diff --git "a/test.json" "b/test.json" deleted file mode 100644--- "a/test.json" +++ /dev/null @@ -1,541 +0,0 @@ -[ - { - "num": 1, - "question": "Consider gamma matrices in the chiral representation using the Pauli matrices\n$$\n\\gamma^\\mu\\equiv\\begin{pmatrix}0&\\sigma^\\mu\\\\\\overline{\\sigma}^\\mu&0\\end{pmatrix},\\quad \\text{with}\\quad\\sigma^\\mu\\equiv(1,\\sigma^i),\\quad\\overline{\\sigma}^\\mu\\equiv(1,-\\sigma^i).\n$$\nWhat is the value of $\\mathrm{tr}(\\overline{\\sigma}^{\\mu}\\sigma^{\\nu}\\overline{\\sigma}^{\\rho}\\sigma^{\\sigma})$? In your answer please take the metric as $\\eta^{\\mu\\nu}$ and take the convention $\\epsilon_{0123}=1$.", - "category": "Classical", - "answer": "$2(\\eta^{\\mu\\nu}\\eta^{\\rho\\sigma}-\\eta^{\\mu\\rho}\\eta^{\\nu\\sigma}+\\eta^{\\mu\\sigma}\\eta^{\\nu\\rho}+i\\epsilon^{\\mu\\nu\\rho\\sigma})$", - "solution": "One can check that\n$$\n\\gamma^\\mu\\gamma^\\nu\\gamma^\\rho\\gamma^\\sigma=\\begin{pmatrix}\\sigma^\\mu\\overline{\\sigma}^\\nu\\sigma^\\rho\\overline{\\sigma}^\\sigma&0\\\\0&\\overline{\\sigma}^\\mu\\sigma^\\nu\\overline{\\sigma}^\\rho\\sigma^\\sigma\\end{pmatrix}.\n$$\nTaking the trace of this equation provides a first identity:\n$$\n\\mathrm{tr}(\\sigma^{\\mu}\\overline{\\sigma}^{\\nu}\\sigma^{\\rho}\\overline{\\sigma}^{\\sigma})+\\mathrm{tr}(\\overline{\\sigma}^{\\mu}\\sigma^{\\nu}\\overline{\\sigma}^{\\rho}\\sigma^{\\sigma})=\\mathrm{tr}(\\gamma^{\\mu}\\gamma^{\\nu}\\gamma^{\\rho}\\gamma^{\\sigma})=4(\\eta^{\\mu\\nu}\\eta^{\\rho\\sigma}-\\eta^{\\mu\\rho}\\eta^{\\nu\\sigma}+\\eta^{\\mu\\sigma}\\eta^{\\nu\\rho}). \\qquad (1)\n$$\nConsider now $\\gamma^5\\equiv i \\gamma^0\\gamma^1\\gamma^2\\gamma^3$. In the above representation of the gamma matrices, an explicit calculation shows that\n$$\n\\gamma^5=\\begin{pmatrix}-1&0\\\\0&1\\end{pmatrix},\\quad\\gamma^5\\gamma^\\mu\\gamma^\\nu\\gamma^\\rho\\gamma^\\sigma=\\begin{pmatrix}-\\sigma^\\mu\\overline{\\sigma}^\\nu\\sigma^\\rho\\overline{\\sigma}^\\sigma&0\\\\0&\\overline{\\sigma}^\\mu\\sigma^\\nu\\overline{\\sigma}^\\rho\\sigma^\\sigma\\end{pmatrix}.\n$$\nTaking the trace of this equation yields a second identity:\n$$\n\\mathrm{tr}(\\overline{\\sigma}^{\\mu}\\sigma^{\\nu}\\overline{\\sigma}^{\\rho}\\sigma^{\\sigma})-\\mathrm{tr}(\\sigma^{\\mu}\\overline{\\sigma}^{\\nu}\\sigma^{\\rho}\\overline{\\sigma}^{\\sigma})=\\mathrm{tr}(\\gamma^{5}\\gamma^{\\mu}\\gamma^{\\nu}\\gamma^{\\rho}\\gamma^{\\sigma})=4\\,i\\,\\epsilon^{\\mu\\nu\\rho\\sigma}. \\qquad (2)\n$$\nHere we used the convention $\\epsilon_{0123}=+1$, i.e. $\\epsilon^{0123}=-1$; as a check, $\\mathrm{tr}(\\gamma^5\\gamma^0\\gamma^1\\gamma^2\\gamma^3)=\\mathrm{tr}(\\gamma^5\\cdot(-i\\gamma^5))=-4i=4i\\,\\epsilon^{0123}$. Adding (1) and (2) and dividing by 2 gives\n$$\n\\mathrm{tr}(\\overline{\\sigma}^{\\mu}\\sigma^{\\nu}\\overline{\\sigma}^{\\rho}\\sigma^{\\sigma})=2(\\eta^{\\mu\\nu}\\eta^{\\rho\\sigma}-\\eta^{\\mu\\rho}\\eta^{\\nu\\sigma}+\\eta^{\\mu\\sigma}\\eta^{\\nu\\rho}+i\\,\\epsilon^{\\mu\\nu\\rho\\sigma}).\n$$\nAs an explicit check, $\\mathrm{tr}(\\overline{\\sigma}^0\\sigma^1\\overline{\\sigma}^2\\sigma^3)=-\\mathrm{tr}(\\sigma^1\\sigma^2\\sigma^3)=-2i=2i\\,\\epsilon^{0123}$.", - "specific_requirement": "When judging the answer of this question, the Levi-Civita term may also be written with lower indices as $-2i\\epsilon_{\\mu\\nu\\rho\\sigma}$, i.e. $2(\\eta^{\\mu\\nu}\\eta^{\\rho\\sigma}-\\eta^{\\mu\\rho}\\eta^{\\nu\\sigma}+\\eta^{\\mu\\sigma}\\eta^{\\nu\\rho}-i\\epsilon_{\\mu\\nu\\rho\\sigma})$ is also correct. This is equivalent to $+2i\\epsilon^{\\mu\\nu\\rho\\sigma}$ because, with $\\epsilon_{0123}=1$, the components satisfy $\\epsilon^{\\mu\\nu\\rho\\sigma}=-\\epsilon_{\\mu\\nu\\rho\\sigma}$. Note that $+2i\\epsilon_{\\mu\\nu\\rho\\sigma}$ (lower indices with a plus sign) is wrong." - }, - { - "num": 2, - "question": "Consider 4-dimensional Yang-Mills theory $\\mathcal{L}_{\\mathrm{YM}}=-\\frac{1}{2}\\operatorname{tr}(F^{2})=-\\frac{1}{4}F^{a\\mu\\nu}F_{\\mu\\nu}^{a}$ in the temporal gauge $A_0=0$. Based on that, calculate the value of\n$$\n\\partial_0 F_a^{0i}+\\left[D_j,F^{ji}\\right]_a .\n$$\nHere $0$ means the time direction and Latin indices mean the space directions.", - "category": "Classical", - "answer": "$0$", - "solution": "By using the temporal gauge $A^0=0$, we circumvent the problem that $A_{a}^{0}$ has a vanishing conjugate momentum, because $A^{0}$ is not a dynamical variable in this gauge. Regarding the other components of the gauge field, the conjugate momentum $\\Pi^{i}$ of $A^{i}$ is given by\n$$\n\\Pi_{a}^{i}\\equiv\\frac{\\partial\\mathcal{L}_{YM}}{\\partial(\\partial_{0}A_{a}^{i})}=\\partial_{0}A_{a}^{i}.\n$$\nIt is convenient to introduce the chromo-electric and chromo-magnetic fields by\n$$\nE_{a}^{i}\\equiv F_{a}^{0i}=\\Pi_{a}^{i}\\ (\\text{in } A^{0}=0 \\text{ gauge}),\\qquad B_{a}^{i}\\equiv\\frac{1}{2}\\epsilon_{ijk}F_{a}^{jk},\n$$\nin terms of which the Lagrangian density can be written as follows:\n$$\n\\begin{aligned}\n-\\frac{1}{4}F_{a}^{\\mu\\nu}F_{\\mu\\nu}^{a}&=-\\frac{1}{4}\\Big(-F_{a}^{0i}F_{a}^{0i}-F_{a}^{i0}F_{a}^{i0}+F_{a}^{ij}F_{a}^{ij}\\Big)=\\frac{1}{4}\\Big(2E_{a}^{i}E_{a}^{i}-\\varepsilon_{ijk}\\varepsilon_{ijl}B_{a}^{k}B_{a}^{l}\\Big)\\\\\n&=\\frac{1}{4}\\Big(2 E_{a}^{i}E_{a}^{i}-(\\delta_{jj}\\delta_{kl}-\\delta_{jk}\\delta_{jl})B_{a}^{k}B_{a}^{l}\\Big)=\\frac{1}{2}\\left(E_{a}^{i}E_{a}^{i}-B_{a}^{i}B_{a}^{i}\\right).\n\\end{aligned}\n$$\nThen, the Hamiltonian is given by\n$$\n\\mathcal{H}=\\underbrace{\\Pi_{a}^{i}(\\partial_{0}A_{a}^{i})}_{E_{a}^{i}E_{a}^{i}}-\\mathcal{L}=\\frac{1}{2}\\big(E_{a}^{i}E_{a}^{i}+B_{a}^{i}B_{a}^{i}\\big)=\\frac{1}{2}\\Pi_{a}^{i}\\Pi_{a}^{i}+\\frac{1}{4}F_{a}^{ij}F_{a}^{ij}.\n$$\nThe Hamilton equations are\n$$\n\\partial_{0}A_{a}^{i}=\\frac{\\partial\\mathcal{H}}{\\partial\\Pi_{a}^{i}}=\\Pi_{a}^{i},\\quad\\partial_{0}\\Pi_{a}^{i}=-\\frac{\\partial\\mathcal{H}}{\\partial A_{a}^{i}}=-\\left[D_j,F^{ji}\\right]_a .\n$$\nThe last step follows because $A^i_a$ enters $\\frac14 F^{ij}_aF^{ij}_a$ both through the derivatives $\\partial_jA^i_a$ (giving, after integration by parts, $\\partial_jF^{ji}_a$) and through the non-Abelian term $f^{abc}A^b_jA^c_i$ (giving the commutator part $\\propto f^{abc}A^b_jF^{ji}_c$); together they form the covariant divergence $[D_j,F^{ji}]_a$.\n\nThe first equation is equivalent to the definition of the conjugate momentum. Using $\\Pi^i_a=E^i_a=F^{0i}_a$, the second one reads\n$$\n\\partial_0 F_a^{0i}+\\left[D_j,F^{ji}\\right]_a=0 .\n$$\nThis is just the spatial component $\\nu=i$ of the Yang-Mills equation $[D_\\mu,F^{\\mu\\nu}]=0$, since $D_0=\\partial_0$ in the temporal gauge. Hence the requested combination vanishes: $\\partial_0 F_a^{0i}+\\left[D_j,F^{ji}\\right]_a=0$." - }, - { - "num": 3, - "question": "Consider a non-Abelian gauge field $A_\\mu^a$ in $d$-dimensional spacetime, with action\n$$\nS = \\int \\mathrm{d}^d x\\, \\left( -\\frac{1}{4} F_{\\mu\\nu}^a F^{a\\,\\mu\\nu} + A_\\mu^a j^{a\\,\\mu} \\right),\n$$\nwhere\n$$\nF_{\\mu\\nu}^a = \\partial_\\mu A_\\nu^a - \\partial_\\nu A_\\mu^a + g f^{abc} A_\\mu^b A_\\nu^c\n$$\nand $f^{abc}$ are the structure constants of the gauge group, and $g$ is the gauge coupling. What is the mass dimension of the coupling constant $g$?", - "category": "Classical", - "answer": "$\\frac{4-d}{2}$", - "solution": "The action is dimensionless and $[\\mathrm{d}^dx]=-d$, so the Lagrangian density has mass dimension $d$. The kinetic term $(\\partial_\\mu A_\\nu^a)^2\\subset F^a_{\\mu\\nu}F^{a\\,\\mu\\nu}$ then fixes the dimension of the gauge field:\n$$\n2\\,(1+[A])=d \\implies [A]=\\frac{d-2}{2}.\n$$\nNow consider the non-Abelian term in $F_{\\mu\\nu}^a$, namely $g f^{abc} A_\\mu^b A_\\nu^c$. The structure constants $f^{abc}$ are dimensionless, and this term must have the same dimension as $\\partial_\\mu A_\\nu^a$, i.e. $1+[A]$:\n$$\n[g]+2[A]=1+[A] \\implies [g]=1-[A]=1-\\frac{d-2}{2}=\\frac{4-d}{2}.\n$$\n(Equivalently, the source term $A^a_\\mu j^{a\\,\\mu}$ fixes $[j]=d-[A]=\\frac{d+2}{2}$, which plays no role for $g$.) In particular $g$ is dimensionless in $d=4$, has positive mass dimension for $d<4$ and negative mass dimension for $d>4$.\n\nFinal answer: $[g]=\\frac{4-d}{2}$." - }, - { - "num": 4, - "question": "Consider the Lagrangian in 4 spacetime dimensions:\n$$\n\\mathcal{L} = -i\\bar{\\psi}\\gamma^\\mu\\partial_\\mu \\psi+\\lambda\\left((\\bar\\psi\\psi)^2-(\\bar\\psi \\gamma_5\\psi)^2\\right)\n$$\nwhere $\\psi$ is a spinor and $\\lambda$ is a constant. Here $\\bar{\\psi} = \\psi^\\dagger \\gamma_0$. Requiring that the Lagrangian is Lorentz invariant, can $\\psi$ be a Dirac spinor (in terms of degrees of freedom)? Your answer should be yes, no or unsure.", - "category": "Classical", - "answer": "yes", - "solution": "Yes. If $\\psi$ transforms as a Dirac spinor, $\\psi\\to\\Lambda_{1/2}\\psi$ with $\\Lambda_{1/2}=\\exp(-\\tfrac{i}{2}\\omega_{\\mu\\nu}S^{\\mu\\nu})$, $S^{\\mu\\nu}=\\tfrac{i}{4}[\\gamma^\\mu,\\gamma^\\nu]$, then $\\bar\\psi\\to\\bar\\psi\\Lambda_{1/2}^{-1}$. The kinetic term is Lorentz invariant, and $\\bar\\psi\\psi$ is a scalar. Since $\\gamma_5$ commutes with the Lorentz generators $S^{\\mu\\nu}$, $\\bar\\psi\\gamma_5\\psi\\to\\bar\\psi\\Lambda_{1/2}^{-1}\\gamma_5\\Lambda_{1/2}\\psi=\\bar\\psi\\gamma_5\\psi$ is also invariant (a pseudoscalar). Hence the whole Lagrangian is Lorentz invariant with $\\psi$ a four-component Dirac spinor." - }, - { - "num": 5, - "question": "Consider the Lagrangian in 4 spacetime dimensions:\n$$\n\\mathcal{L} = -i\\bar{\\psi}\\gamma_\\mu\\partial^\\mu \\psi+\\lambda\\left((\\bar\\psi\\psi)^2-(\\bar\\psi \\gamma_5\\psi)^2\\right)\n$$\nwhere $\\psi$ is a spinor and $\\lambda$ is a constant. Here $\\bar{\\psi} = \\psi^\\dagger \\gamma_0$. Consider the $\\psi\\psi\\to\\psi\\psi$ scattering process. Derive the spin-averaged amplitude squared $|\\bar{\\mathcal{M}}|^2$ at leading order in $\\lambda$. Express the result with the Mandelstam variables $\\{s,t,u\\}$.", - "category": "Classical", - "answer": "$8\\lambda^2(t^2+u^2)$", - "solution": "Consider the interaction Hamiltonian $H_{\\text{int}} = -\\lambda[(\\bar{\\psi}\\psi)(\\bar{\\psi}\\psi)-(\\bar{\\psi}\\gamma_5\\psi)(\\bar{\\psi}\\gamma_5\\psi)]$. The following contractions will contribute to the process:\n$$\n\\langle e(p_3)e(p_4)|:\\bar{\\psi}\\psi\\bar{\\psi}\\psi:|e(p_1)e(p_2)\\rangle-\\langle e(p_3)e(p_4)|:\\bar{\\psi}\\gamma_5\\psi\\bar{\\psi}\\gamma_5\\psi:|e(p_1)e(p_2)\\rangle\n$$\nThere is an additional set of contractions if we interchange the contractions of $\\psi$ into the initial states, which gives a minus sign. We can also interchange the contractions of $\\bar{\\psi}$, but this is equivalent to interchanging the bilinears $\\bar{\\psi}\\psi$, so it will contribute a factor $2!$ to the vertex. With this, we find the Feynman rule for the four-point vertex (with spinor indices $a,b,c,d$ on the four legs):\n$$\n[\\text{4-fermion vertex}] = 2i\\lambda\\left[\\delta_{ac} \\delta_{bd}-\\delta_{ad}\\delta_{bc}-(\\gamma_5)_{ac}(\\gamma_5)_{bd}+(\\gamma_5)_{ad}(\\gamma_5)_{bc}\\right]\n$$\nwhere Latin letters denote spinor indices. So, for the process $\\psi(p_1)\\psi(p_2)\\to\\psi(p_3)\\psi(p_4)$, the amplitude is\n$$\ni\\mathcal{M} = 2i\\lambda\\left[\\bar{u}_3u_1\\bar{u}_4u_2-\\bar{u}_4u_1\\bar{u}_3u_2-\\bar{u}_3\\gamma_5u_1\\bar{u}_4\\gamma_5u_2+\\bar{u}_4\\gamma_5u_1\\bar{u}_3\\gamma_5u_2\\right]\n$$\nwhere we use $u_i\\equiv u(p_i,s_i)$. We can split the amplitude as $\\mathcal{M}=\\mathcal{M}_1+\\mathcal{M}_2$ with\n$$\n\\mathcal{M}_1 = 2\\lambda(\\bar{u}_3u_1\\bar{u}_4u_2-\\bar{u}_3\\gamma_5u_1\\bar{u}_4\\gamma_5u_2), \\qquad \\mathcal{M}_2 = -\\mathcal{M}_1(3 \\leftrightarrow 4)\n$$\n$$\n|\\mathcal{M}|^2=|\\mathcal{M}_1|^2+|\\mathcal{M}_2|^2+(\\mathcal{M}_1\\mathcal{M}_2^*+\\text{c.c.})\n$$\n$$\n\\begin{aligned}\n|\\mathcal{M}_1|^2 &= 4\\lambda^2\\{\\text{Tr}(u_4\\bar{u}_4u_2\\bar{u}_2)\\text{Tr}(u_1\\bar{u}_1u_3\\bar{u}_3)+\\text{Tr}(u_4\\bar{u}_4\\gamma_5u_2\\bar{u}_2\\gamma_5)\\text{Tr}(u_1\\bar{u}_1\\gamma_5u_3\\bar{u}_3\\gamma_5)\\\\\n&\\quad-[\\text{Tr}(u_1\\bar{u}_1\\gamma_5u_3\\bar{u}_3)\\text{Tr}(u_4\\bar{u}_4u_2\\bar{u}_2\\gamma_5)+\\text{c.c.}]\\}\n\\end{aligned}\n$$\nThe Lagrangian contains no mass term, so the fermions are massless and $p_i^2=0$. Using $\\sum_{s_i}u_i\\bar{u}_i = \\slashed{p}_i$, $\\mathrm{tr}(\\slashed{p}_4\\gamma_5\\slashed{p}_2\\gamma_5)=-\\mathrm{tr}(\\slashed{p}_4\\slashed{p}_2)$ and $\\mathrm{tr}(\\slashed{p}_1\\gamma_5\\slashed{p}_3)=0$, we can derive\n$$\n\\begin{aligned}\n|\\bar{\\mathcal{M}}_1|^2 & = \\frac{1}{4}\\sum_{\\{s\\}} |\\mathcal{M}_1|^2 \\\\\n&=\\lambda^2[\\text{Tr}(\\slashed{p}_4 \\slashed{p}_2)\\text{Tr}(\\slashed{p}_1 \\slashed{p}_3)+\\text{Tr}(\\slashed{p}_4 \\slashed{p}_2)\\text{Tr}(\\slashed{p}_1 \\slashed{p}_3)+\\text{Tr}(\\slashed{p}_1 \\gamma_5 \\slashed{p}_3)\\text{Tr}(\\slashed{p}_4 \\gamma_5 \\slashed{p}_2)+\\text{c.c.}] \\\\\n&=32\\lambda^2(p_1p_3)(p_2p_4) =8\\lambda^2 t^2\n\\end{aligned}\n$$\n$$\n|\\bar{\\mathcal{M}}_2|^2 =|\\bar{\\mathcal{M}}_1|^2(3 \\leftrightarrow 4) =32\\lambda^2(p_1p_4)(p_2p_3) =8\\lambda^2 u^2\n$$\nwhere $t=(p_1-p_3)^2=-2p_1\\cdot p_3=-2p_2\\cdot p_4$ and $u=(p_1-p_4)^2=-2p_1\\cdot p_4=-2p_2\\cdot p_3$. For the interference term, note that $(\\bar\\psi\\psi)^2-(\\bar\\psi\\gamma_5\\psi)^2=4(\\bar\\psi P_R\\psi)(\\bar\\psi P_L\\psi)$ with $P_{R,L}=\\frac{1\\pm\\gamma_5}{2}$. In the spin sum, every interference trace then contains a chiral structure of the form $\\slashed{p}_a P_R\\slashed{p}_b P_R=\\slashed{p}_a\\slashed{p}_bP_LP_R=0$ (or the same with $L\\leftrightarrow R$). Therefore\n$$\n\\sum_{\\{s\\}}\\left(\\mathcal{M}_1 \\mathcal{M}_2^*+\\text{c.c.}\\right)=0\n$$\nand\n$$\n|\\bar{\\mathcal{M}}|^2 = 8\\lambda^2(t^2+u^2)\n$$", - "specific_requirement": "When judging the answer of this question, equivalent forms obtained using $s+t+u=0$ are also correct." - }, - { - "num": 6, - "question": "Consider the action of a free massive complex scalar field in four-dimensional spacetime with the Lorentzian metric\n$$\nS=\\int \\mathrm{d}^{4} x\\left[\\partial_{\\mu} \\phi^{*} \\partial^{\\mu} \\phi-m^{2} \\phi^{*} \\phi\\right].\n$$\nThe mode decomposition of the field operator reads\n$$\n\\hat{\\phi}(x)=\\int \\frac{\\mathrm{d}^{3} \\vec{p}}{(2 \\pi)^{3} 2 \\omega_{\\vec{p}}}\\left[\\hat{a}(\\vec{p}) e^{-i p \\cdot x}+\\hat{b}(\\vec{p})^\\dagger e^{i p \\cdot x}\\right].\n$$\nFind the mode decomposition of the conjugate momentum operator (for $\\phi$) $\\hat{\\Pi}(x)$. Here we take the metric as $(+,-,-,-)$.", - "category": "Quantization", - "answer": "$$\n\\hat{\\Pi}(x)=-\\frac{i}{2} \\int \\frac{\\mathrm{d}^3 \\vec{p}}{(2\\pi)^3} \\left[\\hat{b}(\\vec{p}) e^{-ip \\cdot x}-\\hat{a}^{\\dagger}(\\vec{p}) e^{i p \\cdot x} \\right]\n$$", - "solution": "The mode decomposition of the field operator reads\n$$\n\\hat{\\phi}(x)=\\int \\frac{\\mathrm{d}^3 \\vec p}{(2\\pi)^3 2\\omega_{\\vec p}} \\left[\\hat{a} (\\vec p) e^{-i p \\cdot x} + \\hat{b}^{\\dagger} (\\vec p) e^{i p \\cdot x} \\right], \\quad \\text{where} \\quad p \\cdot x= \\omega_{\\vec p}\\, t - \\vec p \\cdot \\vec x\n$$\nThen it immediately follows that\n$$\n\\hat{\\phi}^{*}(x)=\\int \\frac{\\mathrm{d}^3 \\vec p}{(2\\pi)^3 2 \\omega_{\\vec p}} \\left[\\hat{a}^{\\dagger} (\\vec p) e^{i p \\cdot x} + \\hat{b} (\\vec p) e^{-i p \\cdot x} \\right]\n$$\nEquivalently, we could work in terms of $\\hat{\\phi}_1(x), \\hat{\\phi}_2(x)$:\n$$\n\\hat{\\phi}_1(x)=\\int \\frac{\\mathrm{d}^3 \\vec p}{(2\\pi)^3 2 \\omega_{\\vec p}} \\left[\\hat{a}_1(\\vec p) e^{-i p \\cdot x}+\\hat{a}_1^{\\dagger}(\\vec p) e^{i p \\cdot x} \\right]\n$$\n$$\n\\hat{\\phi}_2(x)=\\int \\frac{\\mathrm{d}^3 \\vec p}{(2\\pi)^3 2 \\omega_{\\vec p}} \\left[\\hat{a}_2(\\vec p) e^{-i p \\cdot x}+\\hat{a}_2^{\\dagger}(\\vec p) e^{i p \\cdot x} \\right]\n$$\nUsing that $\\hat{\\phi}(x)=\\frac{\\hat{\\phi}_1(x)+i \\hat{\\phi}_{2}(x)}{\\sqrt{2}}$ we obtain a relation between $\\hat{a},\\hat{b}$ and $\\hat{a}_1,\\hat{a}_2$. Reading off the coefficients of $e^{-ip \\cdot x}$ and $e^{ip \\cdot x}$:\n$$\n\\hat{a}(\\vec p)=\\frac{\\hat{a}_1 (\\vec p)+i \\hat{a}_2 (\\vec p)}{\\sqrt{2}}, \\qquad \\hat{b}^{\\dagger}(\\vec p)=\\frac{\\hat{a}_1^{\\dagger}(\\vec p)+i\\hat{a}_2^{\\dagger}(\\vec p)}{\\sqrt{2}}.\n$$\nOr if we want to express $\\hat{b}(\\vec p)$, we just take the adjoint of the second relation:\n$$\n\\hat{b}(\\vec p)=\\frac{\\hat{a}_1 (\\vec p)-i \\hat{a}_{2}(\\vec p)}{\\sqrt{2}}.\n$$\nTo find the conjugate momentum of $\\hat{\\phi}(x)$, we just use the definition:\n$$\n\\hat{\\Pi}(x)=\\frac{\\partial \\mathcal{L}}{\\partial\\left(\\partial_0 \\hat{\\phi}(x) \\right)}=\\partial^0 \\hat{\\phi}^*(x).\n$$\nUsing that $\\partial^0=\\partial_0$ from the Minkowski metric leads to\n$$\n\\hat{\\Pi}(x)=\\partial_0 \\hat{\\phi}^*(x)=\\dot{\\hat{\\phi}}^*(x),\n$$\nwhich can be expressed using the mode decompositions of the fields:\n$$\n\\hat{\\Pi}(x)=\\dot{\\hat{\\phi}}^*(x)=\\int \\frac{\\mathrm{d}^3 \\vec p}{(2\\pi)^3 2 \\omega_{\\vec p}}\\left[i \\omega_{\\vec p}\\, \\hat{a}^{\\dagger} (\\vec p) e^{i p \\cdot x}-i \\omega_{\\vec p}\\, \\hat{b} (\\vec p) e^{-i p \\cdot x} \\right].\n$$\nThis can be simplified to\n$$\n\\hat{\\Pi}(x)=-\\frac{i}{2} \\int \\frac{\\mathrm{d}^3 \\vec p}{(2\\pi)^3} \\left[\\hat{b}(\\vec p) e^{-ip \\cdot x}-\\hat{a}^{\\dagger}(\\vec p) e^{i p \\cdot x} \\right].\n$$" - }, - { - "num": 7, - "question": "Consider the action of a free massive complex scalar field in four-dimensional spacetime with the Lorentzian metric\n$$\nS=\\int \\mathrm{d}^{4} x\\left[\\partial_{\\mu} \\phi^{*} \\partial^{\\mu} \\phi-m^{2} \\phi^{*} \\phi\\right].\n$$\nThe mode decomposition of the field operator reads\n$$\n\\hat{\\phi}(x)=\\int \\frac{\\mathrm{d}^{3} \\vec{p}}{(2 \\pi)^{3} 2 \\omega_{\\vec{p}}}\\left[\\hat{a}(\\vec{p}) e^{-i p \\cdot x}+\\hat{b}(\\vec{p})^\\dagger e^{i p \\cdot x}\\right].\n$$\nCalculate the commutator $\\left[\\hat{a}(\\vec p_1), \\hat{a}^{\\dagger} (\\vec p_2) \\right]$.", - "category": "Quantization", - "answer": "$(2\\pi)^3 2 \\omega_{\\vec p_1} \\delta^{(3)} (\\vec p_1 - \\vec p_2)$", - "solution": "We write $\\hat\\phi=(\\hat\\phi_1+i\\hat\\phi_2)/\\sqrt2$ in terms of two independent real scalar fields, as in the previous problem, so that the ladder operators are related by\n$$\n\\hat{a}(\\vec p)=\\frac{\\hat{a}_1 (\\vec p)+i \\hat{a}_2 (\\vec p)}{\\sqrt{2}}, \\qquad \\hat{b}(\\vec p)=\\frac{\\hat{a}_1 (\\vec p)-i \\hat{a}_{2}(\\vec p)}{\\sqrt{2}}.\n$$\nWith the relativistic normalization used in the mode expansion, the real-field operators satisfy\n$$\n\\left[\\hat a_i(\\vec p_1),\\hat a_j^\\dagger(\\vec p_2)\\right]=(2\\pi)^3\\,2\\omega_{\\vec p_1}\\,\\delta_{ij}\\,\\delta^{(3)}(\\vec p_1-\\vec p_2),\\qquad \\left[\\hat a_i(\\vec p_1),\\hat a_j(\\vec p_2)\\right]=0,\\qquad i,j=1,2 .\n$$\nThen we have\n$$\n\\left[\\hat{a}(\\vec p_1),\\hat{a}(\\vec p_2) \\right]=\\left[\\frac{\\hat{a}_1 (\\vec p_1)+i \\hat{a}_2(\\vec p_1)}{\\sqrt{2}},\\frac{\\hat{a}_1 (\\vec p_2)+i \\hat{a}_2(\\vec p_2)}{\\sqrt{2}} \\right]\n$$\nUsing the commutator relation\n$$\n\\left[\\hat A +\\hat B, \\hat C +\\hat D \\right]=\\left[\\hat{A},\\hat{C} \\right]+\\left[\\hat{A},\\hat{D} \\right]+\\left[\\hat{B},\\hat{C} \\right]+\\left[\\hat{B},\\hat{D} \\right],\n$$\nwe have\n$$\n\\left[\\hat{a}(\\vec p_1),\\hat{a}(\\vec p_2) \\right]=\\frac{1}{2} \\left[\\hat{a}_1 (\\vec p_1), \\hat{a}_1 (\\vec p_2) \\right]+ \\frac{i}{2} \\left[\\hat{a}_1 (\\vec p_1), \\hat{a}_{2} (\\vec p_2) \\right]+\\frac{i}{2} \\left[\\hat{a}_{2} (\\vec p_1), \\hat{a}_{1}(\\vec p_2) \\right] - \\frac{1}{2} \\left[\\hat{a}_2 (\\vec p_1), \\hat{a}_{2}(\\vec p_2) \\right]\n$$\nUsing the commutation relations for the real scalar field, it immediately follows that $\\left[\\hat{a}(\\vec p_1),\\hat{a}(\\vec p_2) \\right]=0$. The next one we compute is\n$$\n\\left[\\hat{a}(\\vec p_1), \\hat{b}(\\vec p_2) \\right]= \\left[\\frac{\\hat{a}_1 (\\vec p_1)+i \\hat{a}_2(\\vec p_1)}{\\sqrt{2}},\\frac{\\hat{a}_1 (\\vec p_2)-i \\hat{a}_{2}(\\vec p_2)}{\\sqrt{2}} \\right]\n$$\n$$\n= \\frac{1}{2}\\left[\\hat{a}_1 (\\vec p_1),\\hat{a}_1 (\\vec p_2) \\right] -\\frac{i}{2} \\left[\\hat{a}_1 (\\vec p_1),\\hat{a}_2( \\vec p_2) \\right] +\\frac{i}{2} \\left[\\hat{a}_2(\\vec p_1), \\hat{a}_1(\\vec p_2) \\right] + \\frac{1}{2} \\left[\\hat{a}_2 (\\vec p_1), \\hat{a}_2 (\\vec p_2) \\right]=0 .\n$$\nIt can be shown similarly that\n$$\n\\left[\\hat{a}(\\vec p_1),\\hat{b}^{\\dagger} (\\vec p_2) \\right] =0,\\quad\n\\left[\\hat{a}^{\\dagger} (\\vec p_1),\\hat{b} (\\vec p_2) \\right]=0, \\quad\n\\left[\\hat{a}^{\\dagger} (\\vec p_1),\\hat{b}^{\\dagger} (\\vec p_2) \\right]=0,\n$$\n$$\n\\left[\\hat{b}(\\vec p_1), \\hat{b}(\\vec p_2) \\right]=0,\\quad\n\\left[\\hat{b}^{\\dagger}(\\vec p_1), \\hat{b}^{\\dagger}(\\vec p_2) \\right]=0,\\quad\n\\left[\\hat{a}^{\\dagger}(\\vec p_1), \\hat{a}^{\\dagger}(\\vec p_2) \\right]=0 .