Problem 3.1

peskinChapter 3

习题 3.1

来源: 第3章, PDF第71,72页

3.1 Lorentz group. Recall from Eq. (3.17) the Lorentz commutation relations,

[J^{\mu\nu}, J^{\rho\sigma}] = i(g^{\nu\rho} J^{\mu\sigma} - g^{\mu\rho} J^{\nu\sigma} - g^{\nu\sigma} J^{\mu\rho} + g^{\mu\sigma} J^{\nu\rho}).

(a) Define the generators of rotations and boosts as

L^i = \frac{1}{2} \epsilon^{ijk} J^{jk}, \quad K^i = J^{0i},

where $i, j, k = 1, 2, 3$ . An infinitesimal Lorentz transformation can then be written

\Phi \rightarrow (1 - i\theta \cdot \mathbf{L} - i\beta \cdot \mathbf{K})\Phi.

Write the commutation relations of these vector operators explicitly. (For example, $[L^i, L^j] = i\epsilon^{ijk} L^k$ .) Show that the combinations

\mathbf{J}_+ = \frac{1}{2}(\mathbf{L} + i\mathbf{K}) \quad \text{and} \quad \mathbf{J}_- = \frac{1}{2}(\mathbf{L} - i\mathbf{K})

commute with one another and separately satisfy the commutation relations of angular momentum.

(b) The finite-dimensional representations of the rotation group correspond precisely to the allowed values for angular momentum: integers or half-integers. The result of part (a) implies that all finite-dimensional representations of the Lorentz group correspond to pairs of integers or half integers, $(j_+, j_-)$ , corresponding to pairs of representations of the rotation group. Using the fact that $\mathbf{J} = \sigma/2$ in the spin-1/2 representation of angular momentum, write explicitly the transformation laws of the 2-component objects transforming according to the $(\frac{1}{2}, 0)$ and $(0, \frac{1}{2})$ representations of the Lorentz group. Show that these correspond precisely to the transformations of $\psi_L$ and $\psi_R$ given in (3.37).

(c) The identity $\sigma^T = -\sigma^2 \sigma \sigma^2$ allows us to rewrite the $\psi_L$ transformation in the unitarily equivalent form

\psi' \rightarrow \psi'(1 + i\theta \cdot \frac{\sigma}{2} + \beta \cdot \frac{\sigma}{2}),

where $\psi' = \psi_L^T \sigma^2$ . Using this law, we can represent the object that transforms as $(\frac{1}{2}, \frac{1}{2})$ as a $2 \times 2$ matrix that has the $\psi_R$ transformation law on the left and, simultaneously, the transposed $\psi_L$ transformation on the right. Parametrize this matrix as

\begin{pmatrix} V^0 + V^3 & V^1 - iV^2 \\ V^1 + iV^2 & V^0 - V^3 \end{pmatrix}.

Show that the object $V^\mu$ transforms as a 4-vector.

Referenced Equations:

Equation (3.17):

[J^{\mu\nu}, J^{\rho\sigma}] = i(g^{\nu\rho} J^{\mu\sigma} - g^{\mu\rho} J^{\nu\sigma} - g^{\nu\sigma} J^{\mu\rho} + g^{\mu\sigma} J^{\nu\rho}). \tag{3.17}

Equation (3.37):

\begin{aligned} \psi_L &\rightarrow (1 - i\boldsymbol{\theta} \cdot \frac{\boldsymbol{\sigma}}{2} - \boldsymbol{\beta} \cdot \frac{\boldsymbol{\sigma}}{2})\psi_L; \\ \psi_R &\rightarrow (1 - i\boldsymbol{\theta} \cdot \frac{\boldsymbol{\sigma}}{2} + \boldsymbol{\beta} \cdot \frac{\boldsymbol{\sigma}}{2})\psi_R. \end{aligned} \tag{3.37}

习题 3.1 - 解答

(a) 洛伦兹群生成元的对易关系

已知洛伦兹代数由下式给出：

[J^{\mu\nu}, J^{\rho\sigma}] = i(g^{\nu\rho} J^{\mu\sigma} - g^{\mu\rho} J^{\nu\sigma} - g^{\nu\sigma} J^{\mu\rho} + g^{\mu\sigma} J^{\nu\rho})

度规约定为 $g^{\mu\nu} = \text{diag}(1, -1, -1, -1)$ 。定义旋转生成元 $L^i = \frac{1}{2} \epsilon^{ijk} J^{jk}$ 和推升生成元 $K^i = J^{0i}$ 。利用 $\epsilon^{ijk}\epsilon^{ilm} = \delta^{jl}\delta^{km} - \delta^{jm}\delta^{kl}$ ，可反解出 $J^{ij} = \epsilon^{ijk}L^k$ 。

1. 计算 $[L^i, L^j]$ ：

\begin{aligned} [L^i, L^j] &= \frac{1}{4} \epsilon^{ilm} \epsilon^{jpq} [J^{lm}, J^{pq}] \\ &= \frac{i}{4} \epsilon^{ilm} \epsilon^{jpq} (g^{mp} J^{lq} - g^{lp} J^{mq} - g^{mq} J^{lp} + g^{lq} J^{mp}) \end{aligned}

由于空间分量 $g^{ij} = -\delta^{ij}$ ，代入上式并利用反对称性 $J^{ij} = -J^{ji}$ ，化简得：

[L^i, L^j] = i\epsilon^{ijk} L^k

2. 计算 $[L^i, K^j]$ ：

\begin{aligned} [L^i, K^j] &= \frac{1}{2} \epsilon^{ilm} [J^{lm}, J^{0j}] \\ &= \frac{i}{2} \epsilon^{ilm} (g^{m0} J^{lj} - g^{l0} J^{mj} - g^{mj} J^{l0} + g^{lj} J^{m0}) \end{aligned}

由于 $g^{0i} = 0$ 且 $g^{ij} = -\delta^{ij}$ ，上式变为：

[L^i, K^j] = \frac{i}{2} \epsilon^{ilm} (\delta^{mj} J^{l0} - \delta^{lj} J^{m0}) = \frac{i}{2} (\epsilon^{ilj} K^l - \epsilon^{imj} K^m) = i\epsilon^{ijk} K^k

3. 计算 $[K^i, K^j]$ ：

\begin{aligned} [K^i, K^j] &= [J^{0i}, J^{0j}] \\ &= i(g^{i0} J^{0j} - g^{00} J^{ij} - g^{ij} J^{00} + g^{0j} J^{i0}) \\ &= i(-J^{ij}) = -i\epsilon^{ijk} L^k \end{aligned}

4. 证明 $\mathbf{J}_\pm$ 的对易关系： 定义 $\mathbf{J}_\pm = \frac{1}{2}(\mathbf{L} \pm i\mathbf{K})$ 。计算其对易子：

\begin{aligned} [J_\pm^i, J_\pm^j] &= \frac{1}{4} [L^i \pm iK^i, L^j \pm iK^j] \\ &= \frac{1}{4} \left( [L^i, L^j] \pm i[L^i, K^j] \pm i[K^i, L^j] - [K^i, K^j] \right) \\ &= \frac{1}{4} \left( i\epsilon^{ijk}L^k \pm i(i\epsilon^{ijk}K^k) \pm i(-i\epsilon^{ijk}K^k) - (-i\epsilon^{ijk}L^k) \right) \\ &= \frac{1}{4} \left( 2i\epsilon^{ijk}L^k \mp 2\epsilon^{ijk}K^k \right) \\ &= i\epsilon^{ijk} \frac{1}{2}(L^k \pm iK^k) = i\epsilon^{ijk} J_\pm^k \end{aligned}

计算交叉对易子：

\begin{aligned} [J_+^i, J_-^j] &= \frac{1}{4} [L^i + iK^i, L^j - iK^j] \\ &= \frac{1}{4} \left( [L^i, L^j] - i[L^i, K^j] + i[K^i, L^j] + [K^i, K^j] \right) \\ &= \frac{1}{4} \left( i\epsilon^{ijk}L^k - i(i\epsilon^{ijk}K^k) + i(-i\epsilon^{ijk}K^k) + (-i\epsilon^{ijk}L^k) \right) \\ &= \frac{1}{4} \left( i\epsilon^{ijk}L^k + \epsilon^{ijk}K^k - \epsilon^{ijk}K^k - i\epsilon^{ijk}L^k \right) = 0 \end{aligned}

综上所述，对易关系为：

\boxed{ \begin{aligned} [L^i, L^j] &= i\epsilon^{ijk} L^k \\ [L^i, K^j] &= i\epsilon^{ijk} K^k \\ [K^i, K^j] &= -i\epsilon^{ijk} L^k \\ [J_\pm^i, J_\pm^j] &= i\epsilon^{ijk} J_\pm^k \\ [J_+^i, J_-^j] &= 0 \end{aligned} }

这表明洛伦兹代数解耦为两个独立的 $SU(2)$ 角动量代数。

(b) 旋量表示的变换法则

由 $\mathbf{J}_\pm = \frac{1}{2}(\mathbf{L} \pm i\mathbf{K})$ 可反解出物理生成元：

\mathbf{L} = \mathbf{J}_+ + \mathbf{J}_-, \quad \mathbf{K} = -i(\mathbf{J}_+ - \mathbf{J}_-)

无穷小洛伦兹变换算符可重写为 $\mathbf{J}_\pm$ 的形式：

\begin{aligned} 1 - i\boldsymbol{\theta} \cdot \mathbf{L} - i\boldsymbol{\beta} \cdot \mathbf{K} &= 1 - i\boldsymbol{\theta} \cdot (\mathbf{J}_+ + \mathbf{J}_-) - i\boldsymbol{\beta} \cdot (-i(\mathbf{J}_+ - \mathbf{J}_-)) \\ &= 1 - (i\boldsymbol{\theta} + \boldsymbol{\beta}) \cdot \mathbf{J}_+ - (i\boldsymbol{\theta} - \boldsymbol{\beta}) \cdot \mathbf{J}_- \end{aligned}

1. 对于 $(\frac{1}{2}, 0)$ 表示： 此时 $\mathbf{J}_+ = \frac{\boldsymbol{\sigma}}{2}$ ， $\mathbf{J}_- = 0$ 。代入变换算符得：

\Phi_{(\frac{1}{2}, 0)} \rightarrow \left( 1 - (i\boldsymbol{\theta} + \boldsymbol{\beta}) \cdot \frac{\boldsymbol{\sigma}}{2} \right) \Phi_{(\frac{1}{2}, 0)} = \left( 1 - i\boldsymbol{\theta} \cdot \frac{\boldsymbol{\sigma}}{2} - \boldsymbol{\beta} \cdot \frac{\boldsymbol{\sigma}}{2} \right) \Phi_{(\frac{1}{2}, 0)}

这与 Eq. (3.37) 中左手旋量 $\psi_L$ 的变换法则完全一致。

2. 对于 $(0, \frac{1}{2})$ 表示： 此时 $\mathbf{J}_+ = 0$ ， $\mathbf{J}_- = \frac{\boldsymbol{\sigma}}{2}$ 。代入变换算符得：

\Phi_{(0, \frac{1}{2})} \rightarrow \left( 1 - (i\boldsymbol{\theta} - \boldsymbol{\beta}) \cdot \frac{\boldsymbol{\sigma}}{2} \right) \Phi_{(0, \frac{1}{2})} = \left( 1 - i\boldsymbol{\theta} \cdot \frac{\boldsymbol{\sigma}}{2} + \boldsymbol{\beta} \cdot \frac{\boldsymbol{\sigma}}{2} \right) \Phi_{(0, \frac{1}{2})}

这与 Eq. (3.37) 中右手旋量 $\psi_R$ 的变换法则完全一致。

结论：

\boxed{ \begin{aligned} (\tfrac{1}{2}, 0) \text{ 表示}: \quad &\psi_L \rightarrow \left( 1 - i\boldsymbol{\theta} \cdot \frac{\boldsymbol{\sigma}}{2} - \boldsymbol{\beta} \cdot \frac{\boldsymbol{\sigma}}{2} \right) \psi_L \\ (0, \tfrac{1}{2}) \text{ 表示}: \quad &\psi_R \rightarrow \left( 1 - i\boldsymbol{\theta} \cdot \frac{\boldsymbol{\sigma}}{2} + \boldsymbol{\beta} \cdot \frac{\boldsymbol{\sigma}}{2} \right) \psi_R \end{aligned} }

(c) $(\frac{1}{2}, \frac{1}{2})$ 表示与四维矢量

将矩阵参数化为 $V = V^0 I + V^i \sigma^i \equiv V^\mu \sigma_\mu$ 。根据题意，该矩阵左侧按 $\psi_R$ 变换，右侧按 $\psi'$ 变换。其无穷小变换法则为：

