Problem 4.1

peskinChapter 4

习题 4.1

来源: 第4章, PDF第126,127页

4.1 Let us return to the problem of the creation of Klein-Gordon particles by a classical source. Recall from Chapter 2 that this process can be described by the Hamiltonian

H = H_0 + \int d^3x (-j(t, \mathbf{x})\phi(x)),

where $H_0$ is the free Klein-Gordon Hamiltonian, $\phi(x)$ is the Klein-Gordon field, and $j(x)$ is a c-number scalar function. We found that, if the system is in the vacuum state before the source is turned on, the source will create a mean number of particles

\langle N \rangle = \int \frac{d^3p}{(2\pi)^3} \frac{1}{2E_{\mathbf{p}}} |\tilde{j}(p)|^2.

In this problem we will verify that statement, and extract more detailed information, by using a perturbation expansion in the strength of the source.

(a) Show that the probability that the source creates no particles is given by

P(0) = \left| \langle 0 | T \left\{ \exp \left[ i \int d^4x j(x) \phi_I(x) \right] \right\} | 0 \rangle \right|^2.

(b) Evaluate the term in $P(0)$ of order $j^2$ , and show that $P(0) = 1 - \lambda + \mathcal{O}(j^4)$ , where $\lambda$ equals the expression given above for $\langle N \rangle$ .

(c) Represent the term computed in part (b) as a Feynman diagram. Now represent the whole perturbation series for $P(0)$ in terms of Feynman diagrams. Show that this series exponentiates, so that it can be summed exactly: $P(0) = \exp(-\lambda)$ .

(d) Compute the probability that the source creates one particle of momentum $k$ . Perform this computation first to $\mathcal{O}(j)$ and then to all orders, using the trick of part (c) to sum the series.

(e) Show that the probability of producing $n$ particles is given by

P(n) = (1/n!) \lambda^n \exp(-\lambda).

This is a Poisson distribution.

(f) Prove the following facts about the Poisson distribution:

\sum_{n=0}^{\infty} P(n) = 1; \quad \langle N \rangle = \sum_{n=0}^{\infty} n P(n) = \lambda.

The first identity says that the $P(n)$ 's are properly normalized probabilities, while the second confirms our proposal for $\langle N \rangle$ . Compute the mean square fluctuation $\langle (N - \langle N \rangle)^2 \rangle$ .

习题 4.1 - 解答

习题 4.1 分析与解答

(a) 证明源不产生粒子的概率为 $P(0) = \left| \langle 0 | T \left\{ \exp \left[ i \int d^4x j(x) \phi_I(x) \right] \right\} | 0 \rangle \right|^2$

分析与推导： 在相互作用绘景中，系统的相互作用哈密顿量为 $H_I(t) = -\int d^3x j(t, \mathbf{x})\phi_I(x)$ 。系统从 $t \to -\infty$ 演化到 $t \to \infty$ 的时间演化算符为：

U_I(\infty, -\infty) = T \exp\left( -i \int_{-\infty}^{\infty} dt H_I(t) \right) = T \exp\left( i \int d^4x j(x) \phi_I(x) \right)

假设系统在源开启前（ $t \to -\infty$ ）处于真空态 $|0\rangle$ ，那么在源关闭后（ $t \to \infty$ ），系统仍然保持在真空态（即不产生任何粒子）的跃迁振幅为 $\langle 0 | U_I(\infty, -\infty) | 0 \rangle$ 。概率是跃迁振幅的绝对值平方，因此：

\boxed{ P(0) = \left| \langle 0 | T \left\{ \exp \left[ i \int d^4x j(x) \phi_I(x) \right] \right\} | 0 \rangle \right|^2 }

(b) 计算 $P(0)$ 中 $\mathcal{O}(j^2)$ 的项，并证明 $P(0) = 1 - \lambda + \mathcal{O}(j^4)$

分析与推导： 将时间演化算符展开到 $j$ 的二阶：

U_I = 1 + i \int d^4x j(x) \phi_I(x) - \frac{1}{2} \int d^4x d^4y j(x) j(y) T\{\phi_I(x) \phi_I(y)\} + \mathcal{O}(j^3)

取真空期望值，由于 $\langle 0 | \phi_I(x) | 0 \rangle = 0$ ，线性项消失。二阶项包含费曼传播子 $D_F(x-y) = \langle 0 | T\{\phi_I(x) \phi_I(y)\} | 0 \rangle$ ：

\langle 0 | U_I | 0 \rangle = 1 - \frac{1}{2} \int d^4x d^4y j(x) j(y) D_F(x-y) + \mathcal{O}(j^4) \equiv 1 + A + \mathcal{O}(j^4)

其中 $A = - \frac{1}{2} \int d^4x d^4y j(x) j(y) D_F(x-y)$ 。概率 $P(0) = |1 + A|^2 = 1 + 2\text{Re}(A) + \mathcal{O}(j^4)$ 。我们来计算 $A$ ：

A = - \frac{1}{2} \int d^4x d^4y j(x) j(y) \int \frac{d^4p}{(2\pi)^4} \frac{i}{p^2 - m^2 + i\epsilon} e^{-ip \cdot (x-y)} = - \frac{i}{2} \int \frac{d^4p}{(2\pi)^4} \frac{\tilde{j}(p)\tilde{j}(-p)}{p^2 - m^2 + i\epsilon}

因为 $j(x)$ 是实函数，所以 $\tilde{j}(-p) = \tilde{j}^*(p)$ ，从而 $\tilde{j}(p)\tilde{j}(-p) = |\tilde{j}(p)|^2$ 。利用恒等式 $\frac{1}{x + i\epsilon} = \mathcal{P}\frac{1}{x} - i\pi \delta(x)$ ，提取 $A$ 的实部：

\text{Re}(A) = - \frac{1}{2} \int \frac{d^4p}{(2\pi)^4} |\tilde{j}(p)|^2 \pi \delta(p^2 - m^2)

利用 $\delta(p^2 - m^2) = \frac{1}{2E_{\mathbf{p}}} \left[ \delta(p^0 - E_{\mathbf{p}}) + \delta(p^0 + E_{\mathbf{p}}) \right]$ ，对 $p^0$ 积分：

\text{Re}(A) = - \frac{1}{4} \int \frac{d^3p}{(2\pi)^3} \frac{1}{2E_{\mathbf{p}}} \left( |\tilde{j}(E_{\mathbf{p}}, \mathbf{p})|^2 + |\tilde{j}(-E_{\mathbf{p}}, \mathbf{p})|^2 \right)

由于 $j(x)$ 为实， $|\tilde{j}(-E_{\mathbf{p}}, \mathbf{p})|^2 = |\tilde{j}(E_{\mathbf{p}}, -\mathbf{p})|^2$ 。在第二项中作变量代换 $\mathbf{p} \to -\mathbf{p}$ ，两项贡献相等，因此：

