逐字来源、作者主张与候选译文分别保留。工程核验不批准 OCR、译文或数学正确性。
§13 自旋–统计回顾与反对易关系
作者据自旋–统计定理要求旋量产生和湮灭算符反对易。
条件与约定：旋量场；关系中的动量归一化和自旋指标按 Schwartz 保留。
逐字作者来源与版本
The creation and annihilation operators for spinors must anticommute by the spin-statistics theorem:$$
\left\{a _ {p} ^ {s \dagger}, a _ {q} ^ {s ^ {\prime} \dagger} \right\} = \left\{a _ {p} ^ {s}, a _ {q} ^ {s ^ {\prime}} \right\} = \left\{b _ {p} ^ {s \dagger}, b _ {q} ^ {s ^ {\prime} \dagger} \right\} = \left\{b _ {p} ^ {s}, b _ {q} ^ {s ^ {\prime}} \right\} = 0 \tag {13.4}
$$and$$
\left\{a _ {p} ^ {s}, a _ {q} ^ {s ^ {\prime} \dagger} \right\} = \left\{b _ {p} ^ {s}, b _ {q} ^ {s ^ {\prime} \dagger} \right\} = \delta_ {s s ^ {\prime}} (2 \pi) ^ {3} \delta^ {3} (p - q). \tag {13.5}
$$PDF 文件页：118；OCR 未因此获得逐页核验。
使用前置陈述：第12章自旋–统计陈述（既有种子）。作者有明确编号回指。
证据：作者编号回指
The creation and annihilation operators for spinors must anticommute by the spin-statistics theorem:
§13 费米子负号的自旋–统计回指
额外负号归因于自旋统计，作者说稍后解释。
条件与约定：这是回指和后文承诺，不在此处补证明。
逐字作者来源与版本
The extra minus sign is due to spin-statistics, as will be explained shortly. Contracting all the spinor indices and replacing $p_1^\mu$ by $p^\mu + k^\mu$ and $p_2^\mu$ by $k^\mu$ :PDF 文件页：121；OCR 未因此获得逐页核验。
§13 无质量电子极限的螺旋度守恒
三枚伽马矩阵的耦合保持左右手态；作者限定无质量电子极限下左右手态解耦。
条件与约定：不能把有质量粒子的手征性与螺旋度普遍等同。
逐字作者来源与版本
In particular, our interaction $\mathcal{M}_t$ has three $\gamma$ -matrices, so it couples $RR$ and $LL$ . Thus,$$
\mathcal {M} _ {t} = - \frac {e ^ {2}}{\omega \phi} \left[ \bar {u} _ {L} \left(p _ {3}\right) \epsilon_ {1} / k \epsilon_ {4} ^ {*} u _ {L} \left(p _ {2}\right) + \bar {u} _ {R} \left(p _ {3}\right) \epsilon_ {1} / k \epsilon_ {4} ^ {*} u _ {R} \left(p _ {2}\right) \right], \tag {13.157}
$$which is helicity conserving. So, $u(p_2)$ and $\bar{u}(p_3)$ should be either both right-handed or both left-handed. This is consistent with a general property of QED, that in the limit of a massless electron, the left- and right-handed states completely decouple.PDF 文件页：139；OCR 未因此获得逐页核验。
§13 光子螺旋度与左右手约定检查
作者用光子和电子的自旋沿动量投影检查左右手符号一致。
条件与约定：动量沿 z 轴，光子本征值 ±1，电子 ±1/2；保留 E 与角动量矩阵约定。
逐字作者来源与版本
For the photons, we need to use the helicity eigenstates:$$
\epsilon_ {R} ^ {\mu} = \frac {1}{\sqrt {2}} (0, 1, i, 0), \quad \epsilon_ {L} ^ {\mu} = \frac {1}{\sqrt {2}} (0, 1, - i, 0). \tag {13.159}
$$To see that the convention for "left" and "right" is being used consistently, it is easy to check that $\frac{\vec{S} \cdot \vec{p}}{E} \epsilon_{L} = -\epsilon_{L}$ and $\frac{\vec{S} \cdot \vec{p}}{E} u_{L} = -\frac{1}{2} u_{L}$ using $p^{\mu} = (E, 0, 0, p_{z})$ and $S_{z} = S_{12}$ or $S_{z} = V_{12}$ in Eqs. (10.117) and (10.113) respectively.PDF 文件页：139；OCR 未因此获得逐页核验。