Peskin 3

Peskin 4

Peskin 5

Peskin 6

Peskin 7

Peskin 8

Peskin & Schroeder. An Introduction to Quantum Field Theory

Chapter 7: Radiative Corrections: Some Formal Developments

7.2 The LSZ Reduction formula

Chapter 10の繰り込みで使用するので，しっかり理解しておく必要がある

(7.45)

直前にある4点相関函数について証明する．LSZ簡約公式(7.42)から

(7.44)とその後の式から

以上から，

Chapter 10: Systematics of Renormalization

Callan-Symanzik方程式を使ってβ函数を求める．counter termの計算は積分の発散項のみ見ればいいので楽

10.3 Renormalization of Quantum Electrodynamics

(10.46)

(10.39)で定義したように，

Chapter 12: The Renormalization Group

12.1 Wilson's Approach to Renormalization Theory

(12.10)

(12.8)から \begin{align*} \braket{ \hat\phi(x) \hat\phi(x) } &= \int \frac{d^dk}{(2\pi)^d} \frac{d^dp}{(2\pi)^d} e^{-i(k+p) \cdot x} \braket{ \hat\phi(k) \hat\phi}(p) \\ % &= \int \frac{d^dk}{(2\pi)^d} \frac{d^dp}{(2\pi)^d} e^{-i(k+p) \cdot x} \frac{(2\pi)^d}{k^2} \mathop{\delta^{(d)}}(k+p) \Theta(k) \\ % &= \int_{b\Lambda \leq \lvert k \rvert < \Lambda} \frac{d^dk}{(2\pi)^d} \frac{1}{k^2} . \end{align*} よって， \begin{align*} - \int d^dx \, \frac{\lambda}{4} \phi(x) \phi(x) \braket{ \hat\phi(x) {\hat\phi}(x) } &= - \int d^dx \, \frac{\lambda}{4} \phi(x) \phi(x) \int_{b\Lambda \leq \lvert k \rvert < \Lambda} \frac{d^dk}{(2\pi)^d} \frac{1}{k^2} \\ % &= - \frac{\mu}{2} \int d^dx \, \phi(x) \phi(x) \\ % &= - \frac{\mu}{2} \int d^dx \, \int \frac{d^dk}{(2\pi)^d} \frac{d^dp}{(2\pi)^d} e^{-i(k+p) \cdot x} \phi(k) \phi(p) \\ % &= - \frac{\mu}{2} \int \frac{d^dk}{(2\pi)^d} \frac{d^dp}{(2\pi)^d} \mathop{\delta^{(d)}}(k+p) \phi(k) \phi(p) \\ &= - \frac{\mu}{2} \int \frac{d^dk}{(2\pi)^d} \phi(k) \phi(-k) . \end{align*}

(12.14)

(12.8)から \begin{align*} \braket{ {\hat\phi}(x) {\hat\phi}(y) } &= \int \frac{d^dk}{(2\pi)^d} \frac{d^dp}{(2\pi)^d} e^{-i(k \cdot x + p \cdot y)} \braket{ {\hat\phi}(k) {\hat\phi}(p) } \\ % &= \int \frac{d^dk}{(2\pi)^d} \frac{d^dp}{(2\pi)^d} e^{-i(k \cdot x + p \cdot y)} \frac{(2\pi)^d}{k^2} \mathop{\delta^{(d)}}(k+p) \Theta(k) \\ % &= \int_{b\Lambda \leq \lvert k \rvert < \Lambda} \frac{d^dk}{(2\pi)^d} e^{-ik \cdot (x-y)} \frac{1}{k^2} . \end{align*} $\exp (-\lambda \phi^2 \hat\phi^2 / 4)$の2次の展開 \[ \frac{1}{2} \int d^dx \, \frac{\lambda}{4} \phi(x) \phi(x) \hat\phi(x) \hat\phi(x) \int d^dy \, \frac{\lambda}{4} \phi(y) \phi(y) \hat\phi(y) \hat\phi(y) \] を考える．$\hat\phi$の縮約には