\text{Re}(A) = - \frac{1}{2} \int \frac{d^3p}{(2\pi)^3} \frac{1}{2E_{\mathbf{p}}} |\tilde{j}(E_{\mathbf{p}}, \mathbf{p})|^2 = - \frac{1}{2} \lambda

代入概率公式得到：

\boxed{ P(0) = 1 - \lambda + \mathcal{O}(j^4) }

(c) 用费曼图表示微扰级数，并证明 $P(0) = \exp(-\lambda)$

分析与推导： (b) 中计算的项 $A$ 可以用一个费曼图表示：两个源（用叉号表示）通过一条标量传播子内线连接。对于真空到真空的跃迁振幅 $\langle 0 | U_I | 0 \rangle$ ，完整的微扰级数是所有真空图的总和。由于相互作用哈密顿量对场 $\phi$ 是线性的，唯一连通的真空图就是 $A$ （两个源相连）。根据不连通图的指数化定理（Linked Cluster Theorem），所有真空图的总和等于连通真空图之和的指数：

\langle 0 | U_I | 0 \rangle = \exp(A)

因此，不产生粒子的概率为：

\boxed{ P(0) = |\exp(A)|^2 = \exp(A + A^*) = \exp(2\text{Re}(A)) = \exp(-\lambda) }

(d) 计算源产生一个动量为 $k$ 的粒子的概率

分析与推导： 产生一个动量为 $k$ 的粒子的跃迁振幅为 $\mathcal{M}(\mathbf{k}) = \langle \mathbf{k} | U_I | 0 \rangle$ 。 到 $\mathcal{O}(j)$ 阶：

U_I \approx 1 + i \int d^4x j(x) \phi_I(x)

利用 $\langle \mathbf{k} | \phi_I(x) | 0 \rangle = e^{ik \cdot x}$ （其中 $k^0 = E_{\mathbf{k}}$ ），振幅为：

\mathcal{M}(\mathbf{k}) = i \int d^4x j(x) e^{ik \cdot x} = i \tilde{j}(k)

对应的微分概率（相空间密度）为：

\boxed{ dP_1(\mathbf{k}) = \frac{d^3k}{(2\pi)^3 2E_{\mathbf{k}}} |\tilde{j}(k)|^2 }

到全阶： 利用 Wick 定理，时间演化算符可以写为正规序与真空图的乘积：

U_I = : \exp\left( i \int d^4x j(x) \phi_I(x) \right) : \exp(A)

由于单粒子态 $\langle \mathbf{k} |$ 只能与正规序中包含一个 $\phi$ 的项发生非零收缩，振幅变为：

\mathcal{M}(\mathbf{k}) = \langle \mathbf{k} | i \int d^4x j(x) \phi_I(x) | 0 \rangle \exp(A) = i \tilde{j}(k) \exp(A)

取绝对值平方并乘上相空间因子，得到全阶的微分概率：

\boxed{ dP_1(\mathbf{k}) = \frac{d^3k}{(2\pi)^3 2E_{\mathbf{k}}} |\tilde{j}(k)|^2 e^{-\lambda} }

(e) 证明产生 $n$ 个粒子的概率为 $P(n) = (1/n!) \lambda^n \exp(-\lambda)$

分析与推导： 产生 $n$ 个动量分别为 $k_1, \dots, k_n$ 的粒子的振幅为：

\mathcal{M}(\mathbf{k}_1, \dots, \mathbf{k}_n) = \langle \mathbf{k}_1, \dots, \mathbf{k}_n | : \exp\left( i \int d^4x j(x) \phi_I(x) \right) : | 0 \rangle \exp(A)

正规序展开中只有包含 $n$ 个 $\phi$ 场的项 $\frac{i^n}{n!} : (\int j \phi)^n :$ 有贡献。由于 $n$ 个全同玻色子态的对称性，收缩会产生 $n!$ 的组合因子，因此：

\mathcal{M}(\mathbf{k}_1, \dots, \mathbf{k}_n) = i^n \tilde{j}(k_1) \dots \tilde{j}(k_n) \exp(A)

产生 $n$ 个粒子的总概率需要对所有粒子的相空间进行积分，并除以 $n!$ （全同粒子相空间因子）：

P(n) = \frac{1}{n!} \int \frac{d^3k_1}{(2\pi)^3 2E_1} \dots \frac{d^3k_n}{(2\pi)^3 2E_n} |\mathcal{M}(\mathbf{k}_1, \dots, \mathbf{k}_n)|^2

将振幅代入并分离积分：

P(n) = \frac{1}{n!} \left( \int \frac{d^3k}{(2\pi)^3 2E_{\mathbf{k}}} |\tilde{j}(k)|^2 \right)^n |\exp(A)|^2

识别出括号内的积分为 $\lambda$ ，且 $|\exp(A)|^2 = \exp(-\lambda)$ ，即得泊松分布：

\boxed{ P(n) = \frac{1}{n!} \lambda^n \exp(-\lambda) }

(f) 证明泊松分布的性质并计算均方涨落

分析与推导：

归一化：

\sum_{n=0}^{\infty} P(n) = \sum_{n=0}^{\infty} \frac{\lambda^n}{n!} e^{-\lambda} = e^{\lambda} e^{-\lambda} = \boxed{ 1 }

平均粒子数 $\langle N \rangle$ ：

\langle N \rangle = \sum_{n=0}^{\infty} n P(n) = \sum_{n=1}^{\infty} n \frac{\lambda^n}{n!} e^{-\lambda} = \lambda e^{-\lambda} \sum_{n=1}^{\infty} \frac{\lambda^{n-1}}{(n-1)!} = \lambda e^{-\lambda} e^{\lambda} = \boxed{ \lambda }

均方涨落 $\langle (N - \langle N \rangle)^2 \rangle$ ： 首先计算 $\langle N^2 \rangle$ ：

\langle N^2 \rangle = \sum_{n=0}^{\infty} n^2 P(n) = \sum_{n=1}^{\infty} (n(n-1) + n) \frac{\lambda^n}{n!} e^{-\lambda}

= \lambda^2 e^{-\lambda} \sum_{n=2}^{\infty} \frac{\lambda^{n-2}}{(n-2)!} + \sum_{n=1}^{\infty} n \frac{\lambda^n}{n!} e^{-\lambda} = \lambda^2 + \lambda

因此，均方涨落为：

\langle (N - \langle N \rangle)^2 \rangle = \langle N^2 \rangle - \langle N \rangle^2 = (\lambda^2 + \lambda) - \lambda^2 = \boxed{ \lambda }

4.2

Problem 4.2

peskinChapter 4

习题 4.2

来源: 第4章, PDF第127页

4.2 Decay of a scalar particle. Consider the following Lagrangian, involving two real scalar fields $\Phi$ and $\phi$ :

\mathcal{L} = \frac{1}{2}(\partial_{\mu} \Phi)^2 - \frac{1}{2} M^2 \Phi^2 + \frac{1}{2}(\partial_{\mu} \phi)^2 - \frac{1}{2} m^2 \phi^2 - \mu \Phi \phi \phi.