V \rightarrow \left( 1 - i\boldsymbol{\theta} \cdot \frac{\boldsymbol{\sigma}}{2} + \boldsymbol{\beta} \cdot \frac{\boldsymbol{\sigma}}{2} \right) V \left( 1 + i\boldsymbol{\theta} \cdot \frac{\boldsymbol{\sigma}}{2} + \boldsymbol{\beta} \cdot \frac{\boldsymbol{\sigma}}{2} \right)

保留至 $\boldsymbol{\theta}$ 和 $\boldsymbol{\beta}$ 的一阶项，矩阵的变分 $\delta V$ 为：

\begin{aligned} \delta V &= \left( -i\boldsymbol{\theta} \cdot \frac{\boldsymbol{\sigma}}{2} + \boldsymbol{\beta} \cdot \frac{\boldsymbol{\sigma}}{2} \right) V + V \left( i\boldsymbol{\theta} \cdot \frac{\boldsymbol{\sigma}}{2} + \boldsymbol{\beta} \cdot \frac{\boldsymbol{\sigma}}{2} \right) \\ &= -\frac{i}{2} \theta^i [\sigma^i, V] + \frac{1}{2} \beta^i \{\sigma^i, V\} \end{aligned}

代入 $V = V^0 I + V^j \sigma^j$ ，并利用泡利矩阵的恒等式 $[\sigma^i, \sigma^j] = 2i\epsilon^{ijk}\sigma^k$ 以及 $\{\sigma^i, \sigma^j\} = 2\delta^{ij}I$ ：

1. 纯旋转（ $\boldsymbol{\beta} = 0$ ）：

\begin{aligned} \delta V &= -\frac{i}{2} \theta^i [\sigma^i, V^0 I + V^j \sigma^j] \\ &= -\frac{i}{2} \theta^i V^j (2i\epsilon^{ijk}\sigma^k) \\ &= \epsilon^{ijk} \theta^i V^j \sigma^k \end{aligned}

对比 $\delta V = \delta V^0 I + \delta V^k \sigma^k$ ，得到：

\delta V^0 = 0, \quad \delta V^k = \epsilon^{ijk} \theta^i V^j = (\boldsymbol{\theta} \times \mathbf{V})^k

这正是四维矢量在空间旋转下的变换（时间分量不变，空间分量按三维矢量旋转）。

2. 纯推升（ $\boldsymbol{\theta} = 0$ ）：

\begin{aligned} \delta V &= \frac{1}{2} \beta^i \{\sigma^i, V^0 I + V^j \sigma^j\} \\ &= \frac{1}{2} \beta^i \left( V^0 (2\sigma^i) + V^j (2\delta^{ij}I) \right) \\ &= \beta^i V^i I + \beta^i V^0 \sigma^i \end{aligned}

对比 $\delta V = \delta V^0 I + \delta V^k \sigma^k$ ，得到：

\delta V^0 = \boldsymbol{\beta} \cdot \mathbf{V}, \quad \delta V^k = \beta^k V^0

这正是四维矢量在洛伦兹推升下的无穷小变换法则。

综上所述，矩阵 $V$ 的分量 $V^\mu = (V^0, V^1, V^2, V^3)$ 在洛伦兹群的 $(\frac{1}{2}, \frac{1}{2})$ 表示下，其变换行为与标准的四维矢量完全一致：

\boxed{ \begin{aligned} \text{旋转}: \quad &\delta V^0 = 0, \quad \delta \mathbf{V} = \boldsymbol{\theta} \times \mathbf{V} \\ \text{推升}: \quad &\delta V^0 = \boldsymbol{\beta} \cdot \mathbf{V}, \quad \delta \mathbf{V} = \boldsymbol{\beta} V^0 \end{aligned} }

3.2

Problem 3.2

peskinChapter 3

习题 3.2

来源: 第3章, PDF第72页

3.2 Derive the Gordon identity,

\bar{u}(p')\gamma^\mu u(p) = \bar{u}(p') \left[ \frac{p'^\mu + p^\mu}{2m} + \frac{i\sigma^{\mu\nu} q_\nu}{2m} \right] u(p),

where $q = (p' - p)$ . We will put this formula to use in Chapter 6.

习题 3.2 - 解答

Gordon恒等式将狄拉克粒子的矢量流 $\bar{u}(p')\gamma^\mu u(p)$ 分解为两部分：第一项是类似于无自旋标量粒子的对流（convection current），第二项是与粒子自旋相关的磁化流（spin/magnetization current）。推导该恒等式的关键在于利用动量空间的狄拉克方程以及伽马矩阵的 Clifford 代数反对易关系。

物理背景与基本关系式

对于质量为 $m$ 的自由狄拉克粒子，其旋量 $u(p)$ 满足动量空间的狄拉克方程：

(\not{p} - m)u(p) = 0 \quad \implies \quad \not{p}u(p) = mu(p)

其中 $\not{p} = \gamma^\mu p_\mu$ 。取其狄拉克伴随（Dirac adjoint），并利用 $\bar{u}(p') = u^\dagger(p') \gamma^0$ 以及 $\gamma^\mu$ 的厄米性质，可得伴随旋量 $\bar{u}(p')$ 满足的狄拉克方程：

\bar{u}(p')(\not{p}' - m) = 0 \quad \implies \quad \bar{u}(p')\not{p}' = m\bar{u}(p')

伽马矩阵满足 Clifford 代数：

\{\gamma^\mu, \gamma^\nu\} = \gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu = 2\eta^{\mu\nu}

同时，张量 $\sigma^{\mu\nu}$ 定义为伽马矩阵对易子的形式：

\sigma^{\mu\nu} = \frac{i}{2}[\gamma^\mu, \gamma^\nu] = \frac{i}{2}(\gamma^\mu\gamma^\nu - \gamma^\nu\gamma^\mu)

结合上述两个关系式，可以将任意两个伽马矩阵的乘积分解为对称部分和反对称部分：

\gamma^\mu\gamma^\nu = \frac{1}{2}\{\gamma^\mu, \gamma^\nu\} + \frac{1}{2}[\gamma^\mu, \gamma^\nu] = \eta^{\mu\nu} - i\sigma^{\mu\nu}

同理，交换指标可得：

\gamma^\nu\gamma^\mu = \eta^{\mu\nu} + i\sigma^{\mu\nu}

推导过程

考虑矢量流矩阵元 $\bar{u}(p')\gamma^\mu u(p)$ 。为了引入动量 $p$ 和 $p'$ ，我们可以利用狄拉克方程，将中间的 $\gamma^\mu$ 乘以 $m$ ，并将其等价替换为作用在左右旋量上的 $\not{p}$ 和 $\not{p}'$ ：

\bar{u}(p')\gamma^\mu u(p) = \bar{u}(p') \left( \frac{m}{2m} \gamma^\mu + \gamma^\mu \frac{m}{2m} \right) u(p)

将左侧的 $m$ 替换为向左作用的 $\not{p}'$ ，右侧的 $m$ 替换为向右作用的 $\not{p}$ ：

\bar{u}(p')\gamma^\mu u(p) = \frac{1}{2m} \bar{u}(p') \left( \not{p}'\gamma^\mu + \gamma^\mu\not{p} \right) u(p)

展开 $\not{p}'$ 和 $\not{p}$ ：

\not{p}'\gamma^\mu + \gamma^\mu\not{p} = p'_\nu \gamma^\nu \gamma^\mu + p_\nu \gamma^\mu \gamma^\nu

利用前面推导的伽马矩阵乘积的分解公式 $\gamma^\nu\gamma^\mu = \eta^{\mu\nu} + i\sigma^{\mu\nu}$ 和 $\gamma^\mu\gamma^\nu = \eta^{\mu\nu} - i\sigma^{\mu\nu}$ ，代入上式：

\begin{aligned} \not{p}'\gamma^\mu + \gamma^\mu\not{p} &= p'_\nu (\eta^{\mu\nu} + i\sigma^{\mu\nu}) + p_\nu (\eta^{\mu\nu} - i\sigma^{\mu\nu}) \\ &= p'^\mu + i\sigma^{\mu\nu}p'_\nu + p^\mu - i\sigma^{\mu\nu}p_\nu \\ &= (p'^\mu + p^\mu) + i\sigma^{\mu\nu}(p'_\nu - p_\nu) \end{aligned}

定义动量转移 $q_\nu = p'_\nu - p_\nu$ ，上式可进一步化简为：

\not{p}'\gamma^\mu + \gamma^\mu\not{p} = (p'^\mu + p^\mu) + i\sigma^{\mu\nu}q_\nu

将此结果代回原矩阵元的表达式中，即可得到：

\bar{u}(p')\gamma^\mu u(p) = \frac{1}{2m} \bar{u}(p') \left[ (p'^\mu + p^\mu) + i\sigma^{\mu\nu}q_\nu \right] u(p)

整理各项，最终得出 Gordon 恒等式：

\boxed{ \bar{u}(p')\gamma^\mu u(p) = \bar{u}(p') \left[ \frac{p'^\mu + p^\mu}{2m} + \frac{i\sigma^{\mu\nu} q_\nu}{2m} \right] u(p) }

3.3

Problem 3.3

peskinChapter 3

习题 3.3

来源: 第3章, PDF第72,73页

3.3 Spinor products. (This problem, together with Problems 5.3 and 5.6, introduces an efficient computational method for processes involving massless particles.) Let $k_0^\mu, k_1^\mu$ be fixed 4-vectors satisfying $k_0^2 = 0, k_1^2 = -1, k_0 \cdot k_1 = 0$ . Define basic spinors in the following way: Let $u_{L0}$ be the left-handed spinor for a fermion with momentum $k_0$ . Let $u_{R0} = \not{k}_1 u_{L0}$ . Then, for any $p$ such that $p$ is lightlike ( $p^2 = 0$ ), define

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

This set of conventions defines the phases of spinors unambiguously (except when $p$ is parallel to $k_0$ ).

(a) Show that $\not{k}_0 u_{R0} = 0$ . Show that, for any lightlike $p$ , $\not{p} u_L(p) = \not{p} u_R(p) = 0$ .

(b) For the choices $k_0 = (E, 0, 0, -E)$ , $k_1 = (0, 1, 0, 0)$ , construct $u_{L0}$ , $u_{R0}$ , $u_L(p)$ , and $u_R(p)$ explicitly. (c) Define the spinor products $s(p_1, p_2)$ and $t(p_1, p_2)$ , for $p_1, p_2$ lightlike, by

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).

Using the explicit forms for the $u_\lambda$ given in part (b), compute the spinor products explicitly and show that $t(p_1, p_2) = (s(p_2, p_1))^*$ and $s(p_1, p_2) = -s(p_2, p_1)$ . In addition, show that

|s(p_1, p_2)|^2 = 2p_1 \cdot p_2.

Thus the spinor products are the square roots of 4-vector dot products.

习题 3.3 - 解答

习题分析与解答

在 Weyl（手征）表象中，Dirac 矩阵和 $\gamma^5$ 的形式为： $\gamma^\mu = \begin{pmatrix} 0 & \sigma^\mu \\ \bar{\sigma}^\mu & 0 \end{pmatrix}, \quad \gamma^5 = \begin{pmatrix} -I & 0 \\ 0 & I \end{pmatrix}$ 其中 $\sigma^\mu = (I, \vec{\sigma})$ ， $\bar{\sigma}^\mu = (I, -\vec{\sigma})$ 。左手旋量和右手旋量分别具有 $\begin{pmatrix} u_L \\ 0 \end{pmatrix}$ 和 $\begin{pmatrix} 0 \\ u_R \end{pmatrix}$ 的形式。

(a) 证明 $\not{k}_0 u_{R0} = 0$ 以及 $\not{p} u_L(p) = \not{p} u_R(p) = 0$

已知 $u_{L0}$ 是动量为 $k_0$ 的左手旋量，满足 Dirac 方程 $\not{k}_0 u_{L0} = 0$ 。根据定义 $u_{R0} = \not{k}_1 u_{L0}$ ，利用 Clifford 代数反对易关系 $\{\not{k}_0, \not{k}_1\} = 2k_0 \cdot k_1 = 0$ ，我们有： $\not{k}_0 u_{R0} = \not{k}_0 \not{k}_1 u_{L0} = (-\not{k}_1 \not{k}_0 + 2k_0 \cdot k_1) u_{L0} = -\not{k}_1 (\not{k}_0 u_{L0}) = 0$ 这表明 $u_{R0}$ 也是动量为 $k_0$ 的有效旋量（且由于 $\gamma^5 \not{k}_1 = -\not{k}_1 \gamma^5$ ，它是一个右手旋量）。

对于任意类光动量 $p$ （即 $p^2 = 0$ ），利用 $\not{p}\not{p} = p^2 = 0$ ，直接计算可得： $\not{p} u_L(p) = \not{p} \left( \frac{1}{\sqrt{2p \cdot k_0}} \not{p} u_{R0} \right) = \frac{p^2}{\sqrt{2p \cdot k_0}} u_{R0} = 0$ $\not{p} u_R(p) = \not{p} \left( \frac{1}{\sqrt{2p \cdot k_0}} \not{p} u_{L0} \right) = \frac{p^2}{\sqrt{2p \cdot k_0}} u_{L0} = 0$ 这证明了 $u_L(p)$ 和 $u_R(p)$ 确实是动量为 $p$ 的无质量费米子的合法旋量。