\n$$\nThe only non-vanishing commutators are $\\left[\\hat{a}(\\vec p_1),\\hat{a}^{\\dagger}(\\vec p_2) \\right]$ and $\\left[\\hat{b}(\\vec p_1),\\hat{b}^{\\dagger}(\\vec p_2) \\right]$. To compute them explicitly,\n$$\n\\left[\\hat{a}(\\vec p_1),\\hat{a}^{\\dagger}(\\vec p_2) \\right]= \\left[\\frac{\\hat{a}_1 (\\vec p_1)+i \\hat{a}_2(\\vec p_1)}{\\sqrt{2}}, \\frac{\\hat{a}_1^{\\dagger}(\\vec p_2) -i \\hat{a}^{\\dagger}_{2} (\\vec p_2)}{\\sqrt{2}} \\right]\n$$\n$$\n= \\frac{1}{2} \\left[\\hat{a}_1(\\vec p_1), \\hat{a}^{\\dagger}_1(\\vec p_2) \\right] - \\frac{i}{2} \\left[\\hat{a}_1 (\\vec p_1),\\hat{a}^{\\dagger}_2(\\vec p_2) \\right] + \\frac{i}{2} \\left[\\hat{a}_2(\\vec p_1), \\hat{a}^{\\dagger}_1(\\vec p_2) \\right] + \\frac{1}{2} \\left[\\hat{a}_2 (\\vec p_1), \\hat{a}^{\\dagger}_2(\\vec p_2) \\right]\n$$\nThe second and the third term vanish, and from the first and the last term we have\n$$\n\\left[\\hat{a}(\\vec p_1),\\hat{a}^{\\dagger}(\\vec p_2) \\right]=\\frac{1}{2} (2\\pi)^3 2 \\omega_{\\vec p_1} \\delta^{(3)} (\\vec p_1 -\\vec p_2)+ \\frac{1}{2} (2\\pi)^3 2 \\omega_{\\vec p_1} \\delta^{(3)} (\\vec p_1 -\\vec p_2),\n$$\nwhich simplifies to\n$$\n\\left[\\hat{a}(\\vec p_1), \\hat{a}^{\\dagger} (\\vec p_2) \\right]=(2\\pi)^3 2 \\omega_{\\vec p_1} \\delta^{(3)} (\\vec p_1 - \\vec p_2).\n$$\nAnd the last commutator is\n$$\n\\left[\\hat{b}(\\vec p_1), \\hat{b}^{\\dagger} (\\vec p_2) \\right]=(2\\pi)^3 2 \\omega_{\\vec p_1} \\delta^{(3)} (\\vec p_1 - \\vec p_2).\n$$" - }, - { - "num": 8, - "question": "Consider the action\n$$\nS= \\int d^4x \\, \\left( -\\frac{1}{4} F_{\\mu\\nu} F^{\\mu \\nu}-\\frac{1}{2}m^2A_\\mu A^\\mu\\right)\n$$\nwhere $F_{\\mu \\nu} = \\partial_\\mu A_\\nu-\\partial_\\nu A_\\mu$ and $m$ is the mass of the vector field. The metric is $\\eta_{\\mu \\nu}=\\mathrm{diag}(-1,+1,+1,+1)$. The momentum-space Feynman propagator $\\tilde{\\Delta}_F^{\\nu\\rho}(p)$ is defined by\n$$\n\\langle 0 |T\\,A^\\nu(x)A^\\rho(y)|0\\rangle=\\int\\frac{d^4p}{(2\\pi)^4}\\,e^{ip\\cdot(x-y)}\\,\\tilde\\Delta_F^{\\nu\\rho}(p)\n$$\nwith $p\\cdot x\\equiv\\eta_{\\mu\\nu}p^\\mu x^\\nu$. Determine $\\tilde{\\Delta}_F^{\\nu\\rho}(p)$. In your final answer, set the infinitesimal $\\epsilon$ in the propagator to zero.", - "category": "Quantization", - "answer": "$-\\frac{i}{p^2+m^2} \\left(\\eta^{\\nu \\rho} +\\frac{p^\\nu p^\\rho}{m^2} \\right)$", - "solution": "Considering $F_{\\mu\\nu} F^{\\mu \\nu} = 2(\\partial_\\mu A_\\nu \\partial^\\mu A^\\nu-\\partial_\\mu A_\\nu \\partial^\\nu A^\\mu)$, the action can be rewritten (after integrating by parts) as\n$$\nS=\\int d^4x \\,\\left[-\\frac{1}{2}(\\partial_\\mu A_\\nu \\partial^\\mu A^\\nu-\\partial_\\mu A_\\nu \\partial^\\nu A^\\mu)-\\frac{1}{2}m^2A^\\mu \\eta_{\\mu \\nu}A^\\nu\\right]=\\int d^4x \\,\\frac{1}{2}A^\\mu K_{\\mu \\nu}A^\\nu\n$$\nwhere $K_{\\mu \\nu} = \\eta_{\\mu \\nu} \\Box -\\partial_\\mu \\partial_\\nu-\\eta_{\\mu \\nu} m^2$. With the Fourier convention $e^{ip\\cdot(x-y)}$ we have $\\partial_\\mu\\to ip_\\mu$, so in momentum space $\\tilde{K}_{\\mu \\nu} = -\\eta_{\\mu \\nu} p^2+p_\\mu p_\\nu-\\eta_{\\mu \\nu}m^2$. For the Gaussian weight $e^{iS}$ with $S=\\frac12\\int A K A$, the two-point function is $\\langle A A\\rangle = i K^{-1}$, i.e.\n$$\n\\tilde{K}_{\\mu \\nu} \\tilde{\\Delta}_F^{\\nu\\rho}(p) = i\\delta^\\rho_\\mu .\n$$\nWe can make the following ansatz:\n$$\n\\tilde{\\Delta}_F^{\\nu\\rho}(p) = f(p^2) \\eta^{\\nu \\rho}+g(p^2) p^\\nu p^\\rho\n$$\nInserting it gives\n$$\n-\\delta_\\mu^\\rho f(p^2)(p^2+m^2)+[f(p^2)-g(p^2)m^2]\\,p_\\mu p^\\rho = i\\delta^\\rho_\\mu .\n$$\nMatching the coefficients of $\\delta_\\mu^\\rho$ and of $p_\\mu p^\\rho$, we find\n$$\nf(p^2) = -\\frac{i}{p^2+m^2}, \\quad g(p^2) = \\frac{f(p^2)}{m^2}=-\\frac{i}{p^2+m^2} \\frac{1}{m^2}\n$$\n(with $p^2+m^2\\to p^2+m^2-i\\epsilon$ before setting $\\epsilon\\to0$). Finally we have\n$$\n\\tilde{\\Delta}_F^{\\nu\\rho}(p)=-\\frac{i}{p^2+m^2} \\left(\\eta^{\\nu \\rho} +\\frac{p^\\nu p^\\rho}{m^2} \\right)\n$$" - }, - { - "num": 9, - "question": "Consider the path integral of a gauge invariant action:\n$$\n\\int \\mathcal{D}A\\, e^{i\\int \\mathrm{d}^4x\\, \\mathcal{L}(A(x),x)}\n$$\nin which $A$ is the gauge field. Let $G(A(x),\\lambda(x))-C(x)$ be a gauge fixing equation and $\\lambda$ be the gauge parameter. We perform the Faddeev-Popov trick and insert\n$$\n1=\\Delta_{\\text{FP}}[A]\\int \\mathrm{d}\\lambda\\, \\delta(G(A(x),\\lambda(x))-C(x)).\n$$\nSuppose now we average the original path integral over a weight function\n$$\nW(C)=e^{-i\\frac{\\xi}{2}\\int \\mathrm{d}^4x\\, C^2(x)}.\n$$\nWrite the resulting gauge-fixed Lagrangian $\\mathcal{L}_{gauge-fixed}$ (omit the ghost term here). The answer should keep $G$ general and express $G$ as $G(A(x))$. Denote the original Lagrangian exactly as it appears above if needed, i.e. as $\\mathcal{L}(A(x),x)$ — do not rename it, abbreviate it, or expand it.", - "category": "Quantization", - "answer": "$\\mathcal{L}_{gauge-fixed} = \\mathcal{L}(A(x),x)-\\frac{\\xi}{2}G(A(x))^2$", - "solution": "Here $G(A(x),\\lambda(x))\\equiv G(A^\\lambda(x))$ denotes the gauge-fixing function evaluated on the gauge-transformed field $A^\\lambda$. Inserting the identity into the path integral gives\n$$\n\\int \\mathrm{d}\\lambda\\int \\mathcal{D}A\\, \\Delta_{\\text{FP}}[A]\\, \\delta\\big(G(A^\\lambda(x))-C(x)\\big)\\, e^{i\\int \\mathrm{d}^4x\\, \\mathcal{L}(A(x),x)} .\n$$\nSince the action and the measure $\\mathcal{D}A$ are gauge invariant, and $\\Delta_{\\text{FP}}$ is gauge invariant as well, we can change variables $A\\to A^{\\lambda^{-1}}$. The integrand then no longer depends on $\\lambda$, and $\\int\\mathrm{d}\\lambda$ factors out as an (infinite) constant, the gauge-group volume:\n$$\n\\Big(\\int \\mathrm{d}\\lambda\\Big)\\int \\mathcal{D}A\\, \\Delta_{\\text{FP}}[A]\\, \\delta\\big(G(A(x))-C(x)\\big)\\, e^{i\\int \\mathrm{d}^4x\\, \\mathcal{L}(A(x),x)} .\n$$\nThis result is independent of $C(x)$. We can therefore average over $C$ with the weight $W(C)$, which only changes the overall normalization:\n$$\n\\int \\mathcal{D}C\\, e^{-i\\frac{\\xi}{2}\\int \\mathrm{d}^4x\\, C^2(x)} \\int \\mathcal{D}A\\, \\Delta_{\\text{FP}}[A]\\, \\delta\\big(G(A(x))-C(x)\\big)\\, e^{i\\int \\mathrm{d}^4x\\, \\mathcal{L}(A(x),x)} .\n$$\nThe delta function sets $C(x)=G(A(x))$, and after integrating over $C$ we obtain\n$$\n\\int \\mathcal{D}A\\, \\Delta_{\\text{FP}}[A]\\, e^{i\\int \\mathrm{d}^4x \\left[\\mathcal{L}(A(x),x) -\\frac{\\xi}{2} G(A(x))^2\\right]}.\n$$\nExponentiating $\\Delta_{\\text{FP}}$ would give the ghost term, which is omitted here. The gauge-fixed Lagrangian is therefore\n$$\n\\mathcal{L}_{gauge-fixed}=\\mathcal{L}(A(x),x) -\\frac{\\xi}{2} G(A(x))^2 .\n$$" - }, - { - "num": 10, - "question": "Consider the action\n$$\nS = \\int d^4x \\,(i \\bar{\\psi} \\gamma^\\mu \\partial_\\mu \\psi -m\\bar{\\psi}\\psi)\n$$\nwhere $\\bar{\\psi}=\\psi^\\dagger\\gamma_0$ and $m$ is the real mass of the fermion. Consider the transformation $\\psi \\to \\psi'=e^{i\\alpha} \\psi$ with $\\alpha$ a constant. Promote $\\alpha \\to \\alpha(x)$; the Noether current is defined by\n$$\n\\delta S=\\int d^4x\\;j^\\mu\\,\\partial_\\mu\\alpha(x).\n$$\nGive the Noether current.", - "category": "QED", - "answer": "$-\\bar{\\psi} \\gamma^\\mu \\psi$", - "solution": "$\\psi \\to \\psi'=e^{i\\alpha} \\psi$ means $\\bar{\\psi} \\to \\bar{\\psi}'=\\bar{\\psi}e^{-i\\alpha}$, so for constant $\\alpha$ the action is invariant. Now promote $\\alpha\\to\\alpha(x)$. The mass term $\\bar\\psi\\psi$ is still invariant, and in the kinetic term the only change comes from the derivative acting on the phase:\n$$\ni\\bar{\\psi}'\\gamma^\\mu\\partial_\\mu\\psi' = i\\bar{\\psi}e^{-i\\alpha}\\gamma^\\mu\\partial_\\mu\\big(e^{i\\alpha}\\psi\\big)=i\\bar{\\psi}\\gamma^\\mu\\partial_\\mu\\psi+i\\bar\\psi\\gamma^\\mu\\big(i\\partial_\\mu\\alpha\\big)\\psi .\n$$\nHence, to first order in $\\alpha$,\n$$\n\\delta S=\\int d^4x\\,\\big(-\\bar{\\psi}\\gamma^\\mu\\psi\\big)\\,\\partial_\\mu\\alpha(x),\n$$\nand comparing with the definition $\\delta S=\\int d^4x\\,j^\\mu\\partial_\\mu\\alpha(x)$ we read off\n$$\nj^\\mu=-\\bar{\\psi}\\gamma^\\mu\\psi .\n$$\nAs a check of conservation, the equations of motion are\n$$\ni\\gamma^\\mu \\partial_\\mu \\psi-m\\psi=0,\\qquad i\\partial_\\mu\\bar{\\psi}\\gamma^\\mu +m\\bar{\\psi}=0,\n$$\nso that\n$$\n\\partial_\\mu j^\\mu = -(\\partial_\\mu \\bar{\\psi}) \\gamma^\\mu \\psi -\\bar{\\psi}\\gamma^\\mu \\partial_\\mu \\psi = -im\\bar\\psi\\psi+im\\bar\\psi\\psi=0 .\n$$\n(The overall sign of $j^\\mu$ is fixed by the definition through $\\delta S$; the standard Noether formula $\\frac{\\partial\\mathcal L}{\\partial(\\partial_\\mu\\psi)}\\delta\\psi$ with $\\delta\\psi=i\\alpha\\psi$ gives the same result $-\\bar\\psi\\gamma^\\mu\\psi$ per unit $\\alpha$.)" - }, - { - "num": 11, - "question": "From the action\n$$\nS = \\int d^4x \\,(i \\bar{\\psi} \\gamma^\\mu \\partial_\\mu \\psi -m\\bar{\\psi}\\psi)\n$$\nwhere $\\bar{\\psi}=\\psi^\\dagger\\gamma_0$ and $m$ is the real mass of the fermion, consider the transformation\n$$\n\\psi \\to \\psi'=e^{i\\alpha\\gamma_5}\\psi\n$$\nwith $\\alpha$ a constant and $\\gamma_5=i\\gamma_0\\gamma_1\\gamma_2\\gamma_3$. What value of $m$ can ensure that the action is invariant under the above transformation?", - "category": "QED", - "answer": "0", - "solution": "From $\\psi \\to \\psi'=e^{i\\alpha\\gamma_5} \\psi$ we have\n$$\n\\psi^\\dagger\\to \\psi^{\\dagger\\prime} = (e^{i\\alpha \\gamma_5}\\psi)^\\dagger=\\psi^\\dagger e^{-i\\alpha\\gamma_5}\n$$\nsince $\\gamma_5=\\gamma_5^\\dagger$. Using $\\gamma_5^2=1$ and $\\{\\gamma_5,\\gamma^0\\}=0$,\n$$\n\\begin{aligned}\n\\bar{\\psi}' &= \\psi^{\\dagger\\prime} \\gamma^0=\\psi^\\dagger e^{-i\\alpha\\gamma_5}\\gamma^0=\\psi^\\dagger \\sum_{n=0}^\\infty \\frac{(-i\\alpha\\gamma_5)^n}{n!}\\gamma^0 \\\\\n&=\\psi^\\dagger(\\cos \\alpha-i\\gamma_5 \\sin \\alpha) \\gamma^0 = \\bar{\\psi}(\\cos \\alpha+i\\gamma_5 \\sin \\alpha) \\\\\n& = \\bar{\\psi}\\sum_{n=0}^\\infty \\frac{(i\\alpha\\gamma_5)^n}{n!} = \\bar{\\psi}e^{i\\alpha \\gamma_5}\n\\end{aligned}\n$$\n$$\n\\begin{aligned}\nS' &= \\int d^4 x \\,(i \\bar{\\psi}' \\gamma^\\mu \\partial_\\mu \\psi' -m\\bar{\\psi}'\\psi') \\\\\n&= \\int d^4x \\, (i \\bar{\\psi} e^{i\\alpha\\gamma_5}\\gamma^\\mu e^{i\\alpha\\gamma_5} \\partial_\\mu \\psi -m\\bar{\\psi}e^{2i\\alpha\\gamma_5}\\psi)\n\\end{aligned}\n$$\nSince $\\{\\gamma_5,\\gamma^\\mu\\}=0$ implies $e^{i\\alpha \\gamma_5}\\gamma^\\mu=\\gamma^\\mu e^{-i\\alpha \\gamma_5}$, we have $e^{i\\alpha \\gamma_5}\\gamma^\\mu e^{i\\alpha \\gamma_5}=\\gamma^\\mu$ and find\n$$\nS' =\\int d^4x \\,(i \\bar{\\psi} \\gamma^\\mu \\partial_\\mu \\psi -m\\bar{\\psi}e^{2i\\alpha\\gamma_5}\\psi)\n$$\nSince $\\bar\\psi e^{2i\\alpha\\gamma_5}\\psi=\\cos(2\\alpha)\\,\\bar\\psi\\psi+i\\sin(2\\alpha)\\,\\bar\\psi\\gamma_5\\psi\\neq\\bar\\psi\\psi$ for generic $\\alpha$, $S'=S$ only if $m=0$." - }, - { - "num": 12, - "question": "Consider the following theory describing the cubic interaction of a real field $\\phi$ with a complex scalar $\\chi$, with action\n$$\nS = \\int d^4x \\,\\left[\\frac{1}{2}(\\partial_\\mu \\phi)^2-\\frac{1}{2}m_\\phi^2 \\phi^2+|\\partial_\\mu \\chi|^2-m^2_\\chi |\\chi|^2-g\\phi |\\chi|^2\\right]\n$$\nwhere $g$ is a real parameter with mass dimension 1. The decomposition of $\\phi$ and $\\chi$ in box normalization reads\n$$\n\\phi(x) = \\sum_{\\vec{p}}\\frac{1}{\\sqrt{2V\\omega_{\\vec{p}}}}\\left(\\hat{c}_{\\vec{p}}\\, e^{-ipx}+\\hat{c}_{\\vec{p}}^\\dagger e^{ipx}\\right),\n$$\n$$\n\\chi(x) = \\sum_{\\vec{p}}\\frac{1}{\\sqrt{2V\\omega_{\\vec{p}}}}\\left(\\hat{a}_{\\vec{p}}\\, e^{-ipx}+\\hat{b}_{\\vec{p}}^\\dagger e^{ipx}\\right).\n$$\nHere $\\omega_{\\vec p}$ denotes the energy of the corresponding particle. Calculate the amplitude $\\mathcal{A}$ of the decay process $\\phi \\to \\chi+\\chi^\\dagger$ at the lowest order in $g$. Here the momentum of the incoming particle $\\phi$ is $\\vec{r}$ and the momenta of the outgoing particles $\\chi$ and $\\chi^\\dagger$ are $\\vec{p}$ and $\\vec{q}$ respectively. In your answer, keep the box normalization factors $\\frac{1}{\\sqrt{2V \\omega_{\\vec p}}}$ and write the energy-momentum conservation factor in the infinite-volume form $(2\\pi)^4\\delta^{(4)}(p+q-r)$.", - "category": "QED", - "answer": "$\\mathcal{A}=-\\frac{ig}{\\sqrt{8V^3\\omega_{\\vec{r}}\\omega_{\\vec{p}}\\omega_{\\vec{q}}}}(2\\pi)^4\\delta^{(4)}(p+q-r)$", - "solution": "The evolution operator at leading order reads\n$$\nU(t,t_0)=1-i\\int_{t_0}^{t} dt' \\, H_{\\text{int}}+\\cdots\n$$\nwith $H_{\\text{int}}=g\\int d^3x\\,\\phi|\\chi|^2$. The transition amplitude from the initial state $|i\\rangle$ to the final state $|f\\rangle$ at first order is therefore $-i\\langle f|\\int dt \\, H_{\\text{int}}|i\\rangle$. In our case,\n$$\n\\mathcal{A} =-ig \\langle \\vec{p},\\vec{q}|\\int d^4x \\, \\phi(x) \\chi^\\dagger(x) \\chi(x) |\\vec{r}\\rangle,\n$$\nwhere $|\\vec{r}\\rangle = \\hat{c}_{\\vec{r}}^\\dagger |0\\rangle$ and $|\\vec{p},\\vec{q}\\rangle=\\hat a^\\dagger_{\\vec p}\\hat b^\\dagger_{\\vec q}|0\\rangle$. Only one term in the mode expansions contributes. The field $\\phi$ must annihilate the incoming $\\phi$-particle (the $\\hat c_{\\vec r'}e^{-ir'x}$ part), $\\chi^\\dagger$ must create the $\\chi$-particle (the $\\hat a^\\dagger_{\\vec p'}e^{ip'x}$ part), and $\\chi$ must create the anti-particle (the $\\hat b^\\dagger_{\\vec q'}e^{iq'x}$ part). The box-normalized commutators $[\\hat c_{\\vec r'},\\hat c^\\dagger_{\\vec r}]=\\delta_{\\vec r'\\vec r}$, etc., collapse the mode sums:\n$$\n\\begin{aligned}\n\\mathcal{A} & = -ig \\sum_{\\vec{r}',\\vec{p}',\\vec{q}'} \\frac{1}{\\sqrt{8V^3\\omega_{\\vec{r}'}\\omega_{\\vec{p}'}\\omega_{\\vec{q}'} }} \\int d^4x \\, \\langle0|\\hat{b}_{\\vec{q}}\\hat{a}_{\\vec{p}}\\,\\hat{a}^\\dagger_{\\vec{p}'}\\hat{b}^\\dagger_{\\vec{q}'}\\hat{c}_{\\vec{r}'}\\,\\hat{c}^\\dagger_{\\vec{r}}|0\\rangle\\, e^{i(p'+q'-r')\\cdot x} \\\\\n&=-\\frac{ig}{\\sqrt{8V^3\\omega_{\\vec{r}}\\omega_{\\vec{p}}\\omega_{\\vec{q}}}}\\int d^4x\\, e^{i(p+q-r)\\cdot x}=-\\frac{ig}{\\sqrt{8V^3\\omega_{\\vec{r}}\\omega_{\\vec{p}}\\omega_{\\vec{q}}}}(2\\pi)^4\\delta^{(4)}(p+q-r),\n\\end{aligned}\n$$\nwhere, as instructed, the space-time integral is written in its infinite-volume form." - }, - { - "num": 13, - "question": "Consider the QED Lagrangian\n$$\n\\mathcal{L} = -\\frac{1}{4}F_{\\mu \\nu}F^{\\mu \\nu} +\\bar{\\psi}(i\\slashed{\\partial}-m)\\psi -e\\bar{\\psi}\\slashed{A}\\psi\n$$\nwhere $\\psi$ is a Dirac fermion with mass $m$ and $A$ is a massless photon, with\n$$\n\\psi(x)=\\int_p \\sum_i \\left(u_i(\\vec{p}) a_i(\\vec{p}) e^{-ipx}+v_i(\\vec{p}) b_i^\\dagger(\\vec{p}) e^{ipx} \\right),\n$$\n$$\nA^\\mu(x)=\\int_k \\sum_s \\epsilon_s^\\mu\\left( c_s(\\vec{k}) e^{-ikx}+c_s^\\dagger(\\vec{k}) e^{ikx}\\right).\n$$\nThe interaction part $H_{\\text{int}}$ of the Hamiltonian reads\n$$\nH_{\\text{int}} = e\\int d^3\\vec{x} \\,\\bar{\\psi}\\slashed{A} \\psi.\n$$\nCompute the matrix element\n$$\n\\langle f|\\int dt\\, H_{\\text{int}}|i\\rangle\n$$\nwhere the initial state comprises a fermion-anti-fermion pair $|i\\rangle = a^\\dagger_{s_1}(\\vec{p})b^\\dagger_{s_2}(\\vec{q}) |0\\rangle$, while the final state is a photon $|f\\rangle =c_r^\\dagger(\\vec{k})|0\\rangle$, $r=1$ or $2$. Here we take $\\bar{u}_i(\\vec p)u_j(\\vec p) = 2m\\delta_{ij}$ and $\\bar{v}_i(\\vec p)v_j(\\vec p) = -2m\\delta_{ij}$.", - "category": "QED", - "answer": "$(2\\pi)^4\\delta^{(4)}(k-p-q)\\, e\\, \\bar{v}_{s_2}(\\vec{q}) \\gamma^\\mu \\epsilon_{\\mu,r}\\, u_{s_1}(\\vec{p})$", - "solution": "We have\n$$\n\\langle f|\\int dt\\, H_{\\text{int}}|i\\rangle = e\\int d^4x \\, \\langle 0| c_r(\\vec{k})\\, \\bar\\psi(x) \\gamma_\\mu A^\\mu(x) \\psi(x)\\, a^\\dagger_{s_1}(\\vec{p})b^\\dagger_{s_2}(\\vec{q})|0\\rangle .\n$$\nHere $\\int_p$ denotes the Lorentz-invariant measure, so that the ladder operators obey $\\{a_i(\\vec p),a_j^\\dagger(\\vec p')\\}=(2\\pi)^3 2E_{\\vec p}\\delta_{ij}\\delta^{(3)}(\\vec p-\\vec p')$ (and similarly for $b$ and $c$), consistent with $\\bar u_iu_j=2m\\delta_{ij}$. The only non-vanishing contractions are\n$$\n\\langle 0|c_r(\\vec{k}) A^\\mu(x)= \\langle 0|\\,\\epsilon_r^\\mu e^{ik\\cdot x},\\qquad\n\\bar{\\psi}(x)b^\\dagger_{s_2}(\\vec{q})|0\\rangle \\to \\bar{v}_{s_2}(\\vec{q})e^{-iq\\cdot x}|0\\rangle,\\qquad\n\\psi(x)a^\\dagger_{s_1}(\\vec{p})|0\\rangle\\to u_{s_1}(\\vec{p}) e^{-ip\\cdot x}|0\\rangle ,\n$$\nwhere the photon polarization vectors are taken real. The fermion operators can be contracted in the order written without any extra sign: $\\psi$ is contracted with the adjacent $a^\\dagger$, and then $\\bar\\psi$ with $b^\\dagger$. The remaining integral is $\\int d^4x\\, e^{i(k-p-q)\\cdot x}=(2\\pi)^4\\delta^{(4)}(k-p-q)$, and we find\n$$\n\\langle f|\\int dt\\, H_{\\text{int}}|i\\rangle=(2\\pi)^4\\delta^{(4)}(k-p-q)\\, e\\, \\bar{v}_{s_2}(\\vec{q}) \\gamma^\\mu \\epsilon_{\\mu,r}u_{s_1}(\\vec{p})\n$$", - "specific_requirement": "When judging the answer of this question, an additional overall factor $\\frac{1}{\\sqrt{8E_pE_q\\omega_k}}$ is also correct, because the question does not specify the measure of $\\int_p$. The question asks for the matrix element, so answering $0$ or stating that it equals $0$ is wrong." - }, - { - "num": 14, - "question": "Compute the differential cross section $\\frac{d\\sigma}{d\\cos\\theta}$ for Bhabha scattering at tree level, $e^+ e^- \\to e^+ e^-$. Here we ignore the electron mass. Express the result with the fine-structure constant $\\alpha$ and the Mandelstam variables $\\{s,t,u\\}$.", - "category": "QED", - "answer": "$\\frac{\\pi \\alpha^2}{s}\\left[\\frac{t^2}{s^2}+\\frac{s^2}{t^2}+u^2\\left(\\frac{1}{s}+\\frac{1}{t}\\right)^2\\right]$", - "solution": "At tree level the process consists of two diagrams: the $s$-channel annihilation diagram and the $t$-channel exchange diagram (entering with a relative minus sign). The minus sign before the $t$-channel diagram comes from the exchange of two fermion field operators when contracting with in and out states. In the high energy limit, we can ignore the mass of the electrons, so the amplitude for the whole scattering process is\n$$\ni \\mathcal{M} = (ie)^2\\left[\\bar{v}(k_2)\\gamma^\\mu u(k_1)\\frac{-i}{s}\\bar{u}(p_1)\\gamma_\\mu v(p_2)-\\bar{u}(p_1)\\gamma^\\mu u(k_1)\\frac{-i}{t}\\bar{v}(k_2)\\gamma_\\mu v(p_2)\\right]\n$$\nwhere the Mandelstam variables $\\{s,t,u\\}$ are defined as\n$$\ns = (k_1+k_2)^2, \\quad t = (p_1-k_1)^2, \\quad u = (p_2-k_1)^2\n$$\nIn the massless case, $k_1^2=k_2^2=p_1^2=p_2^2=0$, thus we have\n$$\ns=2k_1 \\cdot k_2=2p_1 \\cdot p_2, \\quad t=-2p_1 \\cdot k_1=-2p_2 \\cdot k_2, \\quad u=-2p_2 \\cdot k_1=-2p_1 \\cdot k_2\n$$\nWe want the unpolarized cross section, so we must average over the ingoing spins and sum over the outgoing spins. That is,\n$$\n\\begin{aligned}\n\\frac{1}{4} \\sum_{\\text{spin}} |\\mathcal{M}|^2 &= \\frac{e^4}{4s^2}\\sum \\left| \\bar{v}(k_2)\\gamma^\\mu u(k_1)\\bar{u}(p_1)\\gamma_\\mu v(p_2) \\right|^2+ \\frac{e^4}{4t^2}\\sum \\left| \\bar{u}(p_1)\\gamma^\\mu u(k_1)\\bar{v}(k_2)\\gamma_\\mu v(p_2) \\right|^2 \\\\\n&\\quad-\\frac{e^4}{4st} \\sum \\left[ \\bar{v}(p_2) \\gamma_\\mu u(p_1)\\bar{u}(k_1)\\gamma^\\mu v(k_2)\\bar{u}(p_1)\\gamma^\\nu u(k_1)\\bar{v}(k_2)\\gamma_\\nu v(p_2)+\\text{c.c.}\\right] \\\\\n&=\\frac{e^4}{4s^2} \\text{tr}(\\slashed{k}_1 \\gamma^\\mu \\slashed{k}_2 \\gamma^\\nu) \\text{tr}(\\slashed{p}_2 \\gamma_\\mu \\slashed{p}_1 \\gamma_\\nu)+\\frac{e^4}{4t^2} \\text{tr}(\\slashed{k}_1 \\gamma^\\mu \\slashed{p}_1 \\gamma^\\nu) \\text{tr}(\\slashed{p}_2 \\gamma_\\mu \\slashed{k}_2 \\gamma_\\nu) \\\\\n&\\quad-\\frac{e^4}{4st} \\left[\\text{tr}(\\slashed{k}_1 \\gamma^\\nu \\slashed{k}_2 \\gamma_\\mu \\slashed{p}_2 \\gamma_\\nu \\slashed{p}_1 \\gamma^\\mu)+\\text{c.c.} \\right] \\\\\n&=\\frac{2e^4(u^2+t^2)}{s^2}+\\frac{2e^4(u^2+s^2)}{t^2}+\\frac{4e^4 u^2}{st} \\\\\n&=2e^4\\left[\\frac{t^2}{s^2}+\\frac{s^2}{t^2}+u^2\\left(\\frac{1}{s}+\\frac{1}{t}\\right)^2\\right]\n\\end{aligned}\n$$\nIn the center-of-mass frame, we have $k_1^0 =k_2^0 \\equiv k^0$ and $\\vec{k}_1=-\\vec{k}_2$. Thus the total energy satisfies $E_{\\text{CM}}^2 = (k_1^0+k_2^0)^2 = 4(k^0)^2=s$. For a $2\\to2$ process with all four particles of equal mass (here massless), the center-of-mass cross section is\n$$\n\\left(\\frac{d\\sigma}{d \\Omega}\\right)_{\\text{CM}} = \\frac{1}{64\\pi^2 E_{\\text{CM}}^2}\\left(\\frac{1}{4} \\sum |\\mathcal{M}|^2 \\right)\n$$\ntherefore\n$$\n\\left(\\frac{d\\sigma}{d \\Omega}\\right)_{\\text{CM}} = \\frac{\\alpha^2}{2s}\\left[\\frac{t^2}{s^2}+\\frac{s^2}{t^2}+u^2\\left(\\frac{1}{s}+\\frac{1}{t}\\right)^2\\right]\n$$\nwhere $\\alpha = \\frac{e^2}{4\\pi}$ is the fine structure constant, so that $\\frac{2e^4}{64\\pi^2 s}=\\frac{\\alpha^2}{2s}$. Since $d\\Omega=d\\cos\\theta\\,d\\phi$ and nothing depends on the azimuthal angle $\\phi$, integrating over $\\phi$ gives a factor $2\\pi$:\n$$\n\\left( \\frac{d\\sigma}{d\\cos\\theta}\\right)_{\\text{CM}} = \\frac{\\pi \\alpha^2}{s}\\left[\\frac{t^2}{s^2}+\\frac{s^2}{t^2}+u^2\\left(\\frac{1}{s}+\\frac{1}{t}\\right)^2\\right]\n$$" - }, - { - "num": 15, - "question": "Consider the QED Lagrangian\n$$\n\\mathcal{L} = i\\bar{\\psi}\\gamma^\\mu D_\\mu \\psi-m\\bar\\psi\\psi-\\frac{1}{4}F_{\\mu\\nu}F^{\\mu \\nu}\n$$\nwhere\n$$\nD_\\mu= \\partial_\\mu-ieA_\\mu, \\quad F_{\\mu \\nu}=\\partial_\\mu A_\\nu-\\partial_\\nu A_\\mu.\n$$\nHow many degrees of freedom are propagated by $A_\\mu$?", - "category": "QED", - "answer": "2", - "solution": "We start initially with a field with 4 components. However, we have the Gauss law and the gauge redundancy (1 constraint + 1 invariance), giving in total 2 propagating degrees of freedom." - }, - { - "num": 16, - "question": "Consider the Proca Lagrangian\n$$\n\\mathcal{L} = i\\bar{\\psi}\\gamma^\\mu D_\\mu \\psi-m\\bar\\psi\\psi+\\frac{1}{2}m^2_A A_\\mu A^\\mu-\\frac{1}{4}F_{\\mu\\nu}F^{\\mu \\nu}\n$$\nwhere\n$$\nD_\\mu= \\partial_\\mu-ieA_\\mu, \\quad F_{\\mu \\nu}=\\partial_\\mu A_\\nu-\\partial_\\nu A_\\mu.\n$$\nHow many degrees of freedom are propagated by $A_\\mu$?", - "category": "QED", - "answer": "3", - "solution": "The equations of motion for the free massive vector read\n$$\n\\partial^\\mu F_{\\mu \\nu}+m^2_A A_\\nu=0\n$$\nfrom which we obtain the constraint\n$$\n\\partial^\\mu A_\\mu=0\n$$\nThe equations of motion for the massive vector in the presence of a source term read\n$$\n\\partial^\\mu F_{\\mu \\nu}+m^2_A A_\\nu=j_\\nu\n$$\nfrom which we obtain the constraint\n$$\n\\partial^\\mu A_\\mu=\\frac{1}{m_A^2}\\partial^\\mu j_\\mu\n$$\nIn both cases the physical degrees of freedom for the massive vector are 3." - }, - { - "num": 17, - "question": "Consider the pseudoscalar Yukawa Lagrangian:\n$$\n\\mathcal{L} = \\frac{1}{2}(\\partial_\\mu \\phi)^2-\\frac{1}{2}m^2 \\phi^2+\\bar{\\psi} (i\\slashed{\\partial}-M)\\psi-ig\\bar{\\psi} \\gamma^5 \\psi \\phi\n$$\nwhere $\\phi$ is a real scalar and $\\psi$ is a Dirac fermion. Which extra term can we add to make the theory renormalizable?\n\nA. Nothing, it is already renormalizable.\nB. $\\lambda\\phi^3$\nC. $\\lambda \\phi^4$\nD. $i\\lambda \\bar{\\psi}\\gamma^5\\psi$\nE. $\\phi\\bar{\\psi}\\psi$\nF. None of the above are correct.", - "category": "Renormalization", - "answer": "C", - "solution": "Here we consider the superficial degree of divergence $D$. In 4d we have\n$$\nD = 4L -2I_\\phi -I_\\psi\n$$\nwhere $I_\\phi$ is the number of internal lines for the scalar and $I_\\psi$ the number of internal lines for the fermion. In a Feynman diagram we have\n$$\nL = I_\\phi+I_\\psi-V+1\n$$\nwhere $V$ is the number of vertices. For the vertices we have\n$$\n2I_\\phi+E_\\phi = V, \\qquad 2I_\\psi+E_\\psi = 2V\n$$\nwhere $E_\\phi$ is the number of external lines for the scalar and $E_\\psi$ for the fermion. We find\n$$\nD=4-E_\\phi-\\frac{3}{2}E_\\psi\n$$\nThe superficially divergent amplitudes are those with $D\\geq0$:\n- fermion two-point function ($E_\\psi=2$): $D=1$, absorbed by $\\delta_\\psi$ and $\\delta_M$;\n- scalar two-point function ($E_\\phi=2$): $D=2$, absorbed by $\\delta_\\phi$ and $\\delta_m$;\n- Yukawa vertex ($E_\\phi=1,E_\\psi=2$): $D=0$, absorbed by $\\delta_g$;\n- scalar one-point function ($E_\\phi=1$, $D=3$) and three-point function ($E_\\phi=3$, $D=1$);\n- scalar four-point function ($E_\\phi=4$): $D=0$.\n\nThe Lagrangian is invariant under parity with $\\phi$ a pseudoscalar, $\\phi\\to-\\phi$, $\\bar\\psi\\gamma^5\\psi\\to-\\bar\\psi\\gamma^5\\psi$. Amplitudes with an odd number of external scalars therefore vanish, so no $\\phi$ or $\\phi^3$ counterterm is needed. For the same reason the parity-odd terms $i\\bar\\psi\\gamma^5\\psi$ and $\\phi\\bar\\psi\\psi$ are never generated. This rules out options B, D and E. The four-scalar amplitude, however, is logarithmically divergent (the fermion box diagram) and is parity-even, and the Lagrangian has no parameter to absorb it. We must therefore add a $\\lambda\\phi^4$ term (with its counterterm) to make the theory renormalizable. The answer is C." - }, - { - "num": 18, - "question": "The massless Yukawa Lagrangian is written as\n$$\n\\mathcal{L} = \\frac{1}{2}(\\partial_\\mu \\phi)^2-\\frac{\\lambda}{4!} \\phi^4+\\sum_{j=1}^N\\bar{\\psi}_j (i\\slashed{\\partial})\\psi_j-\\sum_{j=1}^N ig\\bar{\\psi}_j \\gamma^5 \\psi_j \\phi\n$$\nwhere $\\phi$ is a real scalar and $\\psi_j$ with $j=1,\\cdots,N$ are $N$ Dirac fermions coupled to $\\phi$ with common Yukawa coupling $g$. Compute the Callan-Symanzik $\\beta$ function for $g$,\n$$\n\\beta_g(\\lambda,g),\n$$\nto leading order in the coupling constants, assuming that $g^2$ and $\\lambda$ are of the same order. You only need to consider 1-loop diagrams for the calculation of the counterterms.", - "category": "Renormalization", - "answer": "$\\beta_g(\\lambda,g)=\\frac{(2N+3)g^3}{(4\\pi)^2}$", - "solution": "The Yukawa vertex factor is $i\\times(-ig\\gamma^5)=g\\gamma^5$, and all $N$ fermions couple identically to $\\phi$. The divergent 1-loop diagrams that enter $\\beta_g$ are the following (in $d=4-\\epsilon$). First, the fermion-loop contribution to the scalar self-energy, for a single flavour $j$:\n$$\n[\\text{fermion loop on scalar line}]=-g^2 \\int \\frac{d^4 k}{(2\\pi)^4} \\, \\text{tr} \\left[ \\gamma^5\\frac{i}{\\slashed{k}} \\gamma^5 \\frac{i}{\\slashed{k}-\\slashed{p}}\\right] \\sim \\frac{4ig^2 p^2}{(4\\pi)^2} \\frac{1}{\\epsilon}\n$$\nEvery flavour $j=1,\\dots,N$ runs in the loop, so the full scalar self-energy is $N$ times this. Second, the scalar-exchange correction to the fermion self-energy (flavour diagonal, no factor of $N$):\n$$\n[\\text{scalar loop on fermion line}]= g^2 \\int \\frac{d^4 k}{(2\\pi)^4} \\, \\gamma^5 \\frac{i}{\\slashed{k} }\\gamma^5 \\frac{i}{(k-p)^2} \\sim \\frac{ig^2 \\slashed{p}}{(4\\pi)^2}\\frac{1}{\\epsilon}\n$$\nThird, the vertex correction from scalar exchange (also flavour diagonal, no factor of $N$):\n$$\n[\\text{Yukawa vertex correction}]=g^3 \\int \\frac{d^4 k}{(2\\pi)^4} \\, \\gamma^5 \\frac{i}{\\slashed{k}}\\gamma^5 \\frac{i}{\\slashed{k}}\\gamma^5 \\frac{i}{k^2} \\sim -\\frac{2g^3 \\gamma^5}{(4\\pi)^2} \\frac{1}{\\epsilon}\n$$\nThe $\\lambda\\phi^4$ vertex does not contribute to the 1-loop renormalization of the Yukawa vertex or of the fermion propagator. The only $\\lambda$-dependent 1-loop diagram with two external scalars is a momentum-independent tadpole, which does not contribute to $\\delta_\\phi$ (and vanishes in dimensional regularization for a massless scalar). In the $\\overline{\\text{MS}}$-like form $\\frac{2}{\\epsilon}-\\ln\\mu^2$, the counterterms cancelling these divergences are\n$$\n\\delta_\\phi = -\\frac{2Ng^2 }{(4\\pi)^2}\\left( \\frac{2}{\\epsilon} -\\ln \\mu^2 \\right)\n$$\n$$\n\\delta_\\psi = -\\frac{g^2}{2(4\\pi)^2} \\left( \\frac{2}{\\epsilon} -\\ln \\mu^2 \\right)\n$$\n$$\n\\delta_g = \\frac{g^3}{(4\\pi)^2} \\left( \\frac{2}{\\epsilon} -\\ln \\mu^2 \\right)\n$$\nwhere $\\mu$ is the renormalization scale. The beta function for $g$ to lowest order is given by\n$$\n\\beta_g(\\lambda,g) = \\mu \\frac{\\partial}{\\partial\\mu} \\Big(-\\delta_g+\\frac{1}{2}g\\,\\delta_\\phi+g \\,\\delta_\\psi\\Big).\n$$\nUsing $\\mu\\frac{\\partial}{\\partial\\mu}\\big(\\frac{2}{\\epsilon}-\\ln\\mu^2\\big)=-2$, we obtain\n$$\n\\beta_g(\\lambda,g) = -2\\left[-\\frac{g^3}{(4\\pi)^2}-\\frac{Ng^3}{(4\\pi)^2}-\\frac{g^3}{2(4\\pi)^2}\\right]=\\frac{(2N+3)g^3}{(4\\pi)^2}.\n$$\nThe three contributions come from the vertex ($2g^3$), the scalar wave-function renormalization ($2Ng^3$, from the $N$ fermion loops) and the fermion wave-function renormalization ($g^3$), all in units of $1/(4\\pi)^2$. For $N=1$ this reduces to the familiar $5g^3/(4\\pi)^2$. At this order $\\beta_g$ does not depend on $\\lambda$." - }, - { - "num": 19, - "question": "Consider massless two-dimensional QED, the so-called Schwinger model. The one-loop 1PI photon two-point function is denoted $-i \\Pi_{\\mu \\nu}(p)$; compute it.", - "category": "Renormalization", - "answer": "$-\\frac{ie^2}{\\pi p^2}(p_\\mu p_\\nu -p^2g_{\\mu \\nu})$", - "solution": "With the QED vertex $-ie\\gamma^\\mu$ and the massless fermion propagator $\\frac{i\\slashed{q}}{q^2}$, the one-loop 1PI two-point function (including the $(-1)$ for the fermion loop) is\n$$\n-i \\Pi_{\\mu \\nu}(p) = -(-ie)^2 \\int \\frac{d^D q}{(2\\pi)^D} \\frac{\\text{tr}[\\gamma_\\mu\\, i\\slashed{q}\\,\\gamma_\\nu\\, i(\\slashed{q}-\\slashed{p})]}{q^2(q-p)^2}\n=-e^2 \\int \\frac{d^D q}{(2\\pi)^D} \\frac{\\text{tr}[\\gamma_\\mu\\slashed{q}\\gamma_\\nu(\\slashed{q}-\\slashed{p})]}{q^2(q-p)^2}.\n$$\nWe regulate dimensionally around $D=2$. In $D$ dimensions\n$$\n\\text{tr}(\\gamma_\\mu \\gamma_\\nu \\gamma_\\rho \\gamma_\\sigma)=f(D)(g_{\\mu \\nu}g_{\\rho \\sigma}-g_{\\mu \\rho}g_{\\nu \\sigma}+g_{\\mu \\sigma}g_{\\rho \\nu}),\n$$\nwhere $f(D)$ is the dimension of the spinor representation, with $f(2)=2$. Introducing a Feynman parameter, $\\frac{1}{q^2(q-p)^2}=\\int_0^1dx\\,\\frac{1}{[(q-xp)^2-\\Delta]^2}$ with $\\Delta=-x(1-x)p^2$, and shifting $l=q-xp$, the numerator becomes (dropping terms odd in $l$)\n$$\nf(D)\\Big[2l_\\mu l_\\nu-g_{\\mu\\nu}l^2-2x(1-x)p_\\mu p_\\nu+x(1-x)p^2g_{\\mu\\nu}\\Big],\\qquad l_\\mu l_\\nu\\to\\frac{g_{\\mu\\nu}}{D}l^2 .\n$$\nThe Minkowski-space integrals are\n$$\n\\int \\frac{d^Dl}{(2\\pi)^D} \\frac{1}{(l^2-\\Delta)^2} = \\frac{i\\,\\Gamma(2-\\frac D2)}{(4\\pi)^{D/2}}\\,\\Delta^{\\frac D2-2},\\qquad\n\\int \\frac{d^Dl}{(2\\pi)^D} \\frac{l^2}{(l^2-\\Delta)^2} = -\\frac{D}{2}\\,\\frac{i\\,\\Gamma(1-\\frac D2)}{(4\\pi)^{D/2}}\\,\\Delta^{\\frac D2-1}.\n$$\nThe $l^2$ term therefore gives $\\big(\\frac2D-1\\big)g_{\\mu\\nu}\\times\\big(-\\frac D2\\big)i\\Gamma(1-\\frac D2)\\Delta^{\\frac D2-1}/(4\\pi)^{D/2}=-i\\,g_{\\mu\\nu}\\Gamma(2-\\frac D2)\\Delta^{\\frac D2-1}/(4\\pi)^{D/2}$. The would-be pole of $\\Gamma(1-\\frac D2)$ at $D=2$ is cancelled by the factor $(1-\\frac D2)$, and it is exactly this finite remainder that makes the result transverse. Using $\\Delta=-x(1-x)p^2$ and collecting terms,\n$$\n-i \\Pi_{\\mu \\nu}(p) = 2ie^2 f(D)\\,(p_\\mu p_\\nu-p^2g_{\\mu\\nu})\\int_0^1dx\\,\\frac{x(1-x)\\,\\Gamma(2-\\frac D2)}{(4\\pi)^{D/2}}\\,\\Delta^{\\frac D2-2}.\n$$\nThis expression is finite at $D=2$. Setting $D=2$, $f(2)=2$ and $\\Gamma(1)=1$:\n$$\n-i \\Pi_{\\mu \\nu}(p) = \\frac{4ie^2}{4\\pi}(p_\\mu p_\\nu-p^2g_{\\mu \\nu})\\int_0^1dx\\,\\frac{x(1-x)}{-x(1-x)p^2} = -\\frac{ie^2}{\\pi p^2}(p_\\mu p_\\nu-p^2g_{\\mu \\nu}).\n$$\nThe result is transverse, $p^\\mu\\Pi_{\\mu\\nu}=0$, as required by gauge invariance." - }, - { - "num": 20, - "question": "Consider massless two-dimensional QED, the so-called Schwinger model. Find the mass of the photon from the full photon propagator. Hint: consider the vacuum polarization at one loop, $-i \\Pi_{\\mu \\nu}(p)$.", - "category": "Renormalization", - "answer": "$\\frac{e}{\\sqrt{\\pi}}$", - "solution": "Write the one-loop 1PI two-point function of the previous problem as\n$$\n-i\\Pi_{\\mu\\nu}(p)= i\\,(p^2g_{\\mu\\nu}-p_\\mu p_\\nu)\\,\\Pi(p^2),\\qquad \\Pi(p^2)=\\frac{e^2}{\\pi p^2},\n$$\nand introduce the transverse projector $P^T_{\\mu\\nu}=g_{\\mu\\nu}-\\frac{p_\\mu p_\\nu}{p^2}$, which satisfies $P^T_{\\mu}{}^{\\rho}P^T_{\\rho\\nu}=P^T_{\\mu\\nu}$ and $P^T_{\\mu\\nu}p^\\nu=0$. The full photon propagator is the geometric series of 1PI insertions, starting from the tree-level propagator $\\frac{-ig_{\\mu\\nu}}{p^2}$ in Feynman gauge:\n$$\n\\begin{aligned}\niD_{\\mu \\nu}(p) &= \\frac{-ig_{\\mu \\nu}}{p^2+i\\epsilon}+\\frac{-ig_{\\mu \\rho}}{p^2+i\\epsilon}\\,\\big[i(p^2 g^{\\rho \\sigma}-p^\\rho p^\\sigma)\\Pi(p^2)\\big]\\,\\frac{-ig_{\\sigma \\nu}}{p^2+i\\epsilon} +\\cdots \\\\\n&= \\frac{-i}{p^2}\\,P^T_{\\mu \\nu}\\big(1+\\Pi(p^2)+\\Pi^2(p^2)+\\cdots\\big)+\\frac{-i}{p^2}\\frac{p_\\mu p_\\nu}{p^2} \\\\\n&=\\frac{-i\\,P^T_{\\mu \\nu}}{p^2\\big(1-\\Pi(p^2)\\big)}-\\frac{i\\,p_\\mu p_\\nu}{p^4}\n=\\frac{-i\\left(g_{\\mu \\nu}-\\frac{p_\\mu p_\\nu}{p^2}\\right)}{p^2-\\frac{e^2}{\\pi}}-\\frac{i\\,p_\\mu p_\\nu}{p^4}.\n\\end{aligned}\n$$\nThe $p_\\mu p_\\nu$ term is gauge dependent and drops out of physical amplitudes. The physical, transverse part has a pole at $p^2=e^2/\\pi$, so the photon acquires the mass\n$$\nm_\\gamma=\\frac{e}{\\sqrt{\\pi}} .\n$$" - }, - { - "num": 21, - "question": "Consider the Lagrangian\n$$\n\\mathcal{L} = -\\frac{1}{2} \\mathrm{Tr}[G_{\\mu \\nu} G^{\\mu \\nu}]+i\\bar{\\psi}_j(\\slashed{D}\\psi)_j-m\\bar{\\psi}_j \\psi_j.\n$$\nHere the fields $\\psi$ form a vector in the fundamental $N$-dimensional representation of $SU(N)$ and the invariant form is normalized such that $\\mathrm{Tr}\\,T_aT_b=\\frac{1}{2}\\delta_{ab}$. The covariant derivative is\n$$\nD_\\mu \\psi = \\partial_\\mu \\psi+igA_\\mu^a T_a \\psi\\equiv \\partial_\\mu \\psi+igA_\\mu \\psi\n$$\nand $G_{\\mu \\nu}^a$ is defined as the curvature tensor:\n$$\n[D_\\mu,D_\\nu] \\psi = igG_{\\mu \\nu}^a T_a \\psi \\equiv igG_{\\mu \\nu} \\psi.\n$$\nCalculate the term $(\\partial_\\mu G^{\\mu \\nu}+ig[A_\\mu, G^{\\mu \\nu}])_a$. For your answer, use the same index conventions as in the problem statement. For any additional summed indices you introduce, take Greek indices starting from $\\mu$ and Latin indices starting from $j$. Hint: use the EOM.", - "category": "Non-Abelian", - "answer": "$g \\bar{\\psi}_j T_a \\gamma^\\nu\\psi_j$", - "solution": "From $[D_\\mu,D_\\nu]\\psi=igG_{\\mu\\nu}\\psi$ we have $G_{\\mu\\nu}=\\partial_\\mu A_\\nu-\\partial_\\nu A_\\mu+ig[A_\\mu,A_\\nu]$. The fermion kinetic term contains the interaction $i\\bar\\psi_j\\gamma^\\mu(igA^a_\\mu T_a\\psi)_j=-gA^a_\\mu\\,\\bar{\\psi}_jT_a\\gamma^\\mu \\psi_j$. The variation with respect to the gauge field is therefore\n$$\n\\delta_AS = \\int d^4x\\Big(- \\mathrm{Tr}[G^{\\mu \\nu}\\delta G_{\\mu \\nu}]-\\delta A_\\mu^a\\, g \\bar{\\psi}_jT_a\\gamma^\\mu \\psi_j\\Big),\n$$\nwith\n$$\n\\delta G_{\\mu \\nu} = \\partial_\\mu \\delta A_\\nu -\\partial_\\nu \\delta A_\\mu+ig[\\delta A_\\mu,A_\\nu ]+ig[ A_\\mu,\\delta A_\\nu ]=D_\\mu\\delta A_\\nu-D_\\nu\\delta A_\\mu ,\n$$\nwhere $D_\\mu X=\\partial_\\mu X+ig[A_\\mu,X]$ in the adjoint. Using the antisymmetry of $G^{\\mu\\nu}$,\n$$\n\\mathrm{Tr}[G^{\\mu \\nu} \\delta G_{\\mu \\nu}] = 2\\,\\mathrm{Tr}\\big[G^{\\mu\\nu}\\big(\\partial_\\mu\\delta A_\\nu+ig[A_\\mu,\\delta A_\\nu]\\big)\\big]\n=-2\\,\\mathrm{Tr}\\big[(\\partial_\\mu G^{\\mu \\nu}+ig[A_\\mu,G^{\\mu \\nu}])\\delta A_\\nu\\big]+2\\partial_\\mu\\mathrm{Tr}(G^{\\mu \\nu} \\delta A_\\nu),\n$$\nwhere we used the cyclicity of the trace, $\\mathrm{Tr}(G[A,\\delta A])=-\\mathrm{Tr}([A,G]\\delta A)$. We drop the total derivative. With $\\delta A_\\nu=\\delta A^a_\\nu T_a$ and $\\mathrm{Tr}(T_aT_b)=\\frac12\\delta_{ab}$, we get\n$$\n\\begin{aligned}\n\\delta_AS &= \\int 2\\mathrm{Tr}[(\\partial_\\mu G^{\\mu \\nu}+ig[A_\\mu,G^{\\mu \\nu}])\\delta A_\\nu]-\\delta A^a_\\nu g \\bar{\\psi}_j T_a \\gamma^\\nu \\psi_j \\\\\n&=\\int \\delta A^a_\\nu\\left[(\\partial_\\mu G^{\\mu \\nu}+ig[A_\\mu, G^{\\mu \\nu}])_a - g \\bar{\\psi}_j T_a \\gamma^\\nu\\psi_j\\right]\n\\end{aligned}\n$$\nThe EOM can be written as\n$$\n(\\partial_\\mu G^{\\mu \\nu}+ig[A_\\mu, G^{\\mu \\nu}])_a = g \\bar{\\psi}_j T_a \\gamma^\\nu\\psi_j\n$$" - }, - { - "num": 22, - "question": "Consider the pure Yang-Mills Lagrangian\n$$\n\\mathcal{L} = -\\frac{1}{4} G_{\\mu \\nu}^a G^{\\mu \\nu}_a\n$$\nwhere, due to our normalization of the gauge fields, we use the metric $\\delta_{ab}$ to raise and lower the indices in the adjoint representation. It is in particular invariant under global gauge transformations. Find the corresponding Noether currents $J^\\mu_b$. Here $T_a$ are the group generators as in the previous question, satisfying $[T_a,T_b] = iC_{ab}^cT_c$.", - "category": "Non-Abelian", - "answer": "$G_a^{\\mu \\nu}A_\\nu^c C_{cb}^a$", - "solution": "For a symmetry of the form\n$$\n\\mathcal{L}(\\phi_j+\\epsilon\\delta\\phi_j,\\partial(\\phi_j+\\epsilon\\delta\\phi_j)) = \\mathcal{L}(\\phi_j,\\partial \\phi_j)+\\epsilon\\partial_\\mu K^\\mu (\\phi_j,\\partial \\phi_j)+ \\mathcal{O}(\\epsilon^2)\n$$\nthere is a conserved current\n$$\nJ^\\mu = \\frac{\\partial \\mathcal{L}}{\\partial(\\partial_\\mu \\phi_j)} \\delta\\phi_j-K^\\mu\n$$\nA global (constant) gauge transformation $A_\\mu\\to e^{-i\\alpha^cT_c}A_\\mu e^{i\\alpha^dT_d}$ leaves $\\mathcal L$ exactly invariant, so $K^\\mu=0$ and we only have to compute $\\frac{\\partial \\mathcal{L}}{\\partial(\\partial_\\mu A_\\nu^a)}$ and $\\delta_b A_\\nu^a$:\n$$\n\\frac{\\partial \\mathcal{L}}{\\partial(\\partial_\\mu A_\\nu^a)} = -\\frac{1}{2}G_b^{\\alpha \\beta} \\frac{\\partial G^b_{\\alpha \\beta}}{\\partial(\\partial_\\mu A_\\nu^a)}=-\\frac{1}{2}G_b^{\\alpha \\beta}(\\delta^\\mu_\\alpha\\delta_\\beta^\\nu \\delta_b^a-\\delta^\\mu_\\beta\\delta_\\alpha^\\nu \\delta_b^a) =-G_a^{\\mu \\nu}\n$$\n$$\n\\delta_bA^a_\\nu T_a = \\frac{d}{d\\alpha^b} \\bigg|_{\\alpha=0} e^{-i\\alpha^cT_c} A_\\nu^a T_a e^{i\\alpha^d T_d} = -i[T_b,T_a]A_\\nu^a=C_{ba}^d T_d A_\\nu^a\n$$\nWe find\n$$\n\\delta_b A_\\nu^a = C^a_{bc} A_\\nu^c\n$$\nThe current is therefore\n$$\nJ^\\mu_b=\\frac{\\partial \\mathcal{L}}{\\partial(\\partial_\\mu A_\\nu^a)}\\delta_bA^a_\\nu=-G_a^{\\mu \\nu}C^a_{bc}A_\\nu^c=G_a^{\\mu \\nu}A_\\nu^c C_{cb}^a ,\n$$\nwhere in the last step we used the antisymmetry $C^a_{bc}=-C^a_{cb}$.", - "reference_answer_alt": [ - "-G^{\\mu\\nu}_a \\, C^a_{bc} \\, A_\\nu^c" - ], - "specific_requirement": "When judging the answer of this question, both overall signs are correct." - }, - { - "num": 23, - "question": "Consider the Lagrangian for the electroweak interactions in the Standard Model:\n$$\n\\begin{aligned}\n\\mathcal{L} &= -\\frac{1}{4}W^a_{\\mu \\nu}W^{\\mu \\nu\\,a}-\\frac{1}{4}B_{\\mu \\nu}B^{\\mu \\nu} +i\\sum_j\\bar{\\psi}_j \\slashed{\\partial}\\psi_j +D_\\mu H^\\dagger D^\\mu H+\\mu^2H^\\dagger H-\\lambda(H^\\dagger H)^2 \\\\\n&\\quad -\\bar{E}_L \\Lambda^{(e)} H e_R- \\bar{Q}_L \\Lambda^{(u)} \\tilde{H} u_R-\\bar{Q}_L \\Lambda^{(d)} H d_R+\\mathrm{h.c.}\n\\end{aligned}\n$$\nwhere $D_\\mu = \\partial_\\mu -igW^a_\\mu \\tau^a-ig'\\frac{Y}{2}B_\\mu$, $\\psi_j=\\{Q_L,E_L,u_R,d_R,e_R\\}$, $\\tilde{H} = i\\sigma_2H^\\star$, and the family indices for the various Yukawa couplings $\\Lambda$ are suppressed.\nFrom the potential around the minimum, identify the Higgs mass $m_h$. Work in the unitary gauge and use the convention\n$$\nH(x) = \\frac{1}{\\sqrt{2}}\\begin{pmatrix} 0 \\\\ v+h(x) \\end{pmatrix}.\n$$\nWhen expressing your final result, please express the result in terms of $\\lambda$ and $v$.", - "category": "Electroweak", - "answer": "$\\sqrt{2\\lambda}\\, v$", - "solution": "$$\n\\begin{aligned}\nV&= -\\mu^2 H^\\dagger H+\\lambda(H^\\dagger H)^2=-\\frac{\\mu^2}{2}(v+h)^2+\\frac{\\lambda}{4}(v+h)^4 \\\\\n&=\\left(-\\mu^2+\\frac{\\lambda v^2}{2}\\right) \\frac{v^2}{2}+(-\\mu^2+\\lambda v^2)vh+(-\\mu^2+3\\lambda v^2)\\frac{h^2}{2}+\\lambda vh^3+\\frac{\\lambda}{4}h^4\n\\end{aligned}\n$$\nConsidering $\\frac{\\partial V}{\\partial h} = 0$, we have $-\\mu^2+\\lambda v^2=0$. Thus we can easily find $m_h = \\sqrt{2\\lambda}\\, v$." - }, - { - "num": 24, - "question": "Consider the Lagrangian for the electroweak interactions in the Standard Model:\n$$\n\\begin{aligned}\n\\mathcal{L} &= -\\frac{1}{4}W^a_{\\mu \\nu}W^{\\mu \\nu\\,a}-\\frac{1}{4}B_{\\mu \\nu}B^{\\mu \\nu} +i\\sum_j\\bar{\\psi}_j \\slashed{\\partial}\\psi_j +D_\\mu H^\\dagger D^\\mu H+\\mu^2H^\\dagger H-\\lambda(H^\\dagger H)^2 \\\\\n&\\quad -\\bar{E}_L \\Lambda^{(e)} H e_R- \\bar{Q}_L \\Lambda^{(u)} \\tilde{H} u_R-\\bar{Q}_L \\Lambda^{(d)} H d_R+\\mathrm{h.c.}\n\\end{aligned}\n$$\nwhere $D_\\mu = \\partial_\\mu -igW^a_\\mu \\tau^a-ig'\\frac{Y}{2}B_\\mu$, $\\psi_j=\\{Q_L,E_L,u_R,d_R,e_R\\}$, $\\tilde{H} = i\\sigma_2H^\\star$, and the family indices for the various Yukawa couplings $\\Lambda$ are suppressed.\nWork in the unitary gauge and use the convention\n$$\nH(x) = \\frac{1}{\\sqrt{2}}\\begin{pmatrix} 0 \\\\ v+h(x) \\end{pmatrix}.\n$$\nCalculate the mass of the $Z^\\mu$ boson. Express your answer with $g$, $g'$ and $v$ only. (Hint: define $Z^\\mu$ first.)", - "category": "Electroweak", - "answer": "$\\frac{\\sqrt{g^2+g'^2}\\,v}{2}$", - "solution": "With $\\tau^a=\\sigma^a/2$ and hypercharge $Y=1$ for the Higgs doublet (so that $\\frac{Y}{2}B_\\mu$ acts as $\\frac12 B_\\mu\\mathbb{1}$), the gauge-covariant derivative of $H$ is\n$$\n\\begin{aligned}\nD_\\mu H &=\n\\partial_\\mu H\n- \\frac{ig}{2}\n\\left[\nW_\\mu^1\n\\begin{pmatrix} 0 & 1 \\\\ 1 & 0 \\end{pmatrix}\n+ W_\\mu^2\n\\begin{pmatrix} 0 & -i \\\\ i & 0 \\end{pmatrix}\n+ W_\\mu^3\n\\begin{pmatrix} 1 & 0 \\\\ 0 & -1 \\end{pmatrix}\n\\right] H\n- \\frac{ig'}{2} B_\\mu\n\\begin{pmatrix} 1 & 0 \\\\ 0 & 1 \\end{pmatrix} H\n\\\\\n&=\n\\begin{pmatrix} 0 \\\\ \\frac{1}{\\sqrt{2}} \\partial_\\mu h \\end{pmatrix}\n+\n\\begin{pmatrix}\n-\\frac{i}{2}(gW_\\mu^3+g'B_\\mu) & -\\frac{ig}{2}(W_\\mu^1-iW_\\mu^2) \\\\\n-\\frac{ig}{2}(W_\\mu^1+iW_\\mu^2) & \\frac{i}{2}(gW_\\mu^3-g'B_\\mu)\n\\end{pmatrix}\n\\begin{pmatrix} 0 \\\\ \\frac{1}{\\sqrt{2}}(v+h) \\end{pmatrix}.\n\\end{aligned}\n$$\nHere we define\n$$\nW_\\mu^\\pm = \\frac{1}{\\sqrt2}\\big(W_\\mu^1 \\mp i W_\\mu^2\\big),\\qquad\nZ_\\mu = \\frac{1}{\\sqrt{g^2+g'^2}}\\big(gW^3_\\mu-g'B_\\mu\\big),\\qquad\nA_\\mu= \\frac{1}{\\sqrt{g^2+g'^2}}\\big(g'W^3_\\mu+gB_\\mu\\big).\n$$\nThe photon $A_\\mu$ is the combination that does not appear in the lower (neutral) component. The covariant derivative then takes the form\n$$\nD_\\mu H = \\begin{pmatrix}\n-\\frac{ig}{2}W_\\mu^+(v+h)\\\\ \\frac{1}{\\sqrt{2}}\\partial_\\mu h+i\\frac{\\sqrt{g^2+g'^2}}{2\\sqrt{2}}Z_\\mu(v+h)\n\\end{pmatrix}.\n$$\nThe $Z_\\mu$ mass term therefore reads\n$$\n(D_\\mu H)^\\dagger D^\\mu H \\supset \\frac{(g^2+g'^2)v^2}{8}\\, Z_\\mu Z^\\mu\\equiv\\frac{1}{2}m_Z^2\\,Z_\\mu Z^\\mu ,\n$$\nso that\n$$\nm_Z = \\frac{\\sqrt{g^2+g'^2}\\,v}{2}\n$$\n(and, from the upper component, $m_W=gv/2$, while the photon stays massless)." - }, - { - "num": 25, - "question": "Compute the partial decay widths of the $W^+$ boson into pairs of fermions (including leptons and quarks) with numerical values. Assume that the top quark mass $m_t$ is larger than $m_W$, and ignore the other quark masses. The decay widths to quarks are enhanced by QCD corrections. Use $\\sin^2\\theta_w = 0.23$, and take the relevant Lagrangian to be\n$$\n\\mathcal{L} = \\frac{1}{\\sqrt{2}}gW_\\mu^+ \\left( \\sum_i \\bar{\\nu}_{iL} \\gamma^\\mu e_{iL} +\\sum_{j,c} \\bar{u}_{jL}^c \\gamma^\\mu d_{jL}^c\\right)\n$$\nwhere $g= \\frac{e}{\\sin\\theta_w}$. We should also consider about the QCD correction factor $(1+\\frac{\\alpha_s}{\\pi})$ with $\\alpha_s = \\frac{g^2}{4\\pi}$. Hint:Give only the total width $\\Gamma_{\\text{tot}}(W^+)$ as your final answer, as a single number in GeV.Use $\\alpha = \\frac{1}{128}$, $m_W = 80.4$ GeV and $m_Z=91.2$ GeV.", - "category": "Electroweak", - "answer": "2.06 GeV", - "solution": "In the Lagrangian, $i$ runs over all three generations of leptons, $j$ over the first two generations of quarks (the decay $W^+\\to t\\bar b$ is kinematically forbidden since $m_t>m_W$), and $c$ over the 3 colours. All final-state fermions are treated as massless. The amplitude of the decay into a pair of fermions is\n$$\ni\\mathcal{M} = \\frac{ig}{\\sqrt{2}} \\epsilon_\\mu (k) \\bar{u}(p_1)\\gamma^\\mu \\left(\\frac{1-\\gamma^5}{2} \\right) v(p_2)\n$$\nwhere $\\epsilon_\\mu$ is the polarization vector for $W_\\mu^+$, $k$ is the incoming $W$ momentum, and $p_1,p_2$ are the outgoing fermion momenta. Thus the squared amplitude with initial polarizations averaged is\n$$\n\\begin{aligned}\n\\frac{1}{3} \\sum_{\\text{spin}} |i\\mathcal{M}|^2 &= \\frac{g^2}{6} \\left(-g_{\\mu \\nu}+\\frac{k_\\mu k_\\nu}{m_W^2}\\right) \\text{tr} \\left[\\slashed{p}_2 \\gamma^\\mu \\left(\\frac{1-\\gamma^5}{2}\\right)\\slashed{p}_1 \\gamma^\\nu \\left(\\frac{1-\\gamma^5}{2}\\right)\\right] \\\\\n&=\\frac{g^2}{3} \\left(p_1 \\cdot p_2+2\\frac{k \\cdot p_1\\, k\\cdot p_2}{m_W^2}\\right)\n\\end{aligned}\n$$\nHere the momenta in the center-of-mass frame can be taken to be\n$$\nk=(m_W,0,0,0), \\quad p_1 = (p,0,0,p),\\quad p_2 = (p,0,0,-p),\n$$\nand energy conservation requires that $p = \\frac{m_W}{2}$. Thus we get\n$$\n\\frac{1}{3} \\sum_{\\text{spin}} |i\\mathcal{M}|^2 = \\frac{1}{3}g^2m_W^2\n$$\nand the decay rate\n$$\n\\int d\\Gamma = \\frac{1}{2m_W} \\int \\frac{d^3p_1 d^3 p_2}{(2\\pi)^6 2E_1 2E_2}\\left( \\frac{1}{3}g^2m_W^2\\right) (2\\pi)^4 \\delta^{(4)}(k-p_1-p_2)=\\frac{\\alpha m_W}{12\\sin^2\\theta_w}\n$$\nwhere $\\alpha=\\frac{e^2}{4\\pi}$ and $g=e/\\sin\\theta_w$ as defined in the question. We used $\\int\\frac{d^3p_1 d^3 p_2}{(2\\pi)^6 2E_1 2E_2}(2\\pi)^4 \\delta^{(4)}(k-p_1-p_2)=\\frac{1}{8\\pi}$ for massless final states, so that $\\Gamma=\\frac{1}{2m_W}\\frac{g^2m_W^2}{3}\\frac{1}{8\\pi}=\\frac{g^2m_W}{48\\pi}=\\frac{\\alpha m_W}{12\\sin^2\\theta_w}$.\n\nFor each quark final state we multiply the result by the number of colours, 3, and by the QCD correction factor $\\left(1+\\frac{\\alpha_s}{\\pi}\\right)$. As specified in the question, $\\alpha_s=\\frac{g^2}{4\\pi}=\\frac{\\alpha}{\\sin^2\\theta_w}=\\frac{1/128}{0.23}\\simeq0.0340$, so $1+\\frac{\\alpha_s}{\\pi}\\simeq1.0108$. With $\\alpha=1/128$, $m_W=80.4$ GeV and $\\sin^2\\theta_w=0.23$, and taking into account 3 generations of leptons and 2 generations of quarks, we get\n$$\n\\Gamma(W^+ \\to \\ell_i^+ \\nu_i) = \\frac{\\alpha m_W}{12\\sin^2\\theta_w} \\simeq 0.228\\ \\text{GeV}\\qquad(\\text{each } i=e,\\mu,\\tau)\n$$\n$$\n\\Gamma(W^+ \\to u_j \\bar{d}_j) = \\frac{\\alpha m_W}{4\\sin^2\\theta_w}\\left( 1+\\frac{\\alpha_s}{\\pi}\\right) \\simeq 0.690\\ \\text{GeV}\\qquad(\\text{each } j=1,2)\n$$\n$$\n\\Gamma_{\\text{tot}}(W^+) = \\frac{\\alpha m_W}{12\\sin^2\\theta_w}\\left( 9+6\\frac{\\alpha_s}{\\pi}\\right) \\simeq 2.06\\ \\text{GeV}\n$$", - "numeric_tolerance": 0.02, - "specific_requirement": "When judging the answer of this question, a relative error of 2% is allowed." - }, - { - "num": 26, - "question": "Compute the partial decay widths of the $Z^0$ boson into pairs of fermions (including hadrons, charged leptons, and the invisible modes $\\nu \\bar{\\nu}$) with numerical values, treating the quarks in the same way as in the $W^+$ case. Take the relevant Lagrangian to be\n$$\n\\mathcal{L} = \\frac{g}{\\cos\\theta_w}Z_\\mu \\sum_i \\bar{f}_{i} \\gamma^\\mu (I_i^3-\\sin^2\\theta_w Q_i)f_i\n$$\nwhere $g= \\frac{e}{\\sin\\theta_w}$, while $I^3_i$ and $Q_i$ are the associated third component of the weak isospin and the electric charge respectively. Also include the QCD correction factor $\\left(1+\\frac{\\alpha_s}{\\pi}\\right)$ with $\\alpha_s = \\frac{g^2}{4\\pi}$.", - "category": "Electroweak", - "answer": "2.47 GeV", - "solution": "The corresponding amplitude is\n$$\ni\\mathcal{M} = \\frac{ig}{\\cos \\theta_w} \\epsilon_\\mu(k)\\bar{u}(p_1)\\gamma^\\mu\\big(I^3P_L-\\sin^2\\theta_wQ\\big) v(p_2)\n=\\frac{ig}{\\cos \\theta_w} \\epsilon_\\mu(k)\\bar{u}(p_1)\\gamma^\\mu\\big(g_LP_L+g_RP_R\\big) v(p_2),\n$$\nwith $P_{L,R}=\\frac{1\\mp\\gamma^5}{2}$ and\n$$\ng_L=I^3-\\sin^2\\theta_wQ,\\qquad g_R=-\\sin^2\\theta_wQ .\n$$\n(Only the left-handed component of $f$ carries $I^3\\neq0$.) This is the same structure as in the $W$ decay of the previous problem, with the coupling $\\frac{g}{\\sqrt2}P_L$ replaced by $\\frac{g}{\\cos\\theta_w}(g_LP_L+g_RP_R)$. Left- and right-handed massless fermions do not interfere, so\n$$\n\\frac{1}{3}\\sum_{\\text{spin}} |\\mathcal{M}|^2 = \\frac{2g^2}{3\\,\\cos^2\\theta_w}\\big(g_L^2+g_R^2\\big) \\left(p_1 \\cdot p_2+\\frac{2k\\cdot p_1\\,k\\cdot p_2}{m_Z^2} \\right)=\\frac{2g^2m_Z^2}{3\\,\\cos^2\\theta_w}\\big(g_L^2+g_R^2\\big),\n$$\nand, with the massless two-body phase space $\\frac{1}{8\\pi}$, the partial decay rate per colour is\n$$\n\\Gamma(Z^0 \\to f\\bar{f}) = \\frac{1}{2m_Z}\\,\\frac{2g^2m_Z^2}{3\\cos^2\\theta_w}(g_L^2+g_R^2)\\,\\frac{1}{8\\pi}=\\frac{\\alpha\\, m_Z}{6\\,\\sin^2 \\theta_w \\cos^2\\theta_w}\\big(g_L^2+g_R^2\\big).\n$$\nThe input values are the same as in the previous problem: $\\alpha=1/128$, $m_Z=91.2$ GeV, $\\sin^2\\theta_w=0.23$. This gives $\\frac{\\alpha m_Z}{6\\sin^2\\theta_w\\cos^2\\theta_w}\\simeq0.6705$ GeV. The quarks are treated as before: $t\\bar t$ is kinematically forbidden, and each quark channel gets a colour factor 3 and the QCD factor $1+\\frac{\\alpha_s}{\\pi}$ with $\\alpha_s=\\frac{g^2}{4\\pi}=\\frac{\\alpha}{\\sin^2\\theta_w}\\simeq0.0340$. We find\n$$\n\\Gamma(Z^0 \\to \\nu_i\\bar{\\nu}_i) \\simeq 0.168\\ \\text{GeV}\\quad\\big(g_L=\\tfrac12,\\ g_R=0\\big),\\qquad\\text{each of } \\nu_e,\\nu_\\mu,\\nu_\\tau\n$$\n$$\n\\Gamma(Z^0 \\to \\ell^- \\ell^+) \\simeq 0.084\\ \\text{GeV}\\quad\\big(g_L=-\\tfrac12+0.23,\\ g_R=0.23\\big),\\qquad\\text{each of } e,\\mu,\\tau\n$$\n$$\n\\Gamma(Z^0 \\to u\\bar{u}) = \\Gamma(Z^0 \\to c\\bar{c})\\simeq 0.292\\ \\text{GeV}\\quad\\big(g_L=\\tfrac12-\\tfrac23 \\cdot 0.23,\\ g_R=-\\tfrac23 \\cdot 0.23\\big)\n$$\n$$\n\\Gamma(Z^0 \\to d\\bar{d}) = \\Gamma(Z^0 \\to s\\bar{s})=\\Gamma(Z^0 \\to b\\bar{b})\\simeq 0.376\\ \\text{GeV}\\quad\\big(g_L=-\\tfrac12+\\tfrac13 \\cdot 0.23,\\ g_R=\\tfrac13 \\cdot 0.23\\big)\n$$\nHence $\\Gamma(Z^0\\to\\text{invisible})\\simeq0.503$ GeV, $\\Gamma(Z^0\\to\\ell^+\\ell^-)\\simeq0.253$ GeV (three charged leptons), and $\\Gamma(Z^0\\to\\text{hadrons})\\simeq0.584+1.129=1.713$ GeV. Finally we obtain\n$$\n\\Gamma_{\\text{tot}}(Z^0 \\to \\text{all fermions}) \\simeq 2.47\\ \\text{GeV}.\n$$", - "numeric_tolerance": 0.02, - "specific_requirement": "When judging the answer of this question, a relative error of 2% is allowed." - }, - { - "num": 27, - "question": "Consider the Abelian Higgs Lagrangian\n$$\n\\mathcal{L}[A_\\mu,H] = -\\frac{1}{4}F_{\\mu\\nu}F^{\\mu \\nu}+(D_\\mu H)^\\dagger D^\\mu H-\\frac{\\lambda^2}{2}\\left(H^\\dagger H-\\frac{v^2}{2}\\right)^2\n$$\nwhere\n$$\nD_\\mu= \\partial_\\mu-igA_\\mu, \\quad F_{\\mu \\nu}=\\partial_\\mu A_\\nu-\\partial_\\nu A_\\mu, \\quad H=\\rho(x)e^{i\\theta(x)/v}.\n$$\nIdentify the massive modes for $A_\\mu$. Hint: expand the theory around the vacuum according to the prescription $\\rho(x) = \\frac{1}{\\sqrt{2}}(v+h(x))$.", - "category": "Electroweak", - "answer": "$m_A= gv$", - "solution": "The Lagrangian has the gauge redundancy\n$$\nA_\\mu \\to A_\\mu+\\frac{1}{g} \\partial_\\mu \\alpha, \\quad H \\to e^{i\\alpha}H \\quad(\\text{i.e. } \\theta\\to\\theta+v\\alpha),\n$$\nunder which $D_\\mu H\\to e^{i\\alpha}D_\\mu H$. The potential is minimized at $H^\\dagger H=\\rho^2=\\frac{v^2}{2}$, so we expand around this vacuum with $\\rho=\\frac{1}{\\sqrt2}(v+h)$:\n$$\nH=\\frac{1}{\\sqrt2}(v+h)\\,e^{i\\theta/v}.\n$$\nThen\n$$\nD_\\mu H = (\\partial_\\mu-igA_\\mu)\\left[\\frac{1}{\\sqrt2}(v+h)e^{i\\theta/v}\\right] =\\frac{e^{i\\theta/v}}{\\sqrt2}\\left[\\partial_\\mu h+i(v+h)\\left(\\frac{\\partial_\\mu\\theta}{v}-gA_\\mu\\right)\\right],\n$$\nand thus\n$$\n(D_\\mu H)^\\dagger D^\\mu H =\\frac{1}{2}\\partial_\\mu h\\, \\partial^\\mu h+\\frac{g^2}{2}(v+h)^2\\left(A_\\mu-\\frac{1}{gv}\\partial_\\mu \\theta \\right)\\left(A^\\mu-\\frac{1}{gv}\\partial^\\mu \\theta \\right).\n$$\nFor the potential, $H^\\dagger H-\\frac{v^2}{2}=vh+\\frac{h^2}{2}$, so\n$$\n-\\frac{\\lambda^2}{2}\\Big(H^\\dagger H-\\frac{v^2}{2}\\Big)^2 =-\\frac{\\lambda^2}{2}\\left(vh+\\frac{h^2}{2} \\right)^2 =-\\frac{\\lambda^2v^2}{2}h^2-\\frac{\\lambda^2 v}{2}h^3-\\frac{\\lambda^2}{8} h^4 .\n$$\nDefine the gauge-invariant combination $\\tilde{A}_\\mu \\equiv A_\\mu-\\frac{1}{gv}\\partial_\\mu \\theta$; note that $\\tilde F_{\\mu\\nu}=F_{\\mu\\nu}$. The Goldstone mode $\\theta$ is then eaten (unitary gauge), and the Lagrangian becomes\n$$\n\\begin{aligned}\n\\mathcal{L}[A_\\mu,H] &= -\\frac{1}{4}\\tilde F_{\\mu\\nu}\\tilde F^{\\mu \\nu}+\\frac{g^2v^2}{2}\\tilde A_\\mu\\tilde A^\\mu+\\frac{1}{2}\\partial_\\mu h \\partial^\\mu h-\\frac{\\lambda^2v^2}{2}h^2 \\\\\n&\\quad+g^2\\left(vh+\\frac{h^2}{2}\\right)\\tilde A_\\mu\\tilde A^\\mu-\\frac{\\lambda^2v}{2}h^3-\\frac{\\lambda^2}{8}h^4 .\n\\end{aligned}\n$$\nComparing the $\\tilde A_\\mu\\tilde A^\\mu$ term with the Proca mass term $\\frac{m_A^2}{2}\\tilde A_\\mu\\tilde A^\\mu$, the vector field is massive with\n$$\nm_A=gv ,\n$$\nwhile the Higgs mode has $m_h=\\lambda v$." - }, - { - "num": 28, - "question": "Consider the Abelian Higgs Lagrangian\n$$\n\\mathcal{L}[A_\\mu,H] = -\\frac{1}{4}F_{\\mu\\nu}F^{\\mu \\nu}+(D_\\mu H)^\\dagger D^\\mu H-\\frac{\\lambda^2}{2}(H^\\dagger H-v^2)^2\n$$\nwhere\n$$\nD_\\mu= \\partial_\\mu-igA_\\mu, \\quad F_{\\mu \\nu}=\\partial_\\mu A_\\nu-\\partial_\\nu A_\\mu, \\quad H=\\rho(x)e^{i\\theta(x)/v}.\n$$\nAssume that the Higgs is heavier than the vector, that is $m_A \\lesssim E \\ll m_h$. One can integrate the former out, and in this limit one obtains the Proca Lagrangian\n$$\n\\mathcal{L}_{\\text{Proca}}=-\\frac{1}{4}\\tilde{F}_{\\mu\\nu} \\tilde{F}^{\\mu\\nu}+\\frac{m^2}{2}\\tilde{A}_\\mu\\tilde{A}^\\mu+\\frac{\\xi^4}{4}(\\tilde{A}_\\mu\\tilde{A}^\\mu)^2 .\n$$\nIdentify the coupling $\\xi^2$ in terms of the massive mode $m_A$, the Higgs mass $m_h$, and the coupling constant.", - "category": "Electroweak", - "answer": "$\\xi^2= \\sqrt{2}g\\frac{m_A}{m_h}$", - "solution": "For the potential $\\frac{\\lambda^2}{2}(H^\\dagger H-v^2)^2$ the vacuum is at $\\rho=v$. We expand with a canonically normalized Higgs field, $\\rho=v+\\frac{h}{\\sqrt2}$, and define $\\tilde A_\\mu\\equiv A_\\mu-\\frac{1}{gv}\\partial_\\mu\\theta$ as in the previous problem. The same steps as there give\n$$\n\\begin{aligned}\n\\mathcal{L} &= -\\frac{1}{4}\\tilde F_{\\mu\\nu}\\tilde F^{\\mu \\nu}+g^2v^2\\,\\tilde A_\\mu\\tilde A^\\mu+\\frac{1}{2}\\partial_\\mu h \\partial^\\mu h-\\lambda^2v^2h^2 \\\\\n&\\quad+g^2\\left(\\sqrt{2}vh+\\frac{h^2}{2}\\right)\\tilde A_\\mu\\tilde A^\\mu-\\frac{\\lambda^2v}{\\sqrt{2}}h^3-\\frac{\\lambda^2}{8}h^4 ,\n\\end{aligned}\n$$\nso that\n$$\nm_A=\\sqrt2\\,gv,\\qquad m_h=\\sqrt2\\,\\lambda v,\\qquad \\mathcal L\\supset \\frac{m_A^2}{2}\\tilde A_\\mu\\tilde A^\\mu-\\frac{m_h^2}{2}h^2+g\\,m_A\\,h\\,\\tilde A_\\mu\\tilde A^\\mu+\\cdots,\n$$\nwhere we used $\\sqrt2 g^2v=g\\,m_A$. The equation of motion of $h$, keeping the lowest-order interaction, is\n$$\n(\\Box+m_h^2)h = g\\,m_A\\,\\tilde{A}_\\mu\\tilde{A}^\\mu .\n$$\nFor $E\\ll m_h$ we can neglect $\\Box$ compared with $m_h^2$, and integrating out $h$ at tree level gives\n$$\nh\\simeq\\frac{g\\,m_A}{m_h^2}\\,\\tilde{A}_\\mu\\tilde{A}^\\mu .\n$$\nInserting this back into the Lagrangian, the terms $-\\frac{m_h^2}{2}h^2+g\\,m_A\\,h\\,\\tilde A^2$ give $\\frac{g^2m_A^2}{2m_h^2}(\\tilde A_\\mu\\tilde A^\\mu)^2$. The $h^2\\tilde A^2$ term and the Higgs self-interactions only contribute at order $\\tilde A^6$ and higher. Hence\n$$\n\\mathcal{L} = -\\frac{1}{4}\\tilde F_{\\mu\\nu}\\tilde F^{\\mu \\nu}+\\frac{m_A^2}{2}\\tilde{A}_\\mu\\tilde{A}^\\mu+\\frac{g^2m_A^2}{2m_h^2}(\\tilde{A}_\\mu\\tilde{A}^\\mu)^2+\\mathcal{O}\\big(\\tilde A^6\\big) .\n$$\nComparing with $\\mathcal{L}_{\\text{Proca}}$ (with $m=m_A$), we have\n$$\n\\frac{\\xi^4}{4} = \\frac{g^2m_A^2}{2m_h^2}\n$$\nThus\n$$\n\\xi^2= \\sqrt{2}\\,g\\frac{m_A}{m_h}.\n$$\nExpressed in terms of the physical masses, this result does not depend on how the potential or $\\rho$ is normalized. With the conventions of the previous problem ($\\rho=\\frac{v+h}{\\sqrt2}$, $m_A=gv$, $m_h=\\lambda v$, cubic coupling $g^2v\\,h\\tilde A^2$), one again finds $\\frac{\\xi^4}{4}=\\frac{g^4v^2}{2m_h^2}=\\frac{g^2m_A^2}{2m_h^2}$.", - "specific_requirement": "When judging the answer of this question, note that here $m_A=\\sqrt{2}gv$ and $m_h=\\sqrt{2}\\lambda v$; equivalent forms obtained using these relations are also correct." - }, - { - "num": 29, - "question": "The Higgs Yukawa coupling to fermions is given by\n$$\n\\mathcal{L}_Y =\\frac{g}{2}\\frac{m_f}{M_W}h\\bar{f}f .\n$$\nFind the decay rate $\\Gamma$ for the Higgs into a fermion-antifermion pair at tree level. Take the mass of $h$ to be $m_h$. Account for the situation $f=\\text{quark}$; the number of colors can be written as $N_c$.", - "category": "Electroweak", - "answer": "$\\Gamma=m_h \\frac{N_c g^2m_f^2}{32\\pi M_W^2}\\left(1-\\frac{4m_f^2}{m_h^2}\\right)^{\\frac{3}{2}}$", - "solution": "To keep the expression short we introduce $y=\\frac{g}{2}\\frac{m_f}{M_W}$; the tree-level amplitude reads\n$$\n\\mathcal{M} = y\\bar{u}_{s_1}(\\vec{p}_1)v_{s_2}(\\vec{p}_2)\n$$\nwith $\\{\\vec{p}_i,s_i\\}$ the momenta and spins of the outgoing fermion and anti-fermion. Then\n$$\n\\begin{aligned}\n\\sum_{\\text{spin}} |\\mathcal{M}|^2 &=y^2 \\sum_{\\text{spin}} \\bar{u}_{s_1}(\\vec{p}_1)\\,v_{s_2}(\\vec{p}_2)\\,\\bar v_{s_2}(\\vec{p}_2)\\,u_{s_1}(\\vec{p}_1) \\\\\n&=y^2\\text{Tr}[(\\slashed{p}_2-m_f)(\\slashed{p}_1+m_f)] =y^2\\big(\\text{Tr}[\\slashed{p}_2 \\slashed{p}_1]-4m_f^2\\big) =4y^2(p_1 \\cdot p_2-m_f^2)\n\\end{aligned}\n$$\nIn the center-of-mass frame,\n$$\nk^\\mu=(m_h,\\vec{0}), \\quad p^\\mu_1 = \\left(\\frac{m_h}{2},\\vec{p}\\right), \\quad p^\\mu_2 = \\left(\\frac{m_h}{2},-\\vec{p}\\right)\n$$\nFrom 4-momentum conservation,\n$$\nm_h=2E_f, \\quad E_f^2 = \\vec{p}^2+m_f^2, \\quad p_1\\cdot p_2=\\frac{m_h^2}{2}-m_f^2 .\n$$\nThen\n$$\n\\sum_{\\text{spin}} |\\mathcal{M}|^2=4y^2\\left(\\frac{m_h^2}{2}-2m_f^2\\right)\n$$\nHaving computed the amplitude, the decay rate (summed over the $N_c$ colours of the final-state quark) is\n$$\n\\Gamma = \\frac{N_c}{8\\pi m_h^2}|\\vec{p}|\\sum_{\\text{spin}}|\\mathcal{M}|^2 .\n$$\nSince $|\\vec{p}|=\\frac{m_h}{2}\\left(1-\\frac{4m_f^2}{m_h^2}\\right)^{\\frac{1}{2}}$ and $\\sum_{\\text{spin}}|\\mathcal M|^2=2y^2m_h^2\\big(1-\\frac{4m_f^2}{m_h^2}\\big)$, we get $\\Gamma=\\frac{N_c\\,y^2m_h}{8\\pi}\\big(1-\\frac{4m_f^2}{m_h^2}\\big)^{3/2}$, and therefore\n$$\n\\Gamma=m_h \\frac{N_c g^2m_f^2}{32\\pi M_W^2}\\left(1-\\frac{4m_f^2}{m_h^2}\\right)^{\\frac{3}{2}}\n$$" - }, - { - "num": 30, - "question": "Assume that the mass matrices for the up- and down-type quarks have the following forms (in the basis of weak interaction eigenstates)\n$$\nM^{(u)}=\\begin{pmatrix}\nm_u & 0 & 0 \\\\\n0 & m_c & 0 \\\\\n0 & 0 & m_t\n\\end{pmatrix},\\quad\\text{and} \\quad M^{(d)}=m\\begin{pmatrix}\n1+a^2 & ab & 0 \\\\\nab & 1+b^2 & 0 \\\\\n0 & 0 & 1\n\\end{pmatrix}\n$$\nrespectively. Here $m_i$, $[i=u,c,t]$, is the mass of the respective quark flavor, $m$ is a parameter with dimensions of mass, and $a,b$ are real. The mixing angle is $\\theta_{\\text{mix}}$. Find the value of $\\sin^2(2\\theta_{\\text{mix}})$ in terms of $a,b$.", - "category": "Electroweak", - "answer": "$\\frac{4a^2b^2}{(a^2+b^2)^2}$", - "solution": "First we consider the CKM matrix $V_{CKM} = (L^u)^\\dagger L^d$, where $L^u$ and $L^d$ are the transformation matrices of $u_L$ and $d_L$ respectively. Since $M^{(u)}$ is already diagonal in the weak eigenbasis, the CKM matrix is just $L^d$ from diagonalizing $M^{(d)}$. Since $M^{(d)}$ is Hermitian, it can be diagonalized with just one unitary matrix, namely $V_{CKM}$. First, we find the eigenvalues of the matrix by considering\n$$\n\\det(M^{(d)}-\\lambda\\mathcal{I}_{3 \\times3})=(m-\\lambda)[(1+a^2+b^2)m-\\lambda](m-\\lambda)=0\n$$\nwhich means\n$$\n\\lambda_1=m, \\quad\\lambda_2 = m(1+a^2+b^2),\\quad\\lambda_3=m\n$$\nNext, we find the normalized eigenvectors corresponding to the different eigenvalues\n$$\n\\vec{e}_1 = (0,0,1)^T, \\quad\\vec{e}_2 = \\frac{1}{\\sqrt{a^2+b^2}}(a,b,0)^T,\\quad \\vec{e}_3 = \\frac{1}{\\sqrt{a^2+b^2}}(-b,a,0)^T\n$$\nFrom $M^{(d)} =V_{CKM}M_{\\text{diag}}^{(d)}V_{CKM}^\\dagger$ with\n$$\nM_{\\text{diag}}^{(d)}=\\begin{pmatrix}\n\\lambda_1 & 0 & 0 \\\\\n0 & \\lambda_2 & 0 \\\\\n0 & 0 & \\lambda_3\n\\end{pmatrix}\n$$\nwe have\n$$\nV_{CKM} = \\begin{pmatrix}\n0 & \\frac{a}{\\sqrt{a^2+b^2}} & \\frac{-b}{\\sqrt{a^2+b^2}} \\\\\n0 & \\frac{b}{\\sqrt{a^2+b^2}} & \\frac{a}{\\sqrt{a^2+b^2}} \\\\\n1 & 0 & 0\n\\end{pmatrix}\n$$\nComparing the above with the following rotation matrix\n$$\n\\begin{pmatrix}\n0 & \\cos\\theta_{\\text{mix}} & -\\sin\\theta_{\\text{mix}} \\\\\n0 & \\sin\\theta_{\\text{mix}} & \\cos\\theta_{\\text{mix}} \\\\\n1 & 0 & 0\n\\end{pmatrix}\n$$\nwe find that the mixing angle is $\\theta_{\\text{mix}}=\\arctan \\frac{b}{a}$, i.e.\n$$\n\\cos\\theta_{\\text{mix}}=\\frac{a}{\\sqrt{a^2+b^2}},\\qquad \\sin\\theta_{\\text{mix}}=\\frac{b}{\\sqrt{a^2+b^2}} .\n$$\nTherefore\n$$\n\\sin(2\\theta_{\\text{mix}})=2\\sin\\theta_{\\text{mix}}\\cos\\theta_{\\text{mix}}=\\frac{2ab}{a^2+b^2},\\qquad\n\\sin^2(2\\theta_{\\text{mix}})=\\frac{4a^2b^2}{(a^2+b^2)^2}.\n$$\nTwo eigenvalues are degenerate ($\\lambda_1=\\lambda_3=m$). As a result, the assignment of the mass eigenstates within the first-two-generation block, and hence the angle itself, is only fixed up to $\\theta_{\\text{mix}}\\to-\\theta_{\\text{mix}}$, $\\theta_{\\text{mix}}\\to\\frac{\\pi}{2}\\pm\\theta_{\\text{mix}}$ (e.g. $\\arctan\\frac ab$ instead of $\\arctan\\frac ba$). The combination $\\sin^2(2\\theta_{\\text{mix}})$ is invariant under all of these, which is why it is the convention-independent quantity asked for." - }, - { - "num": 31, - "question": "Consider an $O(N)$ $\\phi^4$ scalar model in the large $N$ limit. Suppose all two-point functions of said theory are proportional to the propagator and can hence be renormalized into the mass. How many connected Feynman diagrams would contribute to the four-point function $\\langle\\phi^i\\phi^i\\phi^j\\phi^j\\rangle$ at all orders up to two loops?", - "category": "SUSY", - "answer": "3", - "solution": "The $O(N)$ $\\phi^4$ scalar model has Lagrangian\n$$\n\\mathcal{L} = \\frac{1}{2} (\\partial_\\mu \\phi^i)^2 + \\frac{m^2}{2} \\phi^i \\phi^i + \\frac{\\lambda}{2N} (\\phi^i \\phi^i)^2\n$$\nin which the index $i$ is summed over.\n\nAt tree level, only 1 cross diagram contributes to the 4-point function.\n\nAt 1-loop level, only 1 bubble diagram in the $s$ channel contributes to the 4-point function, with the index $k$ of the internal propagators summed over. The bubble diagrams in the $t$ and $u$ channels, however, have their internal propagators fixed with indices $i$ and $j$, and are therefore suppressed in the large $N$ limit. Any other loop diagram with a tadpole attached to the cross diagram is renormalized into the cross.