The last term is an interaction that allows a $\Phi$ particle to decay into two $\phi$ 's, provided that $M > 2m$ . Assuming that this condition is met, calculate the lifetime of the $\Phi$ to lowest order in $\mu$ .

习题 4.2 - 解答

习题分析： 本题要求计算标量粒子 $\Phi$ 衰变为两个相同标量粒子 $\phi$ 的寿命（最低阶近似）。已知拉格朗日量中包含相互作用项 $\mathcal{L}_{\text{int}} = -\mu \Phi \phi^2$ 。我们需要通过费曼规则写出衰变的跃迁矩阵元 $\mathcal{M}$ ，然后利用两体衰变相空间公式计算衰变率 $\Gamma$ ，最后取倒数得到寿命 $\tau$ 。

解题过程：

确定费曼规则与矩阵元 从拉格朗日量中提取相互作用项：

\mathcal{L}_{\text{int}} = -\mu \Phi \phi \phi = -\mu \Phi \phi^2

在计算 $\Phi(p) \to \phi(k_1) + \phi(k_2)$ 的跃迁矩阵元时，由于末态有两个相同的 $\phi$ 场，算符 $\phi^2$ 作用在真空态上会产生两种不同的收缩方式（即产生 2 的组合因子）。因此，该顶点的费曼规则为 $-2i\mu$ 。由此可得最低阶的不变矩阵元 $\mathcal{M}$ 为：

i\mathcal{M} = -2i\mu \implies \mathcal{M} = -2\mu

矩阵元模的平方为：

|\mathcal{M}|^2 = 4\mu^2

两体衰变相空间积分 在 $\Phi$ 粒子的静止参考系中，其四维动量为 $p = (M, \mathbf{0})$ 。末态两个 $\phi$ 粒子的动量分别为 $k_1 = (E_1, \mathbf{k})$ 和 $k_2 = (E_2, -\mathbf{k})$ 。两体相空间积分定义为：

\int d\Pi_2 = \int \frac{d^3k_1}{(2\pi)^3 2E_1} \frac{d^3k_2}{(2\pi)^3 2E_2} (2\pi)^4 \delta^{(4)}(p - k_1 - k_2)

利用三维动量的 $\delta$ 函数积掉 $\mathbf{k}_2$ ，此时 $E_1 = E_2 = \sqrt{|\mathbf{k}|^2 + m^2}$ ：

\int d\Pi_2 = \frac{1}{(2\pi)^2} \int \frac{d^3k}{4E_1^2} \delta(M - 2E_1)

将体积元写为球坐标形式 $d^3k = |\mathbf{k}|^2 d|\mathbf{k}| d\Omega$ ，并利用色散关系 $E_1 dE_1 = |\mathbf{k}| d|\mathbf{k}|$ ，对立体角 $d\Omega$ 积分得到 $4\pi$ ：

\int d\Pi_2 = \frac{4\pi}{4\pi^2} \int \frac{|\mathbf{k}| E_1 dE_1}{4E_1^2} \delta(M - 2E_1) = \frac{1}{\pi} \int \frac{|\mathbf{k}|}{4E_1} \delta(M - 2E_1) dE_1

利用 $\delta$ 函数的性质 $\delta(M - 2E_1) = \frac{1}{2}\delta(E_1 - M/2)$ ，积分得到：

\int d\Pi_2 = \frac{1}{\pi} \frac{|\mathbf{k}|}{4(M/2)} \frac{1}{2} = \frac{|\mathbf{k}|}{4\pi M}

由能量守恒 $E_1 = M/2$ ，可求得末态粒子的动量大小：

|\mathbf{k}| = \sqrt{E_1^2 - m^2} = \sqrt{\frac{M^2}{4} - m^2} = \frac{M}{2}\sqrt{1 - \frac{4m^2}{M^2}}

代入相空间积分式中：

\int d\Pi_2 = \frac{1}{8\pi} \sqrt{1 - \frac{4m^2}{M^2}}

计算衰变率与寿命 对于 $1 \to 2$ 衰变过程，衰变率 $\Gamma$ 的一般公式为：

\Gamma = \frac{1}{2M} \frac{1}{S} \int |\mathcal{M}|^2 d\Pi_2

其中 $S$ 为末态全同粒子的统计对称因子。由于末态是两个相同的 $\phi$ 粒子，在对全立体角积分时会发生重复计数，故需要除以对称因子 $S = 2! = 2$ 。将前面求得的物理量代入公式：

\Gamma = \frac{1}{2M} \cdot \frac{1}{2} \cdot (4\mu^2) \cdot \frac{1}{8\pi} \sqrt{1 - \frac{4m^2}{M^2}} = \frac{\mu^2}{8\pi M} \sqrt{1 - \frac{4m^2}{M^2}}

粒子 $\Phi$ 的寿命 $\tau$ 是衰变率的倒数：

\tau = \frac{1}{\Gamma} = \frac{8\pi M}{\mu^2 \sqrt{1 - \frac{4m^2}{M^2}}}

最终结果为：

\boxed{\tau = \frac{8\pi M}{\mu^2 \sqrt{1 - \frac{4m^2}{M^2}}}}

4.3

Problem 4.3

peskinChapter 4

习题 4.3

来源: 第4章, PDF第127,128,129页

4.3 Linear sigma model. The interactions of pions at low energy can be described by a phenomenological model called the linear sigma model. Essentially, this model consists of $N$ real scalar fields coupled by a $\phi^4$ interaction that is symmetric under rotations of the $N$ fields. More specifically, let $\Phi^i(x)$ , $i = 1, \dots, N$ be a set of $N$ fields, governed by the Hamiltonian

H = \int d^3x \left( \frac{1}{2}(\Pi^i)^2 + \frac{1}{2}(\nabla \Phi^i)^2 + V(\Phi^2) \right),

where $(\Phi^i)^2 = \mathbf{\Phi} \cdot \mathbf{\Phi}$ , and

V(\Phi^2) = \frac{1}{2} m^2 (\Phi^i)^2 + \frac{\lambda}{4} ((\Phi^i)^2)^2

is a function symmetric under rotations of $\mathbf{\Phi}$ . For (classical) field configurations of $\Phi^i(x)$ that are constant in space and time, this term gives the only contribution to $H$ ; hence, $V$ is the field potential energy.

(What does this Hamiltonian have to do with the strong interactions? There are two types of light quarks, $u$ and $d$ . These quarks have identical strong interactions, but different masses. If these quarks are massless, the Hamiltonian of the strong interactions is invariant to unitary transformations of the 2-component object $(u, d)$ :

\begin{pmatrix} u \\ d \end{pmatrix} \rightarrow \exp(i \mathbf{\alpha} \cdot \mathbf{\sigma} / 2) \begin{pmatrix} u \\ d \end{pmatrix}.

