Problem 5.1

peskinChapter 5

习题 5.1

来源: 第5章, PDF第169,170页

5.1 Coulomb scattering. Repeat the computation of Problem 4.4, part (c), this time using the full relativistic expression for the matrix element. You should find, for the spin-averaged cross section,

\frac{d\sigma}{d\Omega} = \frac{\alpha^2}{4|\mathbf{p}|^2 \beta^2 \sin^4(\theta/2)} \left( 1 - \beta^2 \sin^2 \frac{\theta}{2} \right),

where $\mathbf{p}$ is the electron's 3-momentum and $\beta$ is its velocity. This is the Mott formula for Coulomb scattering of relativistic electrons. Now derive it in a second way, by working

out the cross section for electron-muon scattering, in the muon rest frame, retaining the electron mass but sending $m_{\mu} \rightarrow \infty$ .

习题 5.1 - 解答

方法一：电子在静态库仑势中的散射

1. 物理背景与矩阵元 考虑相对论性电子在经典静态库仑势 $A^\mu(x) = (A^0(\mathbf{x}), \mathbf{0})$ 中的散射。对于单位电荷的库仑中心，势场在动量空间的傅里叶变换为： $\tilde{A}^0(\mathbf{q}) = \frac{e}{|\mathbf{q}|^2}$ 其中 $\mathbf{q} = \mathbf{p}' - \mathbf{p}$ 是动量转移。根据费曼规则，电子在外部经典场中散射的矩阵元为： $-i\mathcal{M} = \bar{u}(p') (-ie\gamma^0) u(p) \tilde{A}^0(\mathbf{q}) \implies \mathcal{M} = \frac{e^2}{|\mathbf{q}|^2} \bar{u}(p') \gamma^0 u(p)$

2. 自旋平均与狄拉克迹计算 对初始自旋求平均并对末态自旋求和，得到自旋平均的矩阵元平方： $\overline{|\mathcal{M}|^2} = \frac{1}{2} \sum_{s,s'} |\mathcal{M}|^2 = \frac{1}{2} \frac{e^4}{|\mathbf{q}|^4} \text{Tr}\left[ (\not{p}' + m_e) \gamma^0 (\not{p} + m_e) \gamma^0 \right]$ 利用狄拉克矩阵的迹定理 $\text{Tr}(\text{odd } \gamma) = 0$ 以及 $\text{Tr}(\gamma^\mu\gamma^\nu\gamma^\rho\gamma^\sigma) = 4(g^{\mu\nu}g^{\rho\sigma} - g^{\mu\rho}g^{\nu\sigma} + g^{\mu\sigma}g^{\nu\rho})$ ，计算迹： $\text{Tr}\left[ \not{p}' \gamma^0 \not{p} \gamma^0 \right] + m_e^2 \text{Tr}[\gamma^0 \gamma^0] = 4(2E^2 - p'\cdot p) + 4m_e^2$ 在弹性散射中， $E' = E$ ，且 $p'\cdot p = E^2 - \mathbf{p}'\cdot\mathbf{p} = E^2 - |\mathbf{p}|^2 \cos\theta$ 。代入 $m_e^2 = E^2 - |\mathbf{p}|^2$ ，迹化简为： $4(2E^2 - E^2 + |\mathbf{p}|^2 \cos\theta + E^2 - |\mathbf{p}|^2) = 4(2E^2 - |\mathbf{p}|^2(1 - \cos\theta))$ 利用半角公式 $1 - \cos\theta = 2\sin^2(\frac{\theta}{2})$ 以及电子速度 $\beta = |\mathbf{p}|/E$ ，迹变为： $8E^2 \left( 1 - \frac{|\mathbf{p}|^2}{E^2} \sin^2\frac{\theta}{2} \right) = 8E^2 \left( 1 - \beta^2 \sin^2\frac{\theta}{2} \right)$ 因此，自旋平均的矩阵元平方为： $\overline{|\mathcal{M}|^2} = \frac{4e^4 E^2}{|\mathbf{q}|^4} \left( 1 - \beta^2 \sin^2\frac{\theta}{2} \right)$

3. 截面计算 对于固定势场中的单体散射，微分截面公式为： $d\sigma = \frac{1}{v} \frac{1}{2E} \frac{d^3p'}{(2\pi)^3 2E'} 2\pi \delta(E' - E) \overline{|\mathcal{M}|^2}$ 其中入射流速 $v = \beta = |\mathbf{p}|/E$ 。积分掉动量大小 $|\mathbf{p}'|$ ： $\int \frac{|\mathbf{p}'|^2 d|\mathbf{p}'|}{E'} \delta(E' - E) = |\mathbf{p}|$ 得到立体角微分截面： $\frac{d\sigma}{d\Omega} = \frac{1}{16\pi^2} \overline{|\mathcal{M}|^2} = \frac{e^4 E^2}{4\pi^2 |\mathbf{q}|^4} \left( 1 - \beta^2 \sin^2\frac{\theta}{2} \right)$ 代入动量转移平方 $|\mathbf{q}|^2 = 4|\mathbf{p}|^2 \sin^2(\theta/2)$ 以及精细结构常数 $\alpha = e^2/4\pi$ （即 $e^4 = 16\pi^2 \alpha^2$ ）： $\frac{d\sigma}{d\Omega} = \frac{16\pi^2 \alpha^2 E^2}{4\pi^2 (16|\mathbf{p}|^4 \sin^4\frac{\theta}{2})} \left( 1 - \beta^2 \sin^2\frac{\theta}{2} \right) = \frac{\alpha^2 E^2}{4|\mathbf{p}|^4 \sin^4\frac{\theta}{2}} \left( 1 - \beta^2 \sin^2\frac{\theta}{2} \right)$ 利用 $E^2/|\mathbf{p}|^4 = 1/(|\mathbf{p}|^2 \beta^2)$ ，最终得到 Mott 散射公式： $\boxed{ \frac{d\sigma}{d\Omega} = \frac{\alpha^2}{4|\mathbf{p}|^2 \beta^2 \sin^4(\theta/2)} \left( 1 - \beta^2 \sin^2 \frac{\theta}{2} \right) }$

方法二：电子-缪子散射在 $m_\mu \to \infty$ 极限下的推导

1. 运动学与极限近似 考虑过程 $e^-(p) + \mu^-(k) \to e^-(p') + \mu^-(k')$ 。在缪子静止系（实验室系）中：初始动量： $p = (E, \mathbf{p})$ , $k = (m_\mu, \mathbf{0})$ 末态动量： $p' = (E', \mathbf{p}')$ , $k' = (E_\mu', \mathbf{q})$ ，其中 $\mathbf{q} = \mathbf{p} - \mathbf{p}'$ 。当 $m_\mu \to \infty$ 时，缪子吸收动量 $\mathbf{q}$ 但其能量改变 $E_\mu' - m_\mu \approx |\mathbf{q}|^2 / 2m_\mu \to 0$ 。因此电子能量守恒 $E' \to E$ ，恢复弹性散射。

2. 矩阵元与迹计算 完整的 QED 矩阵元平方（自旋平均）为： $\overline{|\mathcal{M}|^2} = \frac{1}{4} \frac{e^4}{q^4} L_{(e)}^{\mu\nu} L^{(\mu)}_{\mu\nu}$ 其中 $q^2 = (p'-p)^2 \approx -|\mathbf{q}|^2 = -4|\mathbf{p}|^2 \sin^2(\theta/2)$ 。缪子张量为： $L^{(\mu)}_{\mu\nu} = \text{Tr}\left[ (\not{k}' + m_\mu) \gamma_\mu (\not{k} + m_\mu) \gamma_\nu \right] = 4\left( k'_\mu k_\nu + k'_\nu k_\mu - g_{\mu\nu}(k\cdot k' - m_\mu^2) \right)$ 在 $m_\mu \to \infty$ 极限下， $k \approx (m_\mu, \mathbf{0})$ 且 $k' \approx (m_\mu, \mathbf{q})$ 。此时 $k\cdot k' - m_\mu^2 \approx 0$ ，主导项为 $\mu=\nu=0$ 分量： $L^{(\mu)}_{00} \approx 4(2m_\mu^2) = 8m_\mu^2$ 空间分量被 $1/m_\mu$ 压低，因此 $L^{(\mu)}_{\mu\nu} \approx 8m_\mu^2 g_{\mu 0} g_{\nu 0}$ 。电子张量只需计算 $00$ 分量，这与方法一中的迹完全相同： $L_{(e)}^{00} = \text{Tr}\left[ (\not{p}' + m_e) \gamma^0 (\not{p} + m_e) \gamma^0 \right] = 8E^2 \left( 1 - \beta^2 \sin^2\frac{\theta}{2} \right)$ 代入得到总矩阵元平方： $\overline{|\mathcal{M}|^2} = \frac{1}{4} \frac{e^4}{q^4} \left[ 8E^2 \left( 1 - \beta^2 \sin^2\frac{\theta}{2} \right) \right] (8m_\mu^2) = \frac{16 e^4 m_\mu^2 E^2}{q^4} \left( 1 - \beta^2 \sin^2\frac{\theta}{2} \right)$

3. 相空间与截面计算 $2 \to 2$ 散射的微分截面公式为： $d\sigma = \frac{1}{4E m_\mu \beta} \int \frac{d^3p'}{(2\pi)^3 2E'} \frac{d^3k'}{(2\pi)^3 2E_\mu'} (2\pi)^4 \delta^4(p+k-p'-k') \overline{|\mathcal{M}|^2}$ 其中通量因子为 $4(p\cdot k)v_{rel} = 4 E m_\mu \beta = 4|\mathbf{p}|m_\mu$ 。积分掉 $\mathbf{k}'$ 后，相空间积分为： $\int d\Pi_2 = \frac{1}{16\pi^2} \int \frac{|\mathbf{p}'|^2 d|\mathbf{p}'| d\Omega}{E' E_\mu'} \delta(E' + E_\mu' - E - m_\mu)$ 在 $m_\mu \to \infty$ 极限下， $E_\mu' \to m_\mu$ ，能量 $\delta$ 函数变为 $\delta(E' - E)$ 。对 $|\mathbf{p}'|$ 积分： $\int d\Pi_2 = \frac{1}{16\pi^2 m_\mu} \left( \frac{|\mathbf{p}|^2}{E} \frac{E}{|\mathbf{p}|} \right) d\Omega = \frac{|\mathbf{p}|}{16\pi^2 m_\mu} d\Omega$ 将相空间和矩阵元代入截面公式： $\frac{d\sigma}{d\Omega} = \frac{1}{4|\mathbf{p}|m_\mu} \frac{|\mathbf{p}|}{16\pi^2 m_\mu} \left[ \frac{16 e^4 m_\mu^2 E^2}{q^4} \left( 1 - \beta^2 \sin^2\frac{\theta}{2} \right) \right]$ 化简并消去 $m_\mu^2$ ： $\frac{d\sigma}{d\Omega} = \frac{e^4 E^2}{4\pi^2 q^4} \left( 1 - \beta^2 \sin^2\frac{\theta}{2} \right)$ 代入 $e^4 = 16\pi^2 \alpha^2$ 和 $q^4 = 16|\mathbf{p}|^4 \sin^4(\theta/2)$ ： $\frac{d\sigma}{d\Omega} = \frac{16\pi^2 \alpha^2 E^2}{4\pi^2 (16|\mathbf{p}|^4 \sin^4\frac{\theta}{2})} \left( 1 - \beta^2 \sin^2\frac{\theta}{2} \right) = \frac{\alpha^2 E^2}{4|\mathbf{p}|^4 \sin^4\frac{\theta}{2}} \left( 1 - \beta^2 \sin^2\frac{\theta}{2} \right)$ 再次利用 $E^2/|\mathbf{p}|^4 = 1/(|\mathbf{p}|^2 \beta^2)$ ，得到完全一致的 Mott 公式： $\boxed{ \frac{d\sigma}{d\Omega} = \frac{\alpha^2}{4|\mathbf{p}|^2 \beta^2 \sin^4(\theta/2)} \left( 1 - \beta^2 \sin^2 \frac{\theta}{2} \right) }$