\n\nAt 2-loop level, only 1 double bubble diagram in the $s$ channel contributes to the 4-point function, in which $k$ and $l$ are summed over. Similar to the 1-loop case, the $t$ and $u$ channel diagrams are suppressed in the large $N$ limit. Any other diagram with a tadpole attached to the one-loop bubble is again renormalized into the one-loop bubble diagram. In total, $1+1+1=3$ connected diagrams contribute up to two loops at leading order in $1/N$. (Counting powers of $N$: each vertex carries $\\lambda/N$ and each closed index loop gives a factor $N$, so the chain of $L$ $s$-channel bubbles scales as $N^{-(L+1)}N^{L}=N^{-1}$, the same order as the tree diagram, while every other topology is suppressed by further powers of $1/N$.)" - }, - { - "num": 32, - "question": "Suppose the four-point correlator $\\langle O_\\Delta(x_1)O_\\Delta(x_2)O_\\Delta(x_3)O_\\Delta(x_4)\\rangle$ of a conformal field theory can be written in spectral representation as\n$$\n\\int_0^\\infty \\mathrm{d}\\nu\\, B^{\\text{free}}(\\nu)\\, g^{\\Delta,\\Delta,\\Delta,\\Delta}_{\\nu}(x_1,x_2,x_3,x_4)+\\int_0^\\infty \\mathrm{d}\\nu\\, B^{\\text{int}}(\\nu)\\, g^{\\Delta,\\Delta,\\Delta,\\Delta}_{\\nu}(x_1,x_2,x_3,x_4),\n$$\nin which $B^{\\text{free}}(\\nu)$ is the spectral function of the free theory while $B^{\\text{int}}(\\nu)$ accounts for the interaction. $g^{\\Delta,\\Delta,\\Delta,\\Delta}_{\\nu}(x_1,x_2,x_3,x_4)$ is the conformal block with four external conformal weights $\\Delta$ and internal double-trace operator with weight $\\nu$. For each $n\\in\\mathbb{N}$, $B^{\\text{free}}(\\nu)$ has a single pole at $\\nu=\\nu_n$, while $B^{\\text{int}}$ has a double pole at $\\nu=\\nu_n$. $g^{\\Delta,\\Delta,\\Delta,\\Delta}_{\\nu}(x_1,x_2,x_3,x_4)$ is holomorphic. Write down an expression for the anomalous dimension that arises from turning on the interaction on top of the free theory, for a double-trace operator with conformal weight $\\nu_n$. Express the result in terms of $B^{\\text{free}}(\\nu)$ and $B^{\\text{int}}(\\nu)$.", - "category": "SUSY", - "answer": "$\\frac{B^{\\text{int}}_{-2}(\\nu_n)}{\\text{Res}B^{free}(\\nu_n)}$ (it is OK to differ by a factor of any combination of $2\\pi$, $-1$ and $i$)", - "solution": "First we consider only the free theory and compare the spectral representation to the conformal block expansion:\n$$\n\\begin{aligned}\n&\\langle O_\\Delta(x_1)O_\\Delta(x_2)O_\\Delta(x_3)O_\\Delta(x_4)\\rangle^{\\text{free}} \\\\\n=&\\int_0^\\infty \\mathrm{d}\\nu\\, B^{\\text{free}}(\\nu) g^{\\Delta,\\Delta,\\Delta,\\Delta}_{\\nu}(x_1,x_2,x_3,x_4) \\\\\n=&\\sum_{n} 2\\pi i\\,\\text{Res} \\left( B^{\\text{free}}(\\nu) g^{\\Delta,\\Delta,\\Delta,\\Delta}_{\\nu}(x_1,x_2,x_3,x_4),\\nu=\\nu_n \\right) \\\\\n=&\\sum_{\\nu_n} A(\\nu_n)g^{\\Delta,\\Delta,\\Delta,\\Delta}_{\\nu_n}(x_1,x_2,x_3,x_4)\n\\end{aligned}\n$$\nin which the contour was closed to pick up the poles of $B^{\\text{free}}$ (the overall sign and factors of $2\\pi i$ depend on the orientation convention, which is why the answer is only fixed up to such factors), $A(\\nu_n)$ is the OPE coefficient (squared) of the double-trace operator with dimension $\\nu_n$, and $g^{\\Delta,\\Delta,\\Delta,\\Delta}_{\\nu_n}$ is the conformal block of that operator. We therefore read off\n$$\nA(\\nu_n)=2\\pi i\\,\\text{Res}B^{free}(\\nu_n).\n$$\nNow turn on the interaction. To first order, each double-trace operator gets a shifted dimension $\\nu_n+\\gamma(\\nu_n)$ and a shifted OPE coefficient $A(\\nu_n)+A^{(1)}(\\nu_n)$. Expanding the full conformal block expansion to first order,\n$$\n\\sum_n \\big(A+A^{(1)}\\big)\\,g_{\\nu_n+\\gamma}=\\sum_n\\Big(A(\\nu_n)\\,g_{\\nu_n}+A^{(1)}(\\nu_n)\\,g_{\\nu_n}+\\gamma(\\nu_n)A(\\nu_n)\\,g'_{\\nu_n}\\Big)+\\dots,\n$$\nwhere $g'_{\\nu}\\equiv\\partial_\\nu g^{\\Delta,\\Delta,\\Delta,\\Delta}_{\\nu}$. The first term is the free-theory part, so the interaction must account for the remaining two:\n$$\n\\begin{aligned}\n&\\langle O_\\Delta(x_1)O_\\Delta(x_2)O_\\Delta(x_3)O_\\Delta(x_4)\\rangle^{\\text{int}} \\\\\n=&\\int_0^\\infty \\mathrm{d}\\nu\\, B^{\\text{int}}(\\nu) g^{\\Delta,\\Delta,\\Delta,\\Delta}_{\\nu}(x_1,x_2,x_3,x_4) \\\\\n=&\\sum_{\\nu_n} \\left( A^{(1)}(\\nu_n)g^{\\Delta,\\Delta,\\Delta,\\Delta}_{\\nu_n} +\\gamma(\\nu_n)A(\\nu_n)g'^{\\Delta,\\Delta,\\Delta,\\Delta}_{\\nu_n}\\right)\n\\end{aligned}\n$$\nOn the other hand, we can evaluate the spectral integral as Laurent series coefficients of $B^{\\text{int}}(\\nu)$ and $g^{\\Delta,\\Delta,\\Delta,\\Delta}_{\\nu}$:\n$$\n\\begin{aligned}\n&\\sum_{n} 2\\pi i\\,\\text{Res} \\left(B^{\\text{int}}(\\nu) g^{\\Delta,\\Delta,\\Delta,\\Delta}_{\\nu}, \\nu=\\nu_n \\right) \\\\\n=&\\sum_{n} 2\\pi i\\left[ B^{\\text{int}}_{-1}(\\nu_n)\\, g^{\\Delta,\\Delta,\\Delta,\\Delta}_{\\nu_n} + B^{\\text{int}}_{-2}(\\nu_n)\\, g'^{\\Delta,\\Delta,\\Delta,\\Delta}_{\\nu_n}\\right] ,\n\\end{aligned}\n$$\nin which $B^{\\text{int}}_{-k}(\\nu_n)$ is the coefficient of $(\\nu-\\nu_n)^{-k}$ in the Laurent series of $B^{\\text{int}}(\\nu)$ around $\\nu_n$. Since $g$ is holomorphic, the residue of the double pole produces the derivative $g'_{\\nu_n}$. Matching the coefficients of $g_{\\nu_n}$ and $g'_{\\nu_n}$ gives $A^{(1)}(\\nu_n)=2\\pi i\\,B^{\\text{int}}_{-1}(\\nu_n)$ and $\\gamma(\\nu_n)A(\\nu_n)=2\\pi i\\,B^{\\text{int}}_{-2}(\\nu_n)$. Hence\n$$\n\\gamma(\\nu_n) = \\frac{2\\pi i\\, B^{\\text{int}}_{-2}(\\nu_n)}{A(\\nu_n)} = \\frac{B^{\\text{int}}_{-2}(\\nu_n)}{\\text{Res}\\,B^{free}(\\nu_n)}.\n$$", - "specific_requirement": "When judging the answer of this question, the numerator must be the coefficient of the second-order pole; only differences by factors of $2\\pi$, $-1$ and $i$ are allowed. Read the expression literally." - }, - { - "num": 33, - "question": "Consider two anticommuting CFT fields $b(z)$ and $c(z)$,\n$$\nb(z)c(w) = -c(w)b(z), \\quad c(z)c(w) = -c(w)c(z), \\quad b(z)b(w) = -b(w)b(z)\n$$\nwith the following OPE between them\n$$\n\\langle c(z)b(w) \\rangle = \\frac{1}{z-w}+\\cdots, \\qquad \\langle b(z)b(w)\\rangle=\\cdots, \\qquad \\langle c(z)c(w) \\rangle=\\cdots\n$$\nHere the dots indicate regular terms. Furthermore, the energy-momentum tensor of the system reads\n$$\nT^{bc}(z) = -\\lambda:b\\partial c(z):+(\\lambda-1):c\\partial b(z):\n$$\nwhere normal ordering is explicitly indicated. Derive the central charge $c_{\\text{central charge}}$ of the system by considering the OPE between two energy-momentum tensors $T^{bc}$.", - "category": "SUSY", - "answer": "$-3(2\\lambda -1)^2+1$", - "solution": "The central charge is read off from the fourth-order pole, $T(z)T(w)\\sim\\frac{c/2}{(z-w)^4}+\\dots$, which comes from the double contractions. Write $T^{bc}=-\\lambda\\,\\mathcal B+(\\lambda-1)\\,\\mathcal C$ with $\\mathcal B\\equiv:b\\partial c:$ and $\\mathcal C\\equiv:c\\partial b:$. Then\n$$\nT^{bc}(z) T^{bc}(w) = \\lambda^2\\,\\mathcal B(z)\\mathcal B(w)-\\lambda(\\lambda-1)\\big[\\mathcal B(z)\\mathcal C(w)+\\mathcal C(z)\\mathcal B(w)\\big]+(\\lambda-1)^2\\,\\mathcal C(z)\\mathcal C(w).\n$$\nFrom $\\langle c(z)b(w) \\rangle = \\frac{1}{z-w}$ and the anticommutation, $\\langle b(z)c(w) \\rangle = -\\langle c(w)b(z)\\rangle=\\frac{1}{z-w}$ as well. Taking derivatives gives the needed contractions:\n$$\n\\langle \\partial c(z)b(w) \\rangle =\\langle \\partial b(z)c(w) \\rangle=-\\frac{1}{(z-w)^2},\\qquad\n\\langle c(z)\\partial b(w) \\rangle =\\langle b(z)\\partial c(w) \\rangle=\\frac{1}{(z-w)^2},\n$$\n$$\n\\langle \\partial c(z)\\partial b(w) \\rangle =\\langle \\partial b(z)\\partial c(w) \\rangle=\\partial_z\\partial_w \\frac{1}{z-w}=-\\frac{2}{(z-w)^3}.\n$$\nFor fermionic fields the full contraction of $:A_1A_2(z)::A_3A_4(w):$ is $-\\langle A_1A_3\\rangle\\langle A_2A_4\\rangle+\\langle A_1A_4\\rangle\\langle A_2A_3\\rangle$, where only $b$-$c$ pairs contract. Hence\n$$\n\\begin{aligned}\n\\mathcal B(z)\\mathcal B(w) &\\sim +\\langle b(z)\\partial c(w)\\rangle\\langle \\partial c(z)b(w)\\rangle=-\\frac{1}{(z-w)^4},\\\\\n\\mathcal C(z)\\mathcal C(w) &\\sim +\\langle c(z)\\partial b(w)\\rangle\\langle \\partial b(z)c(w)\\rangle=-\\frac{1}{(z-w)^4},\\\\\n\\mathcal B(z)\\mathcal C(w) &\\sim -\\langle b(z)c(w)\\rangle\\langle \\partial c(z)\\partial b(w)\\rangle=\\frac{2}{(z-w)^4},\\\\\n\\mathcal C(z)\\mathcal B(w) &\\sim -\\langle c(z)b(w)\\rangle\\langle \\partial b(z)\\partial c(w)\\rangle=\\frac{2}{(z-w)^4}.\n\\end{aligned}\n$$\nCollecting the terms,\n$$\nT^{bc}(z)T^{bc}(w) \\sim \\frac{-\\lambda^2-4\\lambda(\\lambda-1)-(\\lambda-1)^2}{(z-w)^4}+\\dots=\\frac{-6\\lambda^2+6\\lambda-1}{(z-w)^4}+\\dots ,\n$$\nso that\n$$\nc_{\\text{central charge}}=2(-6\\lambda^2+6\\lambda-1)=-3(2\\lambda-1)^2+1 .\n$$\nAs checks, $\\lambda=2$ (the reparametrization ghosts) gives $c=-26$, and $\\lambda=\\frac12$ gives $c=1$." - }, - { - "num": 34, - "question": "Consider two anticommuting CFT fields $b(z)$ and $c(z)$,\n$$\nb(z)c(w) = -c(w)b(z), \\quad c(z)c(w) = -c(w)c(z), \\quad b(z)b(w) = -b(w)b(z)\n$$\nwith the following OPE between them\n$$\n\\langle c(z)b(w) \\rangle = \\frac{1}{z-w}+\\cdots, \\qquad \\langle b(z)b(w)\\rangle=\\cdots, \\qquad \\langle c(z)c(w) \\rangle=\\cdots\n$$\nHere the dots indicate regular terms. Furthermore, the energy-momentum tensor of the system reads\n$$\nT^{bc}(z) = -2:b\\partial c(z):+:c\\partial b(z):\n$$\nwhere normal ordering is explicitly indicated. The modes of the fields $c(z)$ and $b(z)$ are given by\n$$\nc_n = \\oint_{C_0} \\frac{dz}{2\\pi i} z^{n-2}c(z), \\quad b_n = \\oint_{C_0} \\frac{dz}{2\\pi i} z^{n+1}b(z)\n$$\nwhere $C_0$ is a counter-clockwise contour around the origin. Determine the anticommutator between the modes $\\{c_n,b_m\\}$ using the OPE.", - "category": "SUSY", - "answer": "$\\{c_n,b_m\\} = \\delta_{m+n,0}$", - "solution": "$$\n\\begin{aligned}\n\\{c_n,b_m\\} &= \\left\\{\\oint_{C_0} \\frac{dz}{2\\pi i} z^{n-2}c(z),\\oint_{C_0} \\frac{dw}{2\\pi i} w^{m+1}b(w)\\right\\} \\\\\n&=\\oint_{C_0} \\frac{dw}{2\\pi i} w^{m+1}\\oint_{C_0} \\frac{dz}{2\\pi i} z^{n-2} \\{c(z),b(w)\\} \\\\\n&= \\oint_{C_0} \\frac{dw}{2\\pi i} w^{m+1}\\left( \\oint_{|z|>|w|}\\frac{dz}{2\\pi i} z^{n-2} c(z)b(w)+\\oint_{|z|<|w|}\\frac{dz}{2\\pi i} z^{n-2} b(w)c(z) \\right) \\\\\n&= \\oint_{C_0} \\frac{dw}{2\\pi i} w^{m+1}\\left( \\oint_{|z|>|w|}\\frac{dz}{2\\pi i} z^{n-2} c(z)b(w)-\\oint_{|z|<|w|}\\frac{dz}{2\\pi i} z^{n-2} c(z)b(w) \\right) \\\\\n&= \\oint_{C_0} \\frac{dw}{2\\pi i} w^{m+1} \\oint_{C_w}\\frac{dz}{2\\pi i} z^{n-2} c(z)b(w) \\\\\n&=\\oint_{C_0} \\frac{dw}{2\\pi i} w^{m+1} \\oint_{C_w}\\frac{dz}{2\\pi i} z^{n-2} \\left( \\frac{1}{z-w}+\\cdots\\right) \\\\\n&=\\oint_{C_0} \\frac{dw}{2\\pi i} w^{m+1} w^{n-2} = \\oint_{C_0} \\frac{dw}{2\\pi i} w^{m+n-1} = \\delta_{m+n,0}\n\\end{aligned}\n$$" - }, - { - "num": 35, - "question": "Consider a supermanifold $M$ with a sheaf that is locally isomorphic to $C^\\infty(\\mathbb{R}) \\otimes \\Lambda^\\cdot (\\mathbb{R}^2)$, with commutative coordinate $x$ and Grassmannian coordinates $\\theta_1$ and $\\theta_2$. Let $f(x,\\theta_1,\\theta_2)$ be a function on $M$. Make use of the Dirac delta function and write down the general expression of an integrable form that is associated with the function $f(x,\\theta_1,\\theta_2)$. Please put your final answer in a box and state it directly (without an equation). Omit any wedge product sign. Write the differentials of $x$ and $\\theta_i$ as $\\mathrm{d}x$ and $\\mathrm{d}\\theta_i$.", - "category": "SUSY", - "answer": "$f(x,\\theta_1,\\theta_2) \\text{d}x\\delta(d\\theta_1)\\delta(d\\theta_2)$", - "solution": "On a supermanifold the parity of the differentials is opposite to that of the coordinates. $\\mathrm{d}x$ is odd, so $(\\mathrm{d}x)^2=0$, while $\\mathrm{d}\\theta_1$ and $\\mathrm{d}\\theta_2$ are even (commuting) variables. A differential form, being polynomial in the even variables $\\mathrm{d}\\theta_i$, therefore cannot be integrated over them, and there is no top-degree differential form. The objects that can be integrated are integral forms. In these, the dependence on each even differential $\\mathrm{d}\\theta_i$ is through the Dirac delta distribution $\\delta(\\mathrm{d}\\theta_i)$, so that the integral over $\\mathrm{d}\\theta_i$ localizes; this also implies $\\mathrm{d}\\theta_i\\,\\omega=0$, since $\\mathrm{d}\\theta_i\\,\\delta(\\mathrm{d}\\theta_i)=0$. Combining this with the top power of the odd differential $\\mathrm{d}x$, the general integral form of top degree on the $(1|2)$-dimensional supermanifold associated with $f$ is\n$$\n\\omega=f(x,\\theta_1,\\theta_2)\\, \\mathrm{d}x\\,\\delta(\\mathrm{d}\\theta_1)\\,\\delta(\\mathrm{d}\\theta_2),\n$$\nand its integral reduces to the Berezin integral $\\int \\mathrm{d}x\\,\\mathrm{d}\\theta_1\\mathrm{d}\\theta_2\\, f$ (up to orientation conventions)." - }, - { - "num": 36, - "question": "Consider a primary field $\\phi(z,\\bar{z})$ of conformal weight $(h,\\bar{h})$. By considering infinitesimal conformal transformations $f(z)=z+\\epsilon(z)$ and $\\bar{f}(\\bar{z})=\\bar{z}+\\bar{\\epsilon}(\\bar{z})$ with $\\epsilon(z), \\bar{\\epsilon}(\\bar{z}) \\ll 1$, compute the corresponding infinitesimal transformation $\\delta_{\\epsilon,\\bar{\\epsilon}}\\phi(z,\\bar{z})$ of the primary field. Adopt the sign convention $\\delta_{a,\\bar{a}}\\phi=(a\\partial+\\bar{a}\\bar{\\partial})\\phi$ for constant translations $a, \\bar{a}$. Write your answer in fully expanded form and explicitly indicate the function on which each derivative acts.", - "category": "CFT", - "answer": "$$\n\\delta_{\\epsilon,\\bar{\\epsilon}}\\phi(z,\\bar{z})\n=\\epsilon(z)\\,\\partial_z\\phi(z,\\bar{z})\n+h\\,\\partial_z\\epsilon(z)\\,\\phi(z,\\bar{z})+\\bar{\\epsilon}(\\bar{z})\\,\\partial_{\\bar{z}}\\phi(z,\\bar{z})\n+\\bar{h}\\,\\partial_{\\bar{z}}\\bar{\\epsilon}(\\bar{z})\\,\\phi(z,\\bar{z}).\n$$", - "solution": "A primary field $\\phi(z, \\bar{z})$ with conformal weight $(h,\\bar{h})$ is defined by its transformation under a conformal map $z\\mapsto f(z), \\bar{z}\\mapsto \\bar{f}(\\bar{z})$ as\n$$\n\\phi(z,\\bar{z}) \\mapsto \\phi'(z,\\bar{z})=\\left(\\frac{\\partial f}{\\partial z}\\right)^{h} \\left(\\frac{\\partial \\bar{f}}{\\partial \\bar{z}}\\right)^{\\bar{h}}\\phi(f(z),\\bar{f}(\\bar{z}))\n$$\nUsing the Taylor expansions\n$$\n\\left(\\frac{\\partial f}{\\partial z}\\right)^{h}=(1+\\partial\\epsilon)^{h} \\approx 1 + h\\partial\\epsilon, \\qquad \\left(\\frac{\\partial \\bar{f}}{\\partial \\bar{z}}\\right)^{\\bar{h}}=(1+\\bar{\\partial}\\bar{\\epsilon})^{\\bar{h}} \\approx 1 + \\bar{h}\\bar{\\partial}\\bar{\\epsilon},\n$$\nand\n$$\n\\phi(f(z), \\bar{f}(\\bar{z})) \\approx \\phi(z,\\bar{z})+\\epsilon(z)\\partial\\phi(z,\\bar{z})+\\bar{\\epsilon}(\\bar{z})\\bar{\\partial}\\phi(z,\\bar{z}),\n$$\nthen, we get\n$$\n\\begin{aligned}\n\\delta_{\\epsilon,\\bar{\\epsilon}}\\phi(z,\\bar{z})&= \\phi'(z,\\bar{z})-\\phi(z,\\bar{z})=\\left(h\\partial\\epsilon(z)+\\bar{h}\\bar{\\partial}\\bar{\\epsilon}(\\bar{z})+\\epsilon(z)\\partial+\\bar{\\epsilon}(\\bar{z})\\bar{\\partial}\\right)\\phi(z,\\bar{z})\\\\\n&=\\epsilon(z)\\,\\partial_z\\phi(z,\\bar{z})\n+h\\,\\partial_z\\epsilon(z)\\,\\phi(z,\\bar{z})+\\bar{\\epsilon}(\\bar{z})\\,\\partial_{\\bar{z}}\\phi(z,\\bar{z})\n+\\bar{h}\\,\\partial_{\\bar{z}}\\bar{\\epsilon}(\\bar{z})\\,\\phi(z,\\bar{z}).\n\\end{aligned}\n$$", - "specific_requirement": "When judging the answer of this question, an answer written as $(\\epsilon\\partial+h\\partial\\epsilon+\\cdots)\\phi$ is wrong, because it cannot be determined which function each derivative acts on." - }, - { - "num": 37, - "question": "Consider a conformal field theory on flat spacetime in the absence of conformal anomalies. Let $T_{\\mu\\nu}$ be its energy-momentum tensor. Determine its trace $T^\\mu{}_{\\mu}$.", - "category": "CFT", - "answer": "$T^\\mu{}_{\\mu}=0$", - "solution": "For every continuous symmetry in a field theory, there exists a conserved current $j^{\\mu}$ satisfying $\\partial_{\\mu}j^{\\mu}=0$. In particular, for a theory invariant under conformal transformations $x^{\\mu} \\mapsto x^{\\mu}+\\epsilon^{\\mu}(x)$, the conserved current can be written as $j^{\\mu}=T^{\\mu\\nu}\\epsilon_{\\nu}$, where $T^{\\mu\\nu}$ is a symmetric tensor. For constant $\\epsilon^\\mu$, conservation of the current implies\n$$\n\\partial_{\\mu}j^{\\mu}= \\partial_{\\mu}(T^{\\mu\\nu}\\epsilon_{\\nu})=(\\partial_{\\mu}T^{\\mu\\nu})\\epsilon_{\\nu} = 0\n$$\nSince $\\epsilon_{\\nu}$ is constant and arbitrary, it follows that $\\partial_{\\mu}T^{\\mu\\nu}=0$. For more general conformal transformations $\\epsilon^{\\mu}(x)$, we have\n$$\n\\begin{aligned}\n\\partial_{\\mu}j^{\\mu}&= \\partial_{\\mu}(T^{\\mu\\nu}\\epsilon_{\\nu})=(\\partial_{\\mu}T^{\\mu\\nu})\\epsilon_{\\nu}+T^{\\mu\\nu}(\\partial_{\\mu}\\epsilon_{\\nu})\\\\\n&= 0 + T^{\\mu\\nu}(\\partial_{\\mu}\\epsilon_{\\nu})\\\\\n&= \\frac{1}{2}T^{\\mu\\nu}(\\partial_{\\mu}\\epsilon_{\\nu}+\\partial_{\\nu}\\epsilon_{\\mu})\\\\\n&= \\frac{1}{2}\\times\\frac{2}{d} T^{\\mu\\nu}\\eta_{\\mu\\nu}(\\partial_{\\lambda}\\epsilon^{\\lambda})\\\\\n&= \\frac{1}{d}T^{\\mu}{}_{\\mu}(\\partial_{\\lambda}\\epsilon^{\\lambda})=0\n\\end{aligned}\n$$\nWe use the fact that conformal transformations satisfy\n$$\n\\partial_{\\mu}\\epsilon_{\\nu}+\\partial_{\\nu}\\epsilon_{\\mu}=\\frac{2}{d} \\eta_{\\mu\\nu} (\\partial_{\\lambda}\\epsilon^{\\lambda})\n$$\nSince this equation has to be true for arbitrary $\\partial_{\\lambda}\\epsilon^{\\lambda}$, we conclude $T^\\mu{}_{\\mu}=0$." - }, - { - "num": 38, - "question": "Consider the Riemann sphere $\\mathbb{CP}^{1}:=\\mathbb{C}\\cup\\{\\infty\\}$. Möbius transformations act on it by\n$$\nz\\mapsto \\frac{az+b}{cz+d},\n\\qquad ad-bc\\neq 0.\n$$\nWhat is the global conformal group of the Riemann sphere? Select the correct option and answer with the corresponding letter only.\n\nA. $GL_2(\\mathbb{C})$\nB. $SL_2(\\mathbb{C})$\nC. $PGL_2(\\mathbb{C})\\cong PSL_2(\\mathbb{C})$\nD. $PSU(2)$", - "category": "CFT", - "answer": "$\\text{C}$", - "solution": "Every Möbius transformation of the Riemann sphere has the form\n$$\nf_g(z)=\\frac{az+b}{cz+d},\n\\qquad\ng=\n\\begin{pmatrix}\na&b\\\\\nc&d\n\\end{pmatrix}\\in GL_2(\\mathbb C).\n$$\nComposition of Möbius transformations corresponds to matrix multiplication $f_{g_1}\\circ f_{g_2}=f_{g_1g_2}$. Thus we obtain a group homomorphism\n$$\nGL_2(\\mathbb C)\\longrightarrow \\operatorname{Conf}(\\mathbb{CP}^1).\n$$\nThe kernel of this map consists of scalar matrices\n$$\ng=\n\\begin{pmatrix}\n\\lambda&0\\\\\n0&\\lambda\n\\end{pmatrix},\n\\qquad \\lambda\\in \\mathbb C^*.\n$$\nHence the kernel is $\\mathbb C^* I$, and therefore\n$$\n\\operatorname{Conf}(\\mathbb{CP}^1)\n\\simeq\nGL_2(\\mathbb C)/\\mathbb C^*\n=:\nPGL_2(\\mathbb C).\n$$\nMoreover, every class in $PGL_2(\\mathbb C)$ has a representative with determinant $1$, and within $SL_2(\\mathbb C)$ the remaining kernel is $\\{\\pm I\\}$. Thus\n$$\nPGL_2(\\mathbb C)\\simeq SL_2(\\mathbb C)/\\{\\pm I\\}\n=:\nPSL_2(\\mathbb C).\n$$\nTherefore the global holomorphic conformal group of the Riemann sphere is $PGL_2(\\mathbb C)$, or equivalently $PSL_2(\\mathbb C)$." - }, - { - "num": 39, - "question": "Let $|h\\rangle$ be a highest-weight state of the Virasoro algebra, satisfying\n$$\nL_0|h\\rangle=h|h\\rangle,\n\\qquad\nL_n|h\\rangle=0 \\quad (n>0),\n$$\nwith commutation relations\n$$\n[L_m,L_n]=(m-n)L_{m+n} +\\frac{c}{12}m(m^2-1)\\delta_{m+n,0}.\n$$\nSuppose that at level $2$ there is a nonzero singular (null) state of the form $\\left(L_{-1}^2+aL_{-2}\\right)|h\\rangle$. Determine the conformal weight $h$ in terms of the central charge $c$.", - "category": "CFT", - "answer": "$h=\\frac{1}{16}(5-c\\pm \\sqrt{(c-1)(c-25)})$", - "solution": "Let $|\\chi\\rangle=\\left(L_{-1}^2+aL_{-2}\\right)|h\\rangle$ be the null state; it must be annihilated by the positive Virasoro modes. In particular, we require $L_1|\\chi\\rangle=0$, $L_2|\\chi\\rangle=0$. First, acting with $L_1$, we use\n$$\n[L_1,L_{-2}]=3L_{-1},\n\\qquad\n[L_1,L_{-1}]=2L_0.\n$$\nSince $L_1|h\\rangle=0$, we find\n$$\nL_1L_{-2}|h\\rangle=3L_{-1}|h\\rangle, \\qquad L_1L_{-1}^2|h\\rangle=(4h+2)L_{-1}|h\\rangle\n$$\nTherefore,\n$$\nL_1|\\chi\\rangle\n=\n\\left[3a+(4h+2)\\right]L_{-1}|h\\rangle.\n$$\nSimilarly, acting with $L_{2}$, we obtain\n$$\nL_2|\\chi\\rangle\n=\n\\left[6h+\\left(4h+\\frac{c}{2}\\right)a\\right]|h\\rangle\n$$\nRequiring that $L_1|\\chi\\rangle$ and $L_2|\\chi\\rangle$ vanish, we have the relation\n$$\nh=\\frac{1}{16}(5-c\\pm \\sqrt{(c-1)(c-25)})\n$$" - }, - { - "num": 40, - "question": "Let $|h\\rangle$ be a normalized highest-weight state of the Virasoro algebra satisfying\n$$\n\\langle h|h\\rangle=1,\\qquad\nL_0|h\\rangle=h|h\\rangle,\\qquad\nL_n|h\\rangle=0\\quad(n>0).\n$$\nThe Virasoro generators satisfy\n$$\n[L_m,L_n]\n=\n(m-n)L_{m+n}\n+\\frac{c}{12}m(m^2-1)\\delta_{m+n,0},\n\\qquad m,n\\in\\mathbb Z.\n$$\nAssume that $L_n^\\dagger=L_{-n}$. In the ordered level-$2$ basis $|v_1\\rangle=L_{-2}|h\\rangle$, $|v_2\\rangle=L_{-1}^2|h\\rangle$, define the Gram matrix by $(M_2(h,c))_{ij}=\\langle v_i|v_j\\rangle$. Compute the level-$2$ Kac determinant $\\det M_2(h,c)$.", - "category": "CFT", - "answer": "$\\det M_{2}(h,c)=32h\\left(h^2-\\frac{5h}{8}+\\frac{hc}{8}+\\frac{c}{16}\\right)$", - "solution": "At level $2$, using the Virasoro algebra and the highest-weight conditions, we obtain\n$$\n\\langle h|L_2L_{-2}|h\\rangle\n=\n4h+\\frac{c}{2}, \\qquad \\langle h|L_2L_{-1}^2|h\\rangle\n=\n\\langle h|L_1^2L_{-2}|h\\rangle = 6h\n$$\nand\n$$\n\\langle h|L_1^2L_{-1}^2|h\\rangle\n=\n4h+8h^2.\n$$\nTherefore\n$$\nM_2(h,c)\n=\n\\begin{pmatrix}\n4h+\\frac{c}{2} & 6h \\\\\n6h & 4h+8h^2\n\\end{pmatrix}.\n$$\nIts determinant is\n$$\n\\det M_2(h,c)\n=\n\\left(4h+\\frac{c}{2}\\right)(4h+8h^2)-36h^2.\n$$\nSimplifying, we find\n$$\n\\det M_2(h,c)\n=\n32h\\left(\nh^2-\\frac{5h}{8}+\\frac{hc}{8}+\\frac{c}{16}\n\\right)\n$$" - }, - { - "num": 41, - "question": "Let $Q$ be a nilpotent symmetry, $Q^{2}=0$, and suppose the action of a field theory is $Q$-exact, $S_t=t\\{Q,V\\}$, where $t>0$ is a real parameter. Assume the path-integral measure is $Q$-invariant. Compute $\\frac{dZ(t)}{dt}$ for the partition function $Z(t)=\\int \\mathcal{D}X\\, e^{-S_t}$.", - "category": "Quantization", - "answer": "$\\frac{dZ(t)}{dt}=0$", - "solution": "For the partition function $Z(t)$, differentiating with respect to $t$ gives\n$$\n\\frac{dZ(t)}{dt}\n=\n-\\int \\mathcal{D}X\\, \\{Q,V\\}\\, e^{-t\\{Q,V\\}}=\\langle QV+VQ \\rangle =0 .\n$$" - }, - { - "num": 42, - "question": "Consider the path integral for a free particle of mass $m$ on the interval $0\\le t\\le T$, with fixed endpoints $x(0)=x(T)=0$. The propagator can be written as a Gaussian path integral\n$$\nK(0,T;0,0)\n=\n\\int_{x(0)=0}^{x(T)=0}\\mathcal{D}x(t)\\,\n\\exp\\left[\n\\frac{i}{2}\\int_0^T dt\\,dt'\\,\nx(t')A(t',t)x(t)\n\\right].