および

の2通りがある．2つ目の縮約は2通りあるので， \begin{align*} & \frac{\lambda^2}{16} \int d^dx \,d^dy \, \phi(x) \phi(x) \phi(y) \phi(y) \braket{ {\hat\phi}(x) {\hat\phi}(y) } \braket{ {\hat\phi}(x) {\hat\phi}(y) } \\ % &= \frac{\lambda^2}{16} \int d^dx \, d^dy \, \phi^2(x) \phi^2(y) \int\limits_{\substack{b\Lambda \leq \lvert k \rvert < \Lambda \\ b\Lambda \leq \lvert p \rvert < \Lambda}} \frac{d^dk}{(2\pi)^d} \frac{d^dp}{(2\pi)^d} e^{-i(k+p) \cdot (x-y)} \frac{1}{k^2} \frac{1}{p^2} \\ % &= \frac{\lambda^2}{16} \int\limits_{\substack{b\Lambda \leq \lvert k \rvert < \Lambda \\ b\Lambda \leq \lvert p \rvert < \Lambda}} \frac{d^dk}{(2\pi)^d} \frac{d^dp}{(2\pi)^d} \frac{1}{k^2} \frac{1}{p^2} \int d^dx \, \phi^2(x) e^{-i(k+p) \cdot x} \int d^dy \, \phi^2(y) e^{i(k+p) \cdot y} \\ % &= \frac{\lambda^2}{16} \int\limits_{\substack{b\Lambda \leq \lvert k \rvert < \Lambda \\ b\Lambda \leq \lvert p \rvert < \Lambda}} \frac{d^dk}{(2\pi)^d} \frac{d^dp}{(2\pi)^d} \frac{1}{k^2} \frac{1}{p^2} \mathcal{F}[\phi^2](-k-p) \mathcal{F}[\phi^2](k+p) \\ % &= \frac{\lambda^2}{16} \int\limits_{\substack{b\Lambda \leq \lvert k \rvert < \Lambda \\ b\Lambda \leq \lvert p \rvert < \Lambda}} \frac{d^dk}{(2\pi)^d} \frac{d^dp}{(2\pi)^d} \frac{1}{k^2} \frac{1}{p^2} \left\lvert \mathcal{F}[\phi^2](k+p) \right\rvert^2 . \end{align*} $\phi$の運動量に関する条件から，$\mathcal{F}[\phi^2](k+p) \approx \mathcal{F}[\phi^2](0) \mathop{\delta^{(d)}}(k+p)$．

よって， \begin{align*} &\approx \frac{1}{(2\pi)^d} \frac{\lambda^2}{16} \int\limits_{b\Lambda \leq \lvert k \rvert < \Lambda} \frac{d^dk}{(2\pi)^d} \left( \frac{1}{k^2} \right)^2 \left\lvert \mathcal{F}[\phi^2](0) \right\rvert^2 \\ % &\approx \frac{\lambda^2}{16} \int\limits_{b\Lambda \leq \lvert k \rvert < \Lambda} \frac{d^dk}{(2\pi)^d} \left( \frac{1}{k^2} \right)^2 \int_{-\infty}^\infty \frac{d^dp}{(2\pi)^d} \, \left\lvert \mathcal{F}[\phi^2](p) \right\rvert^2 \\ % &= \frac{\lambda^2}{16} \int\limits_{b\Lambda \leq \lvert k \rvert < \Lambda} \frac{d^dk}{(2\pi)^d} \left( \frac{1}{k^2} \right)^2 \left\lVert \mathcal{F}[\phi^2] \right\rVert^2 \\ % &= \frac{\lambda^2}{16} \int\limits_{b\Lambda \leq \lvert k \rvert < \Lambda} \frac{d^dk}{(2\pi)^d} \left( \frac{1}{k^2} \right)^2 \left\lVert \phi^2 \right\rVert^2 \\ % &= \frac{\lambda^2}{16} \int\limits_{b\Lambda \leq \lvert k \rvert < \Lambda} \frac{d^dk}{(2\pi)^d} \left( \frac{1}{k^2} \right)^2 \int d^dx \, \phi^4(x) \\ % &= - \frac{1}{4!} \int d^dx \, \zeta \phi^4 . \end{align*}