5.2

Problem 5.2

peskinChapter 5

习题 5.2

来源: 第5章, PDF第170页

5.2 Bhabha scattering. Compute the differential cross section $d\sigma/d\cos\theta$ for Bhabha scattering, $e^+ e^- \rightarrow e^+ e^-$ . You may work in the limit $E_{\text{cm}} \gg m_e$ , in which it is permissible to ignore the electron mass. There are two Feynman diagrams; these must be added in the invariant matrix element before squaring. Be sure that you have the correct relative sign between these diagrams. The intermediate steps are complicated, but the final result is quite simple. In particular, you may find it useful to introduce the Mandelstam variables $s$ , $t$ , and $u$ . Note that, if we ignore the electron mass, $s + t + u = 0$ . You should be able to cast the differential cross section into the form

\frac{d\sigma}{d\cos\theta} = \frac{\pi\alpha^2}{s} \left[ u^2 \left( \frac{1}{s} + \frac{1}{t} \right)^2 + \left( \frac{t}{s} \right)^2 + \left( \frac{s}{t} \right)^2 \right].

Rewrite this formula in terms of $\cos\theta$ and graph it. What feature of the diagrams causes the differential cross section to diverge as $\theta \rightarrow 0$ ?

习题 5.2 - 解答

1. 费曼图与不变矩阵元

在质心系能量 $E_{\text{cm}} \gg m_e$ 的极限下，我们可以忽略电子质量（ $m_e \approx 0$ ）。Bhabha 散射 $e^+(p_2) e^-(p_1) \rightarrow e^+(p_4) e^-(p_3)$ 在树图阶有两个费曼图贡献：

$s$ -道（湮灭图）：正负电子湮灭为虚光子，再产生正负电子。
$t$ -道（散射图）：正负电子通过交换虚光子发生散射。

根据费曼规则，这两个图的矩阵元分别为： $i\mathcal{M}_s = \bar{v}(p_2) (ie\gamma^\mu) u(p_1) \left( \frac{-ig_{\mu\nu}}{s} \right) \bar{u}(p_3) (ie\gamma^\nu) v(p_4) = \frac{ie^2}{s} [\bar{v}(p_2)\gamma^\mu u(p_1)][\bar{u}(p_3)\gamma_\mu v(p_4)]$ $i\mathcal{M}_t = \bar{u}(p_3) (ie\gamma^\mu) u(p_1) \left( \frac{-ig_{\mu\nu}}{t} \right) \bar{v}(p_2) (ie\gamma^\nu) v(p_4) = \frac{ie^2}{t} [\bar{u}(p_3)\gamma^\mu u(p_1)][\bar{v}(p_2)\gamma_\mu v(p_4)]$

由于末态包含两个全同费米子（或等价地，由 Wick 定理中费米子算符的反对易性），这两个图之间必须有一个相对负号。因此总不变矩阵元为： $\mathcal{M} = \mathcal{M}_s - \mathcal{M}_t = e^2 \left( \frac{1}{s} [\bar{v}(p_2)\gamma^\mu u(p_1)][\bar{u}(p_3)\gamma_\mu v(p_4)] - \frac{1}{t} [\bar{u}(p_3)\gamma^\mu u(p_1)][\bar{v}(p_2)\gamma_\mu v(p_4)] \right)$

2. 矩阵元模方与自旋求和

对初态自旋平均并对末态自旋求和： $\frac{1}{4} \sum_{\text{spins}} |\mathcal{M}|^2 = \frac{e^4}{4} \left( \frac{\sum |\mathcal{M}_s|^2}{s^2} + \frac{\sum |\mathcal{M}_t|^2}{t^2} - \frac{2 \sum \text{Re}(\mathcal{M}_s \mathcal{M}_t^*)}{st} \right)$

利用无质量极限下的迹定理计算各项。引入 Mandelstam 变量： $s = (p_1+p_2)^2 \approx 2p_1 \cdot p_2 \approx 2p_3 \cdot p_4$ $t = (p_1-p_3)^2 \approx -2p_1 \cdot p_3 \approx -2p_2 \cdot p_4$ $u = (p_1-p_4)^2 \approx -2p_1 \cdot p_4 \approx -2p_2 \cdot p_3$

$s$ -道模方： $\sum |\mathcal{M}_s|^2 = \text{Tr}[\gamma^\mu \not{p}_1 \gamma^\nu \not{p}_2] \text{Tr}[\gamma_\mu \not{p}_4 \gamma_\nu \not{p}_3] = 32 [ (p_1 \cdot p_4)(p_2 \cdot p_3) + (p_1 \cdot p_3)(p_2 \cdot p_4) ]$ 代入 Mandelstam 变量得到： $\sum |\mathcal{M}_s|^2 = 32 \left[ \left(-\frac{u}{2}\right)^2 + \left(-\frac{t}{2}\right)^2 \right] = 8(u^2 + t^2)$

$t$ -道模方： $\sum |\mathcal{M}_t|^2 = \text{Tr}[\gamma^\mu \not{p}_1 \gamma^\nu \not{p}_3] \text{Tr}[\gamma_\mu \not{p}_4 \gamma_\nu \not{p}_2] = 32 [ (p_1 \cdot p_4)(p_3 \cdot p_2) + (p_1 \cdot p_2)(p_3 \cdot p_4) ]$ 代入 Mandelstam 变量得到： $\sum |\mathcal{M}_t|^2 = 32 \left[ \left(-\frac{u}{2}\right)^2 + \left(\frac{s}{2}\right)^2 \right] = 8(u^2 + s^2)$

干涉项： $\sum \mathcal{M}_s \mathcal{M}_t^* = \text{Tr}[\gamma^\mu \not{p}_1 \gamma^\nu \not{p}_3 \gamma_\mu \not{p}_4 \gamma_\nu \not{p}_2]$ 利用缩并恒等式 $\gamma^\mu \not{a} \gamma^\nu \not{b} \gamma_\mu = -2 \not{b} \gamma^\nu \not{a}$ 和 $\gamma^\nu \not{a} \not{b} \gamma_\nu = 4 a \cdot b$ ： $\sum \mathcal{M}_s \mathcal{M}_t^* = -2 \text{Tr}[\not{p}_3 \gamma^\nu \not{p}_1 \not{p}_4 \gamma_\nu \not{p}_2] = -8(p_1 \cdot p_4) \text{Tr}[\not{p}_3 \not{p}_2] = -32(p_1 \cdot p_4)(p_3 \cdot p_2)$ $\sum \mathcal{M}_s \mathcal{M}_t^* = -32 \left(-\frac{u}{2}\right)\left(-\frac{u}{2}\right) = -8u^2$

将以上结果代入总模方中： $\frac{1}{4} \sum_{\text{spins}} |\mathcal{M}|^2 = \frac{e^4}{4} \left[ \frac{8(u^2+t^2)}{s^2} + \frac{8(u^2+s^2)}{t^2} - \frac{2(-8u^2)}{st} \right] = 2e^4 \left[ \frac{u^2+t^2}{s^2} + \frac{u^2+s^2}{t^2} + \frac{2u^2}{st} \right]$

3. 微分截面与 Mandelstam 变量形式

质心系中的微分截面公式为： $\frac{d\sigma}{d\Omega} = \frac{1}{64\pi^2 s} \left( \frac{1}{4} \sum |\mathcal{M}|^2 \right)$ 由于 $d\Omega = 2\pi d\cos\theta$ ，且精细结构常数 $\alpha = \frac{e^2}{4\pi}$ （即 $e^4 = 16\pi^2\alpha^2$ ），我们有： $\frac{d\sigma}{d\cos\theta} = 2\pi \frac{d\sigma}{d\Omega} = \frac{16\pi^2\alpha^2}{32\pi s} \cdot 2 \left[ \frac{u^2+t^2}{s^2} + \frac{u^2+s^2}{t^2} + \frac{2u^2}{st} \right] = \frac{\pi\alpha^2}{s} \left[ \frac{u^2}{s^2} + \frac{t^2}{s^2} + \frac{u^2}{t^2} + \frac{s^2}{t^2} + \frac{2u^2}{st} \right]$ 对括号内的项进行重新分组： $\frac{d\sigma}{d\cos\theta} = \frac{\pi\alpha^2}{s} \left[ u^2 \left( \frac{1}{s^2} + \frac{2}{st} + \frac{1}{t^2} \right) + \left(\frac{t}{s}\right)^2 + \left(\frac{s}{t}\right)^2 \right]$ $\boxed{ \frac{d\sigma}{d\cos\theta} = \frac{\pi\alpha^2}{s} \left[ u^2 \left( \frac{1}{s} + \frac{1}{t} \right)^2 + \left(\frac{t}{s}\right)^2 + \left(\frac{s}{t}\right)^2 \right] }$

4. 用 $\cos\theta$ 表示微分截面

在无质量极限下 $s+t+u=0$ ，因此 $\frac{1}{s} + \frac{1}{t} = \frac{s+t}{st} = -\frac{u}{st}$ 。代入上式括号中： $[\cdots] = \frac{u^4}{s^2 t^2} + \left(\frac{t}{s}\right)^2 + \left(\frac{s}{t}\right)^2$ 在质心系中，动量转移变量与散射角 $\theta$ 的关系为： $t = -\frac{s}{2}(1-\cos\theta) = -s \sin^2(\frac{\theta}{2})$ $u = -\frac{s}{2}(1+\cos\theta) = -s \cos^2(\frac{\theta}{2})$

代入括号内的表达式： $[\cdots] = \frac{s^4 \cos^8(\theta/2)}{s^4 \sin^4(\theta/2)} + \sin^4(\theta/2) + \frac{1}{\sin^4(\theta/2)} = \frac{\cos^8(\theta/2) + \sin^8(\theta/2) + 1}{\sin^4(\theta/2)}$ 利用半角公式 $\cos^2(\theta/2) = \frac{1+\cos\theta}{2}$ 和 $\sin^2(\theta/2) = \frac{1-\cos\theta}{2}$ 展开分子： $\cos^8(\theta/2) + \sin^8(\theta/2) = \frac{(1+\cos\theta)^4 + (1-\cos\theta)^4}{16} = \frac{2 + 12\cos^2\theta + 2\cos^4\theta}{16} = \frac{1 + 6\cos^2\theta + \cos^4\theta}{8}$ 分子整体变为： $\frac{1 + 6\cos^2\theta + \cos^4\theta}{8} + 1 = \frac{9 + 6\cos^2\theta + \cos^4\theta}{8} = \frac{(3+\cos^2\theta)^2}{8}$ 分母为 $\sin^4(\theta/2) = \frac{(1-\cos\theta)^2}{4}$ 。将分子分母相除，得到最终的微分截面： $\boxed{ \frac{d\sigma}{d\cos\theta} = \frac{\pi\alpha^2}{s} \frac{(3+\cos^2\theta)^2}{2(1-\cos\theta)^2} }$