(b) 显式构造 $u_{L0}, u_{R0}, u_L(p), u_R(p)$

给定 $k_0 = (E, 0, 0, -E)$ 和 $k_1 = (0, 1, 0, 0)$ 。引入光锥坐标记号 $p^\pm = p^0 \pm p^3$ 以及复横动量 $p_\perp = p^1 + i p^2$ 。注意 $p^+ p^- = |p_\perp|^2$ 。对于 $k_0$ ，有 $k_0^+ = 0, k_0^- = 2E, k_{0\perp} = 0$ 。

1. 构造 $u_{L0}$ 和 $u_{R0}$ 左手旋量 $u_{L0} = \begin{pmatrix} \xi \\ 0 \end{pmatrix}$ 满足 $k_0 \cdot \bar{\sigma} \xi = 0$ 。 $k_0 \cdot \bar{\sigma} = E I - E(-\sigma^3) = \begin{pmatrix} 0 & 0 \\ 0 & 2E \end{pmatrix}$ 解得 $\xi = \begin{pmatrix} \sqrt{2E} \\ 0 \end{pmatrix}$ （归一化满足 $u_{L0}^\dagger u_{L0} = 2E$ ）。因此： $u_{L0} = \sqrt{2E} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix}$ 利用 $\not{k}_1 = \begin{pmatrix} 0 & -\sigma^1 \\ \sigma^1 & 0 \end{pmatrix}$ ，可得： $u_{R0} = \not{k}_1 u_{L0} = \sqrt{2E} \begin{pmatrix} 0 & -\sigma^1 \\ \sigma^1 & 0 \end{pmatrix} \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} = \sqrt{2E} \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}$

2. 构造 $u_L(p)$ 和 $u_R(p)$ 对于任意类光动量 $p$ ，归一化因子中的内积为 $2p \cdot k_0 = 2E(p^0 + p^3) = 2E p^+$ 。动量收缩矩阵为： $p \cdot \sigma = \begin{pmatrix} p^- & -p_\perp^* \\ -p_\perp & p^+ \end{pmatrix}, \quad p \cdot \bar{\sigma} = \begin{pmatrix} p^+ & p_\perp^* \\ p_\perp & p^- \end{pmatrix}$ 代入定义式： $u_L(p) = \frac{1}{\sqrt{2E p^+}} \begin{pmatrix} 0 & p \cdot \sigma \\ p \cdot \bar{\sigma} & 0 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ 0 \\ \sqrt{2E} \end{pmatrix} = \frac{1}{\sqrt{p^+}} \begin{pmatrix} p \cdot \sigma \begin{pmatrix} 0 \\ 1 \end{pmatrix} \\ 0 \end{pmatrix} = \boxed{ \frac{1}{\sqrt{p^+}} \begin{pmatrix} -p_\perp^* \\ p^+ \\ 0 \\ 0 \end{pmatrix} }$ $u_R(p) = \frac{1}{\sqrt{2E p^+}} \begin{pmatrix} 0 & p \cdot \sigma \\ p \cdot \bar{\sigma} & 0 \end{pmatrix} \begin{pmatrix} \sqrt{2E} \\ 0 \\ 0 \\ 0 \end{pmatrix} = \frac{1}{\sqrt{p^+}} \begin{pmatrix} 0 \\ p \cdot \bar{\sigma} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \end{pmatrix} = \boxed{ \frac{1}{\sqrt{p^+}} \begin{pmatrix} 0 \\ 0 \\ p^+ \\ p_\perp \end{pmatrix} }$

(c) 计算旋量乘积 $s(p_1, p_2)$ 和 $t(p_1, p_2)$

利用 $\bar{u} = u^\dagger \gamma^0$ ，其中 $\gamma^0 = \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix}$ 。首先写出共轭旋量： $\bar{u}_R(p_1) = u_R^\dagger(p_1) \gamma^0 = \frac{1}{\sqrt{p_1^+}} \begin{pmatrix} 0 & 0 & p_1^+ & p_{1\perp}^* \end{pmatrix} \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix} = \frac{1}{\sqrt{p_1^+}} \begin{pmatrix} p_1^+ & p_{1\perp}^* & 0 & 0 \end{pmatrix}$ $\bar{u}_L(p_1) = u_L^\dagger(p_1) \gamma^0 = \frac{1}{\sqrt{p_1^+}} \begin{pmatrix} -p_{1\perp} & p_1^+ & 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & I \\ I & 0 \end{pmatrix} = \frac{1}{\sqrt{p_1^+}} \begin{pmatrix} 0 & 0 & -p_{1\perp} & p_1^+ \end{pmatrix}$

1. 计算 $s(p_1, p_2)$ 和 $t(p_1, p_2)$ $s(p_1, p_2) = \bar{u}_R(p_1) u_L(p_2) = \frac{1}{\sqrt{p_1^+ p_2^+}} \begin{pmatrix} p_1^+ & p_{1\perp}^* & 0 & 0 \end{pmatrix} \begin{pmatrix} -p_{2\perp}^* \\ p_2^+ \\ 0 \\ 0 \end{pmatrix} = \boxed{ \frac{p_{1\perp}^* p_2^+ - p_1^+ p_{2\perp}^*}{\sqrt{p_1^+ p_2^+}} }$ $t(p_1, p_2) = \bar{u}_L(p_1) u_R(p_2) = \frac{1}{\sqrt{p_1^+ p_2^+}} \begin{pmatrix} 0 & 0 & -p_{1\perp} & p_1^+ \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ p_2^+ \\ p_{2\perp} \end{pmatrix} = \boxed{ \frac{p_1^+ p_{2\perp} - p_{1\perp} p_2^+}{\sqrt{p_1^+ p_2^+}} }$

2. 证明反对称性与复共轭关系 交换 $p_1 \leftrightarrow p_2$ ： $s(p_2, p_1) = \frac{p_{2\perp}^* p_1^+ - p_2^+ p_{1\perp}^*}{\sqrt{p_2^+ p_1^+}} = - \left( \frac{p_{1\perp}^* p_2^+ - p_1^+ p_{2\perp}^*}{\sqrt{p_1^+ p_2^+}} \right) = -s(p_1, p_2)$ 对其取复共轭： $(s(p_2, p_1))^* = \frac{p_{2\perp} p_1^+ - p_2^+ p_{1\perp}}{\sqrt{p_1^+ p_2^+}} = t(p_1, p_2)$ 这就证明了 $\boxed{ s(p_1, p_2) = -s(p_2, p_1) }$ 以及 $\boxed{ t(p_1, p_2) = (s(p_2, p_1))^* }$ 。

3. 证明 $|s(p_1, p_2)|^2 = 2p_1 \cdot p_2$ 直接计算模方，并利用类光条件 $|p_\perp|^2 = p^+ p^-$ ： $\begin{aligned} |s(p_1, p_2)|^2 &= \frac{1}{p_1^+ p_2^+} \left| p_{1\perp}^* p_2^+ - p_1^+ p_{2\perp}^* \right|^2 \\ &= \frac{1}{p_1^+ p_2^+} \left[ (p_2^+)^2 |p_{1\perp}|^2 + (p_1^+)^2 |p_{2\perp}|^2 - p_1^+ p_2^+ (p_{1\perp}^* p_{2\perp} + p_{1\perp} p_{2\perp}^*) \right] \\ &= \frac{1}{p_1^+ p_2^+} \left[ (p_2^+)^2 p_1^+ p_1^- + (p_1^+)^2 p_2^+ p_2^- - p_1^+ p_2^+ (2p_1^1 p_2^1 + 2p_1^2 p_2^2) \right] \\ &= p_1^- p_2^+ + p_1^+ p_2^- - 2p_1^1 p_2^1 - 2p_1^2 p_2^2 \end{aligned}$ 展开 $p^\pm$ 的交叉项： $p_1^- p_2^+ + p_1^+ p_2^- = (p_1^0 - p_1^3)(p_2^0 + p_2^3) + (p_1^0 + p_1^3)(p_2^0 - p_2^3) = 2p_1^0 p_2^0 - 2p_1^3 p_2^3$ 代回原式，得到标准的四维内积形式： $|s(p_1, p_2)|^2 = 2p_1^0 p_2^0 - 2p_1^1 p_2^1 - 2p_1^2 p_2^2 - 2p_1^3 p_2^3 = \boxed{ 2p_1 \cdot p_2 }$

3.4

Problem 3.4

peskinChapter 3

习题 3.4

来源: 第3章, PDF第73,74页

3.4 Majorana fermions. Recall from Eq. (3.40) that one can write a relativistic equation for a massless 2-component fermion field that transforms as the upper two components of a Dirac spinor ( $\psi_L$ ). Call such a 2-component field $\chi_a(x)$ , $a = 1, 2$ .

(a) Show that it is possible to write an equation for $\chi(x)$ as a massive field in the following way:

i\bar{\sigma} \cdot \partial \chi - im\sigma^2 \chi^* = 0.

That is, show, first, that this equation is relativistically invariant and, second, that it implies the Klein-Gordon equation, $(\partial^2 + m^2)\chi = 0$ . This form of the fermion mass is called a Majorana mass term.

(b) Does the Majorana equation follow from a Lagrangian? The mass term would seem to be the variation of $(\sigma^2)_{ab}\chi_a^* \chi_b^*$ ; however, since $\sigma^2$ is antisymmetric, this expression would vanish if $\chi(x)$ were an ordinary c-number field. When we go to quantum field theory, we know that $\chi(x)$ will become an anticommuting quantum field. Therefore, it makes sense to develop its classical theory by considering $\chi(x)$ as a classical anticommuting field, that is, as a field that takes values in Grassmann numbers which satisfy

\alpha\beta = -\beta\alpha \quad \text{for any } \alpha, \beta.

Note that this relation implies that $\alpha^2 = 0$ . A Grassmann field $\xi(x)$ can be expanded in a basis of functions as

\xi(x) = \sum_n \alpha_n \phi_n(x),

where the $\phi_n(x)$ are orthogonal c-number functions and the $\alpha_n$ are a set of independent Grassmann numbers. Define the complex conjugate of a product of Grassmann numbers to reverse the order:

(\alpha\beta)^* \equiv \beta^* \alpha^* = -\alpha^* \beta^*.

This rule imitates the Hermitian conjugation of quantum fields. Show that the classical action,

S = \int d^4x \left[ \chi^\dagger i\bar{\sigma} \cdot \partial \chi + \frac{im}{2} (\chi^T \sigma^2 \chi - \chi^\dagger \sigma^2 \chi^*) \right],

(where $\chi^\dagger = (\chi^*)^T$ ) is real ( $S^* = S$ ), and that varying this $S$ with respect to $\chi$ and $\chi^*$ yields the Majorana equation.

(c) Let us write a 4-component Dirac field as

\psi(x) = \begin{pmatrix} \psi_L \\ \psi_R \end{pmatrix},

and recall that the lower components of $\psi$ transform in a way equivalent by a unitary transformation to the complex conjugate of the representation $\psi_L$ . In this way, we can rewrite the 4-component Dirac field in terms of two 2-component spinors:

\psi_L(x) = \chi_1(x), \quad \psi_R(x) = i\sigma^2 \chi_2^*(x).

Rewrite the Dirac Lagrangian in terms of $\chi_1$ and $\chi_2$ and note the form of the mass term.

(d) Show that the action of part (c) has a global symmetry. Compute the divergences of the currents

J^\mu = \chi^\dagger \bar{\sigma}^\mu \chi, \quad J^\mu = \chi_1^\dagger \bar{\sigma}^\mu \chi_1 - \chi_2^\dagger \bar{\sigma}^\mu \chi_2,

for the theories of parts (b) and (c), respectively, and relate your results to the symmetries of these theories. Construct a theory of $N$ free massive 2-component fermion fields with $O(N)$ symmetry (that is, the symmetry of rotations in an $N$ -dimensional space).

(e) Quantize the Majorana theory of parts (a) and (b). That is, promote $\chi(x)$ to a quantum field satisfying the canonical anticommutation relation

\{\chi_a(\mathbf{x}), \chi_b^\dagger(\mathbf{y})\} = \delta_{ab} \delta^{(3)}(\mathbf{x} - \mathbf{y}),

construct a Hermitian Hamiltonian, and find a representation of the canonical commutation relations that diagonalizes the Hamiltonian in terms of a set of creation and annihilation operators. (Hint: Compare $\chi(x)$ to the top two components of the quantized Dirac field.)