12.2 The Callan-Symanzik Equation

(12.52)

$n$頂点のGreen函数を考える．(7.42)(12.35)から

となる（くりこみした量でダイアグラムを計算するので，(7.45)の右辺の$\sqrt{Z}$は不要）． tree-levelは，

1PI-loopは，(10.20)などで計算したように，$\log (-p^2)$の発散を持つ：

vertex countertermは，

運動量$p_i$の外線に対するexternal leg correctionは

となる（$g^1$の項のみ考えるので，$i$について和を取れば良い）．

(12.57)

$G^{(2, 0)}(p)$を求める．電子の自己エネルギー(7.15)を使えば(12.49)と同様に

(12.50)と同様に，Callan-Syamzik方程式から \[ - \frac{i}{\cancel{p}} M \frac{\partial}{\partial M} \delta_2 + 2 \gamma_2 \frac{i}{\cancel{p}} = 0 \] なので \[ \gamma_2 = \frac{1}{2} M \frac{\partial}{\partial M} \delta_2 . \]

$G^{(0, 2)}(q)$を求める．真空偏極(7.71)(7.73)を使えば(12.49)と同様に

(12.50)と同様に，Callan-Syamzik方程式から \[ - M \frac{\partial}{\partial M} \delta_3 + 2 \gamma_3 = 0 \] なので \[ \gamma_3 = \frac{1}{2} M \frac{\partial}{\partial M} \delta_3 . \]

(12.58)

$G^{(2,1)}(p_1, p_2, q)$を求める．頂点補正(6.38)を使えば(12.52)と同様に

(12.53)と同様に，Callan-Syamzik方程式から \[ M \frac{\partial}{\partial M} (\delta_1 - 2\delta_2 - \delta_3) + \frac{\beta}{e} + 2 \gamma_2 + \gamma_3 = 0 . \] (12.57)を代入して \[ \beta = e M \frac{\partial}{\partial M} \left( - \delta_1 + \delta_2 + \frac{\delta_3}{2} \right) . \]

Problem 12.2: Beta function of the Gross-Neveu model

Gross-Neveu模型は \[ \mathcal{L} = \sum_{i=1}^N \bar\psi_i (i\cancel\partial) \psi_i + \frac{g^2}{2} \left( \sum_{i=1}^N \bar\psi_i \psi_i \right) \] で与えられる．$d = 2$のDirac行列は \[ (\gamma^0)_{\alpha\beta} = \begin{pmatrix} & -i \\ i & \end{pmatrix} , \quad (\gamma^1)_{\alpha\beta} = \begin{pmatrix} & i \\ i & \end{pmatrix} \] である（スピノルの添字を$\alpha = 0, 1$などのギリシャ文字で表す）．

伝播函数は

で与えられる．ガンマ行列は対角成分を持たないので，$\alpha = \beta$なら伝播函数は$0$である． 4点相関函数

を求める．(4.31)から，1次の展開は \begin{align*} & \bra{0} T \{ \psi_{i\alpha}(x_1) \psi_{j\beta}(x_2) \bar\psi_{k\gamma}(x_3) \bar\psi_{l\delta}(x_4) \left( i \frac{g^2}{2} \right) \int d^4x \sum_{mn} \sum_{\rho\sigma} \bar\psi_{m\rho}(x) \psi_{m\rho}(x) \bar\psi_{n\sigma}(x) \psi_{n\sigma}(x) \} \ket{0} \end{align*} となる．縮約の方法には4通りある：