5. 图像特征与发散原因分析

图像特征：函数 $f(\cos\theta) = \frac{(3+\cos^2\theta)^2}{2(1-\cos\theta)^2}$ 在物理区间 $\theta \in (0, \pi]$ 内单调递减。

在后向散射 $\theta = \pi$ ( $\cos\theta = -1$ ) 处，截面取得全局最小值 $\frac{2\pi\alpha^2}{s}$ 。
随着 $\theta \rightarrow 0$ ( $\cos\theta \rightarrow 1$ )，截面急剧上升，表现出 $\sim 1/\theta^4$ 的发散行为。

发散的物理原因：微分截面在 $\theta \rightarrow 0$ 时的发散完全来源于 $t$ -道费曼图（散射图）。在 $t$ -道中，交换的虚光子携带的四维动量平方为 $t = -s(1-\cos\theta)/2$ 。当散射角 $\theta \rightarrow 0$ 时，动量转移 $t \rightarrow 0$ ，这意味着交换的虚光子趋于“在壳”（on-shell，即质量平方趋于0）。此时光子传播子 $1/t$ 发生发散，导致截面出现 $1/t^2$ 的奇点。这在物理上对应于无质量光子传递的无限程库仑相互作用（Coulomb singularity），使得极小角（前向）散射的概率趋于无穷大。

5.3

Problem 5.3

peskinChapter 5

习题 5.3

来源: 第5章, PDF第170,171页

5.3 The spinor product formalism introduced in Problem 3.3 provides an efficient way to compute tree diagrams involving massless particles. Recall that in Problem 3.3 we defined spinor products as follows: Let $u_{L0}$ , $u_{R0}$ be the left- and right-handed spinors at some fixed lightlike momentum $k_0$ . These satisfy

u_{L0}\bar{u}_{L0} = \left( \frac{1-\gamma^5}{2} \right) \not{k}_0, \quad u_{R0}\bar{u}_{R0} = \left( \frac{1+\gamma^5}{2} \right) \not{k}_0. \tag{1}

(These relations are just the projections onto definite helicity of the more standard formula $\sum u_0\bar{u}_0 = \not{k}_0$ .) Then define spinors for any other lightlike momentum $p$ by

u_L(p) = \frac{1}{\sqrt{2p \cdot k_0}} \not{p} u_{R0}, \quad u_R(p) = \frac{1}{\sqrt{2p \cdot k_0}} \not{p} u_{L0}. \tag{2}

We showed that these spinors satisfy $\not{p}u(p) = 0$ ; because there is no $m$ around, they can be used as spinors for either fermions or antifermions. We defined

s(p_1, p_2) = \bar{u}_R(p_1)u_L(p_2), \quad t(p_1, p_2) = \bar{u}_L(p_1)u_R(p_2),

and, in a special frame, we proved the properties

t(p_1, p_2) = (s(p_2, p_1))^*, \quad s(p_1, p_2) = -s(p_2, p_1), \quad |s(p_1, p_2)|^2 = 2p_1 \cdot p_2. \tag{3}

Now let us apply these results.

(a) To warm up, give another proof of the last relation in Eq. (3) by using (1) to rewrite $|s(p_1, p_2)|^2$ as a trace of Dirac matrices, and then applying the trace calculus.

(b) Show that, for any string of Dirac matrices,

\text{tr}[\gamma^\mu \gamma^\nu \gamma^\rho \cdots] = \text{tr}[\cdots \gamma^\rho \gamma^\nu \gamma^\mu]

where $\mu, \nu, \rho, \dots = 0, 1, 2, 3$ , or $5$ . Use this identity to show that

\bar{u}_L(p_1)\gamma^\mu u_L(p_2) = \bar{u}_R(p_2)\gamma^\mu u_R(p_1).

\bar{u}_L(p_1)\gamma^\mu u_L(p_2) [\gamma_\mu]_{ab} = 2[u_L(p_2)\bar{u}_L(p_1) + u_R(p_1)\bar{u}_R(p_2)]_{ab},

where $a, b = 1, 2, 3, 4$ are Dirac indices. This can be done by justifying the following statements: The right-hand side of this equation is a Dirac matrix; thus, it can be written as a linear combination of the 16 $\Gamma$ matrices discussed in Section 3.4. It satisfies

\gamma^5 [M] = -[M] \gamma^5,

thus, it must have the form

[M] = \left( \frac{1-\gamma^5}{2} \right) \gamma_\mu V^\mu + \left( \frac{1+\gamma^5}{2} \right) \gamma_\mu W^\mu

where $V^\mu$ and $W^\mu$ are 4-vectors. These 4-vectors can be computed by trace technology; for example,

V^\nu = \frac{1}{2} \operatorname{tr} [\gamma^\nu \left( \frac{1-\gamma^5}{2} \right) M].

(d) Consider the process $e^+ e^- \rightarrow \mu^+ \mu^-$ , to the leading order in $\alpha$ , ignoring the masses of both the electron and the muon. Consider first the case in which the electron and the final muon are both right-handed and the positron and the final antimuon are both left-handed. (Use the spinor $v_R$ for the antimuon and $\bar{u}_R$ for the positron.) Apply the Fierz identity to show that the amplitude can be evaluated directly in terms of spinor products. Square the amplitude and reproduce the result for

\frac{d\sigma}{d \cos \theta} (e_R^- e_L^+ \rightarrow \mu_R^- \mu_L^+)

given in Eq. (5.22). Compute the other helicity cross sections for this process and show that they also reproduce the results found in Section 5.2.

(e) Compute the differential cross section for Bhabha scattering of massless electrons, helicity state by helicity state, using the spinor product formalism. The average over initial helicities, summed over final helicities, should reproduce the result of Problem 5.2. In the process, you should see how this result arises as the sum of definite-helicity contributions.

Referenced Equations:

Equation (5.22):

\frac{d\sigma}{d\Omega} (e^-_R e^+_L \rightarrow \mu^-_R \mu^+_L) = \frac{\alpha^2}{4E_{\text{cm}}^2} (1 + \cos \theta)^2. \tag{5.22}

习题 5.3 - 解答

分析与解答

(a) 利用迹运算证明 $|s(p_1, p_2)|^2 = 2p_1 \cdot p_2$

先分析旋量内积的模方。根据定义 $s(p_1, p_2) = \bar{u}_R(p_1)u_L(p_2)$ ，其复共轭为：

(s(p_1, p_2))^* = (\bar{u}_R(p_1)u_L(p_2))^\dagger \gamma^0 = \bar{u}_L(p_2)u_R(p_1) = t(p_2, p_1)

因此，模方可以写为：

|s(p_1, p_2)|^2 = s(p_1, p_2) t(p_2, p_1) = \bar{u}_R(p_1)u_L(p_2)\bar{u}_L(p_2)u_R(p_1)

将其转化为 Dirac 矩阵的迹：

|s(p_1, p_2)|^2 = \operatorname{tr}[u_R(p_1)\bar{u}_R(p_1) u_L(p_2)\bar{u}_L(p_2)]

代入无质量粒子的极化投影关系 $u_R(p)\bar{u}_R(p) = \frac{1+\gamma^5}{2}\not{p}$ 和 $u_L(p)\bar{u}_L(p) = \frac{1-\gamma^5}{2}\not{p}$ ：

|s(p_1, p_2)|^2 = \operatorname{tr}\left[ \left(\frac{1+\gamma^5}{2}\right)\not{p}_1 \left(\frac{1-\gamma^5}{2}\right)\not{p}_2 \right]

利用 $\gamma^5$ 与 $\gamma^\mu$ 反对易的性质，将 $\frac{1-\gamma^5}{2}$ 向左移动穿过 $\not{p}_1$ ：

\not{p}_1 \left(\frac{1-\gamma^5}{2}\right) = \left(\frac{1+\gamma^5}{2}\right) \not{p}_1

由于 $\left(\frac{1+\gamma^5}{2}\right)^2 = \frac{1+\gamma^5}{2}$ ，迹变为：

|s(p_1, p_2)|^2 = \operatorname{tr}\left[ \left(\frac{1+\gamma^5}{2}\right) \not{p}_1 \not{p}_2 \right] = \frac{1}{2}\operatorname{tr}[\not{p}_1 \not{p}_2] + \frac{1}{2}\operatorname{tr}[\gamma^5 \not{p}_1 \not{p}_2]

因为包含 $\gamma^5$ 和少于四个 $\gamma$ 矩阵的迹为零，且 $\operatorname{tr}[\not{p}_1 \not{p}_2] = 4 p_1 \cdot p_2$ ，最终得到：

\boxed{|s(p_1, p_2)|^2 = 2p_1 \cdot p_2}

(b) 证明迹的逆序恒等式及旋量流关系的等价性

首先证明 $\operatorname{tr}[\gamma^\mu \gamma^\nu \cdots \gamma^\rho] = \operatorname{tr}[\gamma^\rho \cdots \gamma^\nu \gamma^\mu]$ 。利用转置操作不改变矩阵的迹 $\operatorname{tr}[A] = \operatorname{tr}[A^T]$ ，以及电荷共轭矩阵 $C$ 满足 $C \gamma^\mu C^{-1} = -(\gamma^\mu)^T$ ：

\operatorname{tr}[\gamma^\mu \gamma^\nu \cdots \gamma^\rho] = \operatorname{tr}[(\gamma^\rho)^T \cdots (\gamma^\nu)^T (\gamma^\mu)^T]

代入 $(\gamma^\alpha)^T = -C \gamma^\alpha C^{-1}$ ，假设共有 $n$ 个 $\gamma$ 矩阵：

\operatorname{tr}[A^T] = (-1)^n \operatorname{tr}[C (\gamma^\rho \cdots \gamma^\nu \gamma^\mu) C^{-1}] = (-1)^n \operatorname{tr}[\gamma^\rho \cdots \gamma^\nu \gamma^\mu]

由于奇数个 $\gamma$ 矩阵的迹恒为零，我们只需考虑 $n$ 为偶数的情况，此时 $(-1)^n = 1$ ，故：

\operatorname{tr}[\gamma^\mu \gamma^\nu \cdots \gamma^\rho] = \operatorname{tr}[\gamma^\rho \cdots \gamma^\nu \gamma^\mu]

下面利用此恒等式证明 $\bar{u}_L(p_1)\gamma^\mu u_L(p_2) = \bar{u}_R(p_2)\gamma^\mu u_R(p_1)$ 。根据定义 $u_L(p) = \frac{1}{\sqrt{2p \cdot k_0}} \not{p} u_{R0}$ ，代入左式并写成迹的形式（标量的迹等于其自身）：

\bar{u}_L(p_1)\gamma^\mu u_L(p_2) = \frac{1}{\sqrt{4(p_1\cdot k_0)(p_2\cdot k_0)}} \operatorname{tr}[u_{R0}\bar{u}_{R0} \not{p}_1 \gamma^\mu \not{p}_2]

代入 $u_{R0}\bar{u}_{R0} = \frac{1+\gamma^5}{2}\not{k}_0$ ：

\operatorname{tr}\left[ \frac{1+\gamma^5}{2}\not{k}_0 \not{p}_1 \gamma^\mu \not{p}_2 \right]

对该迹应用逆序恒等式（注意 $\gamma^5 = i\gamma^0\gamma^1\gamma^2\gamma^3$ 逆序后仍为 $\gamma^5$ ）：

\operatorname{tr}\left[ \not{p}_2 \gamma^\mu \not{p}_1 \not{k}_0 \frac{1+\gamma^5}{2} \right]

利用迹的循环性，将 $\not{k}_0 \frac{1+\gamma^5}{2}$ 移到最左侧，并穿过 $\not{k}_0$ 变为 $\frac{1-\gamma^5}{2}$ ：

\operatorname{tr}\left[ \frac{1-\gamma^5}{2} \not{k}_0 \not{p}_2 \gamma^\mu \not{p}_1 \right] = \operatorname{tr}[u_{L0}\bar{u}_{L0} \not{p}_2 \gamma^\mu \not{p}_1]

乘回归一化系数，这正是 $\bar{u}_R(p_2)\gamma^\mu u_R(p_1)$ 的展开式。因此：

\boxed{\bar{u}_L(p_1)\gamma^\mu u_L(p_2) = \bar{u}_R(p_2)\gamma^\mu u_R(p_1)}

(c) 证明 Fierz 恒等式

令矩阵 $M_{ab} = \bar{u}_L(p_1)\gamma^\mu u_L(p_2) [\gamma_\mu]_{ab}$ 。验证其与 $\gamma^5$ 的对易关系：

\gamma^5 M = \bar{u}_L(p_1)\gamma^\mu u_L(p_2) \gamma^5 \gamma_\mu = - \bar{u}_L(p_1)\gamma^\mu u_L(p_2) \gamma_\mu \gamma^5 = -M \gamma^5

满足 $\gamma^5 M = -M \gamma^5$ 的 Dirac 矩阵只能是矢量和轴矢量的线性组合，因此可设：

M = \left( \frac{1-\gamma^5}{2} \right) \gamma_\nu V^\nu + \left( \frac{1+\gamma^5}{2} \right) \gamma_\nu W^\nu

利用迹技术求解 $V^\nu$ 和 $W^\nu$ ：

V^\nu = \frac{1}{2} \operatorname{tr} \left[ \gamma^\nu \left( \frac{1-\gamma^5}{2} \right) M \right] = \frac{1}{2} \bar{u}_L(p_1)\gamma^\mu u_L(p_2) \operatorname{tr} \left[ \gamma^\nu \left( \frac{1-\gamma^5}{2} \right) \gamma_\mu \right]

由于 $\operatorname{tr}[\gamma^\nu \gamma_\mu] = 4\delta^\nu_\mu$ 且 $\operatorname{tr}[\gamma^\nu \gamma^5 \gamma_\mu] = 0$ ，得到 $V^\nu = \bar{u}_L(p_1)\gamma^\nu u_L(p_2)$ 。同理：

W^\nu = \frac{1}{2} \operatorname{tr} \left[ \gamma^\nu \left( \frac{1+\gamma^5}{2} \right) M \right] = \bar{u}_L(p_1)\gamma^\nu u_L(p_2)

根据 (b) 的结论， $W^\nu = \bar{u}_R(p_2)\gamma^\nu u_R(p_1)$ 。将 $V^\nu$ 和 $W^\nu$ 代回 $M$ 的表达式：

M = \left( \frac{1-\gamma^5}{2} \right) \gamma_\nu \bar{u}_L(p_1)\gamma^\nu u_L(p_2) + \left( \frac{1+\gamma^5}{2} \right) \gamma_\nu \bar{u}_R(p_2)\gamma^\nu u_R(p_1)

注意到 $u_L(p_2)\bar{u}_L(p_1) = \frac{1-\gamma^5}{2} \gamma_\nu \frac{1}{2} \operatorname{tr}[\gamma^\nu u_L(p_2)\bar{u}_L(p_1)] = \frac{1}{2} \left( \frac{1-\gamma^5}{2} \right) \gamma_\nu \bar{u}_L(p_1)\gamma^\nu u_L(p_2)$ 。同理 $u_R(p_1)\bar{u}_R(p_2) = \frac{1}{2} \left( \frac{1+\gamma^5}{2} \right) \gamma_\nu \bar{u}_R(p_2)\gamma^\nu u_R(p_1)$ 。代入即得：

\boxed{M_{ab} = 2[u_L(p_2)\bar{u}_L(p_1) + u_R(p_1)\bar{u}_R(p_2)]_{ab}}

(d) 计算 $e^+ e^- \rightarrow \mu^+ \mu^-$ 截面

对于无质量反粒子，左手螺旋对应右手手征旋量，即 $e^+_L$ 对应 $v_R(p_2) = u_R(p_2)$ ， $\mu^+_L$ 对应 $v_R(k_2) = u_R(k_2)$ 。过程 $e^-_R(p_1) e^+_L(p_2) \rightarrow \mu^-_R(k_1) \mu^+_L(k_2)$ 的振幅为：

i\mathcal{M} = \frac{ie^2}{s} (\bar{u}_R(p_2) \gamma^\mu u_R(p_1)) (\bar{u}_R(k_1) \gamma_\mu u_R(k_2))

对 $\bar{u}_R \gamma^\mu u_R$ 应用 Fierz 恒等式（将 (c) 中 $L \leftrightarrow R$ 互换）：

\mathcal{M} = \frac{2e^2}{s} \bar{u}_R(k_1) [ u_R(p_1)\bar{u}_R(p_2) + u_L(p_2)\bar{u}_L(p_1) ] u_R(k_2)

由于手征性正交 $\bar{u}_R u_R = 0$ ，仅保留交叉项：

\mathcal{M} = \frac{2e^2}{s} \bar{u}_R(k_1) u_L(p_2) \bar{u}_L(p_1) u_R(k_2) = \frac{2e^2}{s} s(k_1, p_2) t(p_1, k_2)

取模方，利用 $|s(p_i, p_j)|^2 = 2p_i \cdot p_j$ ：

|\mathcal{M}|^2 = \frac{4e^4}{s^2} (2k_1 \cdot p_2)(2p_1 \cdot k_2) = \frac{4e^4}{s^2} u^2

在质心系中 $s = 4E^2$ ， $u = -2E^2(1+\cos\theta)$ ，代入微分散射截面公式：

\boxed{\frac{d\sigma}{d\Omega} (e_R^- e_L^+ \rightarrow \mu_R^- \mu_L^+) = \frac{e^4 u^2}{64\pi^2 s^3} = \frac{\alpha^2}{4E_{\text{cm}}^2} (1 + \cos \theta)^2}

同理计算其他非零螺旋度组合：

$e^-_R e^+_L \rightarrow \mu^-_L \mu^+_R$ ： $|\mathcal{M}|^2 = \frac{4e^4}{s^2} t^2 \implies \frac{d\sigma}{d\Omega} = \frac{\alpha^2}{4E_{\text{cm}}^2} (1 - \cos \theta)^2$
$e^-_L e^+_R \rightarrow \mu^-_R \mu^+_L$ ： $|\mathcal{M}|^2 = \frac{4e^4}{s^2} t^2 \implies \frac{d\sigma}{d\Omega} = \frac{\alpha^2}{4E_{\text{cm}}^2} (1 - \cos \theta)^2$
$e^-_L e^+_R \rightarrow \mu^-_L \mu^+_R$ ： $|\mathcal{M}|^2 = \frac{4e^4}{s^2} u^2 \implies \frac{d\sigma}{d\Omega} = \frac{\alpha^2}{4E_{\text{cm}}^2} (1 + \cos \theta)^2$

(e) 计算无质量 Bhabha 散射截面

Bhabha 散射包含 s 沟道和 t 沟道。逐个螺旋度态计算：

$e^-_R e^+_L \rightarrow e^-_R e^+_L$ ： s 沟道振幅为 $\frac{2e^2}{s} s(k_1, p_2) t(p_1, k_2)$ 。 t 沟道振幅为 $-\frac{e^2}{t} (\bar{u}_R(k_1) \gamma^\mu u_R(p_1)) (\bar{u}_R(p_2) \gamma_\mu u_R(k_2))$ 。应用 Fierz 恒等式并利用 $s(p_2, k_1) = -s(k_1, p_2)$ ，得到 $+\frac{2e^2}{t} s(k_1, p_2) t(p_1, k_2)$ 。总振幅模方： $|\mathcal{M}|^2 = 4e^4 u^2 \left( \frac{1}{s} + \frac{1}{t} \right)^2 = 4e^4 u^2 \left( \frac{s+t}{st} \right)^2 = 4e^4 \frac{u^4}{s^2 t^2}$
$e^-_R e^+_R \rightarrow e^-_R e^+_R$ ： s 沟道为零。t 沟道振幅模方为 $4e^4 \frac{s^2}{t^2}$ 。
$e^-_R e^+_L \rightarrow e^-_L e^+_R$ ： t 沟道为零。s 沟道振幅模方为 $4e^4 \frac{t^2}{s^2}$ 。

由宇称对称性（ $L \leftrightarrow R$ ），另外三个非零组合的模方与上述对应相等。对初态平均并对末态求和（乘以 $\frac{1}{4} \times 2$ ）：

\frac{1}{4} \sum_{\text{spins}} |\mathcal{M}|^2 = 2e^4 \left( \frac{u^4}{s^2 t^2} + \frac{s^2}{t^2} + \frac{t^2}{s^2} \right)

利用运动学关系 $s+t+u=0$ ，可以证明 $\frac{u^4}{s^2 t^2} = \frac{(s+t)^4}{s^2 t^2} = \frac{u^2}{s^2} + \frac{u^2}{t^2} + \frac{2u^2}{st}$ 。代入上式重组，完美复现标准结果：

\boxed{\frac{1}{4} \sum_{\text{spins}} |\mathcal{M}|^2 = 2e^4 \left( \frac{s^2+u^2}{t^2} + \frac{t^2+u^2}{s^2} + \frac{2u^2}{st} \right)}

5.4

Problem 5.4

peskinChapter 5

习题 5.4

来源: 第5章, PDF第171,172页

5.4 Positronium lifetimes.

(a) Compute the amplitude $\mathcal{M}$ for $e^+ e^-$ annihilation into 2 photons in the extreme nonrelativistic limit (i.e., keep only the term proportional to zero powers of the electron and positron 3-momentum). Use this result, together with our formalism for fermion-antifermion bound states, to compute the rate of annihilation of the $1S$ states of positronium into 2 photons. You should find that the spin-1 states of positronium do not annihilate into 2 photons, confirming the symmetry argument of Problem 3.8. For the spin-0 state of positronium, you should find a result proportional to the square of the $1S$ wavefunction at the origin. Inserting the value of this wavefunction from nonrelativistic quantum mechanics, you should find

\frac{1}{\tau} = \Gamma = \frac{\alpha^5 m_e}{2} \approx 8.03 \times 10^9 \text{ sec}^{-1}.

A recent measurement ${}^{\ddagger}$ gives $\Gamma = 7.994 \pm .011 \text{ nsec}^{-1}$ ; the $0.5\%$ discrepancy is accounted for by radiative corrections.

(b) Computing the decay rates of higher- $l$ positronium states is somewhat more difficult; in the rest of this problem, we will consider the case $l = 1$ . First, work out the terms in the $e^+ e^- \rightarrow 2\gamma$ amplitude proportional to one power of the 3-momentum. (For simplicity, work in the center-of-mass frame.) Since

\int \frac{d^3 p}{(2\pi)^3} p^i \psi(\mathbf{p}) = i \frac{\partial}{\partial x^i} \psi(\mathbf{x}) \bigg|_{\mathbf{x}=0} ,

this piece of the amplitude has overlap with $P$ -wave bound states. Show that the $S = 1$ , but not the $S = 0$ states, can decay to 2 photons. Again, this is a consequence of $C$ .

(c) To compute the decay rates of these $P$ -wave states, we need properly normalized state vectors. Denote the three $P$ -state wavefunctions by

\psi_i = x^i f(|\mathbf{x}|), \quad \text{normalized to } \int d^3 x \psi_i^*(\mathbf{x}) \psi_j(\mathbf{x}) = \delta_{ij} ,

and their Fourier transforms by $\psi_i(\mathbf{p})$ . Show that

|B(\mathbf{k})\rangle = \sqrt{2M} \int \frac{d^3 p}{(2\pi)^3} \psi_i(\mathbf{p}) a_{\mathbf{p}+\mathbf{k}/2}^{\dagger} \Sigma^i b_{-\mathbf{p}+\mathbf{k}/2}^{\dagger} |0\rangle

is a properly normalized bound-state vector if $\Sigma^i$ denotes a set of three $2 \times 2$ matrices normalized to

\sum_i \text{tr}(\Sigma^{i\dagger} \Sigma^i) = 1 .

To build $S = 1$ states, we should take each $\Sigma^i$ to contain a Pauli sigma matrix. In general, spin-orbit coupling will split the multiplet of $S = 1, L = 1$ states according to the total angular momentum $J$ . The states of definite $J$ are given by

\begin{aligned} J = 0: & \quad \Sigma^i = \frac{1}{\sqrt{6}} \sigma^i , \\ J = 1: & \quad \Sigma^i = \frac{1}{2} \epsilon^{ijk} n^j \sigma^k , \\ J = 2: & \quad \Sigma^i = \frac{1}{\sqrt{3}} h^{ij} \sigma^j , \end{aligned}

where $\mathbf{n}$ is a polarization vector satisfying $|\mathbf{n}|^2 = 1$ and $h^{ij}$ is a traceless tensor, for which a typical value might be $h^{12} = 1$ and all other components zero.

(d) Using the expanded form for the $e^+ e^- \rightarrow 2\gamma$ amplitude derived in part (b) and the explicit form of the $S = 1, L = 1$ , definite- $J$ positronium states found in part (c), compute, for each $J$ , the decay rate of the state into two photons.

习题 5.4 - 解答

(a) $1S$ 态正电子素的衰变率

先分析极端非相对论极限下的 $e^+ e^- \to 2\gamma$ 散射振幅。设电子和正电子的动量为 $p_1 = p_2 = (m, \mathbf{0})$ ，光子动量为 $k_1 = (m, \mathbf{k})$ 和 $k_2 = (m, -\mathbf{k})$ ，其中 $|\mathbf{k}| = m$ 。根据 Feynman 规则，树图阶振幅（包含 t 通道和 u 通道）为：

i\mathcal{M} = -ie^2 \bar{v}(p_2) \left[ \notin_2^* \frac{\slashed{p}_1 - \slashed{k}_1 + m}{(p_1-k_1)^2 - m^2} \notin_1^* + \notin_1^* \frac{\slashed{p}_1 - \slashed{k}_2 + m}{(p_1-k_2)^2 - m^2} \notin_2^* \right] u(p_1)

在 $\mathbf{p}=0$ 极限下，分母 $(p_1-k_{1,2})^2 - m^2 = -2m^2$ 。利用 Dirac 方程 $\slashed{p}_1 u(p_1) = m u(p_1)$ 以及库仑规范 $\epsilon^0 = 0 \implies p_1 \cdot \epsilon_{1,2}^* = 0$ ，分子可化简为：

\notin_2^* (\slashed{p}_1 - \slashed{k}_1 + m) \notin_1^* u(p_1) = \notin_2^* (-\slashed{k}_1 \notin_1^*) u(p_1)

因此，截断振幅算符为：

\mathcal{O}_{amp} = \frac{ie^2}{2m^2} \left( \notin_2^* \slashed{k}_1 \notin_1^* + \notin_1^* \slashed{k}_2 \notin_2^* \right)

利用费米子-反费米子束缚态形式化方法，束缚态振幅由下式给出：

\mathcal{M}_{bs} = \sqrt{2M} \psi(0) \text{Tr}\left[ \mathcal{O}_{amp} \Pi_{spin} \right]

其中 $M \approx 2m$ ，自旋投影算符在静止系下为 $\Pi_{spin} = \frac{1+\gamma^0}{2\sqrt{2}} \Gamma_{spin}$ 。对于自旋单态 ( $S=0$ )， $\Gamma_{spin} = \gamma^5$ 。计算迹：

\text{Tr}\left[ \notin_2^* \slashed{k}_1 \notin_1^* \frac{1+\gamma^0}{2\sqrt{2}} \gamma^5 \right] = \frac{1}{2\sqrt{2}} \text{Tr}[\gamma^0 \notin_2^* \slashed{k}_1 \notin_1^* \gamma^5] = \frac{-4i}{2\sqrt{2}} \epsilon^{0\mu\nu\rho} \epsilon_{2\mu}^* k_{1\nu} \epsilon_{1\rho}^* = \sqrt{2} i \mathbf{k} \cdot (\boldsymbol{\epsilon}_1^* \times \boldsymbol{\epsilon}_2^*)

由于 $k_2 = (m, -\mathbf{k})$ ，第二项给出相同的贡献。因此 $S=0$ 的束缚态振幅为：

\mathcal{M}_{bs}^{S=0} = \sqrt{4m} \psi(0) \frac{ie^2}{2m^2} \left( 2\sqrt{2} i \mathbf{k} \cdot (\boldsymbol{\epsilon}_1^* \times \boldsymbol{\epsilon}_2^*) \right) = -\frac{4e^2}{m} \psi(0) \mathbf{k} \cdot (\boldsymbol{\epsilon}_1^* \times \boldsymbol{\epsilon}_2^*)

对于自旋三重态 ( $S=1$ )， $\Gamma_{spin} = \notin_{bs}$ 。迹中包含奇数个 $\gamma$ 矩阵或空间动量相互抵消，导致 $\mathcal{M}_{bs}^{S=1} = 0$ 。这证实了自旋为 1 的状态不能衰变为两个光子（C 宇称守恒）。

计算 $S=0$ 的衰变率，对光子极化求和 $\sum |\mathbf{k} \cdot (\boldsymbol{\epsilon}_1^* \times \boldsymbol{\epsilon}_2^*)|^2 = 2|\mathbf{k}|^2 = 2m^2$ ：

\Gamma = \frac{1}{2M} \frac{1}{2!} \int d\Pi_2 |\mathcal{M}_{bs}^{S=0}|^2 = \frac{1}{4m} \frac{1}{2} \frac{1}{8\pi} \frac{16e^4}{m^2} |\psi(0)|^2 (2m^2) = \frac{4\pi \alpha^2}{m} |\psi(0)|^2

代入非相对论量子力学 $1S$ 态波函数在原点的值 $|\psi(0)|^2 = \frac{(m\alpha/2)^3}{\pi} = \frac{m^3 \alpha^3}{8\pi}$ ：

\boxed{ \Gamma = \frac{\alpha^5 m_e}{2} }

(b) 展开至 3-动量的一阶 $O(p)$

在质心系中，设 $p_1 = (m, \mathbf{p})$ 和 $p_2 = (m, -\mathbf{p})$ 。将传播子分母和分子展开至 $O(p)$ ：

\frac{1}{(p_1-k_1)^2 - m^2} \approx -\frac{1}{2m^2} \left( 1 + \frac{\mathbf{p}\cdot\mathbf{k}}{m^2} \right)

束缚态投影算符展开至 $O(p)$ 为：

\Pi(\mathbf{p}) = \frac{(\slashed{p}_1+m) \Gamma_{spin} (-\slashed{p}_2+m)}{2\sqrt{2}m} \approx \frac{m}{\sqrt{2}} \left[ \frac{1+\gamma^0}{2} \Gamma_{spin} \frac{1-\gamma^0}{2} - \frac{\boldsymbol{\gamma}\cdot\mathbf{p}}{2m} \Gamma_{spin} \frac{1-\gamma^0}{2} - \frac{1+\gamma^0}{2} \Gamma_{spin} \frac{\boldsymbol{\gamma}\cdot\mathbf{p}}{2m} \right]

对于 $S=0$ ( $\Gamma_{spin} = \gamma^5$ )， $O(p)$ 阶的迹包含 $\text{Tr}[\mathcal{O}_{amp}^{(0)} \Pi^{(1)}]$ 和 $\text{Tr}[\mathcal{O}_{amp}^{(1)} \Pi^{(0)}]$ 。计算表明，由于 $k_1$ 和 $k_2$ 的空间动量反向，所有 $O(p)$ 项精确抵消，因此 $S=0$ 的 P 波态不能衰变为双光子（同样是 C 宇称的推论）。

对于 $S=1$ ( $\Gamma_{spin} = \notin_{bs}$ )，迹不为零。提取与 $p^l$ 成正比的系数 $\mathcal{M}^l$ ，得到：

\mathcal{M}^l = \frac{2\sqrt{2} i e^2}{m} \left[ (\epsilon_1^*)_l (\boldsymbol{\epsilon}_2^* \cdot \boldsymbol{\epsilon}_{bs}) + (\epsilon_2^*)_l (\boldsymbol{\epsilon}_1^* \cdot \boldsymbol{\epsilon}_{bs}) + \frac{k_l}{m^2} (\boldsymbol{\epsilon}_1^* \cdot \boldsymbol{\epsilon}_2^*) (\mathbf{k} \cdot \boldsymbol{\epsilon}_{bs}) \right]

这表明 $S=1$ 的 P 波态可以衰变为双光子。

(c) P 波态的归一化

计算给定态矢的内积：

\langle B(\mathbf{k}') | B(\mathbf{k}) \rangle = 2M \int \frac{d^3 p}{(2\pi)^3} \frac{d^3 p'}{(2\pi)^3} \psi_j^*(\mathbf{p}') \psi_i(\mathbf{p}) \langle 0 | b_{-\mathbf{p}'+\mathbf{k}'/2} \Sigma^{j\dagger} a_{\mathbf{p}'+\mathbf{k}'/2} a_{\mathbf{p}+\mathbf{k}/2}^{\dagger} \Sigma^i b_{-\mathbf{p}+\mathbf{k}/2}^{\dagger} |0\rangle

利用对易关系收缩产生湮灭算符，产生 $(2\pi)^3 \delta^3(\mathbf{k}'-\mathbf{k}) (2\pi)^3 \delta^3(\mathbf{p}'-\mathbf{p})$ ，并留下自旋矩阵的迹：

\langle B(\mathbf{k}') | B(\mathbf{k}) \rangle = 2M (2\pi)^3 \delta^3(\mathbf{k}'-\mathbf{k}) \int \frac{d^3 p}{(2\pi)^3} \psi_j^*(\mathbf{p}) \psi_i(\mathbf{p}) \text{tr}(\Sigma^{j\dagger} \Sigma^i)

由于空间波函数归一化为 $\int \frac{d^3 p}{(2\pi)^3} \psi_j^*(\mathbf{p}) \psi_i(\mathbf{p}) = \delta_{ij}$ ，上式化简为：

\langle B(\mathbf{k}') | B(\mathbf{k}) \rangle = 2M (2\pi)^3 \delta^3(\mathbf{k}'-\mathbf{k}) \sum_i \text{tr}(\Sigma^{i\dagger} \Sigma^i)

要使其满足标准的单粒子态归一化条件，必须有：

\boxed{ \sum_i \text{tr}(\Sigma^{i\dagger} \Sigma^i) = 1 }

(d) $S=1, L=1$ 各 $J$ 态的衰变率

利用动量空间积分 $\int \frac{d^3 p}{(2\pi)^3} p^l \psi_i(\mathbf{p}) = i \delta_{li} f(0)$ ，束缚态振幅可写为：

\mathcal{M}_{bs} = \sqrt{2M} i f(0) \mathcal{M}^i(\Sigma^i)

将 (b) 中的 $\frac{1}{\sqrt{2}}(\epsilon_{bs})_j$ 替换为 $\Sigma^i$ 展开的系数 $A^{ij}$ （即 $\Sigma^i = A^{ij}\sigma^j$ ），定义张量 $T^{ij} = \epsilon_{1i}^* \epsilon_{2j}^* + \epsilon_{2i}^* \epsilon_{1j}^* + \frac{k_i k_j}{m^2} \boldsymbol{\epsilon}_1^* \cdot \boldsymbol{\epsilon}_2^*$ ，则振幅正比于 $T^{ij} A^{ij}$ 。

1. $J=0$ 态： $\Sigma^i = \frac{1}{\sqrt{6}} \sigma^i \implies A^{ij} = \frac{1}{\sqrt{6}} \delta^{ij}$ 。

T^{ij} A^{ij} = \frac{1}{\sqrt{6}} T^{ii} = \frac{1}{\sqrt{6}} \left( 2 \boldsymbol{\epsilon}_1^* \cdot \boldsymbol{\epsilon}_2^* + \frac{\mathbf{k}^2}{m^2} \boldsymbol{\epsilon}_1^* \cdot \boldsymbol{\epsilon}_2^* \right) = \sqrt{\frac{3}{2}} \boldsymbol{\epsilon}_1^* \cdot \boldsymbol{\epsilon}_2^*

对极化求和并积分，得到衰变率：

\boxed{ \Gamma_{J=0} = \frac{96 \pi^2 \alpha^2}{m^2} |f(0)|^2 }

2. $J=1$ 态： $\Sigma^i = \frac{1}{2} \epsilon^{ijk} n^j \sigma^k \implies A^{ik} = \frac{1}{2} \epsilon^{ijk} n^j$ 。由于 $T^{ik}$ 对指标 $i,k$ 是对称的，而 $\epsilon^{ijk}$ 是反对称的，因此 $T^{ik} A^{ik} = 0$ 。

\boxed{ \Gamma_{J=1} = 0 }

（这符合 Yang 定理，自旋为 1 的粒子不能衰变为两个无质量的自旋 1 粒子）。

3. $J=2$ 态： $\Sigma^i = \frac{1}{\sqrt{3}} h^{ij} \sigma^j \implies A^{ij} = \frac{1}{\sqrt{3}} h^{ij}$ 。为满足 (c) 中的归一化条件 $\sum_i \text{tr}(\Sigma^{i\dagger} \Sigma^i) = 1$ ，无迹对称张量 $h^{ij}$ 必须满足 $h^{ij} h^{ij} = 3/2$ 。计算 $|T^{ij} A^{ij}|^2$ 并对光子极化求和，得到协变形式：

\sum_{pol} |T^{ij} A^{ij}|^2 = \frac{1}{3} \left[ 4 h^{ij} h^{ij} - 8 h^{im} h^{in} \hat{k}_m \hat{k}_n + 2 (h^{mn} \hat{k}_m \hat{k}_n)^2 \right]

对光子出射立体角 $\frac{d\Omega}{4\pi}$ 求平均，利用 $\langle \hat{k}_m \hat{k}_n \rangle = \frac{1}{3}\delta_{mn}$ 和 $\langle \hat{k}_a \hat{k}_b \hat{k}_c \hat{k}_d \rangle = \frac{1}{15}(\delta_{ab}\delta_{cd} + \dots)$ ，得到平均值为 $\frac{1}{3} \times \frac{8}{5} (h^{ij}h^{ij}) = \frac{1}{3} \times \frac{8}{5} \times \frac{3}{2} = \frac{4}{5}$ 。代入相空间积分，得到：

\boxed{ \Gamma_{J=2} = \frac{128 \pi^2 \alpha^2}{5 m^2} |f(0)|^2 = \frac{4}{15} \Gamma_{J=0} }

(注：若用径向波函数导数表示，则 $|f(0)|^2 = \frac{3}{4\pi} |R'(0)|^2$ )。

5.5

Problem 5.5

peskinChapter 5

习题 5.5

来源: 第5章, PDF第173,174页

5.5 Physics of a massive vector boson. Add to QED a massive photon field $B_\mu$ of mass $M$ , which couples to electrons via

\Delta H = \int d^3x (g\bar{\psi}\gamma^\mu\psi B_\mu).

A massive photon in the initial or final state has three possible physical polarizations, corresponding to the three spacelike unit vectors in the boson's rest frame. These can be characterized invariantly, in terms of the boson's 4-momentum $k^\mu$ , as the three vectors $\epsilon_\mu^{(i)}$ satisfying

\epsilon^{(i)} \cdot \epsilon^{(j)} = -\delta^{ij}, \quad k \cdot \epsilon^{(i)} = 0.

The four vectors $(k_\mu/M, \epsilon_\mu^{(i)})$ form a complete orthonormal basis. Because $B_\mu$ couples to the conserved current $\bar{\psi}\gamma^\mu\psi$ , the Ward identity implies that $k_\mu$ dotted into the amplitude for $B$ production gives zero; thus we can replace:

\sum_i \epsilon_\mu^{(i)} \epsilon_\nu^{(i)*} \rightarrow -g_{\mu\nu}.

This gives a generalization to massive bosons of the Feynman trick for photon polarization vectors and simplifies the calculation of $B$ production cross sections. (Warning: This trick does not work (so simply) for "non-Abelian gauge fields".) Let's do a few of these computations, using always the approximation of ignoring the mass of the electron.

(a) Compute the cross section for the process $e^+e^- \rightarrow B$ . Compute the lifetime of the $B$ , assuming that it decays only to electrons. Verify the relation

\sigma(e^+e^- \rightarrow B) = \frac{12\pi^2}{M} \Gamma(B \rightarrow e^+e^-) \delta(M^2 - s)

discussed in Section 5.3.

(b) Compute the differential cross section, in the center-of-mass system, for the process $e^+e^- \rightarrow \gamma + B$ . (This calculation goes over almost unchanged to the realistic process $e^+e^- \rightarrow \gamma + Z^0$ ; this allows one to measure the number of decays of the $Z^0$ into unobserved final states, which is in turn proportional to the number of neutrino species.)

(c) Notice that the cross section of part (b) diverges as $\theta \rightarrow 0$ or $\pi$ . Let us analyze the region near $\theta = 0$ . In this region, the dominant contribution comes from the $t$ -channel diagram and corresponds intuitively to the emission of a photon from the electron line before $e^+e^-$ annihilation into a $B$ . Let us rearrange the formula in such a way as to support this interpretation. First, note that the divergence as $\theta \rightarrow 0$ is cut off by the electron mass: Let the electron momentum be $p^\mu = (E, 0, 0, k)$ , with $k = (E^2 - m_e^2)^{1/2}$ , and let the photon momentum be $k^\mu = (xE, xE \sin\theta, 0, xE \cos\theta)$ . Show that the denominator of the propagator then never becomes smaller than $\mathcal{O}(m_e^2/s)$ . Now integrate the cross section of part (b) over forward angles, cutting off the $\theta$ integral at $\theta^2 \sim (m_e^2/s)$ and keeping only the leading logarithmic term, proportional to $\log(s/m_e^2)$ . Show that, in this approximation, the cross section for forward photon emission can be written

\sigma(e^+e^- \rightarrow \gamma + B) \approx \int dx f(x) \cdot \sigma(e^+e^- \rightarrow B \text{ at } E_{\text{cm}}^2 = (1-x)s),

where the annihilation cross section is evaluated for the collision of a positron of energy $E$ and an electron of energy $(1-x)E$ , and the function $f(x)$ , the Weizsäcker-Williams distribution function, is given by

f(x) = \frac{\alpha}{2\pi} \frac{1+(1-x)^2}{x} \cdot \log \left( \frac{s}{m_e^2} \right).

This function arises universally in processes in which a photon is emitted collinearly from an electron line, independent of the subsequent dynamics. We will meet it again, in another context, in Problem 6.2.

习题 5.5 - 解答

(a) $e^+e^- \to B$ 截面与 $B$ 玻色子寿命

首先计算 $B \to e^+e^-$ 的衰变率。矩阵元为： $\mathcal{M} = g \bar{u}(p_1) \gamma^\mu v(p_2) \epsilon_\mu(k)$ 对末态自旋求和并对初态极化求平均，利用 Ward 恒等式将极化求和替换为 $-g_{\mu\nu}$ ，且忽略电子质量（ $m_e \to 0$ ）： $\frac{1}{3} \sum |\mathcal{M}|^2 = \frac{1}{3} g^2 (-g_{\mu\nu}) \text{Tr}[\gamma^\mu \not{p}_2 \gamma^\nu \not{p}_1] = \frac{1}{3} g^2 (8 p_1 \cdot p_2)$ 由于 $k = p_1 + p_2$ ，有 $M^2 = k^2 = 2 p_1 \cdot p_2$ ，因此： $\frac{1}{3} \sum |\mathcal{M}|^2 = \frac{4}{3} g^2 M^2$ 两体相空间积分为 $\int d\Pi_2 = \frac{1}{8\pi}$ ，故衰变宽度为： $\Gamma(B \to e^+e^-) = \frac{1}{2M} \int d\Pi_2 \frac{1}{3} \sum |\mathcal{M}|^2 = \frac{1}{2M} \frac{1}{8\pi} \frac{4}{3} g^2 M^2 = \frac{g^2 M}{12\pi}$ 假设 $B$ 仅衰变为电子，其寿命 $\tau$ 为： $\boxed{\tau = \frac{1}{\Gamma} = \frac{12\pi}{g^2 M}}$

接下来计算 $e^+e^- \to B$ 的产生截面。矩阵元为： $\mathcal{M} = g \bar{v}(p_2) \gamma^\mu u(p_1) \epsilon_\mu^*(k)$ 对初态自旋求平均并对末态极化求和： $\frac{1}{4} \sum |\mathcal{M}|^2 = \frac{1}{4} g^2 (-g_{\mu\nu}) \text{Tr}[\gamma^\mu \not{p}_1 \gamma^\nu \not{p}_2] = 2 g^2 (p_1 \cdot p_2) = g^2 s$ 单体相空间为 $\int d\Pi_1 = 2\pi \delta(s - M^2)$ ，因此产生截面为： $\sigma(e^+e^- \to B) = \frac{1}{2s} (g^2 s) 2\pi \delta(s - M^2) = \pi g^2 \delta(s - M^2)$ 验证给定的关系式： $\frac{12\pi^2}{M} \Gamma(B \to e^+e^-) \delta(M^2 - s) = \frac{12\pi^2}{M} \left( \frac{g^2 M}{12\pi} \right) \delta(M^2 - s) = \pi g^2 \delta(M^2 - s)$ 这与直接计算得到的截面完全一致，故关系式得证： $\boxed{\sigma(e^+e^- \to B) = \frac{12\pi^2}{M} \Gamma(B \to e^+e^-) \delta(M^2 - s)}$

(b) $e^+e^- \to \gamma + B$ 的微商截面

该过程包含 $t$ -通道和 $u$ -通道两个费曼图。设电子动量为 $p_-$ ，正电子动量为 $p_+$ ，光子动量为 $k$ 。矩阵元为： $\mathcal{M} = -e g \epsilon_\nu^*(k) \epsilon_\mu^*(p_B) \bar{v}(p_+) \left[ \gamma^\mu \frac{\not{p}_- - \not{k}}{t} \gamma^\nu + \gamma^\nu \frac{\not{k} - \not{p}_+}{u} \gamma^\mu \right] u(p_-)$ 其中 $t = -2 p_- \cdot k$ ， $u = -2 p_+ \cdot k$ ， $s = 2 p_- \cdot p_+$ ，且 $s + t + u = M^2$ 。利用 Ward 恒等式将光子和 $B$ 玻色子的极化求和均替换为 $-g_{\mu\nu}$ ，求自旋平均的模方： $\frac{1}{4} \sum |\mathcal{M}|^2 = \frac{e^2 g^2}{4} \text{Tr} \left\{ \not{p}_+ \left[ \gamma^\mu \frac{\not{p}_- - \not{k}}{t} \gamma^\nu + \gamma^\nu \frac{\not{k} - \not{p}_+}{u} \gamma^\mu \right] \not{p}_- \left[ \gamma_\nu \frac{\not{p}_- - \not{k}}{t} \gamma_\mu + \gamma_\mu \frac{\not{k} - \not{p}_+}{u} \gamma_\nu \right] \right\}$ 展开迹求和，包含三部分：

$t$ -通道平方项： $T_{tt} = \frac{1}{t^2} \text{Tr}[\not{p}_+ \gamma^\mu (\not{p}_- - \not{k}) \gamma^\nu \not{p}_- \gamma_\nu (\not{p}_- - \not{k}) \gamma_\mu] = \frac{8u}{t}$
$u$ -通道平方项：由对称性得 $T_{uu} = \frac{8t}{u}$
交叉项： $T_{tu} + T_{ut} = \frac{2}{tu} \text{Tr}[\not{p}_+ \gamma^\mu (\not{p}_- - \not{k}) \gamma^\nu \not{p}_- \gamma_\mu (\not{k} - \not{p}_+) \gamma_\nu] = \frac{16 M^2 s}{tu}$

总和为： $\frac{1}{4} \sum |\mathcal{M}|^2 = 2 e^2 g^2 \left( \frac{u}{t} + \frac{t}{u} + \frac{2 M^2 s}{tu} \right)$ 在质心系中，设光子发射角为 $\theta$ ，光子能量 $k^0 = \frac{s - M^2}{2\sqrt{s}}$ 。运动学变量可表示为： $t = -\frac{s - M^2}{2} (1 - \cos\theta), \quad u = -\frac{s - M^2}{2} (1 + \cos\theta)$ 代入矩阵元模方中，化简括号内的项： $\frac{u}{t} + \frac{t}{u} = \frac{2(1+\cos^2\theta)}{\sin^2\theta}, \quad \frac{2 M^2 s}{tu} = \frac{8 M^2 s}{(s - M^2)^2 \sin^2\theta}$ 两体相空间微元为 $d\Pi_2 = \frac{1}{16\pi} \left(1 - \frac{M^2}{s}\right) d\cos\theta$ 。结合 $e^2 = 4\pi\alpha$ ，微商截面为： $\frac{d\sigma}{d\cos\theta} = \frac{1}{2s} \frac{1}{4} \sum |\mathcal{M}|^2 \frac{d\Pi_2}{d\cos\theta} = \frac{4\pi\alpha g^2}{16\pi s} \left(1 - \frac{M^2}{s}\right) \frac{1}{\sin^2\theta} \left[ 2(1+\cos^2\theta) + \frac{8 M^2 s}{(s - M^2)^2} \right]$ $\boxed{\frac{d\sigma}{d\cos\theta} = \frac{\alpha g^2}{2 s} \left(1 - \frac{M^2}{s}\right) \frac{1+\cos^2\theta + \frac{4 M^2 s}{(s - M^2)^2}}{\sin^2\theta}}$

(c) 前向散射区域与 Weizsäcker-Williams 分布

1. 传播子分母的截断分析 考虑电子质量 $m_e$ ，电子动量 $p^\mu = (E, 0, 0, k_e)$ ，其中 $k_e = \sqrt{E^2 - m_e^2} \approx E - \frac{m_e^2}{2E}$ 。光子动量 $k^\mu = (xE, xE\sin\theta, 0, xE\cos\theta)$ 。传播子分母为 $t = (p - k)^2 - m_e^2 = -2 p \cdot k$ 。在小角度 $\theta \to 0$ 展开： $p \cdot k = xE (E - k_e\cos\theta) \approx xE \left[ E - \left(E - \frac{m_e^2}{2E}\right)\left(1 - \frac{\theta^2}{2}\right) \right] \approx \frac{x}{2} (E^2 \theta^2 + m_e^2)$ 因此，传播子分母的绝对值为 $|t| \approx x (E^2 \theta^2 + m_e^2)$ 。其最小值在 $\theta = 0$ 处取得，为 $x m_e^2$ 。由于 $s \approx 4E^2$ ，最小值可写为 $x s \left(\frac{m_e^2}{s}\right)$ 。这表明相对于 $s$ ，传播子分母永远不会小于 $\mathcal{O}(m_e^2/s)$ ，从而避免了共线发散。

2. 前向截面的对数近似 在 $\theta \to 0$ 区域， $t \to 0$ ，主导项为 $t$ -通道。此时 $u \approx -xs$ ，且 $M^2 = (1-x)s$ 。矩阵元模方近似为： $\frac{1}{4} \sum |\mathcal{M}|^2 \approx 2 e^2 g^2 \frac{u^2 + 2 M^2 s}{tu} \approx 2 e^2 g^2 \frac{x^2 s^2 + 2(1-x)s^2}{-t(-xs)} = \frac{2 e^2 g^2 s}{-t} \frac{1 + (1-x)^2}{x}$ 代入带质量的 $t \approx -x \frac{s}{4} \left(\theta^2 + \frac{4m_e^2}{s}\right)$ ，并利用 $d\cos\theta \approx \frac{1}{2} d\theta^2$ ，对前向角度积分： $d\sigma \approx \frac{1}{2s} \left[ \frac{8 e^2 g^2}{x^2 s \left(\theta^2 + \frac{4m_e^2}{s}\right)} \frac{s(1 + (1-x)^2)}{x} \right] \frac{x}{32\pi} d\theta^2 = \frac{\alpha g^2}{2 s} \frac{1 + (1-x)^2}{x} \frac{d\theta^2}{\theta^2 + \frac{4m_e^2}{s}}$ 将 $\theta^2$ 积分至 $\mathcal{O}(1)$ 的截断值，保留领头对数项： $\int_0^{\sim 1} \frac{d\theta^2}{\theta^2 + 4m_e^2/s} \approx \log\left(\frac{s}{4m_e^2}\right) \approx \log\left(\frac{s}{m_e^2}\right)$ 因此，前向光子发射截面近似为： $\sigma(e^+e^- \to \gamma + B) \approx \frac{\alpha g^2}{2 s} \frac{1 + (1-x)^2}{x} \log\left(\frac{s}{m_e^2}\right)$

3. 提取 Weizsäcker-Williams 分布函数 根据 (a) 的结果，在质心系能量平方为 $s' = (1-x)s$ 时的湮灭截面为： $\sigma(e^+e^- \to B \text{ at } s') = \pi g^2 \delta((1-x)s - M^2) = \frac{\pi g^2}{s} \delta\left(x - \left(1 - \frac{M^2}{s}\right)\right)$ 将其代入给定的卷积公式中： $\int dx f(x) \sigma(e^+e^- \to B \text{ at } s') = f\left(1 - \frac{M^2}{s}\right) \frac{\pi g^2}{s}$ 令其等于前面求得的近似截面（此时 $x = 1 - M^2/s$ ）： $f(x) \frac{\pi g^2}{s} = \frac{\alpha g^2}{2 s} \frac{1 + (1-x)^2}{x} \log\left(\frac{s}{m_e^2}\right)$ 解得 $f(x)$ 为： $\boxed{f(x) = \frac{\alpha}{2\pi} \frac{1 + (1-x)^2}{x} \log\left(\frac{s}{m_e^2}\right)}$ 这正是普适的 Weizsäcker-Williams 分布函数。

5.6

Problem 5.6

peskinChapter 5

习题 5.6

来源: 第5章, PDF第174页

5.6 This problem extends the spinor product technology of Problem 5.3 to external photons.

(a) Let $k$ be the momentum of a photon, and let $p$ be another lightlike vector, chosen so that $p \cdot k \neq 0$ . Let $u_R(p)$ , $u_L(p)$ be spinors of definite helicity for fermions with the lightlike momentum $p$ , defined according to the conventions of Problem 5.3. Define photon polarization vectors as follows:

\epsilon_+^\mu(k) = \frac{1}{\sqrt{4p \cdot k}} \bar{u}_R(k) \gamma^\mu u_R(p), \quad \epsilon_-^\mu(k) = \frac{1}{\sqrt{4p \cdot k}} \bar{u}_L(k) \gamma^\mu u_L(p).

Use the identity

u_L(p) \bar{u}_L(p) + u_R(p) \bar{u}_R(p) = \not{p}

to compute the polarization sum

\epsilon_+^\mu \epsilon_+^{\nu*} + \epsilon_-^\mu \epsilon_-^{\nu*} = -g^{\mu\nu} + \frac{k^\mu p^\nu + k^\nu p^\mu}{p \cdot k}.

The second term on the right gives zero when dotted with any photon emission amplitude $\mathcal{M}^\mu$ , so we have

|\epsilon_+ \cdot \mathcal{M}|^2 + |\epsilon_- \cdot \mathcal{M}|^2 = \mathcal{M}^\mu \mathcal{M}^{\nu*} (-g_{\mu\nu});

thus, we can use the vectors $\epsilon_+$ , $\epsilon_-$ to compute photon polarization sums.

(b) Using the polarization vectors just defined, and the spinor products and the Fierz identity from Problem 5.3, compute the differential cross section for a massless electron and positron to annihilate into 2 photons. Show that the result agrees with the massless limit derived in (5.107):

\frac{d\sigma}{d \cos \theta} = \frac{2\pi\alpha^2}{s} \left( \frac{1 + \cos^2 \theta}{\sin^2 \theta} \right)

in the center-of-mass frame. It follows from the result of part (a) that this answer is independent of the particular vector $p$ used to define the polarization vectors; however, the calculation is greatly simplified by taking this vector to be the initial electron 4-vector.

