Sen, Abhijit, and Zurab K. Silagadze. "Two-photon decay of P-wave positronium: a tutorial." Canadian Journal of Physics 97.7 (2019): 693-700. の内容を簡略化した． Peskinの本の中で，このProblem 5.4が一番大変だった．
$e^-e^+ \to 2\gamma$の過程を考える．
束縛状態の式(5.43) \[ \ket{B} = \sqrt{2M} \int \frac{d^3k}{(2\pi)^3} \tilde\psi(\boldsymbol{k}) \frac{1}{\sqrt{2m}} \frac{1}{\sqrt{2m}} \ket{\boldsymbol{k}\uparrow, -\boldsymbol{k}\uparrow} \] を角運動量を考慮した形に拡張する． すなわち，スピン$S$，軌道角運動量$l$，全角運動量$J$，全角運動量の射影$M$の状態$\ket{^{2S+1}l_J; M}$を構成しよう（スピンは粒子の内在的な量なので，電子のスピン，陽電子のスピンを足す．それに対し軌道角運動量は粒子の相対運動に起因するので，電子と陽電子について和を取ることは行わない）． これには，スピンの射影が$S_z$で軌道角運動量の射影が$M-S_Z=:m$の状態をClebash-Gordan係数$\braket{lS; mS_z | lS; JM}$によって足せば良い： \begin{align} \ket{^{2S+1}l_J; M} = \sqrt{2M} \int \frac{d^3p}{(2\pi)^3} \sum_{S_z=-S}^S \braket{lS; mS_z | lS; JM} \tilde{\psi}_{lm}(\boldsymbol{p}) \ket{S, S_z}_{\boldsymbol{p}} \label{prob5_4a_Ps_wf} \end{align} （ただし，慣例に従い$l=0$は$S$，$l=1$は$P$などと表記する）． なお，電子の運動量は$\boldsymbol{p}$，陽電子の運動量は$-\boldsymbol{p}$とする． 換算質量は$m/2$なので，相対運動量は$\boldsymbol{p}$である．
不変振幅$\mathcal{M}$の定義(4.73)に注意して，\eqref{prob5_4a_iM_cal}\eqref{prob5_4a_spinor}\eqref{prob5_4a_1S_0;0}から \begin{align} \begin{split} i\mathcal{M}(^1S_0 \to 2\gamma) &= 2\sqrt{m} \int \frac{d^3p}{(2\pi)^3} \tilde{\psi}_{00}(\boldsymbol{p}) \frac{1}{2m} \frac{i\mathcal{M}(e^-_\uparrow e^+_\downarrow \to 2\gamma) - i\mathcal{M}(e^-_\downarrow e^+_\uparrow \to 2\gamma)}{\sqrt{2}} \\ &= 2\sqrt{m} \int \frac{d^3p}{(2\pi)^3} \tilde{\psi}_{00}(\boldsymbol{p}) \frac{1}{2m} \left( -\frac{2e^2}{m} \right) (\boldsymbol{\epsilon}_1^\ast \times \boldsymbol{\epsilon}_2^\ast) \cdot \boldsymbol{k} \frac{ \eta^{\downarrow\dagger}\xi^\uparrow - \eta^{\uparrow\dagger}\xi^\downarrow}{\sqrt{2}} \\ &= 2\sqrt{m} \int \frac{d^3p}{(2\pi)^3} \tilde{\psi}_{00}(\boldsymbol{p}) \frac{1}{2m} \left( -\frac{2e^2}{m} \right) (\boldsymbol{\epsilon}_1^\ast \times \boldsymbol{\epsilon}_2^\ast) \cdot \boldsymbol{k} \frac{-2}{\sqrt{2}} \\ &= 2\sqrt{m} \int \frac{d^3p}{(2\pi)^3} \tilde{\psi}_{00}(\boldsymbol{p}) \frac{1}{2m} \left( -\frac{2e^2}{m} \right) (\boldsymbol{\epsilon}_1^\ast \times \boldsymbol{\epsilon}_2^\ast) \cdot \boldsymbol{k} \frac{-2}{\sqrt{2}} \\ &= 2\sqrt{2} \frac{e^2}{m\sqrt{m}} (\boldsymbol{\epsilon}_1^\ast \times \boldsymbol{\epsilon}_2^\ast) \cdot \boldsymbol{k} \int \frac{d^3p}{(2\pi)^3} \tilde{\psi}_{00}(\boldsymbol{p}) \\ &= 2\sqrt{2} \frac{e^2}{m\sqrt{m}} (\boldsymbol{\epsilon}_1^\ast \times \boldsymbol{\epsilon}_2^\ast) \cdot \boldsymbol{k} \psi_{00}(0) . \end{split} \label{prob5_4a_1S_0_2hv} \end{align} (A.26)から \[ \sum_\text{polarization} \epsilon^{\mu\ast}\epsilon^\nu \to -g^{\mu\nu} \] なので， \begin{align} \begin{split} \lvert (\boldsymbol{\epsilon}_1^\ast \times \boldsymbol{\epsilon}_2^\ast) \cdot \boldsymbol{k} \rvert^2 &= \sum_\text{pol 1}\sum_\text{pol 2}\sum_{ijklmn} \epsilon^{ijk} \epsilon_1^{i\ast}\epsilon_2^{j\ast} k^k \epsilon^{lmn} \epsilon_1^l\epsilon_2^m k^n \\ &\to \sum_{ijklmn} \epsilon^{ijk}\epsilon^{lmn} g^{il} g^{jm} k^k k^n \\ &= \sum_{klmn} \epsilon^{lmk}\epsilon^{lmn} k^k k^n \\ &= \sum_{lmn} \epsilon^{lmn}\epsilon^{lmn} k^n k^n \\ &= 2\lvert\boldsymbol{k}\rvert^2 = 2E^2 \approx 2m^2 \end{split} \label{prob5_4a_pol_sum} \end{align} を得る． $n=1$の場合（換算質量は$m/2$なので，Bohr半径$a_0 = 2/\alpha m$）を考える． \[ \psi_{100} = R_{10}(r) Y_{00}(\theta, \phi) = \frac{1}{\sqrt{4\pi}} \frac{2}{a_0{}^{3/2}} \exp \left( - \frac{r}{2a_0} \right) \] なので， \begin{align} \lvert \psi_{100}(0) \rvert^2 = \frac{m^3\alpha^3}{8\pi} . \label{prob5_4a_wf_100} \end{align} \eqref{prob5_4a_1S_0_2hv}\eqref{prob5_4a_pol_sum}\eqref{prob5_4a_wf_100}から \begin{align} \sum_\text{polarization} \lvert \mathcal{M}(1^1S_0 \to 2\gamma) \rvert^2 &= 8 \frac{e^4}{m^3} 2m^2 \frac{m^3\alpha^3}{8\pi} = 32\pi m^2 \alpha^5 . \end{align} (4.86)から \begin{align*} \Gamma(1^1S_0 \to 2\gamma) &= \frac{1}{2} \frac{1}{4m} \int \frac{d^3k_1}{(2\pi)^3}\frac{d^3k_2}{(2\pi)^3} \frac{1}{4E_1E_2} \lvert \mathcal{M}(^1S_0 \to 2\gamma) \rvert^2 (2\pi)^4 \mathop{\delta^{(4)}}(k_1+k_2-p_1-p_2) \\ &= \frac{1}{8m} \int \frac{d^3k_1}{(2\pi)^2} \frac{1}{4\lvert \boldsymbol{k}_1 \rvert^2} 32\pi m^2 \alpha^5 \mathop\delta(2E_1 - 2m) \\ &= \pi m\alpha^5 4\pi \int \frac{d\lvert \boldsymbol{k}_1 \rvert}{(2\pi)^2} \frac{1}{2} \mathop\delta(\lvert \boldsymbol{k}_1 \rvert - m) \\ &= \frac{m\alpha^5}{2} \end{align*} （出てくる光子は区別できないので，$1/2$倍する）． \eqref{prob5_4a_1S_0_2hv}と同様に考えれば， \begin{align*} \eta^{\uparrow\dagger}\xi^\uparrow = 0 & , & \frac{\eta^{\uparrow\dagger}\xi^\downarrow + \eta^{\downarrow\dagger}\xi^\uparrow}{\sqrt{2}} = 0 & , & \eta^{\downarrow\dagger}\xi^\downarrow = 0 \end{align*} なので，$\mathcal{M}(^3S_1 \to 2\gamma) = 0$であることが分かる． すなわち，スピン$1$の$1^3S$は2光子に崩壊しない．
対消滅の不変振幅を運動量$p$の$1$次までの精度で求める． \begin{align*} p_1^\mu = (E, \boldsymbol{p}) & , & p_2^\mu = (E, -\boldsymbol{p}) & , & k_1^\mu = (E, \boldsymbol{k}) & , & k_2^\mu = (E, -\boldsymbol{k}) . \end{align*} 光子の偏極は次のようにおく： \begin{align*} \epsilon_{\pm}^\mu(k_1) = \epsilon_{1\pm}^\mu = (0, \boldsymbol{\epsilon}_1) & , & \boldsymbol{\epsilon}_1 \cdot \boldsymbol{k}_1 = 0 & , & \epsilon_{\pm}^\mu(k_2) = \epsilon_{2\pm}^\mu = (0, \boldsymbol{\epsilon}_2) & , & \boldsymbol{\epsilon}_2 \cdot \boldsymbol{k}_2 = 0 . \end{align*} スピノルは次のように近似される： \begin{align} u(p_1) = \begin{pmatrix} \sqrt{\sigma \cdot p_1} \xi \\ \sqrt{\overline\sigma \cdot p_1} \xi \end{pmatrix} \approx \sqrt{m} \begin{pmatrix} \left( 1 - \dfrac{\boldsymbol\sigma \cdot \boldsymbol{p}}{2m} \right) \xi \\[10pt] \left( 1 + \dfrac{\boldsymbol\sigma \cdot \boldsymbol{p}}{2m} \right) \xi \end{pmatrix} & , & v(p_2) = \begin{pmatrix} \sqrt{\sigma \cdot p_2} \eta \\ -\sqrt{\overline\sigma \cdot p_2} \eta \end{pmatrix} \approx \sqrt{m} \begin{pmatrix} \left( 1 + \dfrac{\boldsymbol\sigma \cdot \boldsymbol{p}}{2m} \right) \eta \\[10pt] -\left( 1 - \dfrac{\boldsymbol\sigma \cdot \boldsymbol{p}}{2m} \right) \eta \end{pmatrix} . \label{prob5_4b_spinor_approx} \end{align} さらに， \begin{align*} p_1 \cdot k_1 &= E^2 - \boldsymbol{p} \cdot \boldsymbol{k} \approx m^2 - \boldsymbol{p} \cdot \boldsymbol{k}, \\ p_1 \cdot k_2 &= E^2 + \boldsymbol{p} \cdot \boldsymbol{k} \approx m^2 + \boldsymbol{p} \cdot \boldsymbol{k}, \\ (p_1 \cdot k_1)(p_1 \cdot k_2) &\approx m^4 . \end{align*} Dirac方程式から次を得る： \begin{align*} (\cancel{p}_1 + m) \cancel{\epsilon}_1^\ast u(p_1) &= 2 (p_1 \cdot \epsilon_1^\ast) u(p_1) - \cancel{\epsilon}_1^\ast (\cancel{p}_1 - m) u(p_1) = 2 (p_1 \cdot \epsilon_1^\ast) u(p_1) , \\ (\cancel{p}_1 + m) \cancel{\epsilon}_2^\ast u(p_1) &= 2 (p_1 \cdot \epsilon_2^\ast) u(p_1) - \cancel{\epsilon}_2^\ast (\cancel{p}_1 - m) u(p_1) = 2 (p_1 \cdot \epsilon_2^\ast) u(p_1) . \end{align*} 従って，不変振幅は \begin{align} \begin{split} i\mathcal{M} &= -ie^2 \overline{v}(p_2) \left[ \cancel{\epsilon}_2^\ast \frac{\cancel{p}_1 - \cancel{k}_1 + m}{(p_1 - k_1)^2 - m^2} \cancel{\epsilon}_1^\ast + \cancel{\epsilon}_1^\ast \frac{\cancel{p}_1 - \cancel{k}_2 + m}{(p_1 - k_2)^2 - m^2} \cancel{\epsilon}_2^\ast \right] u(p_1) \\ % &= -ie^2 \overline{v}(p_2) \left[ \cancel{\epsilon}_2^\ast \frac{-\cancel{p}_1 + \cancel{k}_1 - m}{2p_1 \cdot k_1} \cancel{\epsilon}_1^\ast + \cancel{\epsilon}_1^\ast \frac{-\cancel{p}_1 + \cancel{k}_2 - m}{2p_1 \cdot k_2} \cancel{\epsilon}_2^\ast \right] u(p_1) \\ % &= -i\frac{e^2}{2} \overline{v}(p_2) \left[ \frac{\cancel{\epsilon}_2^\ast\cancel{k}_1\cancel{\epsilon}_1^\ast - 2(p_1 \cdot \epsilon_1^\ast)\cancel{\epsilon}_2^\ast}{m^2 - \boldsymbol{p} \cdot \boldsymbol{k}} + \frac{\cancel{\epsilon}_1^\ast\cancel{k}_2\cancel{\epsilon}_2^\ast - 2(p_1 \cdot \epsilon_2^\ast)\cancel{\epsilon}_1^\ast}{m^2 + \boldsymbol{p} \cdot \boldsymbol{k}} \right] u(p_1) \\ % &= -i\frac{e^2}{2m^4} \overline{v}(p_2) \left[ (m^2 + \boldsymbol{p} \cdot \boldsymbol{k})(\cancel{\epsilon}_2^\ast\cancel{k}_1\cancel{\epsilon}_1^\ast - 2(p_1 \cdot \epsilon_1^\ast)\cancel{\epsilon}_2^\ast) + (m^2 - \boldsymbol{p} \cdot \boldsymbol{k})(\cancel{\epsilon}_1^\ast\cancel{k}_2\cancel{\epsilon}_2^\ast - 2(p_1 \cdot \epsilon_2^\ast)\cancel{\epsilon}_1^\ast) \right] u(p_1) \end{split} \label{prob5_4b_iM} \end{align} となる．$[\cdots]$を計算するが，まず次の式を示しておく： \begin{align*} \cancel{a}\cancel{b}\cancel{c} &= a_\mu b_\nu c_\rho \begin{pmatrix} 0 & \sigma^\mu \\ \overline{\sigma}^\mu & 0 \end{pmatrix} \begin{pmatrix} 0 & \sigma^\nu \\ \overline{\sigma}^\nu & 0 \end{pmatrix} \begin{pmatrix} 0 & \sigma^\rho \\ \overline{\sigma}^\rho & 0 \end{pmatrix} \\ &= a_\mu b_\nu c_\rho \begin{pmatrix} 0 & \sigma^\mu \overline{\sigma}^\nu \sigma^\rho \\ \overline{\sigma}^\mu \sigma^\nu \overline{\sigma}^\rho & 0 \end{pmatrix} \\ &= \begin{pmatrix} 0 & (a \cdot \sigma)(b \cdot \overline\sigma)(c \cdot \sigma) \\ (a \cdot \overline\sigma)(b \cdot \sigma)(c \cdot \overline\sigma) & 0 \end{pmatrix} . \end{align*} \eqref{prob5_4b_iM}の$[\cdots]$のうち$m^2$と1つのガンマ行列を含む項は \begin{align*} & -2m^2 (p_1 \cdot \epsilon_1^\ast) \cancel{\epsilon}_2^\ast -2m^2 (p_1 \cdot \epsilon_2^\ast) \cancel{\epsilon}_1^\ast \\ &= -2m^2 (p_1 \cdot \epsilon_1^\ast) \begin{pmatrix} 0 & \epsilon_2^\ast \cdot \sigma \\ \epsilon_2^\ast \cdot \overline\sigma & 0 \end{pmatrix} -2m^2 (p_1 \cdot \epsilon_2^\ast) \begin{pmatrix} 0 & \epsilon_1^\ast \cdot \sigma \\ \epsilon_1^\ast \cdot \overline\sigma & 0 \end{pmatrix} \\ % &= -2m^2 (\boldsymbol{p} \cdot \boldsymbol{\epsilon}_1^\ast) \begin{pmatrix} 0 & \boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma \\ -\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma & 0 \end{pmatrix} -2m^2 (\boldsymbol{p} \cdot \boldsymbol{\epsilon}_2^\ast) \begin{pmatrix} 0 & \boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma \\ -\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma & 0 \end{pmatrix} \\ % &= -2m^2 \begin{pmatrix} 0 & (\boldsymbol{p} \cdot \boldsymbol{\epsilon}_1^\ast)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + (\boldsymbol{p} \cdot \boldsymbol{\epsilon}_2^\ast)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) \\ -(\boldsymbol{p} \cdot \boldsymbol{\epsilon}_1^\ast)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) - (\boldsymbol{p} \cdot \boldsymbol{\epsilon}_2^\ast)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) & 0 \end{pmatrix} \\ & =: \begin{pmatrix} 0 & -2m^2 A \\ 2m^2 A & 0 \end{pmatrix} . \end{align*} \eqref{prob5_4b_iM}の$[\cdots]$のうち$m^2$と3つのガンマ行列を含む項は \begin{align*} & m^2\left[ \cancel{\epsilon}_2^\ast\cancel{k}_1\cancel{\epsilon}_1^\ast + \cancel{\epsilon}_1^\ast\cancel{k}_2\cancel{\epsilon}_2^\ast \right] \\ &\quad = m^2 \begin{pmatrix} 0 & (\epsilon_2^\ast \cdot \sigma)(k_1 \cdot \overline\sigma)(\epsilon_1^\ast \cdot \sigma) \\ (\epsilon_2^\ast \cdot \overline\sigma)(k_1 \cdot \sigma)(\epsilon_1^\ast \cdot \overline\sigma) & 0 \end{pmatrix} \\ & \quad + m^2 \begin{pmatrix} 0 & (\epsilon_1^\ast \cdot \sigma)(k_2 \cdot \overline\sigma)(\epsilon_2^\ast \cdot \sigma) \\ (\epsilon_1^\ast \cdot \overline\sigma)(k_2 \cdot \sigma)(\epsilon_2^\ast \cdot \overline\sigma) & 0 \end{pmatrix} \\ &\quad = m^2 \begin{pmatrix} 0 & (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(k_1 \cdot \overline\sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) \\ (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(k_1 \cdot \sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) & 0 \end{pmatrix} \\ & \quad + m^2 \begin{pmatrix} 0 & (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(k_2 \cdot \overline\sigma)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) \\ (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(k_2 \cdot \sigma)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) & 0 \end{pmatrix} \\ &=: \begin{pmatrix} 0 & m^2\overline{B}_+ \\ m^2 B_+ & 0 \end{pmatrix} . \end{align*} \eqref{prob5_4b_iM}の$[\cdots]$のうち$\boldsymbol{p}\cdot\boldsymbol{k}$と3つのガンマ行列を含む項は \begin{align*} & \boldsymbol{p} \cdot \boldsymbol{k}\left[ \cancel{\epsilon}_2^\ast\cancel{k}_1\cancel{\epsilon}_1^\ast - \cancel{\epsilon}_1^\ast\cancel{k}_2\cancel{\epsilon}_2^\ast \right] \\ &\quad = \boldsymbol{p} \cdot \boldsymbol{k} \begin{pmatrix} 0 & (\epsilon_2^\ast \cdot \sigma)(k_1 \cdot \overline\sigma)(\epsilon_1^\ast \cdot \sigma) \\ (\epsilon_2^\ast \cdot \overline\sigma)(k_1 \cdot \sigma)(\epsilon_1^\ast \cdot \overline\sigma) & 0 \end{pmatrix} \\ & \quad - \boldsymbol{p} \cdot \boldsymbol{k} \begin{pmatrix} 0 & (\epsilon_1^\ast \cdot \sigma)(k_2 \cdot \overline\sigma)(\epsilon_2^\ast \cdot \sigma) \\ (\epsilon_1^\ast \cdot \overline\sigma)(k_2 \cdot \sigma)(\epsilon_2^\ast \cdot \overline\sigma) & 0 \end{pmatrix} \\ &\quad = \boldsymbol{p} \cdot \boldsymbol{k} \begin{pmatrix} 0 & (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(k_1 \cdot \overline\sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) \\ (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(k_1 \cdot \sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) & 0 \end{pmatrix} \\ & \quad - \boldsymbol{p} \cdot \boldsymbol{k} \begin{pmatrix} 0 & (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(k_2 \cdot \overline\sigma)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) \\ (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(k_2 \cdot \sigma)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) & 0 \end{pmatrix} \\ &=: \begin{pmatrix} 0 & -(\boldsymbol{p} \cdot \boldsymbol{k})\overline{B}_- \\ -(\boldsymbol{p} \cdot \boldsymbol{k}) B_- & 0 \end{pmatrix} . \end{align*} \eqref{prob5_4b_iM}の残りの項は運動量の$2$乗なので無視する： \begin{align*} & -2 (\boldsymbol{p}\cdot\boldsymbol{k}) (p_1 \cdot \epsilon_1^\ast) \cancel{\epsilon}_2^\ast + 2(\boldsymbol{p}\cdot\boldsymbol{k}) (p_1 \cdot \epsilon_2^\ast) \cancel{\epsilon}_1^\ast \\ &= -2(\boldsymbol{p}\cdot\boldsymbol{k}) (p_1 \cdot \epsilon_1^\ast) \begin{pmatrix} 0 & \epsilon_2^\ast \cdot \sigma \\ \epsilon_2^\ast \cdot \overline\sigma & 0 \end{pmatrix} + 2(\boldsymbol{p}\cdot\boldsymbol{k}) (p_1 \cdot \epsilon_2^\ast) \begin{pmatrix} 0 & \epsilon_1^\ast \cdot \sigma \\ \epsilon_1^\ast \cdot \overline\sigma & 0 \end{pmatrix} \\ % &= -2(\boldsymbol{p}\cdot\boldsymbol{k}) (\boldsymbol{p} \cdot \boldsymbol{\epsilon}_1^\ast) \begin{pmatrix} 0 & \boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma \\ -\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma & 0 \end{pmatrix} + 2(\boldsymbol{p}\cdot\boldsymbol{k}) (\boldsymbol{p} \cdot \boldsymbol{\epsilon}_2^\ast) \begin{pmatrix} 0 & \boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma \\ -\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma & 0 \end{pmatrix} \\ % &= -2(\boldsymbol{p}\cdot\boldsymbol{k}) \begin{pmatrix} 0 & (\boldsymbol{p} \cdot \boldsymbol{\epsilon}_1^\ast)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) - (\boldsymbol{p} \cdot \boldsymbol{\epsilon}_2^\ast)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) \\ -(\boldsymbol{p} \cdot \boldsymbol{\epsilon}_1^\ast)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + (\boldsymbol{p} \cdot \boldsymbol{\epsilon}_2^\ast)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) & 0 \end{pmatrix} \\ &\approx 0 . \end{align*} 以上から \begin{align} i\mathcal{M}= -i\frac{e^2}{2m^4} \overline{v}(p_2) \begin{pmatrix} 0 & -2m^2A + m^2\overline{B}_+ - (\boldsymbol{p}\cdot\boldsymbol{k}) \overline{B}_- \\ 2m^2A + m^2B_+ - (\boldsymbol{p}\cdot\boldsymbol{k}) B_- & 0 \end{pmatrix} u(p_1) \label{prob5_4b_iM_AB} \end{align} となる．各項の定義は \begin{align} \begin{split} A &= (\boldsymbol{p} \cdot \boldsymbol{\epsilon}_1^\ast)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + (\boldsymbol{p} \cdot \boldsymbol{\epsilon}_2^\ast)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) \\ B_\pm &= (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(k_2 \cdot \sigma)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) \pm (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(k_1 \cdot \sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) \\ \overline{B}_\pm &= (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(k_2 \cdot \overline\sigma)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) \pm (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(k_1 \cdot \overline\sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) . \end{split} \label{prob5_4b_AB_def} \end{align} $(\boldsymbol\sigma \cdot \boldsymbol{a}) (\boldsymbol\sigma \cdot \boldsymbol{b}) = \boldsymbol{a} \cdot \boldsymbol{b} + i \boldsymbol\sigma \cdot (\boldsymbol{a} \times \boldsymbol{b})$から \begin{align*} (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(\boldsymbol{k}\cdot\boldsymbol\sigma)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) &= [\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{k} + i \boldsymbol\sigma \cdot (\boldsymbol{\epsilon}_1^\ast \times \boldsymbol{k})](\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) \\ &= i \boldsymbol\sigma \cdot (\boldsymbol{\epsilon}_1^\ast \times \boldsymbol{k})(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) \\ &= i (\boldsymbol{\epsilon}_1^\ast \times \boldsymbol{k}) \cdot \boldsymbol{\epsilon}_2^\ast - \boldsymbol\sigma \cdot [(\boldsymbol{\epsilon}_1^\ast \times \boldsymbol{k}) \times \boldsymbol{\epsilon}_2^\ast] \\ &= i (\boldsymbol{\epsilon}_2^\ast \times \boldsymbol{\epsilon}_1^\ast) \cdot \boldsymbol{k} - \boldsymbol\sigma \cdot [(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast) \boldsymbol{k} - (\boldsymbol{k} \cdot \boldsymbol{\epsilon}_2^\ast) \boldsymbol{\epsilon}_1^\ast] \\ &= - i (\boldsymbol{\epsilon}_1^\ast \times \boldsymbol{\epsilon}_2^\ast) \cdot \boldsymbol{k} - (\boldsymbol\sigma \cdot \boldsymbol{k})(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast) \\ % (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(\boldsymbol{k}\cdot\boldsymbol\sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) &= i (\boldsymbol{\epsilon}_1^\ast \times \boldsymbol{\epsilon}_2^\ast) \cdot \boldsymbol{k} - (\boldsymbol\sigma \cdot \boldsymbol{k})(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast) \\ % (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) &= 2 \boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast \end{align*} となる．従って， \begin{align} \begin{split} \overline{B}_+ - B_+ &= (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(k_2 \cdot \overline\sigma)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(k_1 \cdot \overline\sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) \\ &- (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(k_2 \cdot \sigma)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) - (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(k_1 \cdot \sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) \\ &= (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(k_2 \cdot (\overline\sigma - \sigma))(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(k_1 \cdot (\overline\sigma - \sigma))(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) \\ &= -2(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(\boldsymbol{k}\cdot\boldsymbol\sigma)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + 2(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(\boldsymbol{k}\cdot\boldsymbol\sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) \\ &= 4i (\boldsymbol{\epsilon}_1^\ast \times \boldsymbol{\epsilon}_2^\ast) \cdot \boldsymbol{k} , \\ % B_- - \overline{B}_- &= (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(k_2 \cdot \sigma)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) - (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(k_1 \cdot \sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) \\ &- (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(k_2 \cdot \overline\sigma)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(k_1 \cdot \overline\sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) \\ &= (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(k_2 \cdot (\sigma - \overline\sigma ))(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) - (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(k_1 \cdot (\sigma - \overline\sigma))(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) \\ &= 2(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(\boldsymbol{k}\cdot\boldsymbol\sigma)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + 2(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(\boldsymbol{k}\cdot\boldsymbol\sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) \\ &= -4(\boldsymbol\sigma \cdot \boldsymbol{k})(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast) , \\ % B_+ + \overline{B}_+ &= (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(k_2 \cdot \sigma)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(k_1 \cdot \sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) \\ &+ (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(k_2 \cdot \overline\sigma)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(k_1 \cdot \overline\sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) \\ &= 2E(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + 2E(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) \\ &\approx 2m (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + 2m(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) \\ &= 4m \boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast . \end{split} \label{prob5_4b_BBB} \end{align} \eqref{prob5_4b_iM_AB}に\eqref{prob5_4b_spinor_approx}を代入して，\eqref{prob5_4b_BBB}を使えば， \begin{align*} & i\mathcal{M}(e^-_se^+_r \to 2\gamma) \\ &= -i\frac{e^2}{2m^4} \overline{v}(p_2) \begin{pmatrix} 0 & -2m^2A + m^2\overline{B}_+ - (\boldsymbol{p}\cdot\boldsymbol{k}) \overline{B}_- \\ 2m^2A + m^2B_+ - (\boldsymbol{p}\cdot\boldsymbol{k}) B_- & 0 \end{pmatrix} u(p_1) \\ % &= -i\frac{e^2}{2m^3} \begin{pmatrix} \eta^{r\dagger} \left( 1 + \dfrac{\boldsymbol\sigma \cdot \boldsymbol{p}}{2m} \right) & - \eta^{r\dagger} \left( 1 - \dfrac{\boldsymbol\sigma \cdot \boldsymbol{p}}{2m} \right) \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \\ & \quad \times \begin{pmatrix} 0 & -2m^2A + m^2\overline{B}_+ - (\boldsymbol{p}\cdot\boldsymbol{k}) \overline{B}_- \\ 2m^2A + m^2B_+ - (\boldsymbol{p}\cdot\boldsymbol{k}) B_- & 0 \end{pmatrix} \begin{pmatrix} \left( 1 - \dfrac{\boldsymbol\sigma \cdot \boldsymbol{p}}{2m} \right) \xi^s \\[10pt] \left( 1 + \dfrac{\boldsymbol\sigma \cdot \boldsymbol{p}}{2m} \right) \xi^s \end{pmatrix} \\ % & = i\frac{e^2}{2m^3} \eta^{r\dagger} \left( 1 - \dfrac{\boldsymbol\sigma \cdot \boldsymbol{p}}{2m} \right) \left[ -2m^2A + m^2\overline{B}_+ - (\boldsymbol{p}\cdot\boldsymbol{k}) \overline{B}_- \right] \left( 1 + \dfrac{\boldsymbol\sigma \cdot \boldsymbol{p}}{2m} \right) \xi^s \\ &\quad -i\frac{e^2}{2m^3} \eta^{r\dagger} \left( 1 + \dfrac{\boldsymbol\sigma \cdot \boldsymbol{p}}{2m} \right) \left[ 2m^2A + m^2B_+ - (\boldsymbol{p}\cdot\boldsymbol{k}) B_- \right] \left( 1 - \dfrac{\boldsymbol\sigma \cdot \boldsymbol{p}}{2m} \right) \xi^s \\ \\ % & \approx i\frac{e^2}{2m^3} \eta^{r\dagger} \left\{ -2m^2A + m^2\overline{B}_+ - (\boldsymbol{p}\cdot\boldsymbol{k}) \overline{B}_- + \frac{m}{2} [\overline{B}_+, \boldsymbol\sigma \cdot \boldsymbol{p}] \right\} \xi^s \\ & \quad -i\frac{e^2}{2m^3} \eta^{r\dagger} \left\{ 2m^2A + m^2B_+ - (\boldsymbol{p}\cdot\boldsymbol{k}) B_- - \frac{m}{2} [B_+, \boldsymbol\sigma \cdot \boldsymbol{p}] \right\} \xi^s \\ &= i\frac{e^2}{2m^3} \eta^{r\dagger} \left\{ -4m^2A + m^2(\overline{B}_+ - B_+) + (\boldsymbol{p}\cdot\boldsymbol{k}) (B_- - \overline{B}_-) + \frac{m}{2} [B_+ + \overline{B}_+, \boldsymbol\sigma \cdot \boldsymbol{p}] \right\} \xi^s \\ &= -i \frac{2e^2}{m} \eta^{r\dagger} \left[ (\boldsymbol{p} \cdot \boldsymbol{\epsilon}_1^\ast)(\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + (\boldsymbol{p} \cdot \boldsymbol{\epsilon}_2^\ast)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) - i (\boldsymbol{\epsilon}_1^\ast \times \boldsymbol{\epsilon}_2^\ast) \cdot \boldsymbol{k} + \frac{1}{m^2} (\boldsymbol{p}\cdot\boldsymbol{k})(\boldsymbol\sigma \cdot \boldsymbol{k})(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast) \right] \xi^s \end{align*} となる． ここで， \begin{align} i\mathcal{M}^{(1)j} = -i \frac{2e^2}{m} \eta^{r\dagger} \left[ \epsilon_1^{j\ast} (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + \epsilon_2^{j\ast} (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) + \frac{k^j}{m^2} (\boldsymbol\sigma \cdot \boldsymbol{k})(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast) \right] \xi^s \label{prob5_4b_iM_1_def} \end{align} とおけば， \begin{align} & i\mathcal{M}(e^-_se^+_r \to 2\gamma) = i\mathcal{M}^{(0)} + \sum_j p^j i\mathcal{M}^{(1)j} \label{prob5_4b_iM_01} \end{align} とかける．
$\ket{^3P_0;0}$は\eqref{prob5_4a_Ps_wf}で$S=1$, $l=1$, $J=0$, $M=0$なので \begin{align*} \ket{^3P_0;0} &= \sqrt{2M} \int \frac{d^3p}{(2\pi)^3} \sum_{m=-1}^1 \braket{1,1; m,-m | 1,1; 00} \tilde{\psi}_{1,m} \frac{1}{\sqrt{2m}}\frac{1}{\sqrt{2m}} \ket{1, -m} \\ &= \frac{1}{\sqrt{3m}} \int \frac{d^3p}{(2\pi)^3} \left[ \tilde{\psi}_{11} \ket{\downarrow\downarrow} - \tilde{\psi}_{10} \frac{\ket{\uparrow\downarrow} + \ket{\downarrow\uparrow}}{\sqrt{2}} + \tilde{\psi}_{1,-1} \ket{\uparrow\uparrow} \right] \end{align*} となる．ここで \begin{align} \tilde{\psi}^1 = \frac{\tilde{\psi}_{1,-1} - \tilde{\psi}_{1,1}}{\sqrt{2}} & , & \tilde{\psi}^2 = i\frac{\tilde{\psi}_{1,-1} + \tilde{\psi}_{1,1}}{\sqrt{2}} & , & \tilde{\psi}^3 = \tilde{\psi}_{1,0} \label{prob5_4b_wave_func_conv} \end{align} とおけば， \begin{align*} \ket{^3P_0;0} &= \frac{1}{\sqrt{6m}} \int \frac{d^3p}{(2\pi)^3} \left[ \tilde{\psi}^1 (\ket{\uparrow\uparrow} - \ket{\downarrow\downarrow}) - i \tilde{\psi}^2 (\ket{\uparrow\uparrow} + \ket{\downarrow\downarrow}) - \tilde{\psi}^3 (\ket{\uparrow\downarrow} + \ket{\downarrow\uparrow}) \right] \end{align*} となる．
$n=2$の場合を考える． \begin{align*} \psi^1(\boldsymbol{x}) &= \frac{1}{4\sqrt{2\pi}} \frac{1}{a_0{}^{5/2}} x \exp\left( -\frac{r}{2a_0} \right) \\ \psi^2(\boldsymbol{x}) &= \frac{1}{4\sqrt{2\pi}} \frac{1}{a_0{}^{5/2}} y \exp\left( -\frac{r}{2a_0} \right) \\ \psi^3(\boldsymbol{x}) &= \frac{1}{4\sqrt{2\pi}} \frac{1}{a_0{}^{5/2}} z \exp\left( -\frac{r}{2a_0} \right) \end{align*} なので， \[ \frac{\partial \psi^i}{\partial x^j}(0) = \frac{\delta^{ij}}{4\sqrt{2\pi} a_0{}^{5/2}} . \] ポジトロニウムの崩壊の不変振幅は \begin{align*} i\mathcal{M}(2^3P_0 \to 2\gamma) &= \frac{1}{\sqrt{6m}} \int \frac{d^3p}{(2\pi)^3} \Bigl[ \tilde{\psi}^1(\boldsymbol{p}) \left\{ i\mathcal{M}(e^-_\uparrow e^+_\uparrow \to 2\gamma) - i\mathcal{M}(e^-_\downarrow e^+_\downarrow \to 2\gamma) \right\} \\ &\qquad\qquad\qquad\quad - i \tilde{\psi}^2(\boldsymbol{p}) \left\{ i\mathcal{M}(e^-_\uparrow e^+_\uparrow \to 2\gamma) + i\mathcal{M}(e^-_\downarrow e^+_\downarrow \to 2\gamma) \right\} \\ &\qquad\qquad\qquad\quad - \tilde{\psi}^3(\boldsymbol{p}) \left\{ i\mathcal{M}(e^-_\uparrow e^+_\downarrow \to 2\gamma) + i\mathcal{M}(e^-_\downarrow e^+_\uparrow \to 2\gamma) \right\} \Bigr] \end{align*} で与えられる．\eqref{prob5_4b_iM_01}のように，不変振幅は$i\mathcal{M} = i\mathcal{M}^{(0)} + \sum ip^j \mathcal{M}^{(1)j}$と表すことができたが， \[ \int \frac{d^3p}{(2\pi)^3} \tilde{\psi}^i i\mathcal{M}^{(0)} = \psi^i(0) i\mathcal{M}^{(0)} = 0 \] である．さらに， \[ \int \frac{d^3p}{(2\pi)^3} \tilde{\psi}^i \sum_j i p^j \mathcal{M}^{(1)j} = i \frac{1}{4\sqrt{2\pi} a_0{}^{5/2}} i\mathcal{M}^{(1)i} \] なので \[ \int \frac{d^3p}{(2\pi)^3} \tilde{\psi}^i i \mathcal{M} = i \frac{1}{4\sqrt{2\pi} a_0{}^{5/2}} i\mathcal{M}^{(1)i} \] となる．従って \begin{align*} i\mathcal{M}(2^3P_0 \to 2\gamma) &= \frac{1}{\sqrt{6m}} \frac{i}{4\sqrt{2\pi} a_0{}^{5/2}} \Bigl[ \left\{ i\mathcal{M}^{(1)1}(e^-_\uparrow e^+_\uparrow \to 2\gamma) - i\mathcal{M}^{(1)1}(e^-_\downarrow e^+_\downarrow \to 2\gamma) \right\} \\ &\qquad\qquad\qquad\qquad - i \left\{ i\mathcal{M}^{(1)2}(e^-_\uparrow e^+_\uparrow \to 2\gamma) + i\mathcal{M}^{(1)2}(e^-_\downarrow e^+_\downarrow \to 2\gamma) \right\} \\ &\qquad\qquad\qquad\qquad\quad - \left\{ i\mathcal{M}^{(1)3}(e^-_\uparrow e^+_\downarrow \to 2\gamma) + i\mathcal{M}^{(1)3}(e^-_\downarrow e^+_\uparrow \to 2\gamma) \right\} \Bigr] \end{align*} となる．\eqref{prob5_4a_spinor}から \begin{align*} \xi^\uparrow\eta^{\uparrow\dagger} - \xi^\downarrow\eta^{\downarrow\dagger} = \sigma^1 & , & -i (\xi^\uparrow\eta^{\uparrow\dagger} + \xi^\downarrow\eta^{\downarrow\dagger}) = \sigma^2 & , & - (\xi^\uparrow\eta^{\downarrow\dagger} + \xi^\downarrow\eta^{\uparrow\dagger}) = \sigma^3 \end{align*} である．\eqref{prob5_4b_iM_1_def}は \begin{align*} \require{physics} i\mathcal{M}^{(1)j} &= -i \frac{2e^2}{m} \eta^{r\dagger} \left[ \epsilon_1^{j\ast} (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + \epsilon_2^{j\ast} (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) + \frac{k^j}{m^2} (\boldsymbol\sigma \cdot \boldsymbol{k})(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast) \right] \xi^s \\ % &= -i \frac{2e^2}{m} \Tr \left[ \xi^s \eta^{r\dagger} \left\{ \epsilon_1^{j\ast} (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + \epsilon_2^{j\ast} (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) + \frac{k^j}{m^2} (\boldsymbol\sigma \cdot \boldsymbol{k})(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast) \right\} \right] \end{align*} と表せるので， \begin{align*} i\mathcal{M}(2^3P_0 \to 2\gamma) &= \frac{1}{\sqrt{6m}} \frac{1}{4\sqrt{2\pi} a_0{}^{5/2}} \frac{2e^2}{m} \\ & \qquad \times \Tr \left[ (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma)(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) + \frac{1}{m^2} (\boldsymbol\sigma \cdot \boldsymbol{k})(\boldsymbol\sigma \cdot \boldsymbol{k})(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast) \right] . \end{align*} $(\boldsymbol\sigma \cdot \boldsymbol{a}) (\boldsymbol\sigma \cdot \boldsymbol{b}) = \boldsymbol{a} \cdot \boldsymbol{b} + i \boldsymbol\sigma \cdot (\boldsymbol{a} \times \boldsymbol{b})$から \begin{align*} i\mathcal{M}(2^3P_0 \to 2\gamma) &= \frac{e^2}{4\sqrt{3\pi m}m a_0{}^{5/2}} \Tr \left[ 2(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast) + \frac{\lvert\boldsymbol{k}\rvert^2}{m^2} (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast) \right] \\ &= \frac{e^2}{2\sqrt{3\pi m}m a_0{}^{5/2}} \frac{2m^2 + \lvert\boldsymbol{k}\rvert^2}{m^2} (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast) . \end{align*} 光子の偏極ベクトルの完全性 \[ \sum_\text{polarization} \epsilon^{i\ast}(k) \epsilon^j(k) = \delta^{ij} - \frac{k^ik^j}{\lvert\boldsymbol{k}\rvert^2} \] から \[ \sum_\text{polarization} \lvert\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast\rvert^2 = \sum_\text{pol 1} \sum_\text{pol 2} \sum_{i=1}^3 \sum_{j=1}^3 \epsilon_1^{i\ast} \epsilon_1^j \epsilon_2^{i\ast} \epsilon_2^j = \sum_{i=1}^3 \sum_{j=1}^3 \left( \delta^{ij} - \frac{k^ik^j}{\lvert\boldsymbol{k}\rvert^2} \right) \left( \delta^{ij} - \frac{k^ik^j}{\lvert\boldsymbol{k}\rvert^2} \right) = 2 \] なので \begin{align*} \sum_\text{polarization} \lvert\mathcal{M}(2^3P_0 \to 2\gamma)\rvert^2 &= \frac{e^4}{12\pi m^3 a_0{}^5} \left( \frac{2m^2 + \lvert\boldsymbol{k}\rvert^2}{m^2} \right)^2 \sum_\text{polarization} \lvert\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast\rvert^2 \\ &= \frac{e^4}{6\pi m^3 a_0{}^5} \left( \frac{2m^2 + \lvert\boldsymbol{k}\rvert^2}{m^2} \right)^2 \\ &= \frac{\pi}{12} m^2\alpha^7 \left( \frac{2m^2 + \lvert\boldsymbol{k}\rvert^2}{m^2} \right)^2 . \end{align*} (4.86)から \begin{align*} \Gamma(2^3P_0 \to 2\gamma) &= \frac{1}{2} \frac{1}{4m} \int \frac{d^3k_1}{(2\pi)^3}\frac{d^3k_2}{(2\pi)^3} \frac{1}{4E_1E_2} \lvert \mathcal{M}(2^3P_0 \to 2\gamma) \rvert^2 (2\pi)^4 \mathop{\delta^{(4)}}(k_1+k_2-p_1-p_2) \\ &= \frac{1}{8m} \int \frac{d^3k_1}{(2\pi)^2} \frac{1}{4\lvert \boldsymbol{k} \rvert^2}\lvert\mathcal{M}(2^3P_2 \to 2\gamma)\rvert^2 \mathop\delta(2E_1 - 2m) \\ &= \frac{1}{256 \pi^2 m} \int d\lvert\boldsymbol{k}\rvert \mathop\delta(\lvert\boldsymbol{k}\rvert - m) \int d\Omega \, \lvert\mathcal{M}(^3P_2 \to 2\gamma)\rvert^2 \\ &= \frac{1}{256 \pi^2 m} \int d\lvert\boldsymbol{k}\rvert \mathop\delta(\lvert\boldsymbol{k}\rvert - m) \times 4\pi \frac{\pi}{12} m^2\alpha^7 \left( \frac{2m^2 + \lvert\boldsymbol{k}\rvert^2}{m^2} \right)^2 \\ &= \frac{3}{256} m\alpha^7 . \end{align*}