なら
なら
なら
なら

以上から，

相殺項は

$\delta_f$を求める．

なので，$\delta_f = 0$．

$\delta_g$を求める．頂点の1ループは

から成る． $\log (-p^2)$の発散項にのみ興味があるので，それ以外の項は無視する． 1つ目は \begin{align*} V_s &= - \frac{(ig^2)^2}{2} \sum_{abcd} \sum_{\mu\nu\rho\sigma} \int \frac{d^dk}{(2\pi)^d} \left( \delta_{kc} \delta_{ld} \delta_{\gamma\rho} \delta_{\delta\sigma} + \delta_{kd} \delta_{lc} \delta_{\gamma\sigma} \delta_{\delta\rho} \right) \\ & \quad\times \left( \frac{i}{\cancel{k} + \cancel{p}} \right)_{\rho\mu} \delta_{ac} \left( \frac{i}{-\cancel{k}} \right)_{\sigma\nu} \delta_{bd} \left( \delta_{ia} \delta_{jb} \delta_{\alpha\mu} \delta_{\beta\nu} + \delta_{ib} \delta_{ja} \delta_{\alpha\nu} \delta_{\beta\mu} \right) . \end{align*} 和を計算すれば \begin{align*} & \sum_{abcd} \sum_{\mu\nu\rho\sigma} \left( \delta_{kc} \delta_{ld} \delta_{\gamma\rho} \delta_{\delta\sigma} + \delta_{kd} \delta_{lc} \delta_{\gamma\sigma} \delta_{\delta\rho} \right) \left( \frac{i}{\cancel{k} + \cancel{p}} \right)_{\rho\mu} \delta_{ac} \left( \frac{i}{-\cancel{k}} \right)_{\sigma\nu} \delta_{bd} \\ & \quad\times \left( \delta_{ia} \delta_{jb} \delta_{\alpha\mu} \delta_{\beta\nu} + \delta_{ib} \delta_{ja} \delta_{\alpha\nu} \delta_{\beta\mu} \right) \\ % % &= \sum_{ab} \sum_{\mu\nu\rho\sigma} \left( \delta_{ka} \delta_{lb} \delta_{\gamma\rho} \delta_{\delta\sigma} + \delta_{kb} \delta_{la} \delta_{\gamma\sigma} \delta_{\delta\rho} \right) \left( \frac{i}{\cancel{k} + \cancel{p}} \right)_{\rho\mu} \left( \frac{i}{-\cancel{k}} \right)_{\sigma\nu} \\ & \quad\times \left( \delta_{ia} \delta_{jb} \delta_{\alpha\mu} \delta_{\beta\nu} + \delta_{ib} \delta_{ja} \delta_{\alpha\nu} \delta_{\beta\mu} \right) \\ % % &= \sum_{ab} \sum_{\mu\nu\rho\sigma} \left( \delta_{ka} \delta_{lb} \delta_{\gamma\rho} \delta_{\delta\sigma} \right) \left( \frac{i}{\cancel{k} + \cancel{p}} \right)_{\rho\mu} \left( \frac{i}{-\cancel{k}} \right)_{\sigma\nu} \left( \delta_{ia} \delta_{jb} \delta_{\alpha\mu} \delta_{\beta\nu} \right) \\ % & \quad + \sum_{ab} \sum_{\mu\nu\rho\sigma} \left( \delta_{ka} \delta_{lb} \delta_{\gamma\rho} \delta_{\delta\sigma} \right) \left( \frac{i}{\cancel{k} + \cancel{p}} \right)_{\rho\mu} \left( \frac{i}{-\cancel{k}} \right)_{\sigma\nu} \left( \delta_{ib} \delta_{ja} \delta_{\alpha\nu} \delta_{\beta\mu} \right) \\ % & \quad + \sum_{ab} \sum_{\mu\nu\rho\sigma} \left( \delta_{kb} \delta_{la} \delta_{\gamma\sigma} \delta_{\delta\rho} \right) \left( \frac{i}{\cancel{k} + \cancel{p}} \right)_{\rho\mu} \left( \frac{i}{-\cancel{k}} \right)_{\sigma\nu} \left( \delta_{ia} \delta_{jb} \delta_{\alpha\mu} \delta_{\beta\nu} \right) \\ % & \quad + \sum_{ab} \sum_{\mu\nu\rho\sigma} \left( \delta_{kb} \delta_{la} \delta_{\gamma\sigma} \delta_{\delta\rho} \right) \left( \frac{i}{\cancel{k} + \cancel{p}} \right)_{\rho\mu} \left( \frac{i}{-\cancel{k}} \right)_{\sigma\nu} \left( \delta_{ib} \delta_{ja} \delta_{\alpha\nu} \delta_{\beta\mu} \right) \\ % % &= \left( \delta_{ik} \delta_{jl} \right) \left( \frac{1}{\cancel{k} + \cancel{p}} \right)_{\gamma\alpha} \left( \frac{1}{\cancel{k}} \right)_{\delta\beta} + \left( \delta_{il} \delta_{jk} \right) \left( \frac{1}{\cancel{k} + \cancel{p}} \right)_{\gamma\beta} \left( \frac{1}{\cancel{k}} \right)_{\delta\alpha} \\ % & \quad + \left( \delta_{il} \delta_{jk} \right) \left( \frac{1}{\cancel{k} + \cancel{p}} \right)_{\delta\alpha} \left( \frac{1}{\cancel{k}} \right)_{\gamma\beta} + \left( \delta_{ik} \delta_{jl} \right) \left( \frac{1}{\cancel{k} + \cancel{p}} \right)_{\delta\beta} \left( \frac{1}{\cancel{k}} \right)_{\gamma\alpha} \end{align*} なので， \begin{align*} V_s &= g^4 \left( \delta_{ik} \delta_{jl} \right) \int \frac{d^dk}{(2\pi)^d} \frac{(\cancel{k} + \cancel{p})_{\gamma\alpha}(\cancel{k})_{\delta\beta} + (\cancel{k})_{\gamma\alpha}(\cancel{k} + \cancel{p})_{\delta\beta}}{2k^2(k+p)^2} \\ & \quad + g^4 \left( \delta_{il} \delta_{jk} \right) \int \frac{d^dk}{(2\pi)^d} \frac{(\cancel{k} + \cancel{p})_{\gamma\beta}(\cancel{k})_{\delta\alpha} + (\cancel{k})_{\gamma\beta}(\cancel{k} + \cancel{p})_{\delta\alpha}}{k^2(k+p)^2} \\ % &\sim g^4 \delta_{ik} \delta_{jl} (\gamma^\mu)_{\gamma\alpha}(\gamma^\nu)_{\delta\beta} \int \frac{d^dk}{(2\pi)^d} \frac{k_\mu k_\nu}{k^2(k+p)^2} + g^4 \delta_{il} \delta_{jk} (\gamma^\mu)_{\gamma\beta}(\gamma^\nu)_{\delta\alpha} \int \frac{d^dk}{(2\pi)^d} \frac{k_\mu k_\nu}{k^2(k+p)^2} \\ % &= g^4 \left[ \delta_{ik} \delta_{jl} (\gamma^\mu)_{\gamma\alpha}(\gamma^\nu)_{\delta\beta} + \delta_{il} \delta_{jk} (\gamma^\mu)_{\gamma\beta}(\gamma^\nu)_{\delta\alpha} \right] \int \frac{d^dk}{(2\pi)^d} \frac{1}{k^2(k+p)^2} \frac{k^2 g_{\mu\nu}}{d} \\ % &= g^4 \left[ \delta_{ik} \delta_{jl} (\gamma^\mu)_{\gamma\alpha}(\gamma_\mu)_{\delta\beta} + \delta_{il} \delta_{jk} (\gamma^\mu)_{\gamma\beta}(\gamma_\mu)_{\delta\alpha} \right] \frac{1}{d} \int \frac{d^dk}{(2\pi)^d} \frac{1}{(k+p)^2} \\ % &= g^4 \left[ \delta_{ik} \delta_{jl} (\gamma^\mu)_{\gamma\alpha}(\gamma_\mu)_{\delta\beta} + \delta_{il} \delta_{jk} (\gamma^\mu)_{\gamma\beta}(\gamma_\mu)_{\delta\alpha} \right] \frac{1}{d} \int \frac{d^dk}{(2\pi)^d} \frac{1}{k^2} , \end{align*} $k_\mu k_\nu$の変換に(A.41)を使った． 