This transformation is called an isospin rotation. If, in addition, the strong interactions are described by a vector "gluon" field (as is true in QCD), the strong interaction Hamiltonian is invariant to the isospin rotations done separately on the left-handed and right-handed components of the quark fields. Thus, the complete symmetry of QCD with two massless quarks is $SU(2) \times SU(2)$ . It happens that $SO(4)$ , the group of rotations in 4 dimensions, is isomorphic to $SU(2) \times SU(2)$ , so for $N = 4$ , the linear sigma model has the same symmetry group as the strong interactions.)

(a) Analyze the linear sigma model for $m^2 > 0$ by noticing that, for $\lambda = 0$ , the Hamiltonian given above is exactly $N$ copies of the Klein-Gordon Hamiltonian. We can then calculate scattering amplitudes as perturbation series in the parameter $\lambda$ . Show that the propagator is

Wick contraction formula for the propagator

where $D_F$ is the standard Klein-Gordon propagator for mass $m$ , and that there is one type of vertex given by

Feynman diagram vertex rule with indices

(That is, the vertex between two $\Phi^1$ s and two $\Phi^2$ s has the value $(-2i\lambda)$ ; that between four $\Phi^1$ s has the value $(-6i\lambda)$ .) Compute, to leading order in $\lambda$ , the differential cross sections $d\sigma/d\Omega$ , in the center-of-mass frame, for the scattering processes

\Phi^1 \Phi^2 \rightarrow \Phi^1 \Phi^2, \quad \Phi^1 \Phi^1 \rightarrow \Phi^2 \Phi^2, \quad \text{and} \quad \Phi^1 \Phi^1 \rightarrow \Phi^1 \Phi^1

as functions of the center-of-mass energy.

(b) Now consider the case $m^2 < 0$ : $m^2 = -\mu^2$ . In this case, $V$ has a local maximum, rather than a minimum, at $\Phi^i = 0$ . Since $V$ is a potential energy, this implies that the ground state of the theory is not near $\Phi^i = 0$ but rather is obtained by shifting $\Phi^i$ toward the minimum of $V$ . By rotational invariance, we can consider this shift to be in the $N$ th direction. Write, then,

\begin{aligned} \Phi^i(x) &= \pi^i(x), \quad i = 1, \dots, N - 1, \\ \Phi^N(x) &= v + \sigma(x), \end{aligned}

where $v$ is a constant chosen to minimize $V$ . (The notation $\pi^i$ suggests a pion field and should not be confused with a canonical momentum.) Show that, in these new coordinates (and substituting for $v$ its expression in terms of $\lambda$ and $\mu$ ), we have a theory of a massive $\sigma$ field and $N - 1$ massless pion fields, interacting through cubic and quartic potential energy terms which all become small as $\lambda \rightarrow 0$ . Construct the Feynman rules by assigning values to the propagators and vertices:

Feynman rules for propagators and vertices in the shifted sigma model

\pi^i(p_1) \pi^j(p_2) \rightarrow \pi^k(p_3) \pi^l(p_4)

to leading order in $\lambda$ . There are now four Feynman diagrams that contribute:

Four Feynman diagrams representing contributions to a scattering process at leading order in lambda, including s, t, and u channel exchanges and a four-point contact interaction.

Show that, at threshold ( $\mathbf{p}_i = 0$ ), these diagrams sum to zero. (Hint: It may be easiest to first consider the specific process $\pi^1 \pi^1 \rightarrow \pi^2 \pi^2$ , for which only the first and fourth diagrams are nonzero, before tackling the general case.) Show that, in the special case $N = 2$ (1 species of pion), the term of $\mathcal{O}(p^2)$ also cancels.

(d) Add to $V$ a symmetry-breaking term,

\Delta V = -a\Phi^N,

where $a$ is a (small) constant. (In QCD, a term of this form is produced if the $u$ and $d$ quarks have the same nonvanishing mass.) Find the new value of $v$ that minimizes $V$ , and work out the content of the theory about that point. Show that the pion acquires a mass such that $m_\pi^2 \sim a$ , and show that the pion scattering amplitude at threshold is now nonvanishing and also proportional to $a$ .

习题 4.3 - 解答

习题 4.3 分析与解答

(a) 对称相 ( $m^2 > 0$ ) 的线性 $\sigma$ 模型

1. 传播子与顶点规则

当 $\lambda = 0$ 时，哈密顿量描述了 $N$ 个独立的、质量为 $m$ 的 Klein-Gordon 场。自由场的拉格朗日密度为： $\mathcal{L}_0 = \frac{1}{2} \partial_\mu \Phi^i \partial^\mu \Phi^i - \frac{1}{2} m^2 \Phi^i \Phi^i$ 由于各分量场相互独立，只有同种场的缩并才非零。因此，传播子为： $\langle 0 | T \Phi^i(x) \Phi^j(y) | 0 \rangle = \delta^{ij} D_F(x-y)$ 其中 $D_F(x-y)$ 是标准的标量场传播子。

相互作用拉格朗日密度为： $\mathcal{L}_{\text{int}} = -V_{\text{int}} = -\frac{\lambda}{4} (\Phi^i \Phi^i)^2 = -\frac{\lambda}{4} \sum_{a,b} \Phi^a \Phi^a \Phi^b \Phi^b$ 为了得到四点顶点 $\Phi^i \Phi^j \Phi^k \Phi^l$ 的 Feynman 规则，我们对作用量求泛函导数： $V^{ijkl} = i \frac{\delta^4 \mathcal{L}_{\text{int}}}{\delta \Phi^i \delta \Phi^j \delta \Phi^k \delta \Phi^l}$ 依次求导： $\frac{\delta \mathcal{L}_{\text{int}}}{\delta \Phi^i} = -\lambda \Phi^i (\Phi^a \Phi^a)$ $\frac{\delta^2 \mathcal{L}_{\text{int}}}{\delta \Phi^j \delta \Phi^i} = -\lambda \left( \delta^{ij} \Phi^a \Phi^a + 2\Phi^i \Phi^j \right)$ $\frac{\delta^3 \mathcal{L}_{\text{int}}}{\delta \Phi^k \delta \Phi^j \delta \Phi^i} = -2\lambda \left( \delta^{ij} \Phi^k + \delta^{ik} \Phi^j + \delta^{jk} \Phi^i \right)$ $\frac{\delta^4 \mathcal{L}_{\text{int}}}{\delta \Phi^l \delta \Phi^k \delta \Phi^j \delta \Phi^i} = -2\lambda \left( \delta^{ij} \delta^{kl} + \delta^{ik} \delta^{jl} + \delta^{il} \delta^{jk} \right)$ 因此，顶点规则为： $\boxed{ -2i\lambda (\delta^{ij} \delta^{kl} + \delta^{ik} \delta^{jl} + \delta^{il} \delta^{jk}) }$