$n=2$の場合を考える． $\ket{2^3P_2;0}$は$S=1$, $l=1$, $J=2$, $M=0$なので \begin{align*} \ket{2^3P_2;0} &= \sqrt{2M} \int \frac{d^3p}{(2\pi)^3} \sum_{m=-1}^1 \braket{1,1; m,-m | 1,1; 20} \tilde{\psi}_{1,m} \frac{1}{\sqrt{2m}}\frac{1}{\sqrt{2m}} \ket{1, -m} \\ &= \frac{1}{\sqrt{6m}} \int \frac{d^3p}{(2\pi)^3} \left[ \tilde{\psi}_{11} \ket{\downarrow\downarrow} + 2 \tilde{\psi}_{10} \frac{\ket{\uparrow\downarrow} + \ket{\downarrow\uparrow}}{\sqrt{2}} + \tilde{\psi}_{1,-1} \ket{\uparrow\uparrow} \right] \\ &= \frac{1}{2\sqrt{3m}} \int \frac{d^3p}{(2\pi)^3} \left[ \tilde{\psi}^1 (\ket{\uparrow\uparrow} - \ket{\downarrow\downarrow}) - i \tilde{\psi}^2 (\ket{\uparrow\uparrow} + \ket{\downarrow\downarrow}) + 2 \tilde{\psi}^3 (\ket{\uparrow\downarrow} + \ket{\downarrow\uparrow}) \right] . \end{align*} $\boldsymbol\sigma' = (\sigma^1, \sigma^2, -2\sigma^3)$とすれば \[ \Tr \left[ (\boldsymbol\sigma' \cdot \boldsymbol{a}) (\boldsymbol\sigma \cdot \boldsymbol{b}) \right] = 2a^1b^1 + 2a^2b^2 - 4a^3b^3 . \] $M=0$の不変振幅は \begin{align*} \def\diag{\operatorname{diag}} i\mathcal{M}(2^3P_2;0 \to 2\gamma) &= \frac{1}{2\sqrt{3m}} \frac{i}{4\sqrt{2\pi} a_0{}^{5/2}} \Bigl[ \left\{ i\mathcal{M}^{(1)1}(e^-_\uparrow e^+_\uparrow \to 2\gamma) - i\mathcal{M}^{(1)1}(e^-_\downarrow e^+_\downarrow \to 2\gamma) \right\} \\ &\qquad\qquad\qquad\qquad - i \left\{ i\mathcal{M}^{(1)2}(e^-_\uparrow e^+_\uparrow \to 2\gamma) + i\mathcal{M}^{(1)2}(e^-_\downarrow e^+_\downarrow \to 2\gamma) \right\} \\ &\qquad\qquad\qquad\qquad\quad + 2 \left\{ i\mathcal{M}^{(1)3}(e^-_\uparrow e^+_\downarrow \to 2\gamma) + i\mathcal{M}^{(1)3}(e^-_\downarrow e^+_\uparrow \to 2\gamma) \right\} \Bigr] \\ &= \frac{1}{2\sqrt{3m}} \frac{1}{4\sqrt{2\pi} a_0{}^{5/2}} \frac{2e^2}{m} \\ & \qquad \times \Tr \left[ (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma') (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma')(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) + \frac{1}{m^2} (\boldsymbol\sigma' \cdot \boldsymbol{k})(\boldsymbol\sigma \cdot \boldsymbol{k})(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast) \right] \\ &= \frac{e^2}{2\sqrt{\pi m}m a_0{}^{5/2}} \left[ 2h_0^{ij} \epsilon_1^{i\ast} \epsilon_2^{j\ast} + \frac{(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast)}{m^2} h_0^{ij} k^ik^j \right] , \quad h_0^{ij} = \frac{1}{\sqrt{6}} \diag(1,1,-2) . \end{align*} $\ket{2^3P_2;1}$は$S=1$, $l=1$, $J=2$, $M=1$なので \begin{align*} \ket{2^3P_2;1} &= \sqrt{2M} \int \frac{d^3p}{(2\pi)^3} \sum_{m=0}^1 \braket{1,1; m,1-m | 1,1; 21} \tilde{\psi}_{1,m} \frac{1}{\sqrt{2m}}\frac{1}{\sqrt{2m}} \ket{1, 1-m} \\ &= \frac{1}{\sqrt{2m}} \int \frac{d^3p}{(2\pi)^3} \left[ \tilde{\psi}_{10} \ket{\uparrow\uparrow} + \tilde{\psi}_{11} \frac{\ket{\uparrow\downarrow} + \ket{\downarrow\uparrow}}{\sqrt{2}} \right] \\ &= \frac{1}{2\sqrt{2m}} \int \frac{d^3p}{(2\pi)^3} \left[ - \tilde\psi^1 (\ket{\uparrow\downarrow} + \ket{\downarrow\uparrow}) - i\tilde\psi^2 (\ket{\uparrow\downarrow} + \ket{\downarrow\uparrow}) + 2 \tilde\psi^3 \ket{\uparrow\uparrow} \right] . \end{align*} $\boldsymbol\sigma' = (\sigma^3, i\sigma^3, 2\sigma^+)$とすれば \[ \Tr \left[ (\boldsymbol\sigma' \cdot \boldsymbol{a}) (\boldsymbol\sigma \cdot \boldsymbol{b}) \right] = 2a^1b^3 + 2i a^2b^3 + 2a^3b^1 + 2i a^3b^2 . \] $M=1$の不変振幅は \begin{align*} i\mathcal{M}(2^3P_2;1 \to 2\gamma) &= \frac{1}{2\sqrt{2m}} \frac{i}{4\sqrt{2\pi} a_0{}^{5/2}} \Bigl[ - \left\{ i\mathcal{M}^{(1)1}(e^-_\uparrow e^+_\downarrow \to 2\gamma) + i\mathcal{M}^{(1)1}(e^-_\downarrow e^+_\uparrow \to 2\gamma) \right\} \\ &\qquad\qquad\qquad\qquad - i \left\{ i\mathcal{M}^{(1)2}(e^-_\uparrow e^+_\downarrow \to 2\gamma) + i\mathcal{M}^{(1)2}(e^-_\downarrow e^+_\uparrow \to 2\gamma) \right\} \\ &\qquad\qquad\qquad\qquad\quad + 2 \left\{ i\mathcal{M}^{(1)3}(e^-_\uparrow e^+_\uparrow \to 2\gamma) \right\} \Bigr] \\ &= \frac{1}{2\sqrt{2m}} \frac{1}{4\sqrt{2\pi} a_0{}^{5/2}} \frac{2e^2}{m} \\ & \qquad \times \Tr \left[ (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma') (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma')(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) + \frac{1}{m^2} (\boldsymbol\sigma' \cdot \boldsymbol{k})(\boldsymbol\sigma \cdot \boldsymbol{k})(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast) \right] \\ &= \frac{e^2}{2\sqrt{\pi m}m a_0{}^{5/2}} \left[ 2h_1^{ij} \epsilon_1^{i\ast} \epsilon_2^{j\ast} + \frac{(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast)}{m^2} h_1^{ij} k^ik^j \right] , \quad h_1^{ij} = \frac{1}{2} \begin{pmatrix} & & 1 \\ & & i \\ 1 & i & \end{pmatrix} . \end{align*} $\ket{2^3P_2;-1}$は$S=1$, $l=1$, $J=2$, $M=-1$なので \begin{align*} \ket{2^3P_2;-1} &= \sqrt{2M} \int \frac{d^3p}{(2\pi)^3} \sum_{m=-1}^0 \braket{1,1; m,-1-m | 1,1; 2, -1} \tilde{\psi}_{1,m} \frac{1}{\sqrt{2m}}\frac{1}{\sqrt{2m}} \ket{1, -1-m} \\ &= \sqrt{2M} \int \frac{d^3p}{(2\pi)^3} \sum_{m=-1}^0 \braket{1,1; -m,1+m | 1,1; 2, 1} \tilde{\psi}_{1,m} \frac{1}{\sqrt{2m}}\frac{1}{\sqrt{2m}} \ket{1, -1-m} \\ &= \frac{1}{\sqrt{2m}} \int \frac{d^3p}{(2\pi)^3} \left[ \tilde{\psi}_{1, -1} \frac{\ket{\uparrow\downarrow} + \ket{\downarrow\uparrow}}{\sqrt{2}} + \tilde{\psi}_{10} \ket{\downarrow\downarrow} \right] \\ &= \frac{1}{2\sqrt{2m}} \int \frac{d^3p}{(2\pi)^3} \left[ \tilde\psi^1 (\ket{\uparrow\downarrow} + \ket{\downarrow\uparrow}) - i\tilde\psi^2 (\ket{\uparrow\downarrow} + \ket{\downarrow\uparrow}) + 2 \tilde\psi^3 \ket{\downarrow\downarrow} \right] . \end{align*} $\boldsymbol\sigma' = (-\sigma^3, i\sigma^3, -2\sigma^-)$とすれば \[ \Tr \left[ (\boldsymbol\sigma' \cdot \boldsymbol{a}) (\boldsymbol\sigma \cdot \boldsymbol{b}) \right] = - 2a^1b^3 + 2i a^2b^3 - 2a^3b^1 + 2i a^3b^2 . \] $M=-1$の不変振幅は \begin{align*} i\mathcal{M}(2^3P_2;1 \to 2\gamma) &= \frac{1}{2\sqrt{2m}} \frac{i}{4\sqrt{2\pi} a_0{}^{5/2}} \Bigl[ \left\{ i\mathcal{M}^{(1)1}(e^-_\uparrow e^+_\downarrow \to 2\gamma) + i\mathcal{M}^{(1)1}(e^-_\downarrow e^+_\uparrow \to 2\gamma) \right\} \\ &\qquad\qquad\qquad\qquad - i \left\{ i\mathcal{M}^{(1)2}(e^-_\uparrow e^+_\downarrow \to 2\gamma) + i\mathcal{M}^{(1)2}(e^-_\downarrow e^+_\uparrow \to 2\gamma) \right\} \\ &\qquad\qquad\qquad\qquad\quad + 2 \left\{ i\mathcal{M}^{(1)3}(e^-_\downarrow e^+_\downarrow \to 2\gamma) \right\} \Bigr] \\ &= \frac{1}{2\sqrt{2m}} \frac{1}{4\sqrt{2\pi} a_0{}^{5/2}} \frac{2e^2}{m} \\ & \qquad \times \Tr \left[ (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma') (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma')(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) + \frac{1}{m^2} (\boldsymbol\sigma' \cdot \boldsymbol{k})(\boldsymbol\sigma \cdot \boldsymbol{k})(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast) \right] \\ &= \frac{e^2}{2\sqrt{\pi m}m a_0{}^{5/2}} \left[ 2h_{-1}^{ij} \epsilon_1^{i\ast} \epsilon_2^{j\ast} + \frac{(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast)}{m^2} h_{-1}^{ij} k^ik^j \right] , \quad h_{-1}^{ij} = \frac{1}{2} \begin{pmatrix} & & -1 \\ & & i \\ -1 & i & \end{pmatrix} . \end{align*} $\ket{2^3P_2;2}$は$S=1$, $l=1$, $J=2$, $M=2$なので（$m=S_z=1$） \begin{align*} \ket{2^3P_2;2} &= \sqrt{2M} \int \frac{d^3p}{(2\pi)^3} \braket{1,1; 1,1 | 1,1; 22} \tilde{\psi}_{1,1} \frac{1}{\sqrt{2m}}\frac{1}{\sqrt{2m}} \ket{1, 1} \\ &= \frac{1}{\sqrt{m}} \int \frac{d^3p}{(2\pi)^3} \tilde{\psi}_{11} \ket{\uparrow\uparrow} \\ &= \frac{1}{\sqrt{2m}} \int \frac{d^3p}{(2\pi)^3} \left[ - \tilde\psi^1 \ket{\uparrow\uparrow} - i \tilde\psi^2 \ket{\uparrow\uparrow} \right] . \end{align*} $\boldsymbol\sigma' = (- \sigma^+, -i\sigma^+, 0)$とすれば \[ \Tr \left[ (\boldsymbol\sigma' \cdot \boldsymbol{a}) (\boldsymbol\sigma \cdot \boldsymbol{b}) \right] = - a^1b^2 - i a^1b^2 - i a^2b^1 + a^2b^2 . \] $M=2$の不変振幅は \begin{align*} i\mathcal{M}(2^3P_2;2 \to 2\gamma) &= \frac{1}{\sqrt{2m}} \frac{i}{4\sqrt{2\pi} a_0{}^{5/2}} \left[ - \left\{ i\mathcal{M}^{(1)1}(e^-_\uparrow e^+_\uparrow \to 2\gamma) \right\} - i \left\{ i\mathcal{M}^{(1)2}(e^-_\uparrow e^+_\uparrow \to 2\gamma) \right\} \right] \\ &= \frac{1}{\sqrt{2m}} \frac{1}{4\sqrt{2\pi} a_0{}^{5/2}} \frac{2e^2}{m} \\ & \qquad \times \Tr \left[ (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma') (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma')(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) + \frac{1}{m^2} (\boldsymbol\sigma' \cdot \boldsymbol{k})(\boldsymbol\sigma \cdot \boldsymbol{k})(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast) \right] \\ &= \frac{e^2}{2\sqrt{\pi m}m a_0{}^{5/2}} \left[ 2h_2^{ij} \epsilon_1^{i\ast} \epsilon_2^{j\ast} + \frac{(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast)}{m^2} h_2^{ij} k^ik^j \right] , \quad h_2^{ij} = \frac{1}{2} \begin{pmatrix} -1 & -i & \\ -i & 1 & \\ & & \end{pmatrix} . \end{align*} $\ket{2^3P_2;-2}$は$S=1$, $l=1$, $J=2$, $M=-2$なので（$m=S_z=-1$） \begin{align*} \ket{2^3P_2;-2} &= \sqrt{2M} \int \frac{d^3p}{(2\pi)^3} \braket{1,1; -1,-1 | 1,1; 2, -2} \tilde{\psi}_{1,-1} \frac{1}{\sqrt{2m}}\frac{1}{\sqrt{2m}} \ket{1, -1} \\ &= \sqrt{2M} \int \frac{d^3p}{(2\pi)^3} \braket{1,1; 1,1 | 1,1; 22} \tilde{\psi}_{1,-1} \frac{1}{\sqrt{2m}}\frac{1}{\sqrt{2m}} \ket{1, -1} \\ &= \frac{1}{\sqrt{m}} \int \frac{d^3p}{(2\pi)^3} \tilde{\psi}_{1,-1} \ket{\downarrow\downarrow} \\ &= \frac{1}{\sqrt{2m}} \int \frac{d^3p}{(2\pi)^3} \left[ \tilde\psi^1 \ket{\downarrow\downarrow} - i \tilde\psi^2 \ket{\downarrow\downarrow} \right] . \end{align*} $\boldsymbol\sigma' = (- \sigma^-, i\sigma^-, 0)$とすれば \[ \Tr \left[ (\boldsymbol\sigma' \cdot \boldsymbol{a}) (\boldsymbol\sigma \cdot \boldsymbol{b}) \right] = - a^1b^2 + i a^1b^2 + i a^2b^1 + a^2b^2 . \] $M=-2$の不変振幅は \begin{align*} i\mathcal{M}(2^3P_2;-2 \to 2\gamma) &= \frac{1}{\sqrt{2m}} \frac{i}{4\sqrt{2\pi} a_0{}^{5/2}} \left[ \left\{ i\mathcal{M}^{(1)1}(e^-_\downarrow e^+_\downarrow \to 2\gamma) \right\} - i \left\{ i\mathcal{M}^{(1)2}(e^-_\downarrow e^+_\downarrow \to 2\gamma) \right\} \right] \\ &= \frac{1}{\sqrt{2m}} \frac{1}{4\sqrt{2\pi} a_0{}^{5/2}} \frac{2e^2}{m} \\ & \qquad \times \Tr \left[ (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma') (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma) + (\boldsymbol{\epsilon}_2^\ast \cdot \boldsymbol\sigma')(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol\sigma) + \frac{1}{m^2} (\boldsymbol\sigma' \cdot \boldsymbol{k})(\boldsymbol\sigma \cdot \boldsymbol{k})(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast) \right] \\ &= \frac{e^2}{2\sqrt{\pi m}m a_0{}^{5/2}} \left[ 2h_{-2}^{ij} \epsilon_1^{i\ast} \epsilon_2^{j\ast} + \frac{(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast)}{m^2} h_{-2}^{ij} k^ik^j \right] , \quad h_{-2}^{ij} = \frac{1}{2} \begin{pmatrix} -1 & i & \\ i & 1 & \\ & & \end{pmatrix} . \end{align*} 以上から， \[ i\mathcal{M}(2^3P_2;M \to 2\gamma) = \frac{e^2}{2\sqrt{\pi m}m a_0{}^{5/2}} \left[ 2h_M^{ij} \epsilon_1^{i\ast} \epsilon_2^{j\ast} + \frac{(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast)}{m^2} h_M^{ij} k^ik^j \right] . \] $h_M$はトレースが$0$の対称行列 \[ h_0 = \frac{1}{\sqrt{6}} \begin{pmatrix} 1 & & \\ & 1 & \\ & & -2 \\ \end{pmatrix} , \quad h_{\pm1} = \frac{1}{2} \begin{pmatrix} & & \pm1 \\ & & i \\ \pm1 & i & \end{pmatrix} , \quad h_{\pm2} = \frac{1}{2} \begin{pmatrix} -1 & \mp i & \\ \mp i & 1 & \\ & & \end{pmatrix} \] で与えられ， \begin{align} \Tr \left( h_M h_{M'}^\dagger \right) = \sum_{ij} h_M^{ij} h_{M'}^{ij\ast} = \delta_{MM'} \label{prob5_4b_h_tens_orth} \end{align} を満たす． 光子の偏極と全角運動量の射影について和を取って， \begin{align*} & \sum_\text{polarization} \sum_M \lvert\mathcal{M}(2^3P_2;M \to 2\gamma)\rvert^2 \\ &= \frac{e^4}{4\pi m^3 a_0{}^5} \sum_\text{polarization} \sum_M \left\lvert 2h_M^{ij} \epsilon_1^{i\ast} \epsilon_2^{j\ast} + \frac{(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast)}{m^2} h_M^{ij} k^ik^j \right\rvert^2 \\ % &= \frac{e^4}{4\pi m^3 a_0{}^5} \sum_\text{polarization} \sum_M \left[ 2h_M^{ij} \epsilon_1^{i\ast} \epsilon_2^{j\ast} + \frac{(\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast)}{m^2} h_M^{ij} k^ik^j \right] \left[ 2h_M^{kl\ast} \epsilon_1^k \epsilon_2^l + \frac{(\boldsymbol{\epsilon}_1 \cdot \boldsymbol{\epsilon}_2)}{m^2} h_M^{kl\ast} k^kk^l \right] \\ % &= \frac{e^4}{4\pi m^3 a_0{}^5} \sum_\text{pol} \sum_M \sum_{ijkl} h_M^{ij} h_M^{kl\ast} \\ & \quad \times \left[ 4 \epsilon_1^{i\ast} \epsilon_1^k \epsilon_2^{j\ast} \epsilon_2^l + \frac{2k^kk^l}{m^2} (\boldsymbol{\epsilon}_1 \cdot \boldsymbol{\epsilon}_2) \epsilon_1^{i\ast} \epsilon_2^{j\ast} + \frac{2k^ik^j}{m^2} (\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast) \epsilon_1^k \epsilon_2^l + \frac{k^ik^jk^kk^l}{m^4} \lvert\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast\rvert^2 \right] \\ % &= \frac{e^4}{4\pi m^3 a_0{}^5} \sum_\text{pol} \sum_M \sum_{ijkl} h_M^{ij} h_M^{kl\ast} \\ & \quad \times \left[ 4 \epsilon_1^{i\ast} \epsilon_1^k \epsilon_2^{j\ast} \epsilon_2^l + \frac{2k^kk^l}{m^2} \sum_m \epsilon_1^{i\ast} \epsilon_1^m \epsilon_2^{j\ast} \epsilon_2^m + \frac{2k^ik^j}{m^2} \sum_m \epsilon_1^{m\ast} \epsilon_1^k \epsilon_2^{m\ast} \epsilon_2^l + \frac{k^ik^jk^kk^l}{m^4} \lvert\boldsymbol{\epsilon}_1^\ast \cdot \boldsymbol{\epsilon}_2^\ast\rvert^2 \right] . \end{align*} $\epsilon^\mu$の完全性から \begin{align} & \lvert \mathcal{M}(2^3P_2 \to 2\gamma) \rvert^2 \notag \\ % &= \frac{e^4}{4\pi m^3 a_0{}^5} 4 \sum_M \sum_{ijkl} h_M^{ij} h_M^{kl\ast} \left( \delta^{ik} - \frac{k^ik^k}{\lvert\boldsymbol{k}\rvert^2} \right) \left( \delta^{jl} - \frac{k^jk^l}{\lvert\boldsymbol{k}\rvert^2} \right) \label{prob5_4b_23P2M2_1} \\ % &\quad + \frac{e^4}{4\pi m^3 a_0{}^5} 2 \sum_M \sum_{ijkl} h_M^{ij} h_M^{kl\ast} \frac{k^kk^l}{m^2} \sum_m \left( \delta^{im} - \frac{k^ik^m}{\lvert\boldsymbol{k}\rvert^2} \right) \left( \delta^{jm} - \frac{k^jk^m}{\lvert\boldsymbol{k}\rvert^2} \right) \label{prob5_4b_23P2M2_2} \\ % &\quad + \frac{e^4}{4\pi m^3 a_0{}^5} 2 \sum_M \sum_{ijkl} h_M^{ij} h_M^{kl\ast} \frac{k^ik^j}{m^2} \sum_m \left( \delta^{mk} - \frac{k^mk^k}{\lvert\boldsymbol{k}\rvert^2} \right) \left( \delta^{ml} - \frac{k^mk^l}{\lvert\boldsymbol{k}\rvert^2} \right) \label{prob5_4b_23P2M2_3} \\ % &\quad + \frac{e^4}{4\pi m^3 a_0{}^5} 2 \sum_M \sum_{ijkl} h_M^{ij} h_M^{kl\ast} \frac{k^ik^jk^kk^l}{m^4} \label{prob5_4b_23P2M2_4} . \end{align} (4.86)から \begin{align*} \Gamma(2^3P_2 \to 2\gamma) &= \frac{1}{2} \frac{1}{4m} \int \frac{d^3k_1}{(2\pi)^3}\frac{d^3k_2}{(2\pi)^3} \frac{1}{4E_1E_2} \lvert \mathcal{M}(2^3P_2 \to 2\gamma) \rvert^2 (2\pi)^4 \mathop{\delta^{(4)}}(k_1+k_2-p_1-p_2) \\ &= \frac{1}{8m} \int \frac{d^3k_1}{(2\pi)^2} \frac{1}{4\lvert \boldsymbol{k} \rvert^2}\lvert\mathcal{M}(2^3P_2 \to 2\gamma)\rvert^2 \mathop\delta(2E_1 - 2m) \\ &= \frac{1}{256 \pi^2 m} \int d\lvert\boldsymbol{k}\rvert \mathop\delta(\lvert\boldsymbol{k}\rvert - m) \int d\Omega \, \lvert\mathcal{M}(2^3P_2 \to 2\gamma)\rvert^2 . \end{align*} $\lvert\mathcal{M}\rvert^2$の中に現れる積分を計算する．まず， \[ \int d\Omega \, \frac{k^ik^j}{\lvert\boldsymbol{k}\rvert^2} \] は$i \neq j$ならば$0$なので，$\delta^{ij}$に比例する： \[ \int d\Omega \, \frac{k^ik^j}{\lvert\boldsymbol{k}\rvert^2} = A \delta^{ij} . \] $i = j = 1, 2, 3$について和を取れば$4\pi = 3A$となるので， \[ \int d\Omega \, \frac{k^ik^j}{\lvert\boldsymbol{k}\rvert^2} = \frac{4\pi}{3} \delta^{ij} . \] 次に， \[ \int d\Omega \, \frac{k^ik^jk^kk^l}{\lvert\boldsymbol{k}\rvert^4} = B ( \delta^{ij}\delta^{kl} + \delta^{ik}\delta^{jl} + \delta^{il}\delta^{jk} ) \] とおく．$i = j = 1, 2, 3$について和を取れば \[ \int d\Omega \, \frac{k^kk^l}{\lvert\boldsymbol{k}\rvert^2} = \frac{4\pi}{3} \delta^{kl} = B ( 3 \delta^{kl} + \delta^{kl} + \delta^{kl} ) = 5B\delta^{kl} \] なので \[ \int d\Omega \, \frac{k^ik^jk^kk^l}{\lvert\boldsymbol{k}\rvert^4} = \frac{4\pi}{15} ( \delta^{ij}\delta^{kl} + \delta^{ik}\delta^{jl} + \delta^{il}\delta^{jk} ) . \] \eqref{prob5_4b_h_tens_orth}に注意して\eqref{prob5_4b_23P2M2_1}を積分すれば \begin{align*} & 4 \sum_M \sum_{ijkl} h_M^{ij} h_M^{kl\ast} \int d\lvert\boldsymbol{k}\rvert \mathop\delta(\lvert\boldsymbol{k}\rvert - m) \int d\Omega \, \left( \delta^{ik} - \frac{k^ik^k}{\lvert\boldsymbol{k}\rvert^2} \right) \left( \delta^{jl} - \frac{k^jk^l}{\lvert\boldsymbol{k}\rvert^2} \right) \\ &= 4 \sum_M \sum_{ijkl} h_M^{ij} h_M^{kl\ast} \left[ 4\pi \delta^{ik} \delta^{jl} - \frac{4\pi}{3} \delta^{ik} \delta^{jl} - \frac{4\pi}{3} \delta^{ik} \delta^{jl} + \frac{4\pi}{15} ( \delta^{ij}\delta^{kl} + \delta^{ik}\delta^{jl} + \delta^{il}\delta^{jk} ) \right] \\ &= 4 \sum_M \sum_{ijkl} h_M^{ij} h_M^{kl\ast} \left[ \frac{4\pi}{15} ( \delta^{ij}\delta^{kl} + \delta^{il}\delta^{jk} ) + \frac{8\pi}{5} \delta^{ik}\delta^{jl} \right] \\ &= 4 \sum_M \sum_{ij} \left[ \frac{4\pi}{15} h_M^{ij} h_M^{ji\ast} + \frac{8\pi}{5} h_M^{ij} h_M^{ij\ast} \right] \\ &= 4 \sum_M \frac{12\pi}{15} \delta_{MM} \\ &= \frac{48\pi}{5} . \end{align*} \eqref{prob5_4b_23P2M2_2}\eqref{prob5_4b_23P2M2_3}を積分すれば \begin{align*} & 2 \sum_M \sum_{ijkl} h_M^{ij} h_M^{kl\ast} \int d\lvert\boldsymbol{k}\rvert \mathop\delta(\lvert\boldsymbol{k}\rvert - m) \int d\Omega \, \frac{k^kk^l}{m^2} \sum_m \left( \delta^{im} - \frac{k^ik^m}{\lvert\boldsymbol{k}\rvert^2} \right) \left( \delta^{jm} - \frac{k^jk^m}{\lvert\boldsymbol{k}\rvert^2} \right) \\ &= 2 \sum_M \sum_{ijkl} h_M^{ij} h_M^{kl\ast} \int d\lvert\boldsymbol{k}\rvert \mathop\delta(\lvert\boldsymbol{k}\rvert - m) \int d\Omega \, \frac{k^kk^l}{m^2} \left( \delta^{ij} - \frac{k^ik^j}{\lvert\boldsymbol{k}\rvert^2} \right) \\ &= 2 \sum_M \sum_{ijkl} h_M^{ij} h_M^{kl\ast} \int d\lvert\boldsymbol{k}\rvert \mathop\delta(\lvert\boldsymbol{k}\rvert - m) \int d\Omega \, \left( \frac{k^kk^l}{\lvert\boldsymbol{k}\rvert^2} \delta^{ij} - \frac{k^ik^jk^kk^l}{\lvert\boldsymbol{k}\rvert^4} \right) \\ &= 2 \sum_M \sum_{ijkl} h_M^{ij} h_M^{kl\ast} \left[ \frac{4\pi}{3} \delta^{ij} \delta^{kl} - \frac{4\pi}{15} ( \delta^{ij}\delta^{kl} + \delta^{ik}\delta^{jl} + \delta^{il}\delta^{jk} ) \right] \\ &= 2 \sum_M \sum_{ijkl} h_M^{ij} h_M^{kl\ast} \left[ \frac{16\pi}{15} \delta^{ij} \delta^{kl} - \frac{4\pi}{15} ( \delta^{ik}\delta^{jl} + \delta^{il}\delta^{jk} ) \right] \\ &= 2 \sum_M \sum_{ij} \left[ - \frac{4\pi}{15} h_M^{ij} h_M^{ij\ast} - \frac{4\pi}{15} h_M^{ij} h_M^{ji\ast} \right] \\ &= - \frac{16\pi}{15} \sum_M \delta_{MM} \\ &= - \frac{16\pi}{5} . \end{align*} \eqref{prob5_4b_23P2M2_4}を積分すれば \begin{align*} & 2 \sum_M \sum_{ijkl} h_M^{ij} h_M^{kl\ast} \int d\lvert\boldsymbol{k}\rvert \mathop\delta(\lvert\boldsymbol{k}\rvert - m) \int d\Omega \, \frac{k^ik^jk^kk^l}{m^4} \\ &= 2 \sum_M \sum_{ijkl} h_M^{ij} h_M^{kl\ast} \frac{4\pi}{15} ( \delta^{ij}\delta^{kl} + \delta^{ik}\delta^{jl} + \delta^{il}\delta^{jk} ) \\ &= 2 \sum_M \sum_{ijkl} \frac{4\pi}{15} ( h_M^{ij} h_M^{ij\ast} + h_M^{ij} h_M^{ji\ast} ) \\ &= 2 \sum_M \sum_{ijkl} \frac{4\pi}{15} ( h_M^{ij} h_M^{ij\ast} + h_M^{ij} h_M^{ji\ast} ) \\ &= \frac{16\pi}{15} \sum_M \delta_{MM} \\ &= \frac{16\pi}{5} . \end{align*} 以上から \[ \Gamma(2^3P_2 \to 2\gamma) = \frac{1}{256 \pi^2 m} \frac{e^4}{4\pi m^3 a_0{}^5} \left( \frac{48\pi}{5} - \frac{16\pi}{5} - \frac{16\pi}{5} + \frac{16\pi}{5}\right) = \frac{m\alpha^7}{320} . \]