2つ目は \begin{align*} V_t &= (ig^2)^2 \sum_{abcd} \sum_{\mu\nu\rho\sigma} \int \frac{d^dk}{(2\pi)^d} \left( \delta_{ik} \delta_{ac} \delta_{\alpha\gamma} \delta_{\mu\rho} + \delta_{ic} \delta_{ak} \delta_{\alpha\rho} \delta_{\mu\gamma} \right) \\ & \quad\times \left( \frac{1}{\cancel{k} + \cancel{p}} \right)_{\sigma\rho} \delta_{cd} \left( \frac{1}{\cancel{k}} \right)_{\mu\nu} \delta_{ab} \left( \delta_{jl} \delta_{db} \delta_{\beta\delta} \delta_{\sigma\nu} + \delta_{jb} \delta_{dl} \delta_{\beta\nu} \delta_{\sigma\delta} \right) . \end{align*} 和を計算すれば \begin{align*} & \sum_{abcd} \sum_{\mu\nu\rho\sigma} \left( \delta_{ik} \delta_{ac} \delta_{\alpha\gamma} \delta_{\mu\rho} + \delta_{ic} \delta_{ak} \delta_{\alpha\rho} \delta_{\mu\gamma} \right) \left( \frac{1}{\cancel{k} + \cancel{p}} \right)_{\sigma\rho} \delta_{cd} \left( \frac{1}{\cancel{k}} \right)_{\mu\nu} \delta_{ab} \\ & \quad\times \left( \delta_{jl} \delta_{db} \delta_{\beta\delta} \delta_{\sigma\nu} + \delta_{jb} \delta_{dl} \delta_{\beta\nu} \delta_{\sigma\delta} \right) \\ % % &= \sum_{ac} \sum_{\mu\nu\rho\sigma} \left( \delta_{ik} \delta_{ac} \delta_{\alpha\gamma} \delta_{\mu\rho} + \delta_{ic} \delta_{ak} \delta_{\alpha\rho} \delta_{\mu\gamma} \right) \left( \frac{1}{\cancel{k} + \cancel{p}} \right)_{\sigma\rho} \left( \frac{1}{\cancel{k}} \right)_{\mu\nu} \\ & \quad\times \left( \delta_{jl} \delta_{ca} \delta_{\beta\delta} \delta_{\sigma\nu} + \delta_{ja} \delta_{cl} \delta_{\beta\nu} \delta_{\sigma\delta} \right) \\ % % &= \sum_{ac} \sum_{\mu\nu\rho\sigma} \left( \delta_{ik} \delta_{ac} \delta_{\alpha\gamma} \delta_{\mu\rho} \right) \left( \frac{1}{\cancel{k} + \cancel{p}} \right)_{\sigma\rho} \left( \frac{1}{\cancel{k}} \right)_{\mu\nu} \left( \delta_{jl} \delta_{ca} \delta_{\beta\delta} \delta_{\sigma\nu} \right) \\ % & \quad + \sum_{ac} \sum_{\mu\nu\rho\sigma} \left( \delta_{ik} \delta_{ac} \delta_{\alpha\gamma} \delta_{\mu\rho} \right) \left( \frac{1}{\cancel{k} + \cancel{p}} \right)_{\sigma\rho} \left( \frac{1}{\cancel{k}} \right)_{\mu\nu} \left( \delta_{ja} \delta_{cl} \delta_{\beta\nu} \delta_{\sigma\delta} \right) \\ % & \quad + \sum_{ac} \sum_{\mu\nu\rho\sigma} \left( \delta_{ic} \delta_{ak} \delta_{\alpha\rho} \delta_{\mu\gamma} \right) \left( \frac{1}{\cancel{k} + \cancel{p}} \right)_{\sigma\rho} \left( \frac{1}{\cancel{k}} \right)_{\mu\nu} \left( \delta_{jl} \delta_{ca} \delta_{\beta\delta} \delta_{\sigma\nu} \right) \\ % & \quad + \sum_{ac} \sum_{\mu\nu\rho\sigma} \left( \delta_{ic} \delta_{ak} \delta_{\alpha\rho} \delta_{\mu\gamma} \right) \left( \frac{1}{\cancel{k} + \cancel{p}} \right)_{\sigma\rho} \left( \frac{1}{\cancel{k}} \right)_{\mu\nu} \left( \delta_{ja} \delta_{cl} \delta_{\beta\nu} \delta_{\sigma\delta} \right) \\ % % &= N \left( \delta_{ik} \delta_{jl} \delta_{\alpha\gamma} \delta_{\beta\delta} \right) \Tr \left( \frac{1}{\cancel{k} + \cancel{p}} \frac{1}{\cancel{k}} \right) + \left( \delta_{ik} \delta_{jl} \delta_{\alpha\gamma} \right) \left( \frac{1}{\cancel{k} + \cancel{p}} \frac{1}{\cancel{k}} \right)_{\delta\beta} \\ % & \quad + \left( \delta_{ik} \delta_{jl} \delta_{\beta\delta} \right) \left( \frac{1}{\cancel{k} + \cancel{p}} \frac{1}{\cancel{k}} \right)_{\gamma\alpha} + \left( \delta_{il} \delta_{jk} \right) \left( \frac{1}{\cancel{k}} \right)_{\delta\alpha} \left( \frac{1}{\cancel{k} + \cancel{p}} \right)_{\gamma\beta} \\ % % &\sim (2N+2) \left( \delta_{ik} \delta_{jl} \delta_{\alpha\gamma} \delta_{\beta\delta} \right) \frac{k^2 + k \cdot p}{k^2(k+p)^2} + \left( \delta_{il} \delta_{jk} \right) \left( \frac{1}{\cancel{k}} \right)_{\delta\alpha} \left( \frac{1}{\cancel{k} + \cancel{p}} \right)_{\gamma\beta} \end{align*} なので \begin{align*} V_t &= - g^4 (2N+2) \delta_{ik} \delta_{jl} \delta_{\alpha\gamma} \delta_{\beta\delta} \int \frac{d^dk}{(2\pi)^d} \frac{k^2 + k \cdot p}{k^2(k+p)^2} - g^4 \delta_{il} \delta_{jk} (\gamma^\mu)_{\gamma\beta}(\gamma^\nu)_{\delta\alpha} \int \frac{d^dk}{(2\pi)^d} \frac{k_\mu k_\nu}{k^4} \\ % &= - g^4 (2N+2) \delta_{ik} \delta_{jl} \delta_{\alpha\gamma} \delta_{\beta\delta} \int \frac{d^dk}{(2\pi)^d} \frac{k^2 + k \cdot p}{k^2(k+p)^2} - g^4 \delta_{il} \delta_{jk} (\gamma^\mu)_{\gamma\beta}(\gamma_\mu)_{\delta\alpha} \frac{1}{d} \int \frac{d^dk}{(2\pi)^d} \frac{1}{k^2} . \end{align*} 3つ目は$V_t$で$k\gamma \leftrightarrow l\delta$を入れ替えたもの： \[ V_u = - g^4 (2N+2) \delta_{il} \delta_{jk} \delta_{\alpha\delta} \delta_{\beta\gamma} \int \frac{d^dk}{(2\pi)^d} \frac{k^2 + k \cdot