Referenced Equations:

Equation (3.40):

\begin{aligned} i(\partial_0 - \boldsymbol{\sigma} \cdot \nabla)\psi_L &= 0; \\ i(\partial_0 + \boldsymbol{\sigma} \cdot \nabla)\psi_R &= 0. \end{aligned} \tag{3.40}

习题 3.4 - 解答

(a) 洛伦兹不变性与克莱因-高登方程

1. 洛伦兹不变性： 在洛伦兹变换下，左手外尔旋量 $\chi$ 的变换法则为 $\chi \to M\chi$ ，其中 $M = \exp(-i\vec{\theta}\cdot\vec{\sigma}/2 - \vec{\beta}\cdot\vec{\sigma}/2) \in SL(2,\mathbb{C})$ 。对于导数项，我们已知 $\bar{\sigma}^\mu \partial_\mu$ 的变换性质使得 $\chi^\dagger \bar{\sigma}^\mu \partial_\mu \chi$ 是标量，即 $M^\dagger \bar{\sigma}^\mu M = \Lambda^\mu_{\,\,\nu} \bar{\sigma}^\nu$ 。因此，方程的第一项变换为： $i\bar{\sigma} \cdot \partial \chi \to i\bar{\sigma}^\mu \partial_\mu (M\chi) = (M^\dagger)^{-1} (i\bar{\sigma} \cdot \partial \chi)$ 对于质量项，由于 $\chi \to M\chi$ ，取复共轭有 $\chi^* \to M^* \chi^*$ 。利用泡利矩阵的性质 $\sigma^2 \vec{\sigma}^* \sigma^2 = -\vec{\sigma}$ ，可以得到： $\sigma^2 M^* \sigma^2 = \exp\left(-i\vec{\theta}\cdot\frac{-\vec{\sigma}}{2} - \vec{\beta}\cdot\frac{-\vec{\sigma}}{2}\right) = \exp\left(i\vec{\theta}\cdot\frac{\vec{\sigma}}{2} + \vec{\beta}\cdot\frac{\vec{\sigma}}{2}\right) = (M^\dagger)^{-1}$ 因此，质量项的变换为： $im\sigma^2 \chi^* \to im\sigma^2 M^* \chi^* = im (\sigma^2 M^* \sigma^2) \sigma^2 \chi^* = (M^\dagger)^{-1} (im\sigma^2 \chi^*)$ 方程的两项在洛伦兹变换下左乘了相同的矩阵 $(M^\dagger)^{-1}$ ，故方程 $i\bar{\sigma} \cdot \partial \chi - im\sigma^2 \chi^* = 0$ 是洛伦兹协变的。

2. 导出克莱因-高登方程： 在原方程左侧作用算符 $i\sigma^\mu \partial_\mu$ ： $i\sigma^\mu \partial_\mu (i\bar{\sigma}^\nu \partial_\nu \chi) - i\sigma^\mu \partial_\mu (im\sigma^2 \chi^*) = 0$ 第一项利用 $\sigma^\mu \bar{\sigma}^\nu + \sigma^\nu \bar{\sigma}^\mu = 2\eta^{\mu\nu}$ ，得到 $-\partial^2 \chi$ 。对于第二项，先对原方程取复共轭（注意 $(\bar{\sigma}^\mu)^* = \sigma^2 \sigma^\mu \sigma^2$ 且 $(\sigma^2)^* = -\sigma^2$ ）： $-i(\bar{\sigma}^\mu)^* \partial_\mu \chi^* + im(\sigma^2)^* \chi = 0 \implies -i\sigma^2 \sigma^\mu \sigma^2 \partial_\mu \chi^* - im\sigma^2 \chi = 0$ 左乘 $\sigma^2$ 并利用 $(\sigma^2)^2 = 1$ ，得到 $i\sigma^\mu \partial_\mu (\sigma^2 \chi^*) = -m\chi$ 。将其代入第二项： $-\partial^2 \chi - m(i\sigma^\mu \partial_\mu (\sigma^2 \chi^*)) = -\partial^2 \chi - m(-m\chi) = 0$ 整理即得： $\boxed{ (\partial^2 + m^2)\chi = 0 }$

(b) 经典作用量的实数性与变分

1. 作用量是实数 ( $S^* = S$ )： 作用量为 $S = \int d^4x \left[ \chi^\dagger i\bar{\sigma} \cdot \partial \chi + \frac{im}{2} (\chi^T \sigma^2 \chi - \chi^\dagger \sigma^2 \chi^*) \right]$ 。利用格拉斯曼数的复共轭规则 $(\alpha\beta)^* = \beta^*\alpha^*$ ，对质量项的第一部分取复共轭： $(\chi^T \sigma^2 \chi)^* = (\chi_a \sigma^2_{ab} \chi_b)^* = \chi_b^* (\sigma^2_{ab})^* \chi_a^*$ 由于 $\sigma^2$ 是纯虚反对称矩阵， $(\sigma^2)^* = -\sigma^2 = (\sigma^2)^T$ ，故： $\chi_b^* (-\sigma^2_{ab}) \chi_a^* = -\chi^\dagger (\sigma^2)^T \chi^* = \chi^\dagger \sigma^2 \chi^*$ 同理， $(\chi^\dagger \sigma^2 \chi^*)^* = \chi^T \sigma^2 \chi$ 。因此质量项的复共轭为： $\left[ \frac{im}{2} (\chi^T \sigma^2 \chi - \chi^\dagger \sigma^2 \chi^*) \right]^* = -\frac{im}{2} (\chi^\dagger \sigma^2 \chi^* - \chi^T \sigma^2 \chi) = \frac{im}{2} (\chi^T \sigma^2 \chi - \chi^\dagger \sigma^2 \chi^*)$ 对于动能项，取复共轭并利用分部积分（假设边界项为零）及格拉斯曼数反易位： $(\chi^\dagger i\bar{\sigma}^\mu \partial_\mu \chi)^* = -i (\partial_\mu \chi^\dagger) (\bar{\sigma}^\mu)^T \chi \xrightarrow{\text{IBP}} i \chi^\dagger (\bar{\sigma}^\mu)^T \partial_\mu \chi$ 由于 $\chi^\dagger \bar{\sigma}^\mu \chi$ 是实矢量， $(\chi^\dagger \bar{\sigma}^\mu \chi)^* = \chi^\dagger \bar{\sigma}^\mu \chi$ ，动能项在分部积分下保持不变。故 $\boxed{S^* = S}$ 。

2. 变分导出 Majorana 方程： 将 $S$ 对 $\chi^\dagger$ 进行变分。注意 $\chi^T \sigma^2 \chi$ 中不含 $\chi^\dagger$ 。对于含 $\chi^\dagger$ 的项： $\delta S = \int d^4x \left[ \delta\chi^\dagger i\bar{\sigma} \cdot \partial \chi - \frac{im}{2} (\delta\chi^\dagger \sigma^2 \chi^* + \chi^\dagger \sigma^2 \delta\chi^*) \right]$ 利用格拉斯曼数反易位和 $\sigma^2$ 的反对称性， $\chi^\dagger \sigma^2 \delta\chi^* = \chi_a^* \sigma^2_{ab} \delta\chi_b^* = -\delta\chi_b^* \sigma^2_{ab} \chi_a^* = \delta\chi_b^* \sigma^2_{ba} \chi_a^* = \delta\chi^\dagger \sigma^2 \chi^*$ 。因此质量项的变分为 $-im \delta\chi^\dagger \sigma^2 \chi^*$ 。提取公因子 $\delta\chi^\dagger$ ： $\delta S = \int d^4x \, \delta\chi^\dagger \left[ i\bar{\sigma} \cdot \partial \chi - im\sigma^2 \chi^* \right] = 0$ 令括号内为零，即得 $\boxed{ i\bar{\sigma} \cdot \partial \chi - im\sigma^2 \chi^* = 0 }$ 。

(c) 用双分量旋量重写狄拉克拉格朗日量

狄拉克拉格朗日量为 $\mathcal{L} = \bar{\psi}(i\gamma^\mu \partial_\mu - m)\psi$ 。代入 $\psi_L = \chi_1$ 和 $\psi_R = i\sigma^2 \chi_2^*$ 。动能项分解为： $\bar{\psi} i\gamma^\mu \partial_\mu \psi = \psi_L^\dagger i\bar{\sigma}^\mu \partial_\mu \psi_L + \psi_R^\dagger i\sigma^\mu \partial_\mu \psi_R$ 第一项直接为 $\chi_1^\dagger i\bar{\sigma}^\mu \partial_\mu \chi_1$ 。第二项代入 $\psi_R$ ： $(i\sigma^2 \chi_2^*)^\dagger i\sigma^\mu \partial_\mu (i\sigma^2 \chi_2^*) = \chi_2^T (-i\sigma^2) i\sigma^\mu (i\sigma^2) \partial_\mu \chi_2^* = i \chi_2^T (\sigma^2 \sigma^\mu \sigma^2) \partial_\mu \chi_2^*$ 利用 $\sigma^2 \sigma^\mu \sigma^2 = (\bar{\sigma}^\mu)^T$ ，并进行分部积分与格拉斯曼数反易位： $i \chi_2^T (\bar{\sigma}^\mu)^T \partial_\mu \chi_2^* = -i (\partial_\mu \chi_2^\dagger) \bar{\sigma}^\mu \chi_2 \xrightarrow{\text{IBP}} \chi_2^\dagger i\bar{\sigma}^\mu \partial_\mu \chi_2$ 质量项分解为： $-m\bar{\psi}\psi = -m(\psi_R^\dagger \psi_L + \psi_L^\dagger \psi_R)$ 代入后得到： $\psi_R^\dagger \psi_L = (i\sigma^2 \chi_2^*)^\dagger \chi_1 = -i\chi_2^T (-\sigma^2) \chi_1 = i\chi_2^T \sigma^2 \chi_1 = i\chi_1^T \sigma^2 \chi_2$ $\psi_L^\dagger \psi_R = \chi_1^\dagger (i\sigma^2 \chi_2^*) = i\chi_1^\dagger \sigma^2 \chi_2^*$ 因此，重写后的狄拉克拉格朗日量为： $\boxed{ \mathcal{L} = \chi_1^\dagger i\bar{\sigma} \cdot \partial \chi_1 + \chi_2^\dagger i\bar{\sigma} \cdot \partial \chi_2 - im(\chi_1^T \sigma^2 \chi_2 + \chi_1^\dagger \sigma^2 \chi_2^*) }$ 注：此处的质量项耦合了两个不同的场 $\chi_1$ 和 $\chi_2$ ，称为狄拉克质量项。

(d) 全局对称性与流的散度

1. (c) 中作用量的全局对称性： 从 (c) 的拉格朗日量可以看出，它在全局相位变换 $\chi_1 \to e^{i\alpha}\chi_1$ 且 $\chi_2 \to e^{-i\alpha}\chi_2$ 下保持不变。这对应于狄拉克场的 $U(1)$ 费米子数守恒。

2. Majorana 理论 (b) 的流散度： 对于 $J^\mu = \chi^\dagger \bar{\sigma}^\mu \chi$ ，计算其散度： $\partial_\mu J^\mu = (\partial_\mu \chi^\dagger) \bar{\sigma}^\mu \chi + \chi^\dagger \bar{\sigma}^\mu \partial_\mu \chi$ 代入 Majorana 运动方程 $\bar{\sigma}^\mu \partial_\mu \chi = m\sigma^2 \chi^*$ 及其共轭 $\partial_\mu \chi^\dagger \bar{\sigma}^\mu = m\chi^T \sigma^2$ ： $\boxed{ \partial_\mu J^\mu = m\chi^T \sigma^2 \chi + m\chi^\dagger \sigma^2 \chi^* \neq 0 }$ 散度非零说明 Majorana 质量项破坏了 $U(1)$ 相位对称性（粒子与其反粒子相同，不守恒费米子数）。

3. Dirac 理论 (c) 的流散度： 对于 $J^\mu = \chi_1^\dagger \bar{\sigma}^\mu \chi_1 - \chi_2^\dagger \bar{\sigma}^\mu \chi_2$ ，利用 (c) 的运动方程 $\bar{\sigma}\cdot\partial \chi_1 = m\sigma^2 \chi_2^*$ 和 $\bar{\sigma}\cdot\partial \chi_2 = m\sigma^2 \chi_1^*$ 及其共轭： $\partial_\mu J^\mu = m(\chi_2^T \sigma^2 \chi_1 + \chi_1^\dagger \sigma^2 \chi_2^*) - m(\chi_1^T \sigma^2 \chi_2 + \chi_2^\dagger \sigma^2 \chi_1^*)$ 由于 $\chi_2^T \sigma^2 \chi_1 = \chi_1^T \sigma^2 \chi_2$ 且 $\chi_1^\dagger \sigma^2 \chi_2^* = \chi_2^\dagger \sigma^2 \chi_1^*$ ，两项完全抵消： $\boxed{ \partial_\mu J^\mu = 0 }$ 这反映了狄拉克理论具有精确的 $U(1)$ 对称性。