2. 散射截面计算

在质心系中，微分截面公式为 $\frac{d\sigma}{d\Omega} = \frac{|\mathcal{M}|^2}{64\pi^2 E_{\text{cm}}^2}$ 。

过程 $\Phi^1 \Phi^2 \rightarrow \Phi^1 \Phi^2$ : 设初态指标为 $i=1, j=2$ ，末态指标为 $k=1, l=2$ 。代入顶点规则： $i\mathcal{M} = -2i\lambda (\delta^{12}\delta^{12} + \delta^{11}\delta^{22} + \delta^{12}\delta^{21}) = -2i\lambda (0 + 1\cdot 1 + 0) = -2i\lambda$ $\boxed{ \frac{d\sigma}{d\Omega}(\Phi^1 \Phi^2 \rightarrow \Phi^1 \Phi^2) = \frac{\lambda^2}{16\pi^2 E_{\text{cm}}^2} }$
过程 $\Phi^1 \Phi^1 \rightarrow \Phi^2 \Phi^2$ : 设初态指标为 $i=1, j=1$ ，末态指标为 $k=2, l=2$ 。 $i\mathcal{M} = -2i\lambda (\delta^{11}\delta^{22} + \delta^{12}\delta^{12} + \delta^{12}\delta^{12}) = -2i\lambda (1\cdot 1 + 0 + 0) = -2i\lambda$ $\boxed{ \frac{d\sigma}{d\Omega}(\Phi^1 \Phi^1 \rightarrow \Phi^2 \Phi^2) = \frac{\lambda^2}{16\pi^2 E_{\text{cm}}^2} }$
过程 $\Phi^1 \Phi^1 \rightarrow \Phi^1 \Phi^1$ : 所有指标均为 1，即 $i=j=k=l=1$ 。 $i\mathcal{M} = -2i\lambda (\delta^{11}\delta^{11} + \delta^{11}\delta^{11} + \delta^{11}\delta^{11}) = -2i\lambda (1 + 1 + 1) = -6i\lambda$ $\boxed{ \frac{d\sigma}{d\Omega}(\Phi^1 \Phi^1 \rightarrow \Phi^1 \Phi^1) = \frac{9\lambda^2}{16\pi^2 E_{\text{cm}}^2} }$

(b) 自发对称性破缺相 ( $m^2 = -\mu^2 < 0$ )

势能变为 $V(\Phi^2) = -\frac{1}{2}\mu^2 \Phi^2 + \frac{\lambda}{4} (\Phi^2)^2$ 。极小值出现在 $\Phi^2 = v^2 = \frac{\mu^2}{\lambda}$ 处。按提示平移场： $\Phi^i = \pi^i$ ( $i=1,\dots,N-1$ )， $\Phi^N = v + \sigma$ 。代入势能并展开： $\Phi^2 = \vec{\pi}^2 + (v+\sigma)^2 = \vec{\pi}^2 + v^2 + 2v\sigma + \sigma^2$ $V = -\frac{1}{2}\mu^2 (\vec{\pi}^2 + v^2 + 2v\sigma + \sigma^2) + \frac{\lambda}{4} (\vec{\pi}^2 + v^2 + 2v\sigma + \sigma^2)^2$ 利用极小值条件 $\mu^2 = \lambda v^2$ ，展开并收集同次幂项：

线性项: $(-\mu^2 v + \lambda v^3)\sigma = 0$ （在极小值处展开，线性项必然消失）。
二次项 (质量项): 对于 $\pi$ 场： $-\frac{1}{2}\mu^2 \vec{\pi}^2 + \frac{\lambda}{4}(2v^2 \vec{\pi}^2) = \frac{1}{2}(-\mu^2 + \lambda v^2)\vec{\pi}^2 = 0$ 。故 $\pi$ 场无质量， $m_\pi = 0$ 。对于 $\sigma$ 场： $-\frac{1}{2}\mu^2 \sigma^2 + \frac{\lambda}{4}(2v^2 \sigma^2 + 4v^2 \sigma^2) = \frac{1}{2}(-\mu^2 + 3\lambda v^2)\sigma^2 = \frac{1}{2}(2\lambda v^2)\sigma^2$ 。故 $m_\sigma^2 = 2\lambda v^2 = 2\mu^2$ 。
相互作用项: $\mathcal{L}_{\text{int}} = - \lambda v \sigma \vec{\pi}^2 - \lambda v \sigma^3 - \frac{\lambda}{4} (\vec{\pi}^2)^2 - \frac{\lambda}{2} \vec{\pi}^2 \sigma^2 - \frac{\lambda}{4} \sigma^4$

由此可读出 Feynman 规则：

$\pi$ 传播子: $\frac{i\delta^{ij}}{p^2 + i\epsilon}$
$\sigma$ 传播子: $\frac{i}{p^2 - m_\sigma^2 + i\epsilon}$
$\sigma \pi^i \pi^j$ 顶点: $\boxed{-2i\lambda v \delta^{ij}}$
$\sigma \sigma \sigma$ 顶点: $\boxed{-6i\lambda v}$
$\pi^i \pi^j \pi^k \pi^l$ 顶点: $\boxed{-2i\lambda (\delta^{ij}\delta^{kl} + \delta^{ik}\delta^{jl} + \delta^{il}\delta^{jk})}$
$\sigma \sigma \pi^i \pi^j$ 顶点: $\boxed{-2i\lambda \delta^{ij}}$
$\sigma \sigma \sigma \sigma$ 顶点: $\boxed{-6i\lambda}$

(c) $\pi$ 介子散射振幅

考虑过程 $\pi^i(p_1) \pi^j(p_2) \rightarrow \pi^k(p_3) \pi^l(p_4)$ 。在 $\mathcal{O}(\lambda)$ 阶，有四个图贡献： $s, t, u$ 沟道的 $\sigma$ 交换图，以及四点接触图。注意顶点因子中包含 $\lambda v$ ，而 $v^2 = \mu^2/\lambda$ ，故 $(\lambda v)^2 \sim \lambda$ ，与接触图同阶。

s-沟道: $i\mathcal{M}_s = (-2i\lambda v \delta^{ij}) \frac{i}{s - m_\sigma^2} (-2i\lambda v \delta^{kl}) = -i \frac{4\lambda^2 v^2}{s - m_\sigma^2} \delta^{ij}\delta^{kl}$ 代入 $2\lambda v^2 = m_\sigma^2$ ，得 $\mathcal{M}_s = -2\lambda \frac{m_\sigma^2}{s - m_\sigma^2} \delta^{ij}\delta^{kl}$ 。
t-沟道 和 u-沟道 同理： $\mathcal{M}_t = -2\lambda \frac{m_\sigma^2}{t - m_\sigma^2} \delta^{ik}\delta^{jl}, \quad \mathcal{M}_u = -2\lambda \frac{m_\sigma^2}{u - m_\sigma^2} \delta^{il}\delta^{jk}$
接触图: $\mathcal{M}_4 = -2\lambda (\delta^{ij}\delta^{kl} + \delta^{ik}\delta^{jl} + \delta^{il}\delta^{jk})$

总振幅为： $\mathcal{M} = -2\lambda \left[ \delta^{ij}\delta^{kl} \left( 1 + \frac{m_\sigma^2}{s - m_\sigma^2} \right) + \delta^{ik}\delta^{jl} \left( 1 + \frac{m_\sigma^2}{t - m_\sigma^2} \right) + \delta^{il}\delta^{jk} \left( 1 + \frac{m_\sigma^2}{u - m_\sigma^2} \right) \right]$ 化简括号内的项 $1 + \frac{m_\sigma^2}{x - m_\sigma^2} = \frac{x}{x - m_\sigma^2}$ ，得到： $\boxed{ \mathcal{M} = -2\lambda \left[ \delta^{ij}\delta^{kl} \frac{s}{s - m_\sigma^2} + \delta^{ik}\delta^{jl} \frac{t}{t - m_\sigma^2} + \delta^{il}\delta^{jk} \frac{u}{u - m_\sigma^2} \right] }$

阈值行为: 在阈值处， $\mathbf{p}_i = 0$ 。由于 $\pi$ 是无质量的 Goldstone 玻色子，其能量 $E_i = |\mathbf{p}_i| = 0$ ，因此四动量 $p_i \rightarrow 0$ 。这导致 Mandelstam 变量 $s, t, u \rightarrow 0$ 。代入上式，显然有 $\boxed{ \mathcal{M} \rightarrow 0 }$ 。这体现了 Goldstone 玻色子的低能定理（Adler zero）。

特殊情况 $N=2$ : 此时只有一种 $\pi$ 介子，所有指标 $i=j=k=l=1$ ，Kronecker delta 均为 1。 $\mathcal{M} = -2\lambda \left( \frac{s}{s - m_\sigma^2} + \frac{t}{t - m_\sigma^2} + \frac{u}{u - m_\sigma^2} \right)$ 在低能下 ( $s,t,u \ll m_\sigma^2$ ) 展开到 $\mathcal{O}(p^2)$ ： $\frac{x}{x - m_\sigma^2} \approx -\frac{x}{m_\sigma^2}$ $\mathcal{M} \approx \frac{2\lambda}{m_\sigma^2} (s + t + u)$ 对于无质量粒子，运动学恒等式为 $s + t + u = \sum m_i^2 = 0$ 。因此， $\mathcal{O}(p^2)$ 项严格抵消，振幅的领头项为 $\mathcal{O}(p^4)$ 。

(d) 显式对称性破缺

加入破缺项 $\Delta V = -a\Phi^N$ 。势能变为： $V = -\frac{1}{2}\mu^2 \Phi^2 + \frac{\lambda}{4} (\Phi^2)^2 - a\Phi^N$ 设新的真空期望值为 $\langle \Phi^N \rangle = v$ ，极小值条件为： $\frac{\partial V}{\partial v} = -\mu^2 v + \lambda v^3 - a = 0 \implies \lambda v^2 - \mu^2 = \frac{a}{v}$ 再次作平移 $\Phi^N = v + \sigma, \Phi^i = \pi^i$ ，展开势能寻找质量项： $V \supset \frac{1}{2} (-\mu^2 + \lambda v^2) \vec{\pi}^2 + \frac{1}{2} (-\mu^2 + 3\lambda v^2) \sigma^2$ 利用新的极小值条件代换 $-\mu^2 + \lambda v^2 = a/v$ ，得到 $\pi$ 介子获得了质量： $\boxed{ m_\pi^2 = \frac{a}{v} \approx \frac{a\sqrt{\lambda}}{\mu} \sim a }$ $\sigma$ 介子的质量变为 $m_\sigma^2 = 2\lambda v^2 + a/v = 2\lambda v^2 + m_\pi^2$ 。

阈值处的散射振幅: 相互作用顶点的形式不变，仍由 $\lambda$ 和新的 $v$ 决定。重新计算 (c) 中的振幅，此时 $2\lambda v^2 = m_\sigma^2 - m_\pi^2$ 。 s-沟道图分子变为 $4\lambda^2 v^2 = 2\lambda(m_\sigma^2 - m_\pi^2)$ 。总振幅化简为： $\mathcal{M} = -2\lambda \left[ \delta^{ij}\delta^{kl} \left( 1 + \frac{m_\sigma^2 - m_\pi^2}{s - m_\sigma^2} \right) + \dots \right] = -2\lambda \left[ \delta^{ij}\delta^{kl} \frac{s - m_\pi^2}{s - m_\sigma^2} + \delta^{ik}\delta^{jl} \frac{t - m_\pi^2}{t - m_\sigma^2} + \delta^{il}\delta^{jk} \frac{u - m_\pi^2}{u - m_\sigma^2} \right]$ 现在 $\pi$ 有质量，阈值条件为 $\mathbf{p}_i = 0 \implies p_i = (m_\pi, \mathbf{0})$ 。此时 $s = (2m_\pi)^2 = 4m_\pi^2$ ，且 $t = 0, u = 0$ 。代入上式： $\mathcal{M}_{\text{thresh}} = -2\lambda \left[ \delta^{ij}\delta^{kl} \frac{3m_\pi^2}{4m_\pi^2 - m_\sigma^2} + \delta^{ik}\delta^{jl} \frac{-m_\pi^2}{-m_\sigma^2} + \delta^{il}\delta^{jk} \frac{-m_\pi^2}{-m_\sigma^2} \right]$ 在小 $a$ 极限下， $m_\pi^2 \ll m_\sigma^2$ ，分母 $4m_\pi^2 - m_\sigma^2 \approx -m_\sigma^2$ 。保留到 $m_\pi^2$ 的领头阶： $\mathcal{M}_{\text{thresh}} \approx \frac{2\lambda m_\pi^2}{m_\sigma^2} \left( 3\delta^{ij}\delta^{kl} - \delta^{ik}\delta^{jl} - \delta^{il}\delta^{jk} \right)$ 由于 $m_\pi^2 = a/v$ ，我们得到阈值振幅非零，且正比于破缺参数 $a$ ： $\boxed{ \mathcal{M}_{\text{thresh}} \propto a }$

4.4

Problem 4.4

peskinChapter 4

习题 4.4

来源: 第4章, PDF第129,130页

4.4 Rutherford scattering. The cross section for scattering of an electron by the Coulomb field of a nucleus can be computed, to lowest order, without quantizing the electromagnetic field. Instead, treat the field as a given, classical potential $A_\mu(x)$ . The interaction Hamiltonian is

H_I = \int d^3x \, e\bar{\psi}\gamma^\mu\psi A_\mu,

where $\psi(x)$ is the usual quantized Dirac field.