p}{k^2(k+p)^2} - g^4 \delta_{ik} \delta_{jl} (\gamma^\mu)_{\delta\beta}(\gamma_\mu)_{\gamma\alpha} \frac{1}{d} \int \frac{d^dk}{(2\pi)^d} \frac{1}{k^2} . \] 以上から \begin{align*} V_s + V_t + V_u &= - g^4 (2N+2) \left( \delta_{ik} \delta_{jl} \delta_{\alpha\gamma} \delta_{\beta\delta} + \delta_{il} \delta_{jk} \delta_{\alpha\delta} \delta_{\beta\gamma} \right) \int \frac{d^dk}{(2\pi)^d} \frac{k^2 + p \cdot k}{k^2(k+p)^2} \\ % &= - g^4 (2N+2) \left( \delta_{ik} \delta_{jl} \delta_{\alpha\gamma} \delta_{\beta\delta} + \delta_{il} \delta_{jk} \delta_{\alpha\delta} \delta_{\beta\gamma} \right) \int_0^1 dx \int \frac{d^d\ell}{(2\pi)^d} \frac{\ell^2 + \Delta}{(\ell^2 - \Delta)^2} \\ % &\approx ig^4 (2N+2) \left( \delta_{ik} \delta_{jl} \delta_{\alpha\gamma} \delta_{\beta\delta} + \delta_{il} \delta_{jk} \delta_{\alpha\delta} \delta_{\beta\gamma} \right) \int_0^1 dx \frac{1}{(4\pi)^{d/2}} \frac{d}{2} \Gamma(1-d/2) \left( \frac{1}{\Delta} \right)^{1-d/2} . \end{align*} $\epsilon = 2-d$とすれば \begin{align*} \frac{1}{(4\pi)^{d/2}} \frac{d}{2} \Gamma(1-d/2) \left( \frac{1}{\Delta} \right)^{1-d/2} &= \frac{1}{4\pi} (1-\epsilon/2) \Gamma(\epsilon/2) \left( \frac{4\pi}{\Delta} \right)^{\epsilon/2} \\ &\approx \frac{1}{4\pi} \left( 1 - \frac{\epsilon}{2} \right) \left( \frac{2}{\epsilon} - \gamma \right) \left( 1 - \frac{\epsilon}{2} \log\frac{\Delta}{4\pi} \right) \\ &\approx \frac{1}{4\pi} \left( \frac{2}{\epsilon} - \gamma - \log\frac{\Delta}{4\pi} \right) \\ &\approx - \frac{1}{4\pi} \log(-p^2) \end{align*} なので， \[ V_s + V_t + V_u = - \frac{ig^4}{4\pi} (2N+2) \left( \delta_{ik} \delta_{jl} \delta_{\alpha\gamma} \delta_{\beta\delta} + \delta_{il} \delta_{jk} \delta_{\alpha\delta} \delta_{\beta\gamma} \right) \log(-p^2) . \] $p^2 = -M^2$で，これと相殺項の和が$0$なので， \[ \delta_g = \frac{g^3}{4\pi} (N+1) \log M^2 = \frac{g^3}{2\pi} (N+1) \log M . \]

Chapter 16: Quantization of Non-Abelian Gauge Theories

16.3 Ghosts and Unitarity

Figure 16.5

(16.32)から \begin{align*} \mathcal{L}_\text{ghost} &= \bar{c}^a \left( -g\partial^\mu f^{abc} A^b_\mu \right) c^c + \cdots \\ &= -gf^{abc} \bar{c}^a (\partial^\mu A^b_\mu) c^c - gf^{abc} \bar{c}^a A^b_\mu (\partial^\mu c^c) + \cdots \\ &= -gf^{abc} \bar{c}^a (-iq^\mu A^b_\mu) c^c - gf^{abc} \bar{c}^a A^b_\mu (-ik^\mu c^c) + \cdots \\ &= -gf^{abc} \bar{c}^a (-ip^\mu A^b_\mu) c^c + \cdots \end{align*} となるので，頂点は \[ i\mathcal{L} \to -gf^{abc}p_\mu . \]