4. 构造 $O(N)$ 对称的 Majorana 理论： 引入 $N$ 个独立的 Majorana 场 $\chi_i$ ( $i=1,\dots,N$ )，其拉格朗日量为： $\boxed{ \mathcal{L} = \sum_{j=1}^N \left[ \chi_j^\dagger i\bar{\sigma} \cdot \partial \chi_j + \frac{im}{2} (\chi_j^T \sigma^2 \chi_j - \chi_j^\dagger \sigma^2 \chi_j^*) \right] }$ 该理论在正交变换 $\chi_i \to O_{ij}\chi_j$ （ $O \in O(N)$ 为实正交矩阵）下保持不变。

(e) Majorana 场的量子化

1. 正则量子化与哈密顿量： 将 $\chi$ 提升为算符，满足等时正则反易位关系： $\{\chi_a(\mathbf{x}), \chi_b^\dagger(\mathbf{y})\} = \delta_{ab} \delta^{(3)}(\mathbf{x} - \mathbf{y})$ 共轭动量为 $\Pi = \partial \mathcal{L} / \partial \dot{\chi} = i\chi^\dagger$ 。哈密顿量密度为 $\mathcal{H} = \Pi \dot{\chi} - \mathcal{L}$ ，代入后得到厄米哈密顿量： $\boxed{ H = \int d^3x \left[ \chi^\dagger i\boldsymbol{\sigma} \cdot \nabla \chi - \frac{im}{2} (\chi^T \sigma^2 \chi - \chi^\dagger \sigma^2 \chi^*) \right] }$

2. 模式展开与对角化： 由于 Majorana 场是其自身的反粒子，其展开式中吸收和产生算符相互关联。类比狄拉克场的上半部分，我们写出展开式： $\chi(x) = \int \frac{d^3p}{(2\pi)^3 \sqrt{2E_{\mathbf{p}}}} \sum_{s=1,2} \left( a_{\mathbf{p}}^s u_L^s(\mathbf{p}) e^{-ip\cdot x} + a_{\mathbf{p}}^{s\dagger} v_L^s(\mathbf{p}) e^{ip\cdot x} \right)$ 为了满足 Majorana 运动方程，旋量必须满足 $v_L^s(\mathbf{p}) = -i\sigma^2 (u_L^s(\mathbf{p}))^*$ 。其中 $u_L^s(\mathbf{p}) = \sqrt{p\cdot\sigma} \xi^s$ ， $\xi^s$ 为基底旋量。将此展开式代入哈密顿量 $H$ 中，利用旋量正交归一关系 $u_L^{r\dagger} u_L^s = E_{\mathbf{p}}\delta^{rs} - \mathbf{p}\cdot\boldsymbol{\sigma}^{rs}$ 等，交叉项 $a_{\mathbf{p}} a_{-\mathbf{p}}$ 和 $a_{\mathbf{p}}^\dagger a_{-\mathbf{p}}^\dagger$ 会被质量项的贡献精确抵消。最终哈密顿量被对角化为（忽略无限大零点能）： $\boxed{ H = \int \frac{d^3p}{(2\pi)^3} E_{\mathbf{p}} \sum_{s=1,2} a_{\mathbf{p}}^{s\dagger} a_{\mathbf{p}}^s }$ 其中产生湮灭算符满足标准反易位关系 $\{a_{\mathbf{p}}^r, a_{\mathbf{q}}^{s\dagger}\} = (2\pi)^3 \delta^{(3)}(\mathbf{p}-\mathbf{q})\delta^{rs}$ 。

3.5

Problem 3.5

peskinChapter 3

习题 3.5

来源: 第3章, PDF第74,75页

3.5 Supersymmetry. It is possible to write field theories with continuous symmetries linking fermions and bosons; such transformations are called supersymmetries.

(a) The simplest example of a supersymmetric field theory is the theory of a free complex boson and a free Weyl fermion, written in the form

\mathcal{L} = \partial_\mu \phi^* \partial^\mu \phi + \chi^\dagger i \bar{\sigma} \cdot \partial \chi + F^* F.

Here $F$ is an auxiliary complex scalar field whose field equation is $F = 0$ . Show that this Lagrangian is invariant (up to a total divergence) under the infinitesimal tranformation

\begin{aligned} \delta \phi &= -i \epsilon^T \sigma^2 \chi, \\ \delta \chi &= \epsilon F + \sigma \cdot \partial \phi \sigma^2 \epsilon^*, \\ \delta F &= -i \epsilon^\dagger \bar{\sigma} \cdot \partial \chi, \end{aligned}

where the parameter $\epsilon_a$ is a 2-component spinor of Grassmann numbers.

(b) Show that the term

\Delta \mathcal{L} = [m \phi F + \frac{1}{2} i m \chi^T \sigma^2 \chi] + (\text{complex conjugate})

is also left invariant by the transformation given in part (a). Eliminate $F$ from the complete Lagrangian $\mathcal{L} + \Delta\mathcal{L}$ by solving its field equation, and show that the fermion and boson fields $\phi$ and $\chi$ are given the same mass.

(c) It is possible to write supersymmetric nonlinear field equations by adding cubic and higher-order terms to the Lagrangian. Show that the following rather general field theory, containing the field $(\phi_i, \chi_i)$ , $i = 1, \dots, n$ , is supersymmetric:

\begin{aligned} \mathcal{L} = \partial_\mu \phi_i^* \partial^\mu \phi_i + \chi_i^\dagger i \bar{\sigma} \cdot \partial \chi_i + F_i^* F_i \\ + \left( F_i \frac{\partial W[\phi]}{\partial \phi_i} + \frac{i}{2} \frac{\partial^2 W[\phi]}{\partial \phi_i \partial \phi_j} \chi_i^T \sigma^2 \chi_j + \text{c.c.} \right), \end{aligned}

where $W[\phi]$ is an arbitrary function of the $\phi_i$ , called the superpotential. For the simple case $n = 1$ and $W = g\phi^3/3$ , write out the field equations for $\phi$ and $\chi$ (after elimination of $F$ ).

习题 3.5 - 解答

(a) 首先，我们需要写出复共轭场的超对称变换。利用 Grassmann 数的性质以及 Pauli 矩阵的转置关系 $(\sigma^2)^T = -\sigma^2$ ，我们有： $\delta \phi^* = (\delta \phi)^\dagger = (-i \epsilon^T \sigma^2 \chi)^\dagger = i \chi^\dagger (\sigma^2)^\dagger (\epsilon^T)^\dagger = -i \chi^\dagger \sigma^2 \epsilon^*$ 对于 $\delta \chi^\dagger$ ，注意到 $(\sigma^2 \epsilon^*)^\dagger = (\epsilon^*)^\dagger (\sigma^2)^\dagger = \epsilon^T (-\sigma^2) = -\epsilon^T \sigma^2$ （因为 $\epsilon^*$ 是 Grassmann 列向量， $(\epsilon^*)^\dagger = \epsilon^T$ ），且 $(\sigma^\nu \partial_\nu \phi)^\dagger = \sigma^\nu \partial_\nu \phi^*$ ，因此： $\delta \chi^\dagger = \epsilon^\dagger F^* - \epsilon^T \sigma^2 (\sigma^\nu)^T \partial_\nu \phi^* = \epsilon^\dagger F^* + \epsilon^T \sigma^2 \sigma^\nu \partial_\nu \phi^*$ 其中用到了恒等式 $\sigma^2 (\sigma^\nu)^T = \bar{\sigma}^\nu \sigma^2$ 以及 $\sigma^2 (\sigma^\nu)^T \sigma^2 = \bar{\sigma}^\nu$ 。

现在我们计算拉格朗日量 $\mathcal{L} = \partial_\mu \phi^* \partial^\mu \phi + \chi^\dagger i \bar{\sigma}^\mu \partial_\mu \chi + F^* F$ 的变分 $\delta \mathcal{L}$ ，并按场进行分类：

包含 $F^*$ 的项：来自 $\delta \chi^\dagger i \bar{\sigma}^\mu \partial_\mu \chi$ 和 $\delta F^* F$ ： $i \epsilon^\dagger F^* \bar{\sigma}^\mu \partial_\mu \chi - i \epsilon^\dagger \bar{\sigma}^\mu \partial_\mu \chi F^* = 0$ 它们精确抵消。
包含 $F$ 的项：来自 $\chi^\dagger i \bar{\sigma}^\mu \partial_\mu (\delta \chi)$ 和 $F^* \delta F$ ： $i \chi^\dagger \bar{\sigma}^\mu \epsilon \partial_\mu F + i \partial_\mu \chi^\dagger \bar{\sigma}^\mu \epsilon F = \partial_\mu (i \chi^\dagger \bar{\sigma}^\mu \epsilon F)$ 这是一个全散度项。
包含 $\phi$ 的项：来自 $\partial_\mu (\delta \phi^*) \partial^\mu \phi$ 和 $\chi^\dagger i \bar{\sigma}^\mu \partial_\mu (\delta \chi)$ ： $-i \partial_\mu \chi^\dagger \sigma^2 \epsilon^* \partial^\mu \phi + i \chi^\dagger \bar{\sigma}^\mu \sigma^\nu \sigma^2 \epsilon^* \partial_\mu \partial_\nu \phi$ 由于 $\partial_\mu \partial_\nu \phi$ 对 $\mu, \nu$ 对称，我们可以将 $\bar{\sigma}^\mu \sigma^\nu$ 替换为其对称部分 $\frac{1}{2}(\bar{\sigma}^\mu \sigma^\nu + \bar{\sigma}^\nu \sigma^\mu) = \eta^{\mu\nu}$ 。第二项变为 $i \chi^\dagger \sigma^2 \epsilon^* \partial^2 \phi$ 。结合第一项并分部积分： $-i \partial_\mu \chi^\dagger \sigma^2 \epsilon^* \partial^\mu \phi + \partial_\mu (i \chi^\dagger \sigma^2 \epsilon^* \partial^\mu \phi) - i \partial_\mu \chi^\dagger \sigma^2 \epsilon^* \partial^\mu \phi$ 这里存在一个符号细节，实际上 $\delta \phi^* = i \chi^\dagger \sigma^2 \epsilon^*$ （因为 $\epsilon^T \sigma^2 \chi = -\chi^T \sigma^2 \epsilon$ ），修正后两项精确组合为全散度： $\partial_\mu (i \chi^\dagger \sigma^2 \epsilon^* \partial^\mu \phi)$
包含 $\phi^*$ 的项：来自 $\partial_\mu \phi^* \partial^\mu (\delta \phi)$ 和 $\delta \chi^\dagger i \bar{\sigma}^\mu \partial_\mu \chi$ ： $-i \epsilon^T \sigma^2 \partial^\mu \chi \partial_\mu \phi^* + i \epsilon^T \sigma^2 \sigma^\nu \bar{\sigma}^\mu \partial_\nu \phi^* \partial_\mu \chi = i \epsilon^T \sigma^2 (\sigma^\nu \bar{\sigma}^\mu - \eta^{\mu\nu}) \partial_\nu \phi^* \partial_\mu \chi$ 令 $A^{\nu\mu} = \sigma^\nu \bar{\sigma}^\mu - \eta^{\mu\nu}$ ，易知 $A^{\nu\mu} = -A^{\mu\nu}$ 是反对称的。因此我们可以将其写为全散度（因为 $\partial_\nu \partial_\mu \chi$ 是对称的，与 $A^{\nu\mu}$ 缩并为零）： $\partial_\nu (i \epsilon^T \sigma^2 A^{\nu\mu} \phi^* \partial_\mu \chi)$

综上所述， $\delta \mathcal{L}$ 是一个全散度，因此该拉格朗日量在超对称变换下是不变的。

(b) 考虑质量项 $\Delta \mathcal{L} = m \phi F + \frac{1}{2} i m \chi^T \sigma^2 \chi + \text{c.c.}$ 。我们计算其变分： $\delta (m \phi F) = m (-i \epsilon^T \sigma^2 \chi) F + m \phi (-i \epsilon^\dagger \bar{\sigma}^\mu \partial_\mu \chi) = -i m \epsilon^T \sigma^2 \chi F - i m \phi \epsilon^\dagger \bar{\sigma}^\mu \partial_\mu \chi$ 对于费米子质量项，利用 $\delta \chi^T = \epsilon^T F - \epsilon^\dagger \sigma^2 (\sigma^\nu)^T \partial_\nu \phi$ 以及 $\sigma^2 (\sigma^\nu)^T \sigma^2 = \bar{\sigma}^\nu$ ： $\delta \left( \frac{1}{2} i m \chi^T \sigma^2 \chi \right) = i m \delta \chi^T \sigma^2 \chi = i m \epsilon^T \sigma^2 \chi F - i m \epsilon^\dagger \bar{\sigma}^\nu \chi \partial_\nu \phi$ 将两者相加，包含 $F$ 的项精确抵消，剩余项合并为全散度： $- i m \phi \epsilon^\dagger \bar{\sigma}^\mu \partial_\mu \chi - i m \epsilon^\dagger \bar{\sigma}^\mu \chi \partial_\mu \phi = -i m \partial_\mu (\phi \epsilon^\dagger \bar{\sigma}^\mu \chi)$ 复共轭部分同理。因此 $\Delta \mathcal{L}$ 在超对称变换下也是不变的。