\n$$\nStarting from the free-particle action\n$$\nS[x]=\\frac{m}{2}\\int_0^T dt\\,\\dot x^2(t),\n$$\nuse integration by parts to identify the kernel $A(t',t)$, or equivalently the corresponding differential operator $A$, and use its Dirichlet spectrum to compute the formal determinant $\\det A$. Report only the final expression for the formal determinant $\\det A$ in terms of $m$ and $T$, written as a product over the positive integer modes $n$.", - "category": "Quantization", - "answer": "$\\det A = \\prod_{n=1}^{\\infty}\\frac{m n^2\\pi^2}{T^2}$", - "solution": "For fixed endpoints $x(0)=x(T)=0$, integration by parts gives\n$$\nS[x]\n=\n\\frac{m}{2}\\int_0^T dt\\,\\dot x^2\n=\n-\\frac{m}{2}\\int_0^T dt\\,x(t)\\frac{d^2}{dt^2}x(t).\n$$\nThus the quadratic operator appearing in the Gaussian path integral is $A=-m\\frac{d^2}{dt^2}$. Equivalently, in kernel notation,\n$$\nA(t',t)=-m\\,\\delta(t'-t)\\frac{d^2}{dt^2}.\n$$\nWe now solve the eigenvalue problem\n$$\nAf_n(t)=\\lambda_n f_n(t),\n\\qquad\nf_n(0)=f_n(T)=0.\n$$\nThe normalized eigenfunctions are\n$$\nf_n(t)=\\sqrt{\\frac{2}{T}}\\sin\\left(\\frac{n\\pi t}{T}\\right),\n\\qquad n=1,2,\\ldots\n$$\nSince $-\\frac{d^2}{dt^2}f_n(t)=\\frac{n^2\\pi^2}{T^2}f_n(t)$, we get $\\lambda_n=\\frac{m n^2\\pi^2}{T^2}$. Therefore the formal determinant is the product of all eigenvalues\n$$\n\\det A\n=\n\\prod_{n=1}^{\\infty}\n\\frac{m n^2\\pi^2}{T^2}.\n$$" - }, - { - "num": 43, - "question": "Consider a theory described by the Lagrangian\n$$\n\\mathcal{L}=-\\frac{1}{4}F_{\\mu\\nu}F^{\\mu\\nu}+\\frac{m^{2}}{2}A_{\\mu}A^{\\mu}.\n$$\nFor a plane-wave solution of the form $A^{\\mu}=\\epsilon^{\\mu}e^{\\pm i p\\cdot x}$, where $\\epsilon^{\\mu}$ is the polarization vector, use the equation of motion to determine the constraint relating the momentum $p_\\mu$ and the polarization vector $\\epsilon^\\mu$.", - "category": "Classical", - "answer": "$p_\\mu\\epsilon^\\mu=0$", - "solution": "The equation of motion following from the Lagrangian is\n$$\n\\partial_\\mu\\partial^\\mu A^\\nu -\\partial^\\nu(\\partial_\\mu A^\\mu) +m^2 A^\\nu=0\n$$\nor equivalently,\n$$\n\\left[\\eta^{\\mu\\nu}(\\Box+m^2)-\\partial^\\mu\\partial^\\nu\\right]A_\\nu=0.\n$$\nApplying $\\partial_{\\mu}$ to the equation of motion and contracting the free Lorentz index $\\mu$, we obtain $m^2\\partial_\\nu A^\\nu=0$. For $m\\neq 0$, this implies the condition $\\partial_\\nu A^\\nu=0$. For a plane-wave solution $A^\\mu(x)=\\epsilon^\\mu e^{\\pm ip\\cdot x}$, this condition becomes $p_\\mu\\epsilon^\\mu=0$." - }, - { - "num": 44, - "question": "Consider a complex scalar field $\\phi$ with a global $U(1)$ phase symmetry generated by\n$$\nU_\\alpha=e^{-i\\alpha Q},\n\\qquad\nQ=i\\int d^3x\\,(\\pi_{\\phi^*}\\phi^*-\\pi_{\\phi}\\phi).\n$$\nHere, $\\pi_\\phi$ and $\\pi_{\\phi^*}$ are the canonical momenta conjugate to $\\phi$ and $\\phi^*$, respectively. Assume that the vacuum is invariant under this symmetry, $U_\\alpha|0\\rangle=|0\\rangle$. Determine the value of the correlation function $\\langle 0|\\phi(x)\\phi(x')|0\\rangle$.", - "category": "Quantization", - "answer": "$\\langle 0|\\phi(x)\\phi(x')|0\\rangle=0$", - "solution": "The charge $Q$ generates the global $U(1)$ symmetry. Using the canonical commutation relations, we find $[Q,\\phi(x)]=-\\phi(x)$. The Baker-Campbell-Hausdorff formula then gives\n$$\nU_\\alpha^\\dagger\\phi(x)U_\\alpha\n=\ne^{-i\\alpha}\\phi(x).\n$$\nSince the vacuum is invariant,\n$$\nU_\\alpha|0\\rangle=|0\\rangle,\n\\qquad\n\\langle0|U_\\alpha^\\dagger=\\langle0|,\n$$\nwe have\n$$\n\\begin{aligned}\n\\langle0|\\phi(x)\\phi(x')|0\\rangle\n&=\n\\langle0|\nU_\\alpha^\\dagger\\phi(x)\\phi(x')U_\\alpha\n|0\\rangle\\\\\n&=\ne^{-2i\\alpha}\n\\langle0|\\phi(x)\\phi(x')|0\\rangle.\n\\end{aligned}\n$$\nBecause this must hold for arbitrary $\\alpha$, it follows that $\\langle 0|\\phi(x)\\phi(x')|0\\rangle=0$." - }, - { - "num": 45, - "question": "Consider the one-loop renormalization group equation for the Higgs quartic coupling\n$$\nQ^2\\frac{d\\lambda}{dQ^2}\n=\n\\frac{3}{4\\pi^2}\\lambda^2 .\n$$\nAssume the boundary condition\n$$\n\\lambda(v^2)=\\frac{M_H^2}{2v^2}.\n$$\nDetermine the Landau pole scale $\\Lambda_c$, defined by the value of $Q$ at which $\\lambda(Q^2)$ diverges.", - "category": "Renormalization", - "answer": "$\\Lambda_{c}=v\\exp\\left(\\frac{4\\pi^2v^2}{3M_H^2}\\right)$", - "solution": "The renormalization group equation is\n$$\nQ^2\\frac{d\\lambda}{dQ^2}\n=\n\\frac{3}{4\\pi^2}\\lambda^2 .\n$$\nSeparating variables and integrating from $Q=v$ to $Q$, we get\n$$\n\\int_{\\lambda(v^2)}^{\\lambda(Q^2)}\n\\frac{d\\lambda}{\\lambda^2}\n=\n\\frac{3}{4\\pi^2}\n\\int_{v^2}^{Q^2}\\frac{dQ^2}{Q^2}.\n$$\nThus\n$$\n-\\frac{1}{\\lambda(Q^2)}\n+\n\\frac{1}{\\lambda(v^2)}\n=\n\\frac{3}{4\\pi^2}\\ln\\frac{Q^2}{v^2}.\n$$\nUsing the boundary condition $\\lambda(v^2)=\\frac{M_H^2}{2v^2}$, we obtain\n$$\n\\lambda(Q^2)\n=\n\\frac{1}{\n\\dfrac{2v^2}{M_H^2}\n-\n\\dfrac{3}{4\\pi^2}\\ln\\dfrac{Q^2}{v^2}\n}\n$$\nThe Landau pole occurs when the denominator vanishes\n$$\n\\frac{2v^2}{M_H^2}\n-\n\\frac{3}{4\\pi^2}\\ln\\frac{\\Lambda_c^2}{v^2}\n=0.\n$$\nHence\n$$\n\\Lambda_c\n=\nv\\exp\\left(\n\\frac{4\\pi^2v^2}{3M_H^2}\n\\right)\n$$" - }, - { - "num": 46, - "question": "In four-dimensional $\\mathcal{N}=1$ superspace, let $V$ be a real vector superfield in Wess-Zumino gauge and define the gauge field-strength superfield by $W_\\alpha:=-\\frac14 \\overline D^2 D_\\alpha V$, $\\overline D^2:=\\overline D^{\\dot\\beta}\\overline D_{\\dot\\beta}$, where\n$$\nV(x,\\theta,\\bar\\theta)\n=\n(\\theta\\sigma^\\mu\\bar\\theta)\\,V_\\mu(x)\n+\n(\\theta\\theta)\\,\\bar\\theta\\,\\bar\\lambda(x)\n+\n(\\bar\\theta\\bar\\theta)\\,\\theta\\,\\lambda(x)\n+\n\\frac{1}{2}(\\theta\\theta)(\\bar\\theta\\bar\\theta)\\,D(x),\n$$\n$$\nD_\\alpha\n:=\n\\partial_\\alpha\n+\ni(\\sigma^\\mu)_{\\alpha\\dot\\beta}\\,\n\\bar\\theta^{\\dot\\beta}\\partial_\\mu,\n\\qquad\n\\overline D_{\\dot\\alpha}\n:=\n-\\overline\\partial_{\\dot\\alpha}\n-\ni\\theta^\\beta(\\sigma^\\mu)_{\\beta\\dot\\alpha}\\partial_\\mu .\n$$\nCompute the derivative $\\overline D_{\\dot\\alpha}W_\\alpha$.", - "category": "SUSY", - "answer": "$\\overline D_{\\dot\\alpha}W_\\alpha=0$", - "solution": "Using $\\overline D^2=\\overline D_{\\dot\\beta}\\overline D^{\\dot\\beta}$, we have\n$$\n\\overline D_{\\dot\\alpha}W_\\alpha\n=\n-\\frac14\n\\overline D_{\\dot\\alpha}\n\\overline D_{\\dot\\beta}\n\\overline D^{\\dot\\beta}\nD_\\alpha V .\n$$\nThe operators $\\overline D_{\\dot\\alpha}$ anticommute:\n$$\n\\{\\overline D_{\\dot\\alpha},\\overline D_{\\dot\\beta}\\}=0 .\n$$\nHence the product of three $\\overline D$'s is totally antisymmetric in three dotted spinor indices. However, dotted spinor indices only take two values, so any completely antisymmetric object with three such indices must vanish. Therefore $\\overline D_{\\dot\\alpha}\\overline D_{\\dot\\beta}\\overline D^{\\dot\\beta}=0$. It follows that $\\overline D_{\\dot\\alpha}W_\\alpha=0$." - }, - { - "num": 47, - "question": "Consider an $SU(2)$ Yang-Mills theory whose Lie algebra $\\mathfrak{su}(2)$ has structure constants $f^{a}{}_{bc}=\\epsilon^{a}{}_{bc}$. Compute the quadratic Casimir $C_2(\\text{adj})$ in the adjoint representation, defined by $f^{acd}f^{b}{}_{cd}=C_2(\\text{adj})\\delta^{ab}$.", - "category": "Non-Abelian", - "answer": "$C_2(\\text{adj})=2$", - "solution": "For the Lie algebra $\\mathfrak{su}(2)$, the structure constants are $f^a{}_{bc}=\\epsilon^a{}_{bc}$. Therefore,\n$$\nf^{acd}f^b{}_{cd}\n=\n\\epsilon^{acd}\\epsilon^b{}_{cd}\n=\n2\\delta^{ab}.\n$$\nComparing this with the definition $f^{acd}f^b{}_{cd}=C_2(\\mathrm{adj})\\delta^{ab}$, we obtain $C_2(\\mathrm{adj})=2$." - }, - { - "num": 48, - "question": "Consider a non-Abelian Yang-Mills theory in Minkowski spacetime with metric signature $(-,+,+,+)$. Impose the axial gauge $A_3^a=0$ by adding the gauge-fixing term $\\mathcal{L}_{\\mathrm{gf}}=-\\frac{1}{2\\xi}(A_3^a)^2$ to the Yang-Mills Lagrangian $\\mathcal{L}=-\\frac14(F_{\\mu\\nu}^a)^2+\\cdots$. Derive the tree-level momentum-space gauge-field propagator $D_{\\mu\\nu}^{ab}(k)$ in the strict axial-gauge limit $\\xi\\to0$, assuming $k_3\\neq0$. Report only the final result, expressed explicitly in terms of $k_\\mu$, $k_3$, $\\eta_{\\mu\\nu}$, and $\\delta^{ab}$.", - "category": "Quantization", - "answer": "$D_{\\mu\\nu}^{ab}(k)=-\\frac{i}{k^2-i\\epsilon} \\left[ \\eta_{\\mu\\nu} + \\frac{k_\\mu k_\\nu}{k_3^2} - \\frac{k_\\mu \\delta_\\nu^3+k_\\nu \\delta_\\mu^3}{k_3} \\right]\\delta^{ab}$", - "solution": "To implement the axial gauge, we use a slightly generalized gauge condition\n$$\nA_3^a(x)=\\omega^a(x),\n$$\nwhere $\\omega^a(x)$ is an arbitrary auxiliary function. This condition is imposed in the path integral by inserting\n$$\n\\delta\\!\\left(A_3^a(x)-\\omega^a(x)\\right).\n$$\nWe then average over $\\omega^a$ with a Gaussian weight,\n$$\n\\int \\mathcal{D}\\omega^a\\,\n\\exp\\left[\n-\\frac{i}{2\\xi}\\int d^4x\\,(\\omega^a(x))^2\n\\right].\n$$\nAfter performing this average, the path integral acquires the gauge-fixing term\n$$\n\\mathcal{L}_{\\mathrm{gf}}\n=\n-\\frac{1}{2\\xi}(A_3^a)^2 .\n$$\nUsing the metric signature $(-,+,+,+)$, the quadratic part of the gauge-field Lagrangian in momentum space is\n$$\n\\mathcal{L}_2 = \\frac12 A_\\sigma^a \\left[ -k^2\\eta^{\\sigma\\mu} +k^\\sigma k^\\mu -\\frac{1}{\\xi}\\delta_3^\\sigma\\delta_3^\\mu \\right] \\delta^{ab} A_\\mu^b .\n$$\nThe propagator is obtained by inverting the operator in brackets. One finds\n$$\nD_{\\mu\\nu}^{ab}(k) = -\\frac{i}{k^2-i\\epsilon} \\left[ \\eta_{\\mu\\nu} + \\frac{(1+\\xi k^2)k_\\mu k_\\nu}{k_3^2} - \\frac{k_\\mu k_3\\delta_\\nu^3+k_\\nu k_3\\delta_\\mu^3}{k_3^2} \\right]\\delta^{ab}.\n$$\nIn the limit $\\xi\\to 0$, the Gaussian weight localizes the integral around $\\omega^a=0$. Therefore the gauge condition becomes $A^{a}_{3}=0$. In this strict axial-gauge limit, the propagator reduces to\n$$\nD_{\\mu\\nu}^{ab}(k) = -\\frac{i}{k^2-i\\epsilon} \\left[ \\eta_{\\mu\\nu} + \\frac{k_\\mu k_\\nu}{k_3^2} - \\frac{k_\\mu \\delta_\\nu^3+k_\\nu \\delta_\\mu^3}{k_3} \\right]\\delta^{ab}.\n$$" - }, - { - "num": 49, - "question": "Consider a non-Abelian Yang-Mills theory with gauge field $A_{\\mu}^{a}$ and field strength\n$$\nF_{\\mu\\nu}^a\n=\n\\partial_\\mu A_\\nu^a-\\partial_\\nu A_\\mu^a\n+g f^{abc}A_\\mu^b A_\\nu^c .\n$$\nChoose the convention $\\epsilon^{0123}=1$. The $\\theta$-term is given by $\\mathcal{L}_\\theta=\\epsilon^{\\mu\\nu\\lambda\\rho}F_{\\mu\\nu}^a F_{\\lambda\\rho}^a$. Show that $\\mathcal{L}_\\theta$ can be written locally as a total derivative $\\mathcal{L}_\\theta=\\partial_{\\mu}K^{\\mu}$, and determine the standard current $K^{\\mu}$ explicitly in terms of the gauge field $A^{a}_{\\mu}$ and field strength $F^{a}_{\\mu\\nu}$.", - "category": "Non-Abelian", - "answer": "$2\\epsilon^{\\mu\\nu\\lambda\\rho}\\left(A_\\nu^aF_{\\lambda\\rho}^a-\\frac{g}{3}f^{abc}A_\\nu^aA_\\lambda^bA_\\rho^c\\right)$", - "solution": "We want to write $\\epsilon^{\\mu\\nu\\lambda\\rho}F_{\\mu\\nu}^aF_{\\lambda\\rho}^a$ as a total derivative. A natural first guess is\n$$\nK_0^\\mu\n=\n2\\epsilon^{\\mu\\nu\\lambda\\rho}A_\\nu^aF_{\\lambda\\rho}^a .\n$$\nTaking the divergence gives\n$$\n\\partial_\\mu K_0^\\mu\n=\n2\\epsilon^{\\mu\\nu\\lambda\\rho}\n(\\partial_\\mu A_\\nu^a)F_{\\lambda\\rho}^a\n+\n2\\epsilon^{\\mu\\nu\\lambda\\rho}\nA_\\nu^a\\partial_\\mu F_{\\lambda\\rho}^a .\n$$\nAfter using\n$$\nF_{\\mu\\nu}^a\n=\n\\partial_\\mu A_\\nu^a-\\partial_\\nu A_\\mu^a\n+\ng f^{abc}A_\\mu^bA_\\nu^c\n$$\nand relabelling dummy indices, this becomes\n$$\n\\partial_\\mu K_0^\\mu\n=\n\\epsilon^{\\mu\\nu\\lambda\\rho}\nF_{\\mu\\nu}^aF_{\\lambda\\rho}^a\n+\n2g\\epsilon^{\\mu\\nu\\lambda\\rho}\nf^{abc}\nA_\\mu^aA_\\nu^b\\partial_\\lambda A_\\rho^c .\n$$\nThus the naive current gives the desired term plus an extra contribution. This extra contribution is itself a total derivative:\n$$\n2g\\epsilon^{\\mu\\nu\\lambda\\rho}\nf^{abc}\nA_\\mu^aA_\\nu^b\\partial_\\lambda A_\\rho^c\n=\n\\frac{2g}{3}\\partial_\\mu\n\\left(\n\\epsilon^{\\mu\\nu\\lambda\\rho}\nf^{abc}A_\\nu^aA_\\lambda^bA_\\rho^c\n\\right).\n$$\nTherefore the correct current is obtained by subtracting this extra piece\n$$\nK^\\mu\n=\n2\\epsilon^{\\mu\\nu\\lambda\\rho}\nA_\\nu^aF_{\\lambda\\rho}^a\n-\n\\frac{2g}{3}\n\\epsilon^{\\mu\\nu\\lambda\\rho}\nf^{abc}A_\\nu^aA_\\lambda^bA_\\rho^c.\n$$\nEquivalently,\n$$\nK^\\mu\n=\n2\\epsilon^{\\mu\\nu\\lambda\\rho}\n\\left(\nA_\\nu^aF_{\\lambda\\rho}^a\n-\n\\frac{g}{3}f^{abc}A_\\nu^aA_\\lambda^bA_\\rho^c\n\\right).\n$$\nWith this choice, $\\mathcal{L}_\\theta$ is a total derivative." - }, - { - "num": 50, - "question": "Consider a theory in $d$ spacetime dimensions with a Dirac fermion $\\psi$ coupled to a massive vector field $A_\\mu$ through the interaction term $-e A_\\mu \\bar{\\psi}\\gamma^\\mu \\psi$. At large momentum, the massive vector propagator scales as $\\mathcal{O}(1)$. Derive the superficial degree of divergence $D$ for a connected Feynman diagram with $E_{e}$ external fermion lines, $E_{A}$ external $A_{\\mu}$ lines, and $V$ vertices. Express $D$ in terms of $d, V, E_{e}$, and $E_{A}$.", - "category": "Renormalization", - "answer": "$D = d+\\frac{d-2}{2}V-\\frac{d-1}{2}E_e-\\frac{d}{2}E_A$", - "solution": "Let $E$ denote external lines and $I$ internal lines; we have $D=dL-I_e$ with $d$ the number of spacetime dimensions and $L$ the number of loops. For a connected diagram,\n$$\nL=I_e+I_A-V+1 .\n$$\nWe use $2V=2I_e+E_e$ and $V=2I_A+E_A$ to obtain\n$$\nI_e=V-\\frac{E_e}{2},\n\\qquad\nI_A=\\frac{V-E_A}{2}.\n$$\nSubstituting into $L$, we get\n$$\nL=1+\\frac{V}{2}-\\frac{E_e}{2}-\\frac{E_A}{2}.\n$$\nTherefore,\n$$\nD\n=dL-I_e\n=d\\left(1+\\frac{V}{2}-\\frac{E_e}{2}-\\frac{E_A}{2}\\right)\n-\\left(V-\\frac{E_e}{2}\\right).\n$$\nHence\n$$\nD\n=\nd+\\frac{d-2}{2}V\n-\\frac{d-1}{2}E_e\n-\\frac{d}{2}E_A\n$$" - }, - { - "num": 51, - "question": "Consider a connected vacuum diagram in a large-$N$ matrix gauge theory. Let $E$ be the number of propagators, $V$ the number of vertices, and $F$ the number of faces. Suppose that the diagram contributes a factor\n$$\nA \\sim (g^2)^E (g^2)^{-V} N^F .\n$$\nIntroduce the 't Hooft coupling\n$$\n\\lambda=g^2N,\n$$\nand suppose that the diagram can be drawn on a closed orientable surface of genus $h$. Since $\\log Z$ is given by the sum of all connected vacuum diagrams, determine the large-$N$ genus expansion of $\\log Z$ at fixed $\\lambda$, denoting the contribution from diagrams of genus $h$ as a function $f_{h}(\\lambda)$.", - "category": "Non-Abelian", - "answer": "$\\log Z = \\sum_{h=0}^{\\infty}N^{2-2h}f_h(\\lambda)$", - "solution": "The vacuum diagram has\n$$\nA \\sim (g^2)^E (g^2)^{-V} N^F=\\lambda^{L-1}N^{\\chi}=\\lambda^{L-1}N^{2-2h}\n$$\nwhere $L$ is the number of loops and $\\chi$ is the Euler characteristic. Since $\\log Z$ is the sum over connected vacuum diagrams, we sum over all genera $h$ and all loop numbers $L$:\n$$\n\\log Z =\n\\sum_{h=0}^{\\infty}\\sum_{L} c_{h,L}\\,\nN^{2-2h}\\lambda^{L-1},\n$$\nwhere $c_{h,L}$ denotes the total coefficient obtained by summing all connected diagrams of genus $h$ and loop number $L$. For fixed genus $h$, all dependence on the 't Hooft coupling $\\lambda$ can be collected into a function\n$$\nf_h(\\lambda)\n=\n\\sum_L c_{h,L}\\lambda^{L-1}.\n$$\nTherefore,\n$$\n\\log Z\n=\n\\sum_{h=0}^{\\infty}N^{2-2h}f_h(\\lambda).\n$$" - }, - { - "num": 52, - "question": "Consider a four-dimensional $\\mathcal{N}=1$ Wess-Zumino model with one chiral superfield $\\Phi$ and superpotential\n$$\nW(\\Phi)=\\frac{m}{2}\\Phi^2+\\frac{\\lambda}{3}\\Phi^3 ,\n$$\nwhere $m\\neq0$ and $\\lambda\\neq0$. Determine the values of the scalar field $\\phi$ corresponding to all classical supersymmetric vacua of the theory.", - "category": "SUSY", - "answer": "$\\phi=0, \\,\\, \\phi=-\\frac{m}{\\lambda}$", - "solution": "For a theory of chiral superfields, the scalar potential is determined by the superpotential:\n$$\nV(\\phi)=\\left|W'(\\phi)\\right|^2 .\n$$\nA supersymmetric vacuum must have zero energy, so it must satisfy $V(\\phi)=0$. For the given superpotential\n$$\nW(\\phi)=\\frac{m}{2}\\phi^2+\\frac{\\lambda}{3}\\phi^3,\n$$\nwe find\n$$\nW'(\\phi)=m\\phi+\\lambda\\phi^2\n=\\phi(m+\\lambda\\phi).\n$$\nThus the supersymmetric vacua are given by\n$$\n\\phi(m+\\lambda\\phi)=0.\n$$\nHence there are two solutions:\n$$\n\\phi=0,\n\\qquad\n\\phi=-\\frac{m}{\\lambda}.\n$$\nBoth vacua have zero energy and are degenerate supersymmetric vacua." - }, - { - "num": 53, - "question": "Consider a finite-dimensional supersymmetric multiplet of states with fixed four-momentum $p_\\mu$ and positive energy $p_0>0$. Let the supercharges satisfy\n$$\n\\{Q_\\alpha,\\overline Q_{\\dot\\beta}\\}\n=\n2(\\sigma^\\mu)_{\\alpha\\dot\\beta}P_\\mu .\n$$\nLet $(-1)^F$ be the fermion number operator, acting as\n$$\n(-1)^F|B\\rangle=|B\\rangle,\n\\qquad\n(-1)^F|F\\rangle=-|F\\rangle.\n$$\nDetermine the relation between the number of bosonic states $n_B$ and the number of fermionic states $n_F$ in a supersymmetric multiplet.", - "category": "SUSY", - "answer": "$n_B=n_F$", - "solution": "Since the supercharges map bosonic states to fermionic states and vice versa, they anticommute with the fermion number operator $(-1)^F$:\n$$\n\\{(-1)^F,Q_\\alpha\\}=0,\n\\qquad\n\\{(-1)^F,\\overline Q_{\\dot\\beta}\\}=0.\n$$\nNow consider the trace over a supersymmetric multiplet $\\operatorname{Tr}\\left[(-1)^F\\{Q_\\alpha,\\overline Q_{\\dot\\beta}\\}\\right]$. Using cyclicity of the trace and the anticommutation of $(-1)^F$ with the supercharges, we find $\\operatorname{Tr}\\left[(-1)^F\\{Q_\\alpha,\\overline Q_{\\dot\\beta}\\}\\right]=0$. On the other hand, the supersymmetry algebra gives\n$$\n\\{Q_\\alpha,\\overline Q_{\\dot\\beta}\\}\n=\n2(\\sigma^\\mu)_{\\alpha\\dot\\beta}P_\\mu .\n$$\nFor a multiplet of states with fixed momentum eigenvalue $p_\\mu$, this implies\n$$\n2(\\sigma^\\mu)_{\\alpha\\dot\\beta}p_\\mu\n\\operatorname{Tr}\\left[(-1)^F\\right]=0.\n$$\nThus $\\operatorname{Tr}\\left[(-1)^F\\right]=0$. We obtain\n$$\n\\operatorname{Tr}\\left[(-1)^F\\right]\n=\n\\sum_{\\text{bosons}}1-\\sum_{\\text{fermions}}1\n=\nn_B-n_F=0.\n$$\nTherefore, $n_B=n_F$." - }, - { - "num": 54, - "question": "Let $S_1$ and $S_2$ be superfields. A superfield transforms under supersymmetry as $\\delta S=i[\\epsilon Q+\\bar\\epsilon \\bar Q,S]$. Is the product $S_1S_2$ also a superfield? Justify your answer by computing $\\delta(S_1S_2)$. Please give the final answer as either \"yes\" or \"no\".", - "category": "SUSY", - "answer": "yes", - "solution": "Let the supersymmetry variation be $\\delta S=i[\\epsilon Q+\\bar\\epsilon \\bar Q,S]$. Then $\\delta(S_1S_2)=i[\\epsilon Q+\\bar\\epsilon\\bar Q,S_1S_2]$. Using the Leibniz rule for the supersymmetry generator, we get\n$$\n\\delta(S_1S_2)\n=\n(\\delta S_1)S_2+S_1(\\delta S_2).\n$$\nEquivalently,\n$$\n\\delta(S_1S_2)\n=\ni(\\epsilon Q+\\bar\\epsilon\\bar Q)(S_1S_2).\n$$\nThus $S_1S_2$ transforms in the same way as a superfield. Therefore, $S_{1}S_{2}$ is also a superfield." - }, - { - "num": 55, - "question": "Consider a renormalizable four-dimensional $\\mathcal{N}=1$ theory with a single chiral superfield\n$$\n\\Phi=\\phi+\\sqrt{2}\\theta\\psi+\\theta\\theta F .\n$$\nThe supersymmetric Lagrangian can be written in terms of a Kähler potential $K(\\Phi,\\Phi^\\dagger)$ and a superpotential $W(\\Phi)$ as\n$$\n\\mathcal{L}\n=\n\\left. K(\\Phi,\\Phi^\\dagger)\\right|_D\n+\n\\left(\n\\left. W(\\Phi)\\right|_F\n+\\text{h.c.}\n\\right).\n$$\nSuppose that $K(\\Phi,\\Phi^\\dagger)=\\Phi^\\dagger\\Phi$ and that no additional symmetries constrain the superpotential. Determine the most general holomorphic superpotential $W(\\Phi)$ allowed by power-counting renormalizability. Write your answer in fully expanded form using independent arbitrary complex coefficients $c_{k}$ for $\\Phi^{k}$, including the constant term $c_{0}$, with all numerical factors absorbed into the coefficients.", - "category": "SUSY", - "answer": "$W(\\Phi) = c_{0}+c_{1}\\Phi+c_{2}\\Phi^2+c_{3}\\Phi^3$", - "solution": "For a renormalizable theory in four dimensions, the Lagrangian must have mass dimension $4$. The lowest component of the chiral superfield is the scalar field. Therefore, from $\\Phi=\\phi+\\sqrt2\\,\\theta\\psi+\\theta\\theta F$, we assign $[\\Phi]=[\\phi]=1$. In four dimensions, the fermion has the standard mass dimension $[\\psi]=\\frac{3}{2}$. Since every term in the superfield expansion must have the same dimension, the term $\\theta\\psi$ implies that $\\theta$ has dimension $-1/2$. Thus\n$$\n[\\Phi]=[\\phi]=1,\\qquad\n[\\psi]=\\frac32,\\qquad\n[\\theta]=-\\frac12,\\qquad\n[F]=2.\n$$\nThe $D$-term requires $[K]\\leq 2$, while the $F$-term requires $[W]\\leq 3$. Therefore, the most general renormalizable superpotential is $W(\\Phi)=c_0+c_1\\Phi+c_2\\Phi^2+c_3\\Phi^3$, where $c_0,c_1,c_2,c_3$ are independent complex coefficients. The constant term $c_0$ is included although it does not affect the Lagrangian in global supersymmetry." - }, - { - "num": 56, - "question": "For a supersymmetric gauge theory, define the one-loop coefficient $b$ by\n$$\nb=\\frac{11}{6}T(\\mathrm{adj})\n-\\frac{1}{3}\\sum_f T(\\mathbf R_f)\n-\\frac{1}{6}\\sum_b T(\\mathbf R_b),\n$$\nwhere the sums run over the Weyl fermions and the complex bosons. In an $\\mathcal{N}=1$ supersymmetric gauge theory, the vector multiplet contains a gauge boson and a gaugino in the adjoint representation, while each chiral multiplet contains a Weyl fermion and a complex scalar transforming in the same representation $\\mathbf R_c$. Let $c$ label the chiral multiplets, so that $\\sum_{c}$ denotes the sum over all chiral multiplets. Determine the coefficient $b$ in terms of $T(\\mathrm{adj})$ and $T(\\mathbf R_c)$.", - "category": "SUSY", - "answer": "$b=\\frac{3}{2}T(\\mathrm{adj})-\\frac{1}{2}\\sum_c T(\\mathbf R_c)$", - "solution": "For an $\\mathcal{N}=1$ SUSY gauge theory, the vector multiplet contains a gauge boson $A_{\\mu}$ and a gaugino $\\lambda_{\\alpha}$, both transforming in the adjoint representation. Therefore its contribution is\n$$\nb_{\\text{vect}}=\\frac{11}{6}T(\\mathrm{adj})\n-\\frac{1}{3} T(\\mathrm{adj})=\\frac{3}{2}T(\\mathrm{adj})\n$$\nAn $\\mathcal{N}=1$ chiral multiplet consists of a complex scalar $Q$ and a Weyl fermion $\\psi_{\\alpha}$, both in the same representation $\\mathbf R_c$ of the gauge group. Hence each chiral multiplet contributes\n$$\n-\\frac{1}{3}T(\\mathbf R_c)\n-\\frac{1}{6} T(\\mathbf R_c)=-\\frac{1}{2}T(\\mathbf R_c).\n$$\nSumming over all chiral multiplets, we obtain\n$$\nb_{\\text{chi}}=-\\frac{1}{2}\\sum_c T(\\mathbf R_c).\n$$\nAdding all contributions, we get\n$$\nb=b_{\\text{vect}}+b_{\\text{chi}}=\\frac{3}{2}T(\\mathrm{adj})-\\frac{1}{2}\\sum_c T(\\mathbf R_c).\n$$" - }, - { - "num": 57, - "question": "For $\\mathcal{N}=1$ SQCD with gauge group $SU(N)$ and $N_f$ flavors $(\\mathcal{Q}_i,\\widetilde{\\mathcal{Q}}^{\\,i})$, assume that the IR theory is superconformal. Define the anomalous dimension by $\\Delta(\\mathcal{Q})=1+\\gamma(\\mathcal{Q})$. Determine $\\gamma(\\mathcal{Q})$.", - "category": "SUSY", - "answer": "$\\gamma(\\mathcal{Q})=\\frac{N_{f}-3N}{2N_{f}}$", - "solution": "We use the convention that the superspace coordinate $\\theta^{\\alpha}$ carries $R$-charge $+1$. We have the superfield $W_{\\alpha}$ with the expansion\n$$\nW_{\\alpha}=\\lambda_{\\alpha}+F_{\\alpha\\beta}\\theta^{\\beta}+\\cdots,\n$$\nWith this convention, we are forced to assign $R$-charge $1$ to the gaugino $\\lambda_{\\alpha}$ and $R$-charge $0$ to $F_{\\alpha\\beta}$. To make the $SU(N)^2U(1)_R$ anomaly vanish, the $R$-charge $r=R(\\psi_{\\alpha})=R(\\tilde{\\psi}_{\\alpha})$ of the fermion components $\\psi_{\\alpha},\\tilde{\\psi}_{\\alpha}$ of $\\mathcal{Q}, \\tilde{\\mathcal{Q}}$ should satisfy\n$$\nN+ r\\cdot 2 \\cdot \\frac{1}{2} \\cdot N_{f}=0,\n$$\nmeaning that $r=-\\frac{N}{N_{f}}$. Recall that the chiral superfield has the expansion\n$$\n\\mathcal{Q}=Q+ \\psi_{\\alpha}\\theta^{\\alpha}+\\cdots,\n$$\nso we obtain\n$$\nR(\\mathcal{Q})=R(\\tilde{\\mathcal{Q}})=r+1=1-\\frac{N}{N_{f}}.\n$$\nFor a chiral scalar operator in a four-dimensional $\\mathcal{N}=1$ SCFT, the scaling dimension and the superconformal $R$-charge satisfy $\\Delta=\\frac{3}{2}R$. Therefore, we have the scaling dimension\n$$\n\\Delta(\\mathcal{Q})=\\frac{3}{2}-\\frac{3N}{2N_{f}}.\n$$\nThen, by subtracting the classical dimension $\\Delta_{\\text{cl}}(\\mathcal{Q})=1$, we obtain the anomalous dimension\n$$\n\\gamma(\\mathcal{Q})=\\Delta(\\mathcal{Q})-\\Delta_{\\text{cl}}(\\mathcal{Q})=\\frac{N_{f}-3N}{2N_{f}}.\n$$" - }, - { - "num": 58, - "question": "Consider an $\\mathcal{N}=1$ supersymmetric gauge theory. Let the microscopic holomorphic coupling be\n$$\n\\tau_0=\\frac{4\\pi i}{g^2}+\\frac{\\theta_{\\mathrm{YM}}}{2\\pi},\n$$\nand define the dynamical scale by\n$$\n\\Lambda=\\mu e^{2\\pi i\\tau_0/b},\n\\qquad b\\in\\mathbb{Z}_{>0}.\n$$\nSuppose that the effective holomorphic coupling has the form\n$$\n\\tau_{\\mathrm{eff}}(\\Lambda,\\mu)\n=\n\\frac{b}{2\\pi i}\\log\\left(\\frac{\\Lambda}{\\mu}\\right)\n+f(\\Lambda,\\mu).\n$$\nUsing the theta-angle periodicity $\\theta_{\\mathrm{YM}}\\sim\\theta_{\\mathrm{YM}}+2\\pi$, and assuming that $f$ is holomorphic at $\\Lambda/\\mu=0$, determine the most general form of $\\tau_{\\mathrm{eff}}(\\Lambda,\\mu)$, using $(a_{0},a_{1},\\ldots)$ to denote any independent arbitrary coefficients.", - "category": "SUSY", - "answer": "$\\tau_{\\mathrm{eff}}(\\Lambda,\\mu)=\\frac{b}{2\\pi i }\\text{log}\\left(\\frac{\\Lambda}{\\mu}\\right)+\\sum_{k\\geq 0}a_{k}\\left(\\frac{\\Lambda}{\\mu}\\right)^{bk}$", - "solution": "The periodicity of the theta angle implies $\\tau_{0} \\rightarrow \\tau_{0} +1$. Under this transformation the dynamical scale transforms as $\\Lambda \\rightarrow e^{2\\pi i/b}\\Lambda$, and the logarithmic term changes as\n$$\n\\frac{b}{2\\pi i }\\log\\left(\\frac{\\Lambda}{\\mu}\\right) \\mapsto \\frac{b}{2\\pi i }\\log\\left(\\frac{e^{2\\pi i /b}\\Lambda}{\\mu}\\right)=\\frac{b}{2\\pi i }\\log\\left(\\frac{\\Lambda}{\\mu}\\right)+1.\n$$\nHence the logarithmic term already reproduces the required shift $\\tau_{\\mathrm{eff}} \\mapsto \\tau_{\\mathrm{eff}} + 1$. Therefore the remaining contribution must be invariant\n$$\nf(e^{2\\pi i /b}\\Lambda,\\mu)=f(\\Lambda,\\mu).\n$$\nSince $f$ is holomorphic at $\\frac{\\Lambda}{\\mu}=0$, it admits a power-series expansion\n$$\nf(\\Lambda,\\mu)=\\sum_{n \\geq 0}a_{n}\\left(\\frac{\\Lambda}{\\mu}\\right)^{n}.\n$$\nSubstituting this expansion into the invariance condition gives $a_{n}e^{2\\pi i n /b}=a_{n}$. Thus, whenever $a_{n}\\neq 0$, $e^{2\\pi i n /b}=1$ gives $n = kb$, $k= 0,1,\\cdots$, therefore\n$$\nf(\\Lambda,\\mu)=\\sum_{k\\geq 0}a_{k}\\left(\\frac{\\Lambda}{\\mu}\\right)^{bk}, \\qquad \\left(\\frac{\\Lambda}{\\mu}\\right)^{bk}= e^{2\\pi i \\tau_{0} k}.\n$$\nFor $k\\geq 1$, these terms have the characteristic form of $k$-instanton contributions. Since they are exponentially suppressed as $e^{-8\\pi^{2}k/g^{2}}$, they are non-perturbative and cannot arise from a perturbative loop expansion. We conclude that the most general effective holomorphic coupling is\n$$\n\\tau_{\\mathrm{eff}}(\\Lambda,\\mu)=\\frac{b}{2\\pi i }\\log\\left(\\frac{\\Lambda}{\\mu}\\right)+\\sum_{k\\geq 0}a_{k}\\left(\\frac{\\Lambda}{\\mu}\\right)^{bk}.\n$$" - }, - { - "num": 59, - "question": "Consider $\\mathcal{N}=1$ SQCD with gauge group $SU(N)$ and $N_{f}$ flavors. The NSVZ beta function is\n$$\n\\beta(g)=\\frac{-g^{3}}{16\\pi^{2}}\\frac{3N-N_{F}(1-\\gamma(g^{2}))}{1-\\frac{Ng^{2}}{8\\pi^{2}}},\n$$\nwith\n$$\n\\gamma(g^{2})=-\\frac{g^{2}}{8\\pi^{2}}\\frac{N^{2}-1}{N}+\\mathcal{O}(g^{4}).\n$$\nAssume\n$$\n\\frac{N_{F}}{N}=3-\\epsilon,\n\\qquad\n0<\\epsilon\\ll1.\n$$\nExpand $\\beta(g)$ through order $g^5$, keep only the terms needed at leading order in $\\epsilon$, and determine the leading-order expression for $g_*^2$, neglecting terms of order $\\epsilon^2$ and higher.", - "category": "SUSY", - "answer": "$g_*^2=\\frac{8\\pi^{2}}{3}\\frac{N}{N^{2}-1}\\epsilon$", - "solution": "Using $N_f=N(3-\\epsilon)$, we first expand the numerator of the NSVZ beta function:\n$$\n\\begin{aligned}\n3N-N_f\\bigl(1-\\gamma(g^2)\\bigr)\n&=\n3N-N_f\n-\\frac{N_f(N^2-1)}{8\\pi^2N}g^2\n+\\mathcal O(g^4)\\\\\n&=\nN\\epsilon\n-\\frac{(3-\\epsilon)(N^2-1)}{8\\pi^2}g^2\n+\\mathcal O(g^4).\n\\end{aligned}\n$$\nThe denominator gives\n$$\n\\frac{1}{1-\\frac{Ng^2}{8\\pi^2}}\n=\n1+\\frac{Ng^2}{8\\pi^2}\n+\\mathcal O(g^4).\n$$\nCombining these expansions, we obtain\n$$\n16\\pi^2\\beta(g)\n=\n-N\\epsilon g^3\n+\n\\frac{g^5}{8\\pi^2}\n\\left[\n3(N^2-1)-\\epsilon(2N^2-1)\n\\right]\n+\\mathcal O(g^7).\n$$\nSince the nonzero fixed point satisfies $g_*^2=\\mathcal O(\\epsilon)$, the term proportional to $\\epsilon g^5$ contributes only beyond leading order in $\\epsilon$. Therefore,\n$$\n16\\pi^2\\beta(g)\n=\n-N\\epsilon g^3\n+\n\\frac{3(N^2-1)}{8\\pi^2}g^5\n+\\mathcal O(g^7,\\epsilon g^5).\n$$\nTo find the nonzero fixed point, we set $\\beta(g_*)=0$, which gives\n$$\n-N\\epsilon\n+\n\\frac{3(N^2-1)}{8\\pi^2}g_*^2\n=0.\n$$\nHence,\n$$\ng_*^2\n=\n\\frac{8\\pi^2}{3}\n\\frac{N}{N^2-1}\\epsilon\n+\\mathcal O(\\epsilon^2).\n$$\nTherefore, $g_*$ is an infrared fixed point, namely the supersymmetric Banks-Zaks fixed point." - }, - { - "num": 60, - "question": "In pure $SU(2)$ Seiberg-Witten theory, suppose $a\\sim \\sqrt{2u}$, $a_{D} \\sim \\frac{2i}{\\pi}a\\log a$. Analytically continue once along the path $u\\mapsto e^{2\\pi i}u$. Compute the monodromy matrix $M_\\infty$ defined by\n$$\n\\begin{pmatrix}a_D\\\\a\\end{pmatrix}\n\\longmapsto\nM_\\infty\n\\begin{pmatrix}a_D\\\\a\\end{pmatrix}.\n$$", - "category": "SUSY", - "answer": "$M_{\\infty}=\\begin{pmatrix} -1 & 2\\\\ 0 & -1 \\end{pmatrix}$", - "solution": "Under a large loop around infinity $u\\mapsto e^{2\\pi i}u$, since $a\\sim \\sqrt{2u}$, then $a\\mapsto -a$. Now use the asymptotic formula $a_{D} \\sim \\frac{2i}{\\pi}a\\log a$; we have\n$$\na_{D}\\mapsto \\frac{2i}{\\pi}(-a)(\\log a+i\\pi)=-\\frac{2i}{\\pi}a\\log a+2a=-a_{D}+2a\n$$\nThus the vector transforms as\n$$\n\\begin{pmatrix}\na_D\\\\\na\n\\end{pmatrix}\n\\longmapsto\n\\begin{pmatrix}\n-a_D+2a\\\\\n-a\n\\end{pmatrix}.\n$$\nSo the monodromy matrix at infinity is\n$$\nM_\\infty=\n\\begin{pmatrix}\n-1 & 2\\\\\n0 & -1\n\\end{pmatrix}.\n$$" - }, - { - "num": 61, - "question": "In pure $SU(2)$ Seiberg-Witten theory, the Coulomb branch has a $\\mathbb{Z}_{2}$ symmetry under $u\\mapsto -u$. The moduli space has three singularities located at\n$$\nu=\\infty, \\qquad u=u_{+}, \\qquad u=u_{-}=-u_{+}\n$$\nAt $u=u_{+}$, a monopole with electromagnetic charge $(n_{m},n_{e})=(1,0)$ becomes massless, while at $u=u_{-}$, a dyon with charge $(n_{m},n_{e})=(1,-1)$ becomes massless. The central charge is $Z=n_{m}a_{D}+n_{e}a$. Using the period basis $(a_{D},a)^{T}$ and counterclockwise monodromy loops, compute the ordered pair of monodromy matrices $(M_{u_{+}},M_{u_{-}})$ associated with the two finite singularities $u_{+}$ and $u_{-}$.", - "category": "SUSY", - "answer": "$(M_{u_{+}},M_{u_{-}})=\\left(\\begin{pmatrix} 1 & 0\\\\ -2 & 1 \\end{pmatrix},\\,\\begin{pmatrix} -1 & 2\\\\ -2 & 3 \\end{pmatrix} \\right)$", - "solution": "If $u=\\infty$ were the only singularity, then the monodromy around infinity could be continuously deformed to a trivial contour in the finite $u$-plane. This would imply that the periods $(a_{D},a)^{T}$ have no nontrivial monodromy, so $a$ would be a good global coordinate on the whole moduli space. However, this is not true: the Seiberg-Witten periods have nontrivial monodromy, and the quantum moduli space must contain finite singularities where extra BPS states become massless. At the monopole singularity $u=u_{+}$, a monopole with charge $(n_{m},n_{e})=(1,0)$ becomes massless. Since the BPS central charge is $Z=n_{m}a_{D}+n_{e}a$, the massless condition implies $a_{D}=0$. Near the singularity $u=u_{+}$, the dual coupling behaves as\n$$\n\\tau_D=\\frac{da}{da_D}\n\\sim\n\\frac{i}{\\pi}\\log a_D.\n$$\nIntegrating, we obtain\n$$\na\n\\sim\na_{+}\n+\\frac{i}{\\pi}a_D\\log a_D,\n$$\nwhere $a_{+}=a(u_{+})$ is the finite value of $a$ at the monopole point. For $u-u_{+}\\mapsto e^{2\\pi i}(u-u_{+})$, we obtain\n$$\na_{D}\\mapsto a_{D}, \\qquad a\\mapsto a-2a_{D}\n$$\nSo the monodromy matrix at $u=u_{+}$ is\n$$\nM_{u_{+}}\n=\n\\begin{pmatrix}\n1 & 0\\\\\n-2 & 1\n\\end{pmatrix}\n$$\nFrom the large-$u$ asymptotic behavior, the monodromy at infinity is\n$$\nM_\\infty=\n\\begin{pmatrix}\n-1 & 2\\\\\n0 & -1\n\\end{pmatrix}.\n$$\nFor the chosen counterclockwise loops, the monodromies satisfy\n$$\nM_\\infty=M_{u_+}M_{u_-}.\n$$\nTherefore,\n$$\nM_{u_-}\n=\nM_{u_+}^{-1}M_\\infty\n=\n\\begin{pmatrix}\n-1 & 2\\\\\n-2 & 3\n\\end{pmatrix}.\n$$" - }, - { - "num": 62, - "question": "In pure $SU(2)$ Seiberg-Witten theory, the BPS central charge is $Z=n_{m}a_{D}+n_{e}a$. Suppose a dyon with charge $(n_{m},n_{e})$ becomes massless at a singularity. Using the period basis $(a_D,a)^T$ and a counterclockwise monodromy loop around the singularity, determine the associated monodromy matrix $M_{(n_m,n_e)}$.", - "category": "SUSY", - "answer": "$M_{(n_{m},n_{e})}=\\begin{pmatrix} 1+2n_{m}n_{e} & 2n_{e}^{2}\\\\ -2n_{m}^{2} & 1-2n_{m}n_{e} \\end{pmatrix}$", - "solution": "We write the period vector as $\\Pi=(a_{D},a)^{T}$. Then the central charge $Z=(n_{m},n_{e})\\Pi$. Suppose that a BPS state with charge $(n_{m},n_{e})$ becomes massless at a singularity. Geometrically, this means that the corresponding cycle is a vanishing cycle. By the Picard-Lefschetz formula, when one goes around such a singularity, the period vector transforms as\n$$\n\\Pi \\mapsto \\Pi+2Z\\begin{pmatrix}\n0 & 1\\\\\n-1 & 0\n\\end{pmatrix}\\begin{pmatrix}\nn_{m}\\\\\nn_{e}\n\\end{pmatrix}.\n$$\nTherefore, we have\n$$\na_{D}\\mapsto (1+2n_{m}n_{e})a_{D}+2n_{e}^{2}a, \\qquad a\\mapsto -2n_{m}^{2}a_{D}+(1-2n_{m}n_{e})a\n$$\nHence the monodromy matrix for a massless dyon of charge $(n_{m},n_{e})$ is\n$$\nM_{(n_{m},n_{e})}=\n\\begin{pmatrix}\n1+2n_{m}n_{e} & 2n_{e}^{2}\\\\\n-2n_{m}^{2} & 1-2n_{m}n_{e}\n\\end{pmatrix}\n$$" - }, - { - "num": 63, - "question": "Consider an $SU(2)$ gauge theory with gauge-field Lagrangian\n$$\n\\mathcal{L}\n=\n-\\frac{1}{4}F_{a}^{\\mu\\nu}F^{a}_{\\mu\\nu}\n-\\frac{\\theta e^{2}}{32\\pi^{2}}\nF_{a}^{\\mu\\nu}\\widetilde{F}^{a}_{\\mu\\nu}\n+\\cdots .\n$$\nUsing the gauge transformation\n$$\n\\delta A_\\mu^a=\\frac{1}{ev}D_\\mu\\phi^a\n$$\nand the conventions\n$$\nF_a^{0i}=-(E_a)_i,\n\\qquad\n\\epsilon^{0ijk}(F_a)_{jk}=-2(B_a)_i,\n$$\ncompute the corresponding Noether charge $\\mathcal{Q}$ in terms of\n$$\nQ\n:=\n\\frac{1}{v}\\int d^3x\\,\nE_i^aD_i\\phi^a,\n\\qquad\ng\n:=\n\\frac{1}{v}\\int d^3x\\,\nB_i^aD_i\\phi^a.\n$$", - "category": "Non-Abelian", - "answer": "$\\mathcal{Q}=\\frac{Q}{e}+\\frac{\\theta e g}{8\\pi^2}$", - "solution": "The Noether charge can be written as $\\mathcal{Q}=\\int d^3x\\,\\frac{\\partial\\mathcal{L}}{\\partial(\\partial_0A_\\mu^a)}\\delta A_\\mu^a$. The gauge-field Lagrangian is\n$$\n\\mathcal{L}\n=\n-\\frac{1}{4}F_{a}^{\\mu\\nu}F^{a}_{\\mu\\nu}\n-\\frac{\\theta e^{2}}{32\\pi^{2}}\nF_{a}^{\\mu\\nu}\\widetilde{F}^{a}_{\\mu\\nu}\n+\\cdots .\n$$\nThen, we have\n$$\n\\frac{\\partial\\mathcal{L}}\n{\\partial(\\partial_0A_\\mu^a)}\n=\n-\\frac{1}{2}F_b^{\\alpha\\beta}\n\\frac{\\partial F_{\\alpha\\beta}^b}\n{\\partial(\\partial_0A_\\mu^a)}\n-\n\\frac{\\theta e^2}{32\\pi^2}\n\\epsilon^{\\rho\\sigma\\alpha\\beta}\n(F_b)_{\\alpha\\beta}\n\\frac{\\partial F_{\\rho\\sigma}^b}\n{\\partial(\\partial_0A_\\mu^a)}.\n$$\nFrom the definition of $F_{\\rho\\sigma}^b$,\n$$\n\\frac{\\partial F_{\\rho\\sigma}^b}\n{\\partial(\\partial_0A_\\mu^a)}\n=\n\\delta_a^b\n\\left(\n\\delta_\\rho^0\\delta_\\sigma^\\mu\n-\n\\delta_\\sigma^0\\delta_\\rho^\\mu\n\\right).\n$$\nSubstituting this expression and simplifying, we obtain\n$$\n\\frac{\\partial\\mathcal{L}}\n{\\partial(\\partial_0A_i^a)}\n=\n-F_a^{0i}\n-\n\\frac{\\theta e^2}{16\\pi^2}\n\\epsilon^{0ijk}(F_a)_{jk}.\n$$\nUsing $F_a^{0i}=-(E_a)_i$ and $\\epsilon^{0ijk}(F_a)_{jk}=-2(B_a)_i$, it follows that\n$$\n\\frac{\\partial\\mathcal{L}}\n{\\partial(\\partial_0A_i^a)}\n=\n(E_a)_i\n+\n\\frac{\\theta e^2}{8\\pi^2}(B_a)_i.\n$$\nTherefore, the Noether charge associated with the transformation $\\delta A_\\mu^a=\\frac{1}{ev}D_\\mu\\phi^a$ is\n$$\n\\mathcal{Q}\n=\n\\int d^3x\\,\\frac{1}{ev}\n\\left[\n(E_i)^a\n+\n\\frac{\\theta e^2}{8\\pi^2}(B_i)^a\n\\right]D_i\\phi^a .\n$$\nUsing\n$$\nQ=\\frac{1}{v}\\int d^3x\\,(E_i)^aD_i\\phi^a,\n\\qquad\ng=\\frac{1}{v}\\int d^3x\\,(B_i)^aD_i\\phi^a,\n$$\nwe obtain\n$$\n\\mathcal{Q}\n=\n\\frac{Q}{e}\n+\n\\frac{\\theta e g}{8\\pi^2}.\n$$" - }, - { - "num": 64, - "question": "Consider a four-dimensional gauge theory in Minkowski spacetime with theta angle $\\theta$ and coupling $\\tau=\\frac{\\theta}{2\\pi}+\\frac{4\\pi i}{e^2}$. Define\n$$\nF^2:=F_{\\mu\\nu}F^{\\mu\\nu},\n\\qquad\nF{*}F:=F_{\\mu\\nu}({*}F)^{\\mu\\nu},\n$$\nand use the convention $({*}F)^2=-F^2$. The gauge kinetic terms are written in the compact form\n$$\n\\mathcal{L}\n=\n-\\frac{1}{32\\pi}\n\\operatorname{Im}\\!\\left[\n\\tau(F+i{*}F)^2\n\\right].\n$$\nRewrite $\\mathcal{L}$ in terms of $F^2$ and $F{*}F$.", - "category": "Classical", - "answer": "$\\mathcal{L}=-\\frac{1}{4e^{2}}F^{2}-\\frac{\\theta}{32\\pi^{2}}F*F$", - "solution": "We can compute\n$$\n(F+i*F)^{2}=F^{2}+2iF*F-(*F)^{2}=2F^{2}+2iF*F\n$$\nwhere we have used the identity $(*F)^{2}=-F^{2}$ in four-dimensional Minkowski spacetime. Taking the imaginary part, we have\n$$\n\\operatorname{Im}(\\tau(F+i*F)^{2})= 2\\operatorname{Im}(\\tau F^{2})+2\\operatorname{Re}(\\tau F*F).\n$$\nSince $\\tau=\\frac{\\theta}{2\\pi}+\\frac{4\\pi i}{e^{2}}$, we obtain\n$$\n\\mathcal{L}=-\\frac{1}{32\\pi}\\left(\\frac{8\\pi}{e^{2}}F^{2}+\\frac{2\\theta}{2\\pi}F*F\\right)=-\\frac{1}{4e^{2}}F^{2}-\\frac{\\theta}{32\\pi^{2}}F*F\n$$\nwhich recovers the standard form of the kinetic terms." - }, - { - "num": 65, - "question": "In a four-dimensional $\\mathcal{N}=1$ SCFT, the central charges are\n$$\na=\\frac{3}{32}\\left(3\\operatorname{Tr}R^3-\\operatorname{Tr}R\\right),\n\\qquad\nc=\\frac{1}{32}\\left(9\\operatorname{Tr}R^3-5\\operatorname{Tr}R\\right),\n$$\nwhere $\\operatorname{Tr}R$ and $\\operatorname{Tr}R^3$ denote sums over all left-handed Weyl fermions. Compute the central charges $(a,c)$ of the free chiral multiplet $\\Phi$ consisting of a complex scalar $\\phi$ and a Weyl fermion $\\psi$.", - "category": "SUSY", - "answer": "$(a,c)=\\left(\\frac{1}{48},\\frac{1}{24}\\right)$", - "solution": "If the theory has a Lagrangian description, then $\\operatorname{Tr}$ denotes a sum over all left-handed Weyl fermions. For the free chiral multiplet $\\Phi$, the scalar has scaling dimension $\\Delta(\\phi)=1$. Since a chiral primary satisfies $\\Delta(\\phi)=\\frac{3}{2}R(\\phi)$, we obtain $R(\\phi)=\\frac{2}{3}$. Since the fermion component satisfies $R(\\psi)=R(\\phi)-1$, we have $R(\\psi)=-1/3$. Therefore\n$$\n(a,c)_{\\text{chiral}}=\\left(\\frac{3}{32}\\left(-3\\times\\frac{1}{27}+\\frac{1}{3}\\right),\\frac{1}{32}\\left(-9\\times \\frac{1}{27}+\\frac{5}{3}\\right)\\right)= \\left(\\frac{1}{48},\\frac{1}{24}\\right).\n$$" - }, - { - "num": 66, - "question": "Consider a primary operator $\\mathcal{O}^{a}$ in a $d$-dimensional Euclidean CFT. It has scaling dimension $\\Delta$ and transforms in an irreducible representation $R$ of $SO(d)$. In radial quantization\n$$\nP^{\\dagger}_{\\mu}=K_{\\mu}, \\qquad [K_{\\mu},P_{\\nu}]=2\\delta_{\\mu\\nu}D-2M_{\\mu\\nu}\n$$\nThe primary state satisfies\n$$\nK_\\mu |\\mathcal O^a\\rangle = 0,\n\\qquad\nD|\\mathcal O^a\\rangle = \\Delta |\\mathcal O^a\\rangle,\n\\qquad\nM_{\\mu\\nu}|\\mathcal O^a\\rangle\n= (S_{\\mu\\nu})^{a}{}_{b} |\\mathcal O^b\\rangle, \\qquad \\langle \\mathcal O_b | \\mathcal O^a\\rangle\n= \\delta_b^a,\n$$\nwhere $\\mu,\\nu$ are $SO(d)$ vector indices and $a,b$ are representation indices of $R$. View $\\mathcal S:=\\bigl((S_{\\mu\\nu})^a{}_b\\bigr)$ as the Hermitian matrix acting on $\\mathbb C^d\\otimes R$, and let $s_R$ denote its largest eigenvalue. Using positivity of the norms of the first descendants $P_\\mu |\\mathcal O^a\\rangle$, determine the resulting lower bound on the scaling dimension $\\Delta$.", - "category": "CFT", - "answer": "$\\Delta \\geq s_{R}$", - "solution": "We compute the nonnegative norm squared of an arbitrary linear combination $|\\Psi\\rangle=\\xi^\\nu{}_aP_\\nu|\\mathcal O^a\\rangle$. Then\n$$\n\\begin{aligned}\n0\n&\\leq \\langle\\Psi|\\Psi\\rangle\\\\\n&=\n(\\xi^\\mu{}_b)^*\\xi^\\nu{}_a\n\\bigl(P_\\mu|\\mathcal O^b\\rangle\\bigr)^\\dagger\nP_\\nu|\\mathcal O^a\\rangle\\\\\n&=\n(\\xi^\\mu{}_b)^*\\xi^\\nu{}_a\n\\langle\\mathcal O_b|K_\\mu P_\\nu|\\mathcal O^a\\rangle\\\\\n&=\n(\\xi^\\mu{}_b)^*\\xi^\\nu{}_a\n\\langle\\mathcal O_b|[K_\\mu,P_\\nu]|\\mathcal O^a\\rangle\\\\\n&=\n2(\\xi^\\mu{}_b)^*\\xi^\\nu{}_a\n\\left(\n\\Delta\\delta_{\\mu\\nu}\\delta_b^a\n-(S_{\\mu\\nu})^a{}_b\n\\right).\n\\end{aligned}\n$$\nHere we used the fact that $K_\\mu$ annihilates primary states, the conformal algebra $[K_\\mu,P_\\nu]=2\\delta_{\\mu\\nu}D-2M_{\\mu\\nu}$, the normalization $\\langle\\mathcal O_b|\\mathcal O^a\\rangle=\\delta_b^a$ and the action of $M_{\\mu\\nu}$ and $D$ on $|\\mathcal O^a\\rangle$. Since this inequality holds for arbitrary $\\xi$, it implies that $\\Delta\\geq s_R$." - }, - { - "num": 67, - "question": "Let $\\mathcal O_{\\mu_1\\cdots\\mu_\\ell}$ be a spin-$\\ell$ symmetric traceless primary operator in a unitary $d$-dimensional CFT with $d\\geq3$, and let its scaling dimension be $\\Delta$. Determine the unitarity bound on $\\Delta$ for $\\ell>0$.", - "category": "CFT", - "answer": "$\\Delta \\geq \\ell+d-2$", - "solution": "To compute the eigenvalues of $(S_{\\mu\\nu})^b{}_a$, we can rewrite it in terms of the generators of $SO(d)$. Let $L^{\\alpha\\beta}$ be the generators acting on the vector representation $V_1$ of $SO(d)$, with\n$$\n(L^{\\alpha\\beta})_{\\mu\\nu}\n=\n\\delta^\\alpha_\\mu \\delta^\\beta_\\nu\n-\n\\delta^\\alpha_\\nu \\delta^\\beta_\\mu, \\qquad (S_{\\mu\\nu})^b{}_a\n=\n\\frac12 (L^{\\alpha\\beta})_{\\mu\\nu}\n(S_{\\alpha\\beta})^b{}_a .\n$$\nUsing the adjoint index notation $I=[\\alpha,\\beta]$, this operator can be written as $L^I S_I$. We use the standard identity\n$$\nL^I S_I\n=\n\\frac12\\left((L+S)^2-L^2-S^2\\right).\n$$\nWith our conventions, the operators $-L^2$, $-S^2$, and $-(L+S)^2$ are the quadratic Casimirs on $V_1$, $R$, and $R\\otimes V_\\ell$, respectively. We now specialize to the case $R=V_\\ell$, where $V_\\ell$ is the rank-$\\ell$ symmetric traceless representation of $SO(d)$. Its quadratic Casimir is\n$$\n\\mathrm{Cas}(V_\\ell)=\\ell(\\ell+d-2).\n$$\nThe eigenvalue of $L^I S_I$ on an irreducible component $V_m\\subset V_1\\otimes V_\\ell$ is therefore\n$$\n\\lambda_m\n=\n\\frac12\\left(\n\\mathrm{Cas}(V_\\ell)+\\mathrm{Cas}(V_1)-\\mathrm{Cas}(V_m)\n\\right).\n$$\nThus the maximal eigenvalue comes from the component with the smallest Casimir. For $\\ell>0$, this component is $V_{\\ell-1}$. Hence\n$$\n\\Delta\n\\geq\n\\lambda_{\\max}\n=\n\\frac12\\left(\n\\mathrm{Cas}(V_\\ell)+\\mathrm{Cas}(V_1)-\\mathrm{Cas}(V_{\\ell-1})\n\\right).\n$$\nUsing $\\mathrm{Cas}(V_1)=d-1$ and $\\mathrm{Cas}(V_{\\ell-1})=(\\ell-1)(\\ell+d-3)$, we obtain\n$$\n\\Delta \\geq \\ell+d-2,\n\\qquad \\ell>0.\n$$" - }, - { - "num": 68, - "question": "Let $\\phi$ be a real scalar primary operator of scaling dimension $\\Delta_\\phi$ in a $d$-dimensional CFT, normalized by\n$$\n\\langle \\phi(x_1)\\phi(x_2)\\rangle\n=\n\\frac{1}{x_{12}^{2\\Delta_\\phi}},\n\\qquad\nx_{ij}:=|x_i-x_j|.\n$$\nThe $\\phi\\times\\phi$ OPE is $\\phi(x_1)\\phi(x_2)=\\sum_{\\mathcal O}f_{\\phi\\phi\\mathcal O}\\,C_a(x_{12},\\partial_2)\\,\\mathcal O^a(x_2)$, where $\\mathcal O$ has scaling dimension $\\Delta_{\\mathcal O}$ and spin $\\ell_{\\mathcal O}$. Assume\n$$\n\\langle\n\\mathcal O^a(x)\\mathcal O'^{\\,b}(0)\n\\rangle\n=\n\\delta_{\\mathcal O\\mathcal O'}\n\\frac{I^{ab}(x)}{x^{2\\Delta_{\\mathcal O}}},\n$$\nwhere $I^{ab}(x)$ is the corresponding two-point tensor structure. Define the conformal block by\n$$\ng_{\\Delta_{\\mathcal O},\\ell_{\\mathcal O}}(x_i)\n:=\nx_{12}^{2\\Delta_\\phi}\nx_{34}^{2\\Delta_\\phi}\nC_a(x_{12},\\partial_2)\nC_b(x_{34},\\partial_4)\n\\frac{I^{ab}(x_{24})}\n{x_{24}^{2\\Delta_{\\mathcal O}}}.\n$$\nBy conformal invariance, $g_{\\Delta,\\ell}(x_i)=g_{\\Delta,\\ell}(u,v)$, where $u$ and $v$ are the conformal cross-ratios. Define $g(u,v):=\\sum_{\\mathcal O}f_{\\phi\\phi\\mathcal O}^{\\,2}g_{\\Delta_{\\mathcal O},\\ell_{\\mathcal O}}(u,v)$. Determine the decomposition of the four-point function $\\langle\\phi(x_1)\\phi(x_2)\\phi(x_3)\\phi(x_4)\\rangle$ in terms of $g(u,v)$.", - "category": "CFT", - "answer": "$\\langle \\phi(x_1)\\phi(x_2)\\phi(x_3)\\phi(x_4)\\rangle=\\frac{g(u,v)}{x_{12}^{2\\Delta_\\phi}x_{34}^{2\\Delta_\\phi}}$", - "solution": "Applying the OPE in the $(12)$ and $(34)$ channels, we obtain\n$$\n\\begin{aligned}\n\\langle \\phi(x_1)\\phi(x_2)\\phi(x_3)\\phi(x_4)\\rangle &= \\sum_{\\mathcal O,\\mathcal O'} f_{\\phi\\phi\\mathcal O} f_{\\phi\\phi\\mathcal O'} C_a(x_{12},\\partial_2) C_b(x_{34},\\partial_4) \\langle \\mathcal O^a(x_2)\\mathcal O'^b(x_4)\\rangle .\n\\end{aligned}\n$$\nChoosing an orthonormal basis of primary operators, the two-point function is diagonal\n$$\n\\langle \\mathcal O^a(x)\\mathcal O'^b(0)\\rangle = \\delta_{\\mathcal O\\mathcal O'} \\frac{I^{ab}(x)}{x^{2\\Delta_{\\mathcal O}}},\n$$\nTherefore,\n$$\n\\langle \\phi(x_1)\\phi(x_2)\\phi(x_3)\\phi(x_4)\\rangle = \\sum_{\\mathcal O} f_{\\phi\\phi\\mathcal O}^{\\,2} C_a(x_{12},\\partial_2) C_b(x_{34},\\partial_4) \\frac{I^{ab}(x_{24})}{x_{24}^{2\\Delta_{\\mathcal O}}}.\n$$\nThe conformal block is defined by\n$$\ng_{\\Delta,\\ell}(x_i)\n\\equiv\nx_{12}^{2\\Delta_\\phi}x_{34}^{2\\Delta_\\phi}\nC_a(x_{12},\\partial_2)\nC_b(x_{34},\\partial_4)\n\\frac{I^{ab}(x_{24})}{x_{24}^{2\\Delta_{\\mathcal O}}} .\n$$\nSince conformal invariance implies that it depends only on the cross-ratios $u,v$, $g_{\\Delta,\\ell}(x_i)=g_{\\Delta,\\ell}(u,v)$, hence the four-point function admits the conformal block decomposition\n$$\n\\langle \\phi(x_1)\\phi(x_2)\\phi(x_3)\\phi(x_4)\\rangle\n=\n\\frac{g(u,v)}{x_{12}^{2\\Delta_\\phi}x_{34}^{2\\Delta_\\phi}}, \\qquad g(u,v)\n=\n\\sum_{\\mathcal O}\nf_{\\phi\\phi\\mathcal O}^{\\,2}\ng_{\\Delta_{\\mathcal O},\\ell_{\\mathcal O}}(u,v).\n$$" - }, - { - "num": 69, - "question": "Let $\\phi$ and $\\mathcal{O}$ be scalar primary operators in a $d$-dimensional CFT with scaling dimensions $\\Delta_\\phi$ and $\\Delta$, respectively. Assume that the two-point function is normalized as\n$$\n\\langle \\mathcal O(x)\\mathcal O(y)\\rangle\n=\n\\frac{1}{|x-y|^{2\\Delta}}.\n$$\nWrite the $\\phi\\times\\phi$ OPE as\n$$\n\\phi(x)\\phi(0)=f_{\\phi\\phi\\mathcal{O}}C(x,\\partial)\\mathcal{O}(0)+\\cdots,\n$$\nwhere the differential operator $C(x,\\partial)$ has the expansion\n$$\nC(x,\\partial)\n=\n|x|^{\\Delta-2\\Delta_\\phi}\n\\left(\n1+\\alpha\\,x\\cdot \\partial\n+\\beta x^\\mu x^\\nu \\partial_\\mu\\partial_\\nu\n+\\gamma x^2\\partial^2+\\cdots\n\\right).\n$$\nAssume that $\\Delta\\neq\\frac{d-2}{2}$. Determine the coefficient tuple $(\\alpha,\\beta,\\gamma)$.", - "category": "CFT", - "answer": "$(\\alpha,\\beta,\\gamma)=\\left(\\frac12,\\frac{\\Delta+2}{8(\\Delta+1)},-\\frac{\\Delta}{16\\left(\\Delta-\\frac{d-2}{2}\\right)(\\Delta+1)}\\right)$", - "solution": "By conformal invariance, the three-point function of scalar operators is fixed to be\n$$\n\\langle \\phi(x)\\phi(0)\\mathcal O(z)\\rangle = \\frac{f_{\\phi\\phi\\mathcal O}} {|x|^{2\\Delta_\\phi-\\Delta}|z|^\\Delta |z-x|^\\Delta} = \\frac{f_{\\phi\\phi\\mathcal O} |x|^{\\Delta-2\\Delta_\\phi}}{|z|^\\Delta |z-x|^\\Delta}.\n$$\nOn the other hand, using the OPE, we obtain\n$$\n\\langle \\phi(x)\\phi(0)\\mathcal O(z)\\rangle = f_{\\phi\\phi\\mathcal O}\\, C(x,\\partial_{y}) \\langle \\mathcal O(y)\\mathcal O(z)\\rangle\\big|_{y=0}\n$$\nWe choose the normalization\n$$\n\\langle \\mathcal O(y)\\mathcal O(z)\\rangle = \\frac{1}{|y-z|^{2\\Delta}}.\n$$\nTherefore, $C(x,\\partial_{y})$ must satisfy\n$$\n\\left.\n\\frac{C(x,\\partial_{y})}{|x|^{\\Delta-2\\Delta_\\phi}}\\frac{1}{|z-y|^{2\\Delta}}\\right|_{y=0}= \\frac{1}{|z|^\\Delta |z-x|^\\Delta}.\n$$\nNow we expand the right-hand side\n$$\n\\frac{1}{|z|^\\Delta |z-x|^\\Delta} = \\frac{1}{|z|^{2\\Delta}} \\left[ 1 +\\Delta\\frac{x\\cdot z}{z^2} -\\frac{\\Delta}{2}\\frac{x^2}{z^2} +\\frac{\\Delta(\\Delta+2)}{2} \\frac{(x\\cdot z)^2}{z^4} +\\cdots \\right].\n$$\nThe left-hand side gives\n$$\n\\frac{1}{|z|^{2\\Delta}}\n\\left[\n1\n+2\\alpha\\Delta \\frac{x\\cdot z}{z^2}\n+4\\beta\\Delta(\\Delta+1)\\frac{(x\\cdot z)^2}{z^4}\n+\\left(\n-2\\beta\\Delta\n+2\\gamma\\Delta(2\\Delta+2-d)\n\\right)\\frac{x^2}{z^2}\n+\\cdots\n\\right].\n$$\nComparing coefficients gives\n$$\n\\alpha=\\frac12,\n\\qquad\n\\beta=\\frac{\\Delta+2}{8(\\Delta+1)},\n\\qquad\n\\gamma=\n-\\frac{\\Delta}\n{16\\left(\\Delta-\\frac{d-2}{2}\\right)(\\Delta+1)}.\n$$" - }, - { - "num": 70, - "question": "Let $\\mathcal{O}_1(x)$ and $\\mathcal{O}_2(x)$ be scalar primary operators in a $d$-dimensional Euclidean conformal field theory, with scaling dimensions $\\Delta_1$ and $\\Delta_2$, respectively. Using conformal invariance, determine the two-point function $\\left\\langle \\mathcal{O}_1(x_1)\\mathcal{O}_2(x_2)\\right\\rangle$ up to an overall constant $C$. Give the answer separately for $\\Delta_1=\\Delta_2$ and $\\Delta_1\\neq\\Delta_2$.", - "category": "CFT", - "answer": "$$\n\\left\\langle \\mathcal{O}_1(x_1)\\mathcal{O}_2(x_2)\\right\\rangle\n=\n\\begin{cases}\n\\dfrac{C}{|x_1-x_2|^{2\\Delta_1}},\n& \\Delta_1=\\Delta_2,\\\\[6pt]\n0,\n& \\Delta_1\\neq\\Delta_2.\n\\end{cases}\n$$", - "solution": "We know that translation, rotation, and scale invariance fix the two-point function of scalars up to a constant\n$$\n\\langle \\mathcal{O}_{1}(x_{1}) \\mathcal{O}_{2}(x_{2}) \\rangle = \\frac{C}{|x_{1}-x_{2}|^{\\Delta_{1}+\\Delta_{2}}}.\n$$\nFor a conformal transformation,\n$$\n(x_{1}-x_{2})^{2}\\Omega(x_{1}')\\Omega(x_{2}')=(x_{1}'-x_{2}')^{2}\n$$\nwhere $\\Omega(x)$ is a finite, position-dependent rescaling factor. Using this transformation, we obtain\n$$\n\\frac{C}{|x_{1}-x_{2}|^{\\Delta_{1}+\\Delta_{2}}}=\\Omega(x_{1}')^{\\frac{\\Delta_{1}+\\Delta_{2}}{2}}\\Omega(x_{2}')^{\\frac{\\Delta_{1}+\\Delta_{2}}{2}}\\frac{C}{|x_{1}'-x_{2}'|^{\\Delta_{1}+\\Delta_{2}}}\n$$\nFor primary scalars in a CFT, the correlators must satisfy\n$$\n\\left\\langle \\mathcal O_1(x_1)\\cdots \\mathcal O_n(x_n)\\right\\rangle=\n\\Omega(x_1')^{\\Delta_1}\\cdots \\Omega(x_n')^{\\Delta_n}\n\\left\\langle\n\\mathcal O_1(x_1')\\cdots \\mathcal O_n(x_n')\n\\right\\rangle.\n$$\nConsistency with conformal covariance requires either $\\Delta_{1}=\\Delta_{2}$ or $C=0$. Therefore,\n$$\n\\left\\langle \\mathcal{O}_1(x_1)\\mathcal{O}_2(x_2)\\right\\rangle\n=\n\\begin{cases}\n\\dfrac{C}{|x_1-x_2|^{2\\Delta_1}},\n& \\Delta_1=\\Delta_2,\\\\[6pt]\n0,\n& \\Delta_1\\neq\\Delta_2.\n\\end{cases}\n$$" - }, - { - "num": 71, - "question": "Consider a two-dimensional CFT with central charge $(c,\\bar{c})=(c,c)$ on the complex plane with stress tensors $T(z)$ and $\\overline{T}(\\bar{z})$. Define the Virasoro modes by\n$$\nT(z)=\\sum_{n\\in\\mathbb Z}L_n z^{-n-2},\n\\qquad\n\\overline T(\\bar z)\n=\n\\sum_{n\\in\\mathbb Z}\\overline L_n\\bar z^{-n-2}.\n$$\nDefine the torus $T^2=\\mathbb{C}/\\bigl(\\mathbb{Z}\\omega_1+\\mathbb{Z}\\omega_2\\bigr)$ with $\\omega_1=L\\in\\mathbb{R}_{>0}$, $\\omega_2=L\\tau$, and $\\tau=\\tau_1+i\\tau_2$, where $\\tau_1\\in\\mathbb{R}$ and $\\tau_2>0$. Using the conformal transformation law of the stress tensor, derive the torus partition function $Z(\\tau,\\bar{\\tau})$ and express it as a trace over the Hilbert space $\\mathcal{H}$ in terms of $L_0$, $\\overline{L}_0$, the central charge $c$, and $q=e^{2\\pi i\\tau}$, $\\bar q=e^{-2\\pi i\\bar{\\tau}}$.", - "category": "CFT", - "answer": "$Z(\\tau,\\bar{\\tau})=\\operatorname{Tr}_{\\mathcal H}\\left(q^{L_0-\\frac{c}{24}}\\overline q^{\\,\\overline L_0-\\frac{c}{24}}\\right)$", - "solution": "Let $w$ be the coordinate on the covering space of the torus. The first identification\n$$\nw\\sim w+\\omega_1,\n\\qquad\n\\omega_1=L,\n$$\ndefines a cylinder with spatial circumference $L$. We map this cylinder to the complex plane by\n$$\nz\n=\ne^{2\\pi i w/\\omega_1}\n=\ne^{2\\pi i w/L}.\n$$\nThe stress tensor transforms as\n$$\nT_{\\mathrm{cyl}}(w)\n=\n\\left(\\frac{dz}{dw}\\right)^2T(z)\n+\\frac{c}{12}\\{z,w\\},\n$$\nwhere $\\{z,w\\}$ is the Schwarzian derivative. We can compute\n$$\nT_{\\mathrm{cyl}}(w)\n=\n-\\left(\\frac{2\\pi}{L}\\right)^2\n\\left(\n\\sum_{n\\in\\mathbb Z}\nL_n e^{-2\\pi i n w/L}\n-\\frac{c}{24}\n\\right),\n$$\nand similarly,\n$$\n\\overline T_{\\mathrm{cyl}}(\\bar w)\n=\n-\\left(\\frac{2\\pi}{L}\\right)^2\n\\left(\n\\sum_{n\\in\\mathbb Z}\n\\overline L_n e^{2\\pi i n\\bar w/L}\n-\\frac{c}{24}\n\\right).\n$$\nTherefore, the zero modes are shifted as\n$$\nL_0\\mapsto L_0-\\frac{c}{24},\n\\qquad\n\\overline L_0\\mapsto\\overline L_0-\\frac{c}{24}.\n$$\nHence, the Hamiltonian and momentum on the cylinder are\n$$\nH\n=\n\\frac{2\\pi}{L}\n\\left(\nL_0+\\overline L_0-\\frac{c}{12}\n\\right),\n\\qquad\nP\n=\n\\frac{2\\pi}{L}\n\\left(\nL_0-\\overline L_0\n\\right).\n$$\nThe second identification\n$$\nw\\sim w+\\omega_2,\n\\qquad\n\\omega_2=L\\tau=\\omega_1\\tau,\n$$\ncorresponds to Euclidean time evolution by $\\operatorname{Im}\\omega_2=L\\tau_2$ together with a spatial translation by $\\operatorname{Re}\\omega_2=L\\tau_1$. Thus,\n$$\n\\begin{aligned}\nZ(\\tau,\\bar\\tau)\n&=\n\\operatorname{Tr}_{\\mathcal H}\n\\left[\n\\exp\\left(\n-\\operatorname{Im}\\omega_2\\,H\n+i\\operatorname{Re}\\omega_2\\,P\n\\right)\n\\right]\\\\\n&=\n\\operatorname{Tr}_{\\mathcal H}\n\\left[\n\\exp\\left(\n-L\\tau_2H+iL\\tau_1P\n\\right)\n\\right]\\\\\n&=\n\\operatorname{Tr}_{\\mathcal H}\n\\left[\n\\exp\\left(\n2\\pi i\\tau\n\\left(L_0-\\frac{c}{24}\\right)\n-\n2\\pi i\\bar\\tau\n\\left(\\overline L_0-\\frac{c}{24}\\right)\n\\right)\n\\right].\n\\end{aligned}\n$$\nDefining\n$$\nq=e^{2\\pi i\\tau},\n\\qquad\n\\bar q=e^{-2\\pi i\\bar\\tau},\n$$\nwe obtain\n$$\nZ(\\tau,\\bar\\tau)\n=\n\\operatorname{Tr}_{\\mathcal H}\n\\left(\nq^{L_0-\\frac{c}{24}}\n\\bar q^{\\,\\overline L_0-\\frac{c}{24}}\n\\right).\n$$" - }, - { - "num": 72, - "question": "Consider the free boson CFT with left- and right-moving central charges $(c,\\bar c)=(1,1)$. It has a continuum of primary operators\n$$\nV_\\alpha(z,\\bar z):=e^{i\\alpha\\phi(z,\\bar z)},\n\\qquad\n\\alpha\\in\\mathbb R,\n$$\nwith corresponding states $|V_\\alpha\\rangle$ satisfying\n$$\nL_0|V_\\alpha\\rangle=\\frac{\\alpha^2}{2}|V_\\alpha\\rangle,\n\\qquad\n\\overline L_0|V_\\alpha\\rangle=\\frac{\\alpha^2}{2}|V_\\alpha\\rangle.\n$$\nIn taking the trace over the Hilbert space $\\mathcal H$, normalize the continuum measure over the primary labels by\n$$\n\\operatorname{Tr}_{\\mathcal{H}}(\\,\\cdots\\,)\n:=\n\\int_{-\\infty}^{\\infty}d\\alpha\\,(\\,\\cdots\\,).\n$$\nLet $\\tau=\\tau_1+i\\tau_2$ with $\\tau_2>0$. Define the Dedekind eta function by\n$$\n\\eta(\\tau)\n=\nq^{1/24}\n\\prod_{n=1}^{\\infty}(1-q^n),\n\\qquad\nq=e^{2\\pi i\\tau}, \\qquad \\overline{\\eta(\\tau)}\n=\n\\overline q^{\\,1/24}\n\\prod_{n=1}^{\\infty}(1-\\overline q^{\\,n}),\n\\qquad\n\\overline q=e^{-2\\pi i\\overline{\\tau}}.\n$$\nCompute the torus partition function $Z_{\\mathrm{Boson}}(\\tau,\\bar{\\tau})$ and express your answer in terms of the Dedekind eta function $\\eta(\\tau)$ and $\\operatorname{Im}\\tau$.", - "category": "CFT", - "answer": "$Z_{\\mathrm{Boson}}(\\tau,\\bar{\\tau})=\\frac{1}{\\sqrt{2\\,\\operatorname{Im}\\tau}\\,|\\eta(\\tau)|^2}$", - "solution": "Since we know the value of the central charge and the conformal weights of the primary states, we can compute the partition function explicitly. Let $k=(k_1,\\ldots,k_r)$ denote a partition with $1\\leq k_1\\leq\\cdots\\leq k_r$, and define $|k|:=\\sum_{i=1}^{r}k_i$. We use the analogous notation for $\\bar{k}$. Then,\n$$\n\\begin{aligned}\nZ_{\\mathrm{Boson}}(\\tau,\\bar{\\tau})\n&=\n\\operatorname{Tr}_{\\mathcal{H}}\n\\left(\nq^{L_0-\\frac{c}{24}}\n\\overline q^{\\,\\overline L_0-\\frac{c}{24}}\n\\right) \\\\[0.5em]\n&=\n\\int_{-\\infty}^{\\infty} d\\alpha\\,\n\\left(\n\\sum_{\\{k\\},\\,k_1\\leq k_2\\leq\\cdots}\nq^{\\frac{\\alpha^2}{2}+|\\vec k|-\\frac{1}{24}}\n\\right)\n\\left(\n\\sum_{\\{\\overline k\\},\\,\\overline k_1\\leq\\overline k_2\\leq\\cdots}\n\\overline q^{\\,\\frac{\\alpha^2}{2}+|\\overline{\\vec k}|-\\frac{1}{24}}\n\\right) \\\\[0.5em]\n&=\n\\left(\n\\int_{-\\infty}^{\\infty}\nd\\alpha\\,\ne^{-2\\pi\\,\\operatorname{Im}(\\tau)\\alpha^2}\n\\right)\nq^{-\\frac{1}{24}}\n\\overline q^{-\\frac{1}{24}}\n\\prod_{n=1}^{\\infty}(1-q^n)^{-1}\n\\prod_{m=1}^{\\infty}(1-\\overline q^{\\,m})^{-1}\\\\\n&=\\frac{1}\n{\\sqrt{2\\,\\operatorname{Im}\\tau}\\,|\\eta(\\tau)|^2}.\n\\end{aligned}\n$$\nWe use the product formula for the Dedekind $\\eta$ function\n$$\n\\eta(\\tau)\n=\nq^{1/24}\n\\prod_{n=1}^{\\infty}(1-q^n),\n\\qquad\nq=e^{2\\pi i\\tau}, \\qquad \\overline{\\eta(\\tau)}\n=\n\\overline q^{\\,1/24}\n\\prod_{n=1}^{\\infty}(1-\\overline q^{\\,n}),\n\\qquad\n\\overline q=e^{-2\\pi i\\overline{\\tau}}.\n$$" - }, - { - "num": 73, - "question": "Consider the free boson CFT with torus partition function\n$$\nZ_{\\mathrm{Boson}}(\\tau,\\bar{\\tau})=\\frac{1}\n{\\sqrt{2\\,\\operatorname{Im}\\tau}\\,|\\eta(\\tau)|^2}, \\qquad \\eta(\\tau)=q^{1/24}\n\\prod_{n=1}^{\\infty}(1-q^n),\n$$\nwhere $\\tau=\\tau_1+i\\tau_2$, $\\tau_2>0$, $q=e^{2\\pi i\\tau}$. Determine the relation between $Z_{\\mathrm{Boson}}(\\tau,\\bar{\\tau})$ and $Z_{\\mathrm{Boson}}\\!\\left(-\\frac{1}{\\tau},-\\frac{1}{\\bar{\\tau}}\\right)$ under the modular transformation $S:\\tau\\mapsto -\\frac{1}{\\tau}$, $\\bar{\\tau}\\mapsto -\\frac{1}{\\bar{\\tau}}$.", - "category": "CFT", - "answer": "$Z_{\\mathrm{Boson}}\\!\\left(-\\frac{1}{\\tau},-\\frac{1}{\\bar{\\tau}}\\right)=Z_{\\mathrm{Boson}}(\\tau,\\bar{\\tau})$", - "solution": "Under the modular transformation $S:\\tau\\mapsto -\\frac{1}{\\tau},\\bar{\\tau}\\mapsto -\\frac{1}{\\bar{\\tau}}$, the Dedekind $\\eta$ function satisfies\n$$\n\\left|\\eta\\!\\left(-\\frac{1}{\\tau}\\right)\\right|^2\n=\n|\\tau|\\,|\\eta(\\tau)|^2,\n$$\nand\n$$\n\\operatorname{Im}\\!\\left(-\\frac{1}{\\tau}\\right)\n=\n\\operatorname{Im}\\!\\left(-\\frac{\\overline{\\tau}}{|\\tau|^2}\\right)\n=\n\\frac{\\operatorname{Im}\\tau}{|\\tau|^2}.\n$$\nTherefore\n$$\n\\sqrt{2\\,\\operatorname{Im}\\!\\left(-\\frac{1}{\\tau}\\right)}\n\\left|\\eta\\!\\left(-\\frac{1}{\\tau}\\right)\\right|^2\n=\n\\sqrt{\\frac{2\\,\\operatorname{Im}\\tau}{|\\tau|^2}}\n\\,|\\tau|\\,|\\eta(\\tau)|^2\n=\n\\sqrt{2\\,\\operatorname{Im}\\tau}\\,|\\eta(\\tau)|^2.\n$$\nThus\n$$\nZ_{\\mathrm{Boson}}\\!\\left(-\\frac{1}{\\tau},-\\frac{1}{\\bar{\\tau}}\\right)\n=\nZ_{\\mathrm{Boson}}(\\tau,\\bar{\\tau}).\n$$\nHence $Z_{\\mathrm{Boson}}(\\tau,\\bar{\\tau})$ is invariant under $S:\\tau\\mapsto -\\frac{1}{\\tau}, \\bar{\\tau}\\mapsto -\\frac{1}{\\bar{\\tau}}$." - }, - { - "num": 74, - "question": "For $n$ left-moving free complex fermions $\\psi_i(z)$ with action\n$$\nS=\\int d^2z\\,\\psi_i\\bar{\\partial}\\psi^i,\n$$\nwhere $\\psi^i=(\\psi_i)^*$, work in the Neveu-Schwarz sector and adopt the oscillator normalization\n$$\n\\{\\psi_{i,r},\\psi_s^j\\}\n=\n\\delta_i^{\\,j}\\delta_{r+s,0},\\qquad \\{\\psi_{i,r},\\psi_{j,s}\\}=0,\n\\qquad\n\\{\\psi^i_r,\\psi^j_s\\}=0.\n$$\nExpand the fields in half-integer oscillator modes and determine the radially ordered two-point function $\\langle\\psi_i(z)\\psi^j(w)\\rangle$.", - "category": "CFT", - "answer": "$\\langle \\psi_i(z)\\psi^j(w)\\rangle=\\frac{\\delta_i^{\\,j}}{z-w}$", - "solution": "The mode expansions are\n$$\n\\psi_i(z)=\\sum_{r\\in\\mathbb Z+\\frac12}\\psi_{i,r}\\,z^{-r-\\frac12},\n\\qquad\n\\psi^i(z)=\\sum_{r\\in\\mathbb Z+\\frac12}\\psi^i_r\\,z^{-r-\\frac12}.\n$$\nThe fermionic modes satisfy the anti-commutation relations\n$$\n\\{\\psi_{i,r},\\psi^j_s\\}\n=\n\\delta_i^{\\,j}\\delta_{r+s,0}, \\qquad \\{\\psi_{i,r},\\psi_{j,s}\\}=0,\n\\qquad\n\\{\\psi^i_r,\\psi^j_s\\}=0.\n$$\nWe choose the vacuum $|0\\rangle$ to be annihilated by all positive modes:\n$$\n\\psi_{i,r}|0\\rangle=0,\n\\qquad\n\\psi^i_r|0\\rangle=0,\n\\qquad r>0.\n$$\nFor $|z|>|w|$, we compute\n$$\n\\langle \\psi_i(z)\\psi^j(w)\\rangle\n=\n\\sum_{r,s\\in\\mathbb Z+\\frac12}\nz^{-r-\\frac12}w^{-s-\\frac12}\n\\langle 0|\\psi_{i,r}\\psi^j_s|0\\rangle .\n$$\nOnly the contraction between $\\psi_{i,r}$ with $r>0$ and $\\psi^j_{-r}$ contributes. Hence\n$$\n\\begin{aligned}\n\\langle \\psi_i(z)\\psi^j(w)\\rangle\n&=\n\\sum_{r>0}\nz^{-r-\\frac12}w^{r-\\frac12}\\delta_i^{\\,j} \\\\\n&=\n\\delta_i^{\\,j}\n\\sum_{k=0}^{\\infty}\nz^{-k-1}w^k,\n\\qquad r=k+\\frac12 \\\\\n&=\n\\frac{\\delta_i^{\\,j}}{z}\n\\sum_{k=0}^{\\infty}\n\\left(\\frac{w}{z}\\right)^k \\\\\n&=\n\\frac{\\delta_i^{\\,j}}{z-w}.\n\\end{aligned}\n$$\nTherefore,\n$$\n\\langle \\psi_i(z)\\psi^j(w)\\rangle\n=\n\\frac{\\delta_i^{\\,j}}{z-w}.\n$$" - }, - { - "num": 75, - "question": "Let $\\psi_i(z)$, $i=1,\\ldots,n$, be $n$ left-moving free complex fermions, and let $\\psi^i(z)$ be their complex conjugates. Let $\\psi_{i,r}$ and $\\psi_r^i$, with $r\\in\\mathbb Z+\\frac12$, denote their oscillator modes in the Neveu-Schwarz sector, normalized by\n$$\n\\{\\psi_{i,r},\\psi_s^j\\}\n=\n\\delta_i^{\\,j}\\delta_{r+s,0},\n\\qquad\n\\{\\psi_{i,r},\\psi_{j,s}\\}\n=\n\\{\\psi_r^i,\\psi_s^j\\}=0.\n$$\nDefine the spin-one currents\n$$\nJ_i^{\\,j}(z):=:\\!\\psi_i(z)\\psi^j(z)\\!:\n=\n\\sum_{m\\in\\mathbb Z}J_{i,m}^{\\,j}z^{-m-1},\n$$\nand impose the Hermiticity condition $\\left(J_{i,m}^{\\,j}\\right)^\\dagger=J_{j,-m}^{\\,i}$. Use the commutation relations of the zero modes $J_{i,0}^{\\,j}$, together with the Hermiticity condition, to determine the real Lie algebra they generate.", - "category": "CFT", - "answer": "$\\mathfrak u(n)$", - "solution": "Using the creation-annihilation normal ordering, the current can be expanded as\n$$\nJ_i^{\\,j}(z)\n=\n:\\psi_i(z)\\psi^j(z):\n=\n\\sum_{r,s\\in\\mathbb Z+\\frac12}\n:\\psi_{i,r}\\psi^j_s:\\,z^{-r-s-1}\n=\n\\sum_{m\\in\\mathbb Z}J_{i,m}^{\\,j}z^{-m-1}.\n$$\nHence $J_{i,m}^{\\,j}=\\sum_{r\\in\\mathbb Z+\\frac12}:\\psi_{i,m-r}\\psi^j_r:$. To compute the operator part of the commutator, we may first drop the normal-ordering constants, since they are constants and hence commute with all modes. The central term must then be recovered when the resulting products are rewritten in normal-ordered form. Using $\\{\\psi_{i,r},\\psi^j_s\\}=\\delta_i^{\\,j}\\delta_{r+s,0}$, we get\n$$\n\\begin{aligned}\n[J_{i,m}^{\\,j},J_{k,n}^{\\,l}]\n&=\n\\sum_{r,s\\in\\mathbb Z+\\frac12}\n[\\psi_{i,m-r}\\psi^j_r,\\psi_{k,n-s}\\psi^l_s] \\\\\n&=\n\\sum_{r,s\\in\\mathbb Z+\\frac12}\n\\left(\n\\psi_{i,m-r}\\{\\psi^j_r,\\psi_{k,n-s}\\}\\psi^l_s\n-\n\\psi_{k,n-s}\\{\\psi_{i,m-r},\\psi^l_s\\}\\psi^j_r\n\\right) \\\\\n&=\n\\sum_{r\\in\\mathbb Z+\\frac12}\n\\left(\n\\delta_k^{\\,j}\\psi_{i,m-r}\\psi^l_{n+r}\n-\n\\delta_i^{\\,l}\\psi_{k,m+n-r}\\psi^j_r\n\\right).\n\\end{aligned}\n$$\nReturning to normal-ordered products gives\n$$\n\\begin{aligned}\n[J_{i,m}^{\\,j},J_{k,n}^{\\,l}]\n&=\n\\delta_k^{\\,j}\n\\sum_{r\\in\\mathbb Z+\\frac12}\n:\\psi_{i,m-r}\\psi^l_{n+r}:\n-\n\\delta_i^{\\,l}\n\\sum_{r\\in\\mathbb Z+\\frac12}\n:\\psi_{k,m+n-r}\\psi^j_r:\n+\nC \\\\\n&=\n\\delta_k^{\\,j}J_{i,m+n}^{\\,l}\n-\n\\delta_i^{\\,l}J_{k,m+n}^{\\,j}\n+\nC .\n\\end{aligned}\n$$\nThe extra constant $C$ comes from the change in normal ordering. It is nonzero only when $m+n=0$, and the difference of the two normal-ordering constants is\n$$\nC\n=\nm\\,\\delta_{m+n,0}\\delta_i^{\\,l}\\delta_k^{\\,j}.\n$$\nTherefore\n$$\n[J_{i,m}^{\\,j},J_{k,n}^{\\,l}]\n=\n\\delta_k^{\\,j}J_{i,m+n}^{\\,l}\n-\n\\delta_i^{\\,l}J_{k,m+n}^{\\,j}\n+\nm\\,\\delta_{m+n,0}\\delta_i^{\\,l}\\delta_k^{\\,j}.\n$$\nEquivalently, in adjoint notation,\n$$\n[J_m^a,J_n^b]\n=\nif^{ab}{}_{c}J_{m+n}^c\n+\nm\\,\\delta^{ab}\\delta_{m+n,0}.\n$$\nFor the zero modes, the central term vanishes:\n$$\n[J_{i,0}^{\\,j},J_{k,0}^{\\,l}]\n=\n\\delta_k^{\\,j}J_{i,0}^{\\,l}\n-\n\\delta_i^{\\,l}J_{k,0}^{\\,j}.\n$$\nThis is the matrix Lie algebra $\\mathfrak{gl}_n$ over $\\mathbb C$, whose compact real form is $\\mathfrak u(n)$." - } -] \ No newline at end of file