(a) Show that the $T$ -matrix element for electron scattering off a localized classical potential is, to lowest order,

\langle p' | iT | p \rangle = -ie \bar{u}(p')\gamma^\mu u(p) \cdot \tilde{A}_\mu(p' - p),

where $\tilde{A}_\mu(q)$ is the four-dimensional Fourier transform of $A_\mu(x)$ .

(b) If $A_\mu(x)$ is time independent, its Fourier transform contains a delta function of energy. It is then natural to define

\langle p' | iT | p \rangle \equiv i\mathcal{M} \cdot (2\pi)\delta(E_f - E_i),

where $E_i$ and $E_f$ are the initial and final energies of the particle, and to adopt a new Feynman rule for computing $\mathcal{M}$ :

where $\tilde{A}_\mu(\mathbf{q})$ is the three-dimensional Fourier transform of $A_\mu(x)$ . Given this definition of $\mathcal{M}$ , show that the cross section for scattering off a time-independent,

localized potential is

d\sigma = \frac{1}{v_i} \frac{1}{2E_i} \frac{d^3 p_f}{(2\pi)^3} \frac{1}{2E_f} |\mathcal{M}(p_i \rightarrow p_f)|^2 (2\pi) \delta(E_f - E_i),

where $v_i$ is the particle's initial velocity. This formula is a natural modification of (4.79). Integrate over $|p_f|$ to find a simple expression for $d\sigma/d\Omega$ .

(c) Specialize to the case of electron scattering from a Coulomb potential ( $A^0 = Ze/4\pi r$ ). Working in the nonrelativistic limit, derive the Rutherford formula,

\frac{d\sigma}{d\Omega} = \frac{\alpha^2 Z^2}{4m^2 v^4 \sin^4(\theta/2)}.

(With a few calculational tricks from Section 5.1, you will have no difficulty evaluating the general cross section in the relativistic case; see Problem 5.1.)

Referenced Equations:

Equation (4.79):

d\sigma = \frac{1}{2E_A 2E_B |v_A - v_B|} \left( \prod_f \frac{d^3 p_f}{(2\pi)^3} \frac{1}{2E_f} \right) \tag{4.79}

习题 4.4 - 解答

(a) 推导 $T$ 矩阵元

在相互作用绘景中，系统的时间演化由 $S$ 矩阵描述： $S = \mathcal{T} \exp\left( -i \int d^4x \, \mathcal{H}_I(x) \right)$ 其中相互作用哈密顿量密度为 $\mathcal{H}_I(x) = e\bar{\psi}(x)\gamma^\mu\psi(x) A_\mu(x)$ 。保留到最低阶（一阶）， $S$ 矩阵展开为： $S \approx 1 - i \int d^4x \, e\bar{\psi}(x)\gamma^\mu\psi(x) A_\mu(x)$ 由 $S = 1 + iT$ 定义 $T$ 矩阵，因此最低阶的 $iT$ 算符为： $iT = -i \int d^4x \, e\bar{\psi}(x)\gamma^\mu\psi(x) A_\mu(x)$ 我们需要计算从初态 $|p\rangle$ 到末态 $|p'\rangle$ 的跃迁矩阵元 $\langle p' | iT | p \rangle$ 。这里 $|p\rangle$ 和 $|p'\rangle$ 是单电子态，采用相对论协变归一化 $\langle p' | p \rangle = (2\pi)^3 2E_p \delta^{(3)}(\mathbf{p}' - \mathbf{p})$ 。将狄拉克场算符 $\psi(x)$ 和 $\bar{\psi}(x)$ 展开为平面波： $\psi(x) = \int \frac{d^3k}{(2\pi)^3 \sqrt{2E_k}} \sum_s \left( a_{\mathbf{k}}^s u^s(k) e^{-ik\cdot x} + b_{\mathbf{k}}^{s\dagger} v^s(k) e^{ik\cdot x} \right)$ 场算符作用于单粒子态给出： $\psi(x) |p\rangle = u(p) e^{-ip\cdot x} |0\rangle$ $\langle p' | \bar{\psi}(x) = \langle 0 | \bar{u}(p') e^{ip'\cdot x}$ 将这些代入矩阵元中： $\langle p' | iT | p \rangle = -i \int d^4x \, e \langle p' | \bar{\psi}(x)\gamma^\mu\psi(x) | p \rangle A_\mu(x)$ $\langle p' | iT | p \rangle = -ie \int d^4x \, \bar{u}(p')\gamma^\mu u(p) e^{i(p'-p)\cdot x} A_\mu(x)$ 将旋量部分提出积分外，并识别出经典势 $A_\mu(x)$ 的四维傅里叶变换 $\tilde{A}_\mu(q) = \int d^4x \, e^{iq\cdot x} A_\mu(x)$ ，令 $q = p' - p$ ，即可得到： $\langle p' | iT | p \rangle = -ie \bar{u}(p')\gamma^\mu u(p) \tilde{A}_\mu(p' - p)$

(b) 推导与时间无关势的散射截面

如果 $A_\mu(x)$ 与时间无关，即 $A_\mu(x) = A_\mu(\mathbf{x})$ ，其四维傅里叶变换可以分离出能量的狄拉克 $\delta$ 函数： $\tilde{A}_\mu(p' - p) = \int dt \, e^{i(E_{p'} - E_p)t} \int d^3x \, e^{-i(\mathbf{p}' - \mathbf{p})\cdot\mathbf{x}} A_\mu(\mathbf{x}) = 2\pi \delta(E_{p'} - E_p) \tilde{A}_\mu(\mathbf{p}' - \mathbf{p})$ 其中 $\tilde{A}_\mu(\mathbf{q})$ 是三维傅里叶变换。代入 (a) 的结果中： $\langle p' | iT | p \rangle = -ie \bar{u}(p')\gamma^\mu u(p) \tilde{A}_\mu(\mathbf{p}' - \mathbf{p}) 2\pi \delta(E_{p'} - E_p)$ 根据定义 $\langle p' | iT | p \rangle \equiv i\mathcal{M} \cdot (2\pi)\delta(E_f - E_i)$ ，可提取出不变矩阵元： $\mathcal{M} = -e \bar{u}(p')\gamma^\mu u(p) \tilde{A}_\mu(\mathbf{q})$ 这正是题目中给出的费曼规则。

接下来推导截面公式。考虑归一化体积 $V$ 中的波包，初末态的归一化因子分别为 $1/\sqrt{2E_i V}$ 和 $1/\sqrt{2E_f V}$ 。跃迁概率为： $P = \frac{|\langle p' | iT | p \rangle|^2}{(2E_i V)(2E_f V)} = \frac{|\mathcal{M}|^2 [2\pi \delta(E_f - E_i)]^2}{4 E_i E_f V^2}$ 利用 $[2\pi \delta(E)]^2 = 2\pi \delta(E) \cdot T$ （ $T$ 为相互作用总时间），单位时间的跃迁速率为： $w = \frac{P}{T} = \frac{|\mathcal{M}|^2 2\pi \delta(E_f - E_i)}{4 E_i E_f V^2}$ 末态相空间密度为 $\frac{V d^3p_f}{(2\pi)^3}$ ，入射粒子流强为 $J = \frac{v_i}{V}$ 。微分散射截面 $d\sigma$ 定义为跃迁速率乘以相空间密度再除以入射流强： $d\sigma = \frac{w \cdot \frac{V d^3p_f}{(2\pi)^3}}{J} = \frac{\frac{|\mathcal{M}|^2 2\pi \delta(E_f - E_i)}{4 E_i E_f V^2} \frac{V d^3p_f}{(2\pi)^3}}{\frac{v_i}{V}}$ 化简后即得题目要求的公式： $d\sigma = \frac{1}{v_i} \frac{1}{2E_i} \frac{d^3 p_f}{(2\pi)^3} \frac{1}{2E_f} |\mathcal{M}(p_i \rightarrow p_f)|^2 (2\pi) \delta(E_f - E_i)$

为了得到 $d\sigma/d\Omega$ ，我们需要对 $|p_f|$ 进行积分。在球坐标下 $d^3p_f = p_f^2 dp_f d\Omega$ 。利用关系式 $E_f^2 = p_f^2 + m^2$ ，微分得 $E_f dE_f = p_f dp_f$ ，因此 $d^3p_f = p_f E_f dE_f d\Omega$ 。代入截面公式： $d\sigma = \frac{1}{v_i 2E_i} \frac{p_f E_f dE_f d\Omega}{(2\pi)^3 2E_f} |\mathcal{M}|^2 2\pi \delta(E_f - E_i) = \frac{1}{v_i 2E_i} \frac{p_f dE_f d\Omega}{8\pi^2} |\mathcal{M}|^2 \delta(E_f - E_i)$ 对 $E_f$ 积分， $\delta$ 函数使得 $E_f = E_i$ ，从而 $p_f = p_i \equiv p$ 且 $v_f = v_i \equiv v$ 。由于 $v = p/E_i$ ，有 $v_i E_i = p$ ： $\frac{d\sigma}{d\Omega} = \frac{1}{2(v_i E_i)} \frac{p}{8\pi^2} |\mathcal{M}|^2 = \frac{1}{2p} \frac{p}{8\pi^2} |\mathcal{M}|^2 = \frac{|\mathcal{M}|^2}{16\pi^2}$

(c) 非相对论极限下的卢瑟福散射公式

对于库仑势 $A^0(x) = \frac{Ze}{4\pi r}$ ， $\mathbf{A}(x) = 0$ 。其三维傅里叶变换为： $\tilde{A}^0(\mathbf{q}) = \int d^3x \, e^{-i\mathbf{q}\cdot\mathbf{x}} \frac{Ze}{4\pi r} = \frac{Ze}{\mathbf{q}^2}$ 代入矩阵元表达式： $\mathcal{M} = -e \bar{u}(p')\gamma^0 u(p) \tilde{A}^0(\mathbf{q}) = -\frac{Ze^2}{\mathbf{q}^2} u^\dagger(p') u(p)$ 对于非极化散射，我们需要对初态自旋求平均并对末态自旋求和： $\frac{1}{2} \sum_{s,s'} |\mathcal{M}|^2 = \frac{1}{2} \left( \frac{Ze^2}{\mathbf{q}^2} \right)^2 \sum_{s,s'} |u^{s'\dagger}(p') u^s(p)|^2$ 利用自旋求和公式 $\sum_s u^s(p)\bar{u}^s(p) = \not{p} + m$ ，计算迹： $\sum_{s,s'} |u^{s'\dagger}(p') u^s(p)|^2 = \text{Tr}[\gamma^0 (\not{p} + m) \gamma^0 (\not{p}' + m)] = \text{Tr}[(E\gamma^0 + \mathbf{p}\cdot\boldsymbol{\gamma} + m)(E\gamma^0 - \mathbf{p}'\cdot\boldsymbol{\gamma} + m)]$ $= 4E^2 + 4\mathbf{p}\cdot\mathbf{p}' + 4m^2$ 在非相对论极限下， $v \ll 1$ ，动量 $p \ll m$ ，能量 $E \approx m$ 。保留到最低阶，迹的结果近似为 $4m^2 + 0 + 4m^2 = 8m^2$ 。因此： $\frac{1}{2} \sum_{s,s'} |\mathcal{M}|^2 \approx \frac{1}{2} \left( \frac{Ze^2}{\mathbf{q}^2} \right)^2 (8m^2) = \frac{4m^2 Z^2 e^4}{\mathbf{q}^4}$ 弹性散射中 $|\mathbf{p}'| = |\mathbf{p}| = p = mv$ ，动量转移的平方为： $\mathbf{q}^2 = (\mathbf{p}' - \mathbf{p})^2 = 2p^2(1 - \cos\theta) = 4p^2 \sin^2(\theta/2) = 4m^2 v^2 \sin^2(\theta/2)$ 将其代入微分散射截面公式 $\frac{d\sigma}{d\Omega} = \frac{|\mathcal{M}|^2}{16\pi^2}$ 中： $\frac{d\sigma}{d\Omega} = \frac{1}{16\pi^2} \frac{4m^2 Z^2 e^4}{[4m^2 v^2 \sin^2(\theta/2)]^2} = \frac{4m^2 Z^2 e^4}{16\pi^2 \cdot 16m^4 v^4 \sin^4(\theta/2)} = \frac{Z^2 e^4}{64\pi^2 m^2 v^4 \sin^4(\theta/2)}$ 引入精细结构常数 $\alpha = \frac{e^2}{4\pi}$ ，即 $e^4 = 16\pi^2 \alpha^2$ ，代入上式化简得到最终的卢瑟福散射公式： $\boxed{ \frac{d\sigma}{d\Omega} = \frac{\alpha^2 Z^2}{4m^2 v^4 \sin^4(\theta/2)} }$

Problem 4.1

习题 4.1

习题 4.1 - 解答

Problem 4.2

习题 4.2

习题 4.2 - 解答

Problem 4.3

习题 4.3

习题 4.3 - 解答

(a) 对称相 (m2>0m^2 > 0m2>0) 的线性 σ\sigmaσ 模型

(b) 自发对称性破缺相 (m2=−μ2<0m^2 = -\mu^2 < 0m2=−μ2<0)

(c) π\piπ 介子散射振幅

(d) 显式对称性破缺

Problem 4.4

习题 4.4

习题 4.4 - 解答

(a) 推导 TTT 矩阵元

(b) 推导与时间无关势的散射截面

(c) 非相对论极限下的卢瑟福散射公式

(a) 对称相 ( $m^2 > 0$ ) 的线性 $\sigma$ 模型

(b) 自发对称性破缺相 ( $m^2 = -\mu^2 < 0$ )

(c) $\pi$ 介子散射振幅

(a) 推导 $T$ 矩阵元