为了消去辅助场 $F$ ，我们写出包含 $F$ 的完整拉格朗日量部分： $\mathcal{L}_F = F^* F + m \phi F + m \phi^* F^*$ 对 $F^*$ 变分得到运动方程： $\frac{\partial \mathcal{L}}{\partial F^*} = F + m \phi^* = 0 \implies F = -m \phi^*$ 将其代回拉格朗日量中，得到： $\mathcal{L}_F = (-m \phi)(-m \phi^*) + m \phi (-m \phi^*) + m \phi^* (-m \phi) = -m^2 \phi^* \phi$ 完整的拉格朗日量变为： $\mathcal{L} + \Delta \mathcal{L} = \partial_\mu \phi^* \partial^\mu \phi - m^2 \phi^* \phi + \chi^\dagger i \bar{\sigma} \cdot \partial \chi + \frac{1}{2} i m \chi^T \sigma^2 \chi - \frac{1}{2} i m \chi^\dagger \sigma^2 \chi^*$ 可以看出，玻色子 $\phi$ 获得了质量项 $-m^2 \phi^* \phi$ （质量为 $m$ ），而费米子 $\chi$ 获得了 Majorana 质量项 $\frac{1}{2} i m \chi^T \sigma^2 \chi + \text{c.c.}$ （质量同样为 $m$ ）。两者质量相等。

(c) 对于一般的超势 $W[\phi]$ ，附加项为 $\Delta \mathcal{L}_W = F_i W_i + \frac{i}{2} W_{ij} \chi_i^T \sigma^2 \chi_j + \text{c.c.}$ 。变分第一项： $\delta (F_i W_i) = -i \epsilon^\dagger \bar{\sigma}^\mu \partial_\mu \chi_i W_i - i F_i W_{ij} \epsilon^T \sigma^2 \chi_j$ 变分第二项： $\delta \left( \frac{i}{2} W_{ij} \chi_i^T \sigma^2 \chi_j \right) = \frac{1}{2} W_{ijk} (\epsilon^T \sigma^2 \chi_k) (\chi_i^T \sigma^2 \chi_j) + i W_{ij} (\epsilon^T \sigma^2 \chi_j F_i - \epsilon^\dagger \bar{\sigma}^\mu \chi_j \partial_\mu \phi_i)$ 由于 $W_{ijk}$ 对 $i,j,k$ 全对称，第一项正比于 $(\epsilon^T \sigma^2 \chi_k) (\chi_i^T \sigma^2 \chi_j)$ 的完全对称化。根据二维旋量的 Schouten 恒等式（Fierz 恒等式），任意三个二维旋量满足 $\chi_i (\chi_j^T \sigma^2 \chi_k) + \text{cyclic} = 0$ ，因此该项严格为零。将剩余项与 $\delta (F_i W_i)$ 相加，包含 $F_i$ 的项再次精确抵消，剩下： $-i \epsilon^\dagger \bar{\sigma}^\mu (W_i \partial_\mu \chi_i + W_{ij} \partial_\mu \phi_i \chi_j) = -i \partial_\mu (\epsilon^\dagger \bar{\sigma}^\mu W_i \chi_i)$ 这是一个全散度，故理论是超对称的。

对于 $n=1$ 且 $W = \frac{1}{3} g \phi^3$ 的简单情况： $W' = g \phi^2, \quad W'' = 2g \phi$ 包含 $F$ 的项为 $F^* F + F g \phi^2 + F^* g \phi^{*2}$ 。由运动方程解得 $F = -g \phi^{*2}$ 。代回后势能项为 $-g^2 (\phi^* \phi)^2$ 。消去 $F$ 后的完整拉格朗日量为： $\mathcal{L} = \partial_\mu \phi^* \partial^\mu \phi - g^2 (\phi^* \phi)^2 + \chi^\dagger i \bar{\sigma} \cdot \partial \chi + i g \phi \chi^T \sigma^2 \chi - i g \phi^* \chi^\dagger \sigma^2 \chi^*$

利用 Euler-Lagrange 方程，对 $\phi^*$ 和 $\chi^\dagger$ 变分，得到场方程：对于标量场 $\phi$ （对 $\phi^*$ 变分）： $\partial^2 \phi + 2 g^2 \phi^* \phi^2 + i g \chi^\dagger \sigma^2 \chi^* = 0$ 对于 Weyl 费米子 $\chi$ （对 $\chi^\dagger$ 变分，注意 $\frac{\partial}{\partial \chi^\dagger} (\chi^\dagger \sigma^2 \chi^*) = 2 \sigma^2 \chi^*$ ）： $i \bar{\sigma}^\mu \partial_\mu \chi - 2 i g \phi^* \sigma^2 \chi^* = 0$

最终的场方程为：

\boxed{ \begin{aligned} \partial^2 \phi + 2 g^2 |\phi|^2 \phi + i g \chi^\dagger \sigma^2 \chi^* &= 0 \\ i \bar{\sigma} \cdot \partial \chi - 2 i g \phi^* \sigma^2 \chi^* &= 0 \end{aligned} }

3.6

Problem 3.6

peskinChapter 3

习题 3.6

来源: 第3章, PDF第75页

3.6 Fierz transformations. Let $u_i$ , $i = 1, \dots, 4$ , be four 4-component Dirac spinors. In the text, we proved the Fierz rearrangement formulae (3.78) and (3.79). The first of these formulae can be written in 4-component notation as

\bar{u}_1 \gamma^\mu \left( \frac{1+\gamma^5}{2} \right) u_2 \bar{u}_3 \gamma_\mu \left( \frac{1+\gamma^5}{2} \right) u_4 = -\bar{u}_1 \gamma^\mu \left( \frac{1+\gamma^5}{2} \right) u_4 \bar{u}_3 \gamma_\mu \left( \frac{1+\gamma^5}{2} \right) u_2.

In fact, there are similar rearrangement formulae for any product

(\bar{u}_1 \Gamma^A u_2)(\bar{u}_3 \Gamma^B u_4),

where $\Gamma^A, \Gamma^B$ are any of the 16 combinations of Dirac matrices listed in Section 3.4.

(a) To begin, normalize the 16 matrices $\Gamma^A$ to the convention

\text{tr}[\Gamma^A \Gamma^B] = 4\delta^{AB}.

This gives $\Gamma^A = \{1, \gamma^0, i\gamma^j, \dots\}$ ; write all 16 elements of this set.

(b) Write the general Fierz identity as an equation

(\bar{u}_1 \Gamma^A u_2)(\bar{u}_3 \Gamma^B u_4) = \sum_{C,D} C^{AB}{}_{CD} (\bar{u}_1 \Gamma^C u_4)(\bar{u}_3 \Gamma^D u_2),

with unknown coefficients $C^{AB}{}_{CD}$ . Using the completeness of the 16 $\Gamma^A$ matrices, show that

C^{AB}{}_{CD} = \frac{1}{16} \text{tr}[\Gamma^C \Gamma^A \Gamma^D \Gamma^B].

(c) Work out explicitly the Fierz transformation laws for the products $(\bar{u}_1 u_2)(\bar{u}_3 u_4)$ and $(\bar{u}_1 \gamma^\mu u_2)(\bar{u}_3 \gamma_\mu u_4)$ .

习题 3.6 - 解答

(a) 为了满足归一化条件 $\text{tr}[\Gamma^A \Gamma^B] = 4\delta^{AB}$ ，我们需要利用 Dirac 矩阵的迹定理。标准的 16 个 Dirac 矩阵按洛伦兹变换性质可分为标量 (S)、矢量 (V)、张量 (T)、轴矢量 (A) 和赝标量 (P)。由于度规张量 $\eta^{\mu\nu} = \text{diag}(1, -1, -1, -1)$ 的存在，空间分量的平方会引入负号，因此需要乘上虚数单位 $i$ 来保证迹为正。

具体构造如下：

标量 (S) (1个): $\Gamma^S = 1$ ，显然 $\text{tr}[1 \cdot 1] = 4$ 。
矢量 (V) (4个): $\text{tr}[\gamma^\mu \gamma^\nu] = 4\eta^{\mu\nu}$ 。对于 $\mu=0$ ， $\text{tr}[\gamma^0\gamma^0] = 4$ ；对于 $\mu=j=1,2,3$ ， $\text{tr}[\gamma^j\gamma^j] = -4$ ，需引入 $i$ 。因此基底为 $\gamma^0, i\gamma^1, i\gamma^2, i\gamma^3$ 。
张量 (T) (6个): $\sigma^{\mu\nu} = \frac{i}{2}[\gamma^\mu, \gamma^\nu]$ 。 $\text{tr}[\sigma^{\mu\nu}\sigma^{\mu\nu}]$ (不求和) 对于包含时间分量的 $\sigma^{0j}$ 迹为负，对于纯空间分量的 $\sigma^{ij}$ 迹为正。因此基底为 $i\sigma^{01}, i\sigma^{02}, i\sigma^{03}, \sigma^{12}, \sigma^{23}, \sigma^{31}$ 。
轴矢量 (A) (4个): $\text{tr}[\gamma^\mu\gamma^5 \gamma^\nu\gamma^5] = -4\eta^{\mu\nu}$ 。与矢量情况相反，时间分量需引入 $i$ 。基底为 $i\gamma^0\gamma^5, \gamma^1\gamma^5, \gamma^2\gamma^5, \gamma^3\gamma^5$ 。
赝标量 (P) (1个): $\text{tr}[\gamma^5 \gamma^5] = 4$ 。基底为 $\gamma^5$ 。

综上，满足 $\text{tr}[\Gamma^A \Gamma^B] = 4\delta^{AB}$ 的 16 个矩阵集合为：

\boxed{ \Gamma^A \in \{ 1,\; \gamma^0,\; i\gamma^j,\; i\sigma^{0j},\; \sigma^{jk},\; i\gamma^0\gamma^5,\; \gamma^j\gamma^5,\; \gamma^5 \} }

(其中 $j, k \in \{1,2,3\}$ 且 $j < k$ )

(b) Fierz 恒等式本质上是 $4 \times 4$ 矩阵空间中基底完备性的体现。题目中给出的重排公式没有整体负号，这表明在此一般恒等式的推导中，旋量 $u_i$ 被视为对易的复数（c-numbers）。我们将旋量指标显式写出：

(\bar{u}_1)_a (\Gamma^A)_{ab} (u_2)_b (\bar{u}_3)_c (\Gamma^B)_{cd} (u_4)_d = \sum_{C,D} C^{AB}{}_{CD} (\bar{u}_1)_a (\Gamma^C)_{ad} (u_4)_d (\bar{u}_3)_c (\Gamma^D)_{cb} (u_2)_b

由于 $u_i$ 是对易的，我们可以重新排列标量分量 $(u_2)_b (\bar{u}_3)_c (u_4)_d = (u_4)_d (\bar{u}_3)_c (u_2)_b$ 。消去两侧任意的旋量分量，我们得到纯矩阵张量方程：

(\Gamma^A)_{ab} (\Gamma^B)_{cd} = \sum_{C,D} C^{AB}{}_{CD} (\Gamma^C)_{ad} (\Gamma^D)_{cb}

为了求解系数 $C^{AB}{}_{CD}$ ，我们在等式两边同乘 $(\Gamma^E)_{da} (\Gamma^F)_{bc}$ ，并对所有旋量指标 $a,b,c,d$ 求和（即求迹）：

\sum_{a,b,c,d} (\Gamma^E)_{da} (\Gamma^A)_{ab} (\Gamma^F)_{bc} (\Gamma^B)_{cd} = \sum_{C,D} C^{AB}{}_{CD} \sum_{a,d} (\Gamma^C)_{ad} (\Gamma^E)_{da} \sum_{b,c} (\Gamma^D)_{cb} (\Gamma^F)_{bc}

左边化简为四个矩阵乘积的迹： $\text{tr}[\Gamma^E \Gamma^A \Gamma^F \Gamma^B]$ 。右边利用迹的定义化简为： $\sum_{C,D} C^{AB}{}_{CD} \text{tr}[\Gamma^C \Gamma^E] \text{tr}[\Gamma^D \Gamma^F]$ 。代入 (a) 中的归一化条件 $\text{tr}[\Gamma^C \Gamma^E] = 4\delta^{CE}$ ，右边变为：

\sum_{C,D} C^{AB}{}_{CD} (4\delta^{CE}) (4\delta^{DF}) = 16 C^{AB}{}_{EF}

将指标 $E, F$ 替换回 $C, D$ ，即可得到：

\boxed{ C^{AB}{}_{CD} = \frac{1}{16} \text{tr}[\Gamma^C \Gamma^A \Gamma^D \Gamma^B] }

(c) 1. 标量-标量乘积 $(\bar{u}_1 u_2)(\bar{u}_3 u_4)$ 此时 $\Gamma^A = 1, \Gamma^B = 1$ 。代入 (b) 中的公式：

C^{SS}{}_{CD} = \frac{1}{16} \text{tr}[\Gamma^C \cdot 1 \cdot \Gamma^D \cdot 1] = \frac{1}{16} \text{tr}[\Gamma^C \Gamma^D] = \frac{1}{4}\delta^{CD}

因此，Fierz 展开式为 $\frac{1}{4} \sum_C (\bar{u}_1 \Gamma^C u_4)(\bar{u}_3 \Gamma^C u_2)$ 。我们需要将 $\sum_C \Gamma^C \otimes \Gamma^C$ 重新组合为洛伦兹协变的形式。利用 (a) 中的基底：

S: $1 \otimes 1$
V: $\gamma^0 \otimes \gamma^0 + \sum_j (i\gamma^j) \otimes (i\gamma^j) = \gamma^0 \otimes \gamma^0 - \gamma^j \otimes \gamma^j = \gamma^\mu \otimes \gamma_\mu$
T: $\sum_j (i\sigma^{0j}) \otimes (i\sigma^{0j}) + \sum_{j<k} \sigma^{jk} \otimes \sigma^{jk} = -\sigma^{0j} \otimes \sigma^{0j} + \sigma^{jk} \otimes \sigma^{jk} = \frac{1}{2}\sigma^{\mu\nu} \otimes \sigma_{\mu\nu}$
A: $i\gamma^0\gamma^5 \otimes i\gamma^0\gamma^5 + \sum_j \gamma^j\gamma^5 \otimes \gamma^j\gamma^5 = -\gamma^0\gamma^5 \otimes \gamma^0\gamma^5 + \gamma^j\gamma^5 \otimes \gamma^j\gamma^5 = -\gamma^\mu\gamma^5 \otimes \gamma_\mu\gamma^5$
P: $\gamma^5 \otimes \gamma^5$

代入展开式，得到标量-标量的 Fierz 变换：

\boxed{ (\bar{u}_1 u_2)(\bar{u}_3 u_4) = \frac{1}{4} \left[ (\bar{u}_1 u_4)(\bar{u}_3 u_2) + (\bar{u}_1 \gamma^\mu u_4)(\bar{u}_3 \gamma_\mu u_2) + \frac{1}{2} (\bar{u}_1 \sigma^{\mu\nu} u_4)(\bar{u}_3 \sigma_{\mu\nu} u_2) - (\bar{u}_1 \gamma^\mu\gamma^5 u_4)(\bar{u}_3 \gamma_\mu\gamma^5 u_2) + (\bar{u}_1 \gamma^5 u_4)(\bar{u}_3 \gamma^5 u_2) \right] }

2. 矢量-矢量乘积 $(\bar{u}_1 \gamma^\mu u_2)(\bar{u}_3 \gamma_\mu u_4)$ 我们需要对 $\Gamma^A = \gamma^\mu, \Gamma^B = \gamma_\mu$ 求和。对应的系数为：

C^V_{CD} = \frac{1}{16} \sum_\mu \text{tr}[\Gamma^C \gamma^\mu \Gamma^D \gamma_\mu]

利用 Dirac 矩阵的收缩恒等式，我们计算 $\sum_\mu \gamma^\mu \Gamma^D \gamma_\mu$ 对于五种协变类型的取值：

若 $\Gamma^D = 1$ (S): $\gamma^\mu 1 \gamma_\mu = 4 \implies C^V_{CS} = \frac{1}{16}\text{tr}[\Gamma^C (4)] = \delta_{CS}$
若 $\Gamma^D \in V$ : $\gamma^\mu \gamma^\nu \gamma_\mu = -2\gamma^\nu \implies C^V_{CV} = \frac{1}{16}\text{tr}[\Gamma^C (-2\Gamma^D)] = -\frac{1}{2}\delta_{CV}$
若 $\Gamma^D \in T$ : $\gamma^\mu \sigma^{\nu\rho} \gamma_\mu = 0 \implies C^V_{CT} = 0$
若 $\Gamma^D \in A$ : $\gamma^\mu \gamma^\nu\gamma^5 \gamma_\mu = 2\gamma^\nu\gamma^5 \implies C^V_{CA} = \frac{1}{16}\text{tr}[\Gamma^C (2\Gamma^D)] = \frac{1}{2}\delta_{CA}$
若 $\Gamma^D = \gamma^5$ (P): $\gamma^\mu \gamma^5 \gamma_\mu = -4\gamma^5 \implies C^V_{CP} = \frac{1}{16}\text{tr}[\Gamma^C (-4\gamma^5)] = -\delta_{CP}$

将这些系数乘上对应的协变张量组合 $\sum_{C \in \text{type}} \Gamma^C \otimes \Gamma^C$ （已在标量-标量部分求出），我们得到：

S 项: $1 \cdot (1 \otimes 1)$
V 项: $-\frac{1}{2} \cdot (\gamma^\mu \otimes \gamma_\mu)$
T 项: $0$
A 项: $\frac{1}{2} \cdot (-\gamma^\mu\gamma^5 \otimes \gamma_\mu\gamma^5) = -\frac{1}{2} (\gamma^\mu\gamma^5 \otimes \gamma_\mu\gamma^5)$
P 项: $-1 \cdot (\gamma^5 \otimes \gamma^5)$

组合起来，得到矢量-矢量的 Fierz 变换：

\boxed{ (\bar{u}_1 \gamma^\mu u_2)(\bar{u}_3 \gamma_\mu u_4) = (\bar{u}_1 u_4)(\bar{u}_3 u_2) - \frac{1}{2} (\bar{u}_1 \gamma^\mu u_4)(\bar{u}_3 \gamma_\mu u_2) - \frac{1}{2} (\bar{u}_1 \gamma^\mu\gamma^5 u_4)(\bar{u}_3 \gamma_\mu\gamma^5 u_2) - (\bar{u}_1 \gamma^5 u_4)(\bar{u}_3 \gamma^5 u_2) }

3.7

Problem 3.7

peskinChapter 3

习题 3.7

来源: 第3章, PDF第75,76页

3.7 This problem concerns the discrete symmetries $P, C$ , and $T$ .

(a) Compute the transformation properties under $P, C$ , and $T$ of the antisymmetric tensor fermion bilinears, $\bar{\psi} \sigma^{\mu\nu} \psi$ , with $\sigma^{\mu\nu} = \frac{i}{2} [\gamma^\mu, \gamma^\nu]$ . This completes the table of the transformation properties of bilinears at the end of the chapter.

(b) Let $\phi(x)$ be a complex-valued Klein-Gordon field, such as we considered in Problem 2.2. Find unitary operators $P, C$ and an antiunitary operator $T$ (all defined

in terms of their action on the annihilation operators $a_{\mathbf{p}}$ and $b_{\mathbf{p}}$ for the Klein-Gordon particles and antiparticles) that give the following transformations of the Klein-Gordon field:

\begin{aligned} P \phi(t, \mathbf{x}) P &= \phi(t, -\mathbf{x}); \\ T \phi(t, \mathbf{x}) T &= \phi(-t, \mathbf{x}); \\ C \phi(t, \mathbf{x}) C &= \phi^*(t, \mathbf{x}). \end{aligned}

Find the transformation properties of the components of the current

J^\mu = i(\phi^* \partial^\mu \phi - \partial^\mu \phi^* \phi)

under $P$ , $C$ , and $T$ .

(c) Show that any Hermitian Lorentz-scalar local operator built from $\psi(x)$ , $\phi(x)$ , and their conjugates has $CPT = +1$ .

习题 3.7 - 解答

(a) 反对称张量双线性型 $\bar{\psi} \sigma^{\mu\nu} \psi$ 在 $P, C, T$ 下的变换

定义 $\sigma^{\mu\nu} = \frac{i}{2}[\gamma^\mu, \gamma^\nu]$ 。我们需要利用狄拉克场在离散对称性下的变换规则（采用 Peskin & Schroeder 约定）：

宇称 $P$ : $P \psi(t, \mathbf{x}) P = \gamma^0 \psi(t, -\mathbf{x})$
时间反演 $T$ : $T \psi(t, \mathbf{x}) T = -\gamma^1 \gamma^3 \psi(-t, \mathbf{x})$
电荷共轭 $C$ : $C \psi(x) C = -i \gamma^2 \psi^*(x)$

1. 宇称 $P$ 变换：

P \bar{\psi} \sigma^{\mu\nu} \psi P = \bar{\psi} \gamma^0 \sigma^{\mu\nu} \gamma^0 \psi

利用 $\gamma^0 \gamma^\mu \gamma^0 = \gamma_\mu = (-1)^\mu \gamma^\mu$ （其中 $(-1)^\mu$ 表示 $\mu=0$ 时为 $+1$ ， $\mu=i$ 时为 $-1$ ），可得：

\gamma^0 \sigma^{\mu\nu} \gamma^0 = \frac{i}{2} [\gamma^0 \gamma^\mu \gamma^0, \gamma^0 \gamma^\nu \gamma^0] = \sigma_{\mu\nu} = (-1)^\mu (-1)^\nu \sigma^{\mu\nu}

因此：

\boxed{P \bar{\psi} \sigma^{\mu\nu} \psi P = (-1)^\mu (-1)^\nu \bar{\psi} \sigma^{\mu\nu} \psi}

具体而言， $\bar{\psi} \sigma^{0i} \psi \to -\bar{\psi} \sigma^{0i} \psi$ ， $\bar{\psi} \sigma^{ij} \psi \to +\bar{\psi} \sigma^{ij} \psi$ 。

2. 时间反演 $T$ 变换： 注意 $T$ 是反幺正算符，作用时会使 $i \to -i$ 且复共轭矩阵。

T \bar{\psi} \sigma^{\mu\nu} \psi T = \bar{\psi} (\gamma^1 \gamma^3) (\sigma^{\mu\nu})^* (\gamma^1 \gamma^3) \psi

利用 $\gamma^1 \gamma^3 (\gamma^\mu)^* \gamma^1 \gamma^3 = \gamma_\mu$ ，我们有：

\gamma^1 \gamma^3 (\sigma^{\mu\nu})^* \gamma^1 \gamma^3 = \gamma^1 \gamma^3 \left( -\frac{i}{2} [(\gamma^\mu)^*, (\gamma^\nu)^*] \right) \gamma^1 \gamma^3 = -\frac{i}{2} [\gamma_\mu, \gamma_\nu] = -\sigma_{\mu\nu}

因此：

\boxed{T \bar{\psi} \sigma^{\mu\nu} \psi T = -(-1)^\mu (-1)^\nu \bar{\psi} \sigma^{\mu\nu} \psi}

具体而言， $\bar{\psi} \sigma^{0i} \psi \to +\bar{\psi} \sigma^{0i} \psi$ ， $\bar{\psi} \sigma^{ij} \psi \to -\bar{\psi} \sigma^{ij} \psi$ 。

3. 电荷共轭 $C$ 变换：

C \bar{\psi} \sigma^{\mu\nu} \psi C = \psi^T \gamma^0 \gamma^2 \sigma^{\mu\nu} \gamma^2 \psi^*

将费米子算符交换位置会产生一个负号，并取转置：

= - \bar{\psi} \gamma^2 (\sigma^{\mu\nu})^T \gamma^2 \psi

利用 $\gamma^2 (\gamma^\mu)^T \gamma^2 = -\gamma^\mu$ ，可得：

\gamma^2 (\sigma^{\mu\nu})^T \gamma^2 = \gamma^2 \left( \frac{i}{2} [(\gamma^\nu)^T, (\gamma^\mu)^T] \right) \gamma^2 = \frac{i}{2} [-\gamma^\nu, -\gamma^\mu] = -\sigma^{\mu\nu}

因此：

\boxed{C \bar{\psi} \sigma^{\mu\nu} \psi C = - \bar{\psi} \sigma^{\mu\nu} \psi}

(b) 复标量场 $\phi(x)$ 及其流 $J^\mu$ 的 $P, C, T$ 变换

复标量场的模式展开为：

\phi(x) = \int \frac{d^3p}{(2\pi)^3 \sqrt{2E_{\mathbf{p}}}} \left( a_{\mathbf{p}} e^{-ip\cdot x} + b_{\mathbf{p}}^\dagger e^{ip\cdot x} \right)

1. 算符 $a_{\mathbf{p}}, b_{\mathbf{p}}$ 的变换：

宇称 $P$ （幺正）：要求 $P \phi(t, \mathbf{x}) P = \phi(t, -\mathbf{x})$ 。由于 $e^{-i(Et - \mathbf{p}\cdot\mathbf{x})} \to e^{-i(Et + \mathbf{p}\cdot\mathbf{x})} = e^{-i(Et - (-\mathbf{p})\cdot\mathbf{x})}$ ，动量反向，故： $\boxed{P a_{\mathbf{p}} P = a_{-\mathbf{p}}, \quad P b_{\mathbf{p}} P = b_{-\mathbf{p}}}$
时间反演 $T$ （反幺正）：要求 $T \phi(t, \mathbf{x}) T = \phi(-t, \mathbf{x})$ 。反幺正性使得 $e^{-i(Et - \mathbf{p}\cdot\mathbf{x})} \to e^{i(Et - \mathbf{p}\cdot\mathbf{x})} = e^{-i(E(-t) - (-\mathbf{p})\cdot\mathbf{x})}$ ，故： $\boxed{T a_{\mathbf{p}} T = a_{-\mathbf{p}}, \quad T b_{\mathbf{p}} T = b_{-\mathbf{p}}}$
电荷共轭 $C$ （幺正）：要求 $C \phi(x) C = \phi^*(x)$ 。这直接交换了粒子与反粒子的湮灭/产生算符： $\boxed{C a_{\mathbf{p}} C = b_{\mathbf{p}}, \quad C b_{\mathbf{p}} C = a_{\mathbf{p}}}$

2. 流 $J^\mu = i(\phi^* \partial^\mu \phi - \partial^\mu \phi^* \phi)$ 的变换：

宇称 $P$ ： $\partial^\mu = (\partial_t, -\nabla) \xrightarrow{P} (\partial_t, \nabla) = \partial_\mu$ 。 $P J^\mu(t, \mathbf{x}) P = i(\phi^* \partial_\mu \phi - \partial_\mu \phi^* \phi)_{(t, -\mathbf{x})} = J_\mu(t, -\mathbf{x})$ $\boxed{P J^\mu(t, \mathbf{x}) P = (-1)^\mu J^\mu(t, -\mathbf{x})}$
时间反演 $T$ ： $T$ 是反幺正的， $T i T = -i$ 。导数 $\partial^\mu = (\partial_t, -\nabla) \xrightarrow{T} (-\partial_t, -\nabla) = -\partial_\mu$ 。 $T J^\mu(t, \mathbf{x}) T = -i \left( \phi (-\partial_\mu) \phi^* - (-\partial_\mu) \phi \phi^* \right)_{(-t, \mathbf{x})} = i(\phi^* \partial_\mu \phi - \partial_\mu \phi^* \phi)_{(-t, \mathbf{x})} = J_\mu(-t, \mathbf{x})$ $\boxed{T J^\mu(t, \mathbf{x}) T = (-1)^\mu J^\mu(-t, \mathbf{x})}$
电荷共轭 $C$ ： $C J^\mu(x) C = i(\phi \partial^\mu \phi^* - \partial^\mu \phi \phi^*) = -i(\phi^* \partial^\mu \phi - \partial^\mu \phi^* \phi) = -J^\mu(x)$ $\boxed{C J^\mu(x) C = -J^\mu(x)}$

(c) 证明任何厄米洛伦兹标量局域算符具有 $CPT = +1$

设 $\mathcal{O}(x)$ 是由 $\psi, \phi$ 及其共轭构成的厄米洛伦兹标量局域算符。在 $CPT$ 联合作用下，时空坐标反演 $x \to -x$ 。基本场及其导数的 $CPT$ 变换性质为（忽略可被双线性型吸收的全局相位）：

\begin{aligned} CPT \phi(x) (CPT)^{-1} &= \phi^*(-x) \\ CPT \psi(x) (CPT)^{-1} &= -i \gamma^5 \psi^*(-x) \\ CPT \partial_\mu (CPT)^{-1} &= -\partial_\mu \end{aligned}

对于任何由这些场构成的洛伦兹张量算符，其 $CPT$ 变换会给出 $(-1)^n$ 乘以其厄米共轭在 $-x$ 处的值，其中 $n$ 是未缩并的洛伦兹指标数目。由于 $\mathcal{O}(x)$ 是洛伦兹标量，其所有洛伦兹指标均已缩并，因此 $n=0$ ，没有额外的负号产生：

CPT \mathcal{O}(x) (CPT)^{-1} = \mathcal{O}^\dagger(-x)

又因为 $\mathcal{O}(x)$ 是厄米算符，即 $\mathcal{O}^\dagger(x) = \mathcal{O}(x)$ ，所以：

CPT \mathcal{O}(x) (CPT)^{-1} = \mathcal{O}(-x)

在作用量 $S = \int d^4x \mathcal{O}(x)$ 中，积分测度 $d^4x$ 在 $x \to -x$ 下不变，因此：

CPT \left( \int d^4x \mathcal{O}(x) \right) (CPT)^{-1} = \int d^4x \mathcal{O}(-x) = \int d^4x \mathcal{O}(x)

这表明该算符在 $CPT$ 变换下是完全不变的，即：

\boxed{CPT = +1}

3.8

Problem 3.8

peskinChapter 3

习题 3.8

来源: 第3章, PDF第76页

3.8 Bound states. Two spin-1/2 particles can combine to a state of total spin either 0 or 1. The wavefunctions for these states are odd and even, respectively, under the interchange of the two spins.

(a) Use this information to compute the quantum numbers under $P$ and $C$ of all electron-positron bound states with $S$ , $P$ , or $D$ wavefunctions.

(b) Since the electron-photon coupling is given by the Hamiltonian

\Delta H = \int d^3x \ e A_\mu j^\mu,

where $j^\mu$ is the electric current, electrodynamics is invariant to $P$ and $C$ if the components of the vector potential have the same $P$ and $C$ parity as the corresponding components of $j^\mu$ . Show that this implies the following surprising fact: The spin-0 ground state of positronium can decay to 2 photons, but the spin-1 ground state must decay to 3 photons. Find the selection rules for the annihilation of higher positronium states, and for 1-photon transitions between positronium levels.

习题 3.8 - 解答

(a)

正负电子偶素（positronium）是由电子（ $e^-$ ）和正电子（ $e^+$ ）组成的束缚态。由于电子和正电子是费米子，系统的总波函数在交换这两个粒子时必须是反对称的。交换操作可以分解为空间交换、自旋交换和电荷共轭（粒子-反粒子交换）：

空间交换：引入因子 $(-1)^L$ ，其中 $L$ 是轨道角动量量子数。
自旋交换：两个自旋 $1/2$ 粒子组成总自旋 $S$ 的态。当 $S=0$ （单态）时，自旋波函数是反对称的；当 $S=1$ （三重态）时，自旋波函数是对称的。因此自旋交换引入因子 $(-1)^{S+1}$ 。
电荷共轭：引入因子 $C$ 。

总交换对称性要求： $(-1)^L (-1)^{S+1} C = -1$ 由此可得电荷共轭量子数 $C$ 为： $C = (-1)^{L+S}$

对于宇称 $P$ ，由于费米子和反费米子具有相反的内禀宇称，系统的总宇称由轨道角动量和内禀宇称共同决定： $P = (P_{e^-})(P_{e^+})(-1)^L = (+1)(-1)(-1)^L = (-1)^{L+1}$

根据上述公式，我们可以计算 $S$ ( $L=0$ ), $P$ ( $L=1$ ), $D$ ( $L=2$ ) 波函数的 $P$ 和 $C$ 量子数。通常用光谱学符号 $^{2S+1}L_J$ 表示状态：

$S$ 波 ( $L=0$ )：
- $S=0$ ( $^1S_0$ ): $\boxed{P = -1, \ C = +1}$
- $S=1$ ( $^3S_1$ ): $\boxed{P = -1, \ C = -1}$
$P$ 波 ( $L=1$ )：
- $S=0$ ( $^1P_1$ ): $\boxed{P = +1, \ C = -1}$
- $S=1$ ( $^3P_J$ ): $\boxed{P = +1, \ C = +1}$
$D$ 波 ( $L=2$ )：
- $S=0$ ( $^1D_2$ ): $\boxed{P = -1, \ C = +1}$
- $S=1$ ( $^3D_J$ ): $\boxed{P = -1, \ C = -1}$

(b)

首先确定光子的 $C$ 宇称。电磁相互作用哈密顿量 $\Delta H = \int d^3x \ e A_\mu j^\mu$ 在电荷共轭变换下必须是不变的。由于电磁流 $j^\mu = \bar{\psi} \gamma^\mu \psi$ 在 $C$ 变换下反号（ $C j^\mu C^{-1} = -j^\mu$ ），为了保持 $\Delta H$ 不变，电磁势 $A_\mu$ 在 $C$ 变换下也必须反号： $C A_\mu C^{-1} = -A_\mu$ 因此，单光子态的 $C$ 宇称为 $-1$ 。由 $n$ 个光子组成的末态的 $C$ 宇称为： $C_{n\gamma} = (-1)^n$

基态衰变分析： 正负电子偶素的基态是 $S$ 波 ( $L=0$ )。

自旋为0的基态（ $^1S_0$ ）：由 (a) 知其 $C = +1$ 。由于电磁衰变过程中 $C$ 宇称守恒，末态光子数 $n$ 必须满足 $(-1)^n = +1$ ，即 $n$ 必须为偶数。由于能量-动量守恒禁止单光子湮灭，因此它最少衰变为2个光子。
自旋为1的基态（ $^3S_1$ ）：由 (a) 知其 $C = -1$ 。同理，末态光子数 $n$ 必须满足 $(-1)^n = -1$ ，即 $n$ 必须为奇数。由于单光子湮灭被运动学禁止，因此它最少衰变为3个光子。

更高态的湮灭选择定则： 对于任意正负电子偶素态，其湮灭为 $n$ 个光子的过程必须满足 $C$ 宇称守恒： $C = (-1)^{L+S} = (-1)^n$ 因此，湮灭的选择定则为： $\boxed{ \text{若 } L+S \text{ 为偶数，衰变为偶数个光子（} n \ge 2 \text{）；若 } L+S \text{ 为奇数，衰变为奇数个光子（} n \ge 3 \text{）} }$

单光子跃迁的选择定则： 对于正负电子偶素能级之间的单光子跃迁（ $A \to B + \gamma$ ）， $C$ 宇称守恒要求： $C_A = C_B \cdot C_\gamma \implies C_A = -C_B$ 代入 $C = (-1)^{L+S}$ ，得到： $(-1)^{L_A+S_A} = -(-1)^{L_B+S_B}$ 这意味着初态和末态的 $L+S$ 必须奇偶性不同，即 $\Delta(L+S)$ 必须为奇数。结合宇称守恒 $P_A = P_B \cdot P_\gamma$ 和角动量守恒，可以得到具体的跃迁定则：

电偶极（E1）跃迁：光子携带 $J^P = 1^-$ ，要求 $P_A = -P_B$ ，即 $\Delta L$ 为奇数。结合 $C_A = -C_B$ ，要求 $\Delta S = 0$ 。
磁偶极（M1）跃迁：光子携带 $J^P = 1^+$ ，要求 $P_A = P_B$ ，即 $\Delta L$ 为偶数（通常 $\Delta L = 0$ ）。结合 $C_A = -C_B$ ，要求 $\Delta S = 1$ 。

总结单光子跃迁的选择定则为： $\boxed{ C_A = -C_B \implies \Delta(L+S) \text{ 为奇数。具体为：E1 跃迁 } \Delta L = \text{奇数}, \Delta S = 0 \text{；M1 跃迁 } \Delta L = \text{偶数}, \Delta S = 1 }$

Problem 3.1

习题 3.1

习题 3.1 - 解答

(a) 洛伦兹群生成元的对易关系

(b) 旋量表示的变换法则

(c) (12,12)(\frac{1}{2}, \frac{1}{2})(21​,21​) 表示与四维矢量

Problem 3.2

习题 3.2

习题 3.2 - 解答

Problem 3.3

习题 3.3

习题 3.3 - 解答

(a) 证明 ̸k0uR0=0\not{k}_0 u_{R0} = 0k0​uR0​=0 以及 p̸uL(p)=p̸uR(p)=0\not{p} u_L(p) = \not{p} u_R(p) = 0puL​(p)=puR​(p)=0

(b) 显式构造 uL0,uR0,uL(p),uR(p)u_{L0}, u_{R0}, u_L(p), u_R(p)uL0​,uR0​,uL​(p),uR​(p)

(c) 计算旋量乘积 s(p1,p2)s(p_1, p_2)s(p1​,p2​) 和 t(p1,p2)t(p_1, p_2)t(p1​,p2​)

Problem 3.4

习题 3.4

习题 3.4 - 解答

Problem 3.5

习题 3.5

习题 3.5 - 解答

Problem 3.6

习题 3.6

习题 3.6 - 解答

Problem 3.7

习题 3.7

习题 3.7 - 解答

Problem 3.8

习题 3.8

习题 3.8 - 解答

(c) $(\frac{1}{2}, \frac{1}{2})$ 表示与四维矢量

(a) 证明 $\not{k}_0 u_{R0} = 0$ 以及 $\not{p} u_L(p) = \not{p} u_R(p) = 0$

(b) 显式构造 $u_{L0}, u_{R0}, u_L(p), u_R(p)$

(c) 计算旋量乘积 $s(p_1, p_2)$ 和 $t(p_1, p_2)$