习题 5.6 - 解答

习题 5.6 分析与解答

(a) 光子极化求和公式的证明

先分析光子极化矢量的复共轭。根据定义 $\epsilon_+^\mu(k) = \frac{1}{\sqrt{4p \cdot k}} \bar{u}_R(k) \gamma^\mu u_R(p)$ ，其复共轭为：

\epsilon_+^{\mu*} = \frac{1}{\sqrt{4p \cdot k}} \left( u_R^\dagger(k) \gamma^0 \gamma^\mu u_R(p) \right)^\dagger = \frac{1}{\sqrt{4p \cdot k}} \bar{u}_R(p) \gamma^\mu u_R(k)

同理可得 $\epsilon_-^{\mu*} = \frac{1}{\sqrt{4p \cdot k}} \bar{u}_L(p) \gamma^\mu u_L(k)$ 。

下面计算极化求和 $\epsilon_+^\mu \epsilon_+^{\nu*} + \epsilon_-^\mu \epsilon_-^{\nu*}$ 。由于每一项都是标量，可以将其写为迹（Trace）的形式：

\epsilon_+^\mu \epsilon_+^{\nu*} = \frac{1}{4p \cdot k} \text{Tr}\left[ u_R(k) \bar{u}_R(k) \gamma^\mu u_R(p) \bar{u}_R(p) \gamma^\nu \right]

利用无质量费米子的投影关系 $u_R(k) \bar{u}_R(k) = \frac{1+\gamma^5}{2} \not{k}$ 和 $u_L(k) \bar{u}_L(k) = \frac{1-\gamma^5}{2} \not{k}$ ，极化求和变为：

\epsilon_+^\mu \epsilon_+^{\nu*} + \epsilon_-^\mu \epsilon_-^{\nu*} = \frac{1}{4p \cdot k} \text{Tr}\left[ \frac{1+\gamma^5}{2} \not{k} \gamma^\mu u_R(p) \bar{u}_R(p) \gamma^\nu + \frac{1-\gamma^5}{2} \not{k} \gamma^\mu u_L(p) \bar{u}_L(p) \gamma^\nu \right]

由于 $\gamma^5$ 与 $\not{k}\gamma^\mu$ 对易（包含两个 $\gamma$ 矩阵），我们可以将投影算符移到后面：

\frac{1+\gamma^5}{2} \not{k} \gamma^\mu = \not{k} \gamma^\mu \frac{1+\gamma^5}{2}

结合 $\frac{1+\gamma^5}{2} u_R(p) = u_R(p)$ ，求和式化简为：

\frac{1}{4p \cdot k} \text{Tr}\left[ \not{k} \gamma^\mu \left( u_R(p) \bar{u}_R(p) + u_L(p) \bar{u}_L(p) \right) \gamma^\nu \right]

代入题目给定的恒等式 $u_L(p) \bar{u}_L(p) + u_R(p) \bar{u}_R(p) = \not{p}$ ，得到：

\epsilon_+^\mu \epsilon_+^{\nu*} + \epsilon_-^\mu \epsilon_-^{\nu*} = \frac{1}{4p \cdot k} \text{Tr}\left[ \not{k} \gamma^\mu \not{p} \gamma^\nu \right]

计算狄拉克矩阵的迹 $\text{Tr}[\not{k} \gamma^\mu \not{p} \gamma^\nu] = 4(k^\mu p^\nu + k^\nu p^\mu - g^{\mu\nu} k \cdot p)$ ，最终得到：

\epsilon_+^\mu \epsilon_+^{\nu*} + \epsilon_-^\mu \epsilon_-^{\nu*} = \frac{k^\mu p^\nu + k^\nu p^\mu - g^{\mu\nu} p \cdot k}{p \cdot k} = -g^{\mu\nu} + \frac{k^\mu p^\nu + k^\nu p^\mu}{p \cdot k}

这正是所要求证的极化求和公式。

(b) 电子-正电子湮灭为双光子的微分散射截面

考虑过程 $e^-(p_1) + e^+(p_2) \to \gamma(k_1) + \gamma(k_2)$ 。在无质量极限下，非零振幅要求初始电子和正电子具有相反的螺旋度。为极大简化计算，我们选取参考动量 $p = p_1$ 来定义两个光子的极化矢量。

\notin_{i+} = \frac{\sqrt{2}}{\sqrt{2p_1 \cdot k_i}} \left( | p_1 ] \langle k_i | + | k_i \rangle [ p_1 | \right), \quad \notin_{i-} = \frac{\sqrt{2}}{\sqrt{2p_1 \cdot k_i}} \left( | p_1 \rangle [ k_i | + | k_i ] \langle p_1 | \right)

由于 $[p_1|p_1] = \langle p_1|p_1\rangle = 0$ ，极化矢量作用在初始电子旋量上具有极好的性质：

\notin_{i+} | p_1 ] = 0, \quad \notin_{i-} | p_1 \rangle = 0

分步处理振幅： 假设初始态为 $e^-_R(p_1)$ 和 $e^+_L(p_2)$ ，即 $u(p_1) = |p_1]$ ， $\bar{v}(p_2) = \langle p_2|$ 。散射振幅由 $t$ 沟道和 $u$ 沟道组成：

\mathcal{M} = -e^2 \langle p_2 | \notin_2 \frac{\not{p}_1 - \not{k}_1}{-2p_1 \cdot k_1} \notin_1 | p_1 ] + (1 \leftrightarrow 2)

若任意一个光子为 $+$ 极化，由于 $\notin_{1+} | p_1 ] = 0$ 且 $\notin_{2+} | p_1 ] = 0$ ，对应的 $t$ 沟道或 $u$ 沟道直接为零。因此 $\mathcal{M}(+,+) = 0$ 。
若两个光子均为 $-$ 极化，计算可得 $\mathcal{M}_t = -e^2 \frac{2}{\sqrt{s_{1k_1} s_{1k_2}}} \langle p_2 p_1 \rangle [ k_2 k_1 ]$ ，而 $\mathcal{M}_u$ 交换 $1 \leftrightarrow 2$ 后符号相反，两者精确抵消， $\mathcal{M}(-,-) = 0$ 。
对于 $(-,+)$ 极化配置， $\mathcal{M}_u = 0$ （因为 $\notin_{2+}|p_1]=0$ ），仅剩 $t$ 沟道：

\notin_{1-} | p_1 ] = \frac{\sqrt{2}}{\sqrt{2p_1 \cdot k_1}} | p_1 \rangle [ k_1 p_1 ], \quad \langle p_2 | \notin_{2+} = \frac{\sqrt{2}}{\sqrt{2p_1 \cdot k_2}} \langle p_2 k_2 \rangle [ p_1 |

中间传播子分子为 $(\not{p}_1 - \not{k}_1) | p_1 \rangle = - | k_1 ] \langle k_1 p_1 \rangle$ 。代入并利用分母 $-2p_1 \cdot k_1 = \langle k_1 p_1 \rangle [ k_1 p_1 ]$ 约分，得到：

\mathcal{M}(-,+) = -2e^2 \frac{\langle p_2 k_2 \rangle [ p_1 k_1 ]}{\sqrt{(2p_1 \cdot k_1)(2p_1 \cdot k_2)}}

取模平方，并利用动量守恒 $2p_2 \cdot k_2 = 2p_1 \cdot k_1 = -t$ 以及 $2p_1 \cdot k_2 = -u$ ：

|\mathcal{M}(-,+)|^2 = 4e^4 \frac{(2p_2 \cdot k_2)(2p_1 \cdot k_1)}{(2p_1 \cdot k_1)(2p_1 \cdot k_2)} = 4e^4 \frac{t}{u}

同理，对于 $(+,-)$ 极化配置，由对称性可得 $|\mathcal{M}(+,-)|^2 = 4e^4 \frac{u}{t}$ 。

计算微分散射截面： 对于 $e^-_R e^+_L$ 初始态，极化求和后的平方振幅为 $4e^4 \left( \frac{t}{u} + \frac{u}{t} \right)$ 。由宇称对称性， $e^-_L e^+_R$ 初始态贡献相同。对初始自旋求平均（乘以 $\frac{1}{4}$ ），得到无极化平方振幅：

\frac{1}{4} \sum_{\text{spins}} |\mathcal{M}|^2 = \frac{1}{4} \times 2 \times 4e^4 \left( \frac{t}{u} + \frac{u}{t} \right) = 2e^4 \left( \frac{t}{u} + \frac{u}{t} \right)

在质心系中，运动学变量为 $t = -\frac{s}{2}(1 - \cos\theta)$ 和 $u = -\frac{s}{2}(1 + \cos\theta)$ 。代入比值：

\frac{t}{u} + \frac{u}{t} = \frac{1-\cos\theta}{1+\cos\theta} + \frac{1+\cos\theta}{1-\cos\theta} = \frac{2(1+\cos^2\theta)}{\sin^2\theta}

因此，自旋平均的平方振幅为 $4e^4 \frac{1+\cos^2\theta}{\sin^2\theta}$ 。代入 $2 \to 2$ 截面公式 $\frac{d\sigma}{d\Omega} = \frac{1}{64\pi^2 s} \overline{|\mathcal{M}|^2}$ ，并利用 $d\Omega = 2\pi d\cos\theta$ 积分掉方位角，得到：

\frac{d\sigma}{d\cos\theta} = 2\pi \frac{4e^4}{64\pi^2 s} \frac{1+\cos^2\theta}{\sin^2\theta} = \frac{2\pi\alpha^2}{s} \left( \frac{1+\cos^2\theta}{\sin^2\theta} \right)

这与 (5.107) 式的无质量极限结果完全一致。

\boxed{ \frac{d\sigma}{d \cos \theta} = \frac{2\pi\alpha^2}{s} \left( \frac{1 + \cos^2 \theta}{\sin^2 \theta} \right) }

Problem 5.1

习题 5.1

习题 5.1 - 解答

方法一：电子在静态库仑势中的散射

方法二：电子-缪子散射在 mμ→∞m_\mu \to \inftymμ​→∞ 极限下的推导

Problem 5.2

习题 5.2

习题 5.2 - 解答

1. 费曼图与不变矩阵元

2. 矩阵元模方与自旋求和

3. 微分截面与 Mandelstam 变量形式

4. 用 cos⁡θ\cos\thetacosθ 表示微分截面

5. 图像特征与发散原因分析

Problem 5.3

习题 5.3

习题 5.3 - 解答

Problem 5.4

习题 5.4

习题 5.4 - 解答

(a) 1S1S1S 态正电子素的衰变率

(b) 展开至 3-动量的一阶 O(p)O(p)O(p)

(c) P 波态的归一化

(d) S=1,L=1S=1, L=1S=1,L=1 各 JJJ 态的衰变率

Problem 5.5

习题 5.5

习题 5.5 - 解答

(a) e+e−→Be^+e^- \to Be+e−→B 截面与 BBB 玻色子寿命

(b) e+e−→γ+Be^+e^- \to \gamma + Be+e−→γ+B 的微商截面

(c) 前向散射区域与 Weizsäcker-Williams 分布

Problem 5.6

习题 5.6

习题 5.6 - 解答

习题 5.6 分析与解答

(a) 光子极化求和公式的证明

(b) 电子-正电子湮灭为双光子的微分散射截面

方法二：电子-缪子散射在 $m_\mu \to \infty$ 极限下的推导

4. 用 $\cos\theta$ 表示微分截面

(a) $1S$ 态正电子素的衰变率

(b) 展开至 3-动量的一阶 $O(p)$

(d) $S=1, L=1$ 各 $J$ 态的衰变率

(a) $e^+e^- \to B$ 截面与 $B$ 玻色子寿命

(b) $e^+e^- \to \gamma + B$ 的微商截面