# 第19章 散射理论

## 19.1 引言

## 19.2 一维散射的回顾与概述

## 19.3 Born 近似（含时描述）

**习题 19.3.1** 证明

$$
\sigma_{\text{Yukawa}}=16 \pi r_0^2\left(\frac{g \mu r_0}{\hbar^2}\right)^2 \frac{1}{1+4 k^2 r_0^2}
$$

其中 $r_0=1 \mu_0$ 是作用程。将 $\sigma$ 与这一作用程对应的几何截面进行比较。



**习题 19.3.2** (1) 证明如果
$V(r)=-V_0 \theta\left(r_0-r\right)$，
那么

$$
\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega}=4 r_0^2\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2 \frac{\left(\sin q r_0-q r_0 \cos q r_0\right)^2}{\left(q r_0\right)^6}
$$

(2) 证明当
$k r_0 \rightarrow 0$
时，散射趋于各向同性，并且

$$
\sigma \cong \frac{16 \pi r_0^2}{9}\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2
$$

**习题 19.3.3** 证明对于 Gaussian 势
$V(r)=V_0 e^{-r^2 / r_0^3}$，
有

$$
\begin{aligned}
\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega} & =\frac{\pi r_0^2}{4}\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2 \mathrm{e}^{-q^2 r_0^2 / 2} \\
\sigma & =\frac{\pi^2}{2 k^2}\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2\left(1-\mathrm{e}^{-2 k^2 r_0^2}\right)
\end{aligned}
$$

[提示：由于
$q^2=2 k^2(1-\cos \theta)$，
因此
$\mathrm{d}(\cos \theta)=-\mathrm{d}(q^2) / 2 k^2$。]

**习题 19.3.4** 对 Gaussian 势验证上面的说法。

## 19.4 再论 Born 近似（时间无关近似）

**习题 19.4.1** 推导不等式 (19.4.44)。

## 19.5 分波展开

**习题 19.5.1** 证明：一个动能为
$100-\mathrm{MeV}$
的中子入射到一个固定原子核上时，
$l_{\max} \cong 2$。
[提示：核力的作用程约为一个 Fermi
$=10^{-5} \text{\AA}$。
另外，
$\hbar c \cong 200 \mathrm{MeV}~\mathrm{F}$
是核物理中更有用的估算口诀。]

**习题 19.5.2** 推导方程 (19.5.18)，并补全导向光学定理，即方程 (19.5.21) 的缺失步骤。

**习题 19.5.3** (1) 证明对于硬球，当
$k\rightarrow0$
时，
$\sigma_0 \rightarrow 4 \pi r_0^2$。

(2) 再考虑另一极限：
$k r_0$
非常大。由方程 (19.5.27) 以及 $j_l$ 和 $n_l$ 的渐近形式证明

$$
\sin ^2 \delta_l \xrightarrow[k r_0 \rightarrow \infty]{ } \sin ^2\left(k r_0-l \pi / 2\right)
$$

因此

$$
\begin{aligned}
\sigma=\sum_{l=0}^{l_{\max }=k r_0} \sigma_l & \cong \frac{4 \pi}{k^2} \int_0^{k r_0}(2 l) \sin ^2 \delta_l \mathrm{d} l \\
& \cong 2 \pi r_0^2
\end{aligned}
$$

这里我们把对 $l$ 的求和近似为积分，把
$2l+1$
近似为
$2l$，
并把振荡函数
$\sin^2\delta$
用其平均值
$1/2$
代替。

**习题 19.5.4** 证明深度为 $V_0$、作用程为 $r_0$ 的方势阱，其 $s$ 波相移为

$$
\delta_0=-k r_0+\tan ^{-1}\left(\frac{k}{k^{\prime}} \tan k^{\prime} r_0\right)
$$

其中 $k^{\prime}$ 和 $k$ 分别是势阱内部和外部的波数。

当 $k$ 很小时，$k r_0$ 是一个很小的数，我们将其忽略。现在考察随着势阱深度变化，也就是改变 $k^{\prime}$ 时，$\delta_0$ 会发生什么。证明每当

$$
k^{\prime} \simeq k_n^{\prime}=(2 n+1) \pi / 2 r_0
$$

时，$\delta_0$ 都具有方程 (19.5.30) 给出的共振形式，其中

$$
\Gamma / 2=\hbar^2 k_n / \mu r_0
$$

而 $k_n$ 是当
$k^{\prime}=k_n^{\prime}$
时 $k$ 的取值。

从一个浅到不足以支持任何束缚态的势阱开始，证明
$k_1^{\prime}$
对应于势阱刚好产生第一个零能束缚态。（见习题 12.6.9。）
（注意：零能束缚态对应于 $k=0$。）
随着势阱继续加深，这个能级向下移动，而很快在
$k_2^{\prime}$
处又会形成另一个零能束缚态，如此继续。

**习题 19.5.5** 证明即使一个势会吸收粒子，我们仍然可以用

$$
S_l(k)=\eta_l(k) \mathrm{e}^{2\mathrm{i} \delta_l}
$$

来描述它，其中
$\eta(<1)$
称为非弹性因子。

(1) 通过考察概率流，证明

$$
\begin{aligned}
\sigma_{\text{inel}} & =\frac{\pi}{k^2} \sum_{l=0}^{\infty}(2 l+1)\left[1-\eta_l^2\right] \\
\sigma_{\text{el}} & =\frac{\pi}{k^2} \sum_{l=0}^{\infty}(2 l+1)\left(1+\eta_l^2-2 \eta_l \cos 2 \delta_l\right)
\end{aligned}
$$

并且仍然有

$$
\sigma_{\mathrm{tot}}=\frac{4 \pi}{k} \operatorname{Im} f(0)
$$

(2) 考虑一个“黑盘”，它对
$r\leqslant r_0$
的所有粒子都完全吸收，而在更远处不起作用。将其理想化为：
对于
$l\leqslant k r_0$，
$\eta=0$；
对于
$l>k r_0$，
$\eta=1$、$\delta=0$。

证明
$\sigma_{\text{el}}=\sigma_{\text{inel}} \simeq \pi r_0^2$。
把求和换成积分，并假设
$k r_0 \gg 1$。
（见习题 19.5.3。）
为什么只要存在
$\sigma_{\text{inel}}$，
总会同时伴随一个
$\sigma_{\text{el}}$？

**习题 19.5.6** （光学定理）

(1) 证明入射波与散射波之间干涉所产生的流密度，其径向分量为

$$
j_r^{\mathrm{int}} \underset{r \rightarrow \infty}{\sim}\left(\frac{\hbar k}{\mu}\right) \frac{1}{r} \operatorname{Im}\left[\mathrm{i} \mathrm{e}^{\mathrm{i} k r(\cos \theta-1)} f^*(\theta) \cos \theta+\mathrm{i} \mathrm{e}^{\mathrm{i} k r(1-\cos \theta)} f(\theta)\right]
$$

(2) 说明只要
$\theta\neq0$，
由于
$r\rightarrow\infty$，
$j_r^{\text{int}}$
在任意小立体角上的平均值都为零。
[假设 $f(\theta)$ 是光滑函数。]

(3) 将
$j_r^{\text{int}}$
在前向方向的一个很小圆锥内积分，并证明（见提示）

$$
\int_{\text{forward cone}} j_r^{\text{int}} r^2 \mathrm{d}\Omega=-\left(\frac{\hbar k}{\mu}\right) \frac{4 \pi}{k} \operatorname{Im} f(0)
$$

因此，如果我们对靶后方区域中的总流进行积分，会发现干涉项（它只在紧邻前向方向、即靶后方显著）会造成粒子数的亏损，从而形成一个“阴影”。阴影区域中每秒缺失的粒子总数由上面的积分通量给出。把这一损失与入射通量
$\hbar k/\mu$
和截面 $\sigma$ 的乘积相等，就重新得到光学定理。

[提示：由于 $\theta$ 很小，可根据需要取
$\sin\theta\approx\theta$，
$\cos\theta=1$
或
$1-\theta^2/2$。
在计算 $\theta$ 积分的上限时，使用第 1 章中引入的思想：当一个函数的宗量趋于无穷时，如果该函数持续振荡，则其极限可以用平均值来处理。]

## 19.6 两粒子散射

**习题 19.6.1** (1) 从方程 (19.6.16) 和 (19.6.17) 出发，证明
$\mathrm{d}\sigma/\mathrm{d}\Omega$
与
$\mathrm{d}\sigma/\mathrm{d}\Omega_L$
之间的关系为

$$
\left.\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega_L}\right|_{\theta_0}=\left.\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega}\right|_{2 \theta_0} 4 \cos \theta_0
$$

(2) 仅利用能量守恒和动量守恒，证明
$\theta_L \leqslant \pi / 2$。

(3) 对不等质量粒子散射，证明

$$
\tan \theta_L=\frac{\sin \theta}{\cos \theta+\left(m_1 / m_2\right)}
$$

其中 $m_2$ 是靶粒子的质量。

**习题 19.6.2** 利用方程 (19.3.16) 中的
$f_c(\theta)$
推导方程 (19.6.21)。

**习题 19.6.3** 假设
$f=f(\theta)$，
证明对于处于三重态的费米子，

$$
(\mathrm{d}\sigma/\mathrm{d}\Omega)_{\pi/2}=0.
$$

# Chapter 19 Scattering Theory
## 19.1 Introduction
## 19.2 Recapitulation of One-Dimensional Scattering and Overview
## 19.3 The Born Approximation (Time-Dependent Description)
**Exercise 19.3.1** Show that
$$
\sigma_{\text{Yukawa}}=16 \pi r_0^2\left(\frac{g \mu r_0}{\hbar^2}\right)^2 \frac{1}{1+4 k^2 r_0^2}
$$
where $r_0=1 \mu_0$ is the range. Compare $\sigma$ to the geometrical cross section associated with this range.



**Exercise 19.3.2** (1) Show that if $V(r)=-V_0 \theta\left(r_0-r\right)$,
$$
\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega}=4 r_0^2\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2 \frac{\left(\sin q r_0-q r_0 \cos q r_0\right)^2}{\left(q r_0\right)^6}
$$

(2) Show that as $k r_0 \rightarrow 0$, the scattering becomes isotropic and
$$
\sigma \cong \frac{16 \pi r_0^2}{9}\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2
$$

**Exercise 19.3.3** Show that for the Gaussian potential, $V(r)=V_0 e^{-r^2 / r_0^3}$,
$$
\begin{aligned}
\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega} & =\frac{\pi r_0^2}{4}\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2 \mathrm{e}^{-q^2 r_0^2 / 2} \\
\sigma & =\frac{\pi^2}{2 k^2}\left(\frac{\mu V_0 r_0^2}{\hbar^2}\right)^2\left(1-\mathrm{e}^{-2 k^2 r_0^2}\right)
\end{aligned}
$$
[Hint: Since $q^2=2 k^2(1-\cos \theta)$, $\mathrm{d}(\cos \theta)=-\mathrm{d}(q^2) / 2 k^2$.]

**Exercise 19.3.4** Verify the above claim for the Gaussian potential.

## 19.4 Born Again (The Time-Independent Approximation)
**Exercise 19.4.1** Derive the inequality (19.4.44).
## 19.5 The Partial Wave Expansion
**Exercise 19.5.1** Show that for a $100-\mathrm{MeV}$ (kinetic energy) neutron incident on a fixed nucleus, $l_{\max} \cong 2$. [Hint: The range of the nuclear force is roughly a Fermi$=10^{-5} \text{\AA}$. Also $\hbar c \cong 200 \mathrm{MeV}~\mathrm{F}$ is a more useful mnemonic for nuclear physics.]

**Exercise 19.5.2** Derive Eq. (19.5.18) and provide the missing steps leading to the optical theorem, Eq. (19.5.21).

**Exercise 19.5.3** (1) Show that $\sigma_0 \rightarrow 4 \pi r_0^2$ for a hard sphere as $k \rightarrow 0$.

(2) Consider the other extreme of $k r_0$ very large. From Eq. (19.5.27) and the asymptotic forms of $j_l$ and $n_l$ show that
$$
\sin ^2 \delta_l \xrightarrow[k r_0 \rightarrow \infty]{ } \sin ^2\left(k r_0-l \pi / 2\right)
$$
so that
$$
\begin{aligned}
\sigma=\sum_{l=0}^{l_{\max }=k r_0} \sigma_l & \cong \frac{4 \pi}{k^2} \int_0^{k r_0}(2 l) \sin ^2 \delta_l \mathrm{d} l \\
& \cong 2 \pi r_0^2
\end{aligned}
$$
if we approximate the sum over $l$ by an integral, $2 l+1$ by $2 l$, and the oscillating function $\sin ^2 \delta$ by its mean value of $1 / 2$.

**Exercise 19.5.4** Show that the $s$-wave phase shift for a square well of depth $V_0$ and range $r_0$ is
$$
\delta_0=-k r_0+\tan ^{-1}\left(\frac{k}{k^{\prime}} \tan k^{\prime} r_0\right)
$$
where $k^{\prime}$ and $k$ are the wave numbers inside and outside the well. For $k$ small, $k r_0$ is some small number and we ignore it. Let us see what happens to $\delta_0$ as we vary the depth of the well, i.e., change $k^{\prime}$. Show that whenever $k^{\prime} \simeq k_n^{\prime}=(2 n+1) \pi / 2 r_0$, $\delta_0$ takes on the resonant form Eq. (19.5.30) with $\Gamma / 2=\hbar^2 k_n / \mu r_0$, where $k_n$ is the value of $k$ when $k^{\prime}=k_n^{\prime}$. Starting with a well that is too shallow to have any bound state, show $k_1^{\prime}$ corresponds to the well developing its first bound state, at zero energy. (See Exercise 12.6.9.) (Note: A zero-energy bound state corresponds to $k=0$.) As the well is deepened further, this level moves down, and soon, at $k_2^{\prime}$, another zero-energy bound state is formed, and so on.

**Exercise 19.5.5** Show that even if a potential absorbs particles, we can describe it by
$$
S_l(k)=\eta_l(k) \mathrm{e}^{2\mathrm{i} \delta_l}
$$
where $\eta(<1)$, is called the inelasticity factor.

(1) By considering probability currents, show that
$$
\begin{aligned}
\sigma_{\text{inel}} & =\frac{\pi}{k^2} \sum_{l=0}^{\infty}(2 l+1)\left[1-\eta_l^2\right] \\
\sigma_{\mathrm{el}} & =\frac{\pi}{k^2} \sum_{l=0}^{\infty}(2 l+1)\left(1+\eta_l^2-2 \eta_l \cos 2 \delta_l\right)
\end{aligned}
$$
and that once again
$$
\sigma_{\mathrm{tot}}=\frac{4 \pi}{k} \operatorname{Im} f(0)
$$

(2) Consider a “black disk” which absorbs everything for $r \leqslant r_0$ and is ineffective beyond. Idealize it by $\eta=0$ for $l \leqslant k r_0$; $\eta=1$, $\delta=0$ for $l>k r_0$. Show that $\sigma_{\text{el}}=\sigma_{\text{inel}} \simeq \pi r_0^2$. Replace the sum by an integral and assume $k r_0 \gg 1$. (See Exercise 19.5.3.) Why is $\sigma_{\text{inel}}$ always accompanied by $\sigma_{\mathrm{el}}$?

**Exercise 19.5.6** (The Optical Theorem) (1) Show that the radial component of the current density due to interference between the incident and scattered waves is
$$
j_r^{\mathrm{int}} \underset{r \rightarrow \infty}{\sim}\left(\frac{\hbar k}{\mu}\right) \frac{1}{r} \operatorname{Im}\left[\mathrm{i} \mathrm{e}^{\mathrm{i} k r(\cos \theta-1)} f^*(\theta) \cos \theta+\mathrm{i} \mathrm{e}^{\mathrm{i} k r(1-\cos \theta)} f(\theta)\right]
$$

(2) Argue that as long as $\theta \neq 0$, the average of $j_r^{\text {int }}$ over any small solid angle is zero because $r \rightarrow \infty$. [Assume $f(\theta)$ is a smooth function.]

(3) Integrate $j_r^{\text {int }}$ over a tiny cone in the forward direction and show that (see hint)
$$
\int_{\text{forward cone}} j_r^{\text{int}} r^2 \mathrm{d}\Omega=-\left(\frac{\hbar k}{\mu}\right) \frac{4 \pi}{k} \operatorname{Im} f(0)
$$
Thus, if we integrate the total current in the region behind the target, we find that the interference term (important only in the near-forward direction, behind the target) produces a depletion of particles, casting a “shadow”. The total number of particles (per second) missing in the shadow region is given by the above expression for the integrated flux. Equating this loss to the product of the incident flux $\hbar k / \mu$ and the cross section $\sigma$, we regain the optical theorem. [Hint: Since $\theta$ is small, set $\sin \theta \approx \theta$, $\cos \theta=1$ or $1-\theta^2 / 2$ using the judgment. In evaluating the upper limit in the $\theta$ integration, use the idea introduced in Chapter 1, namely, that the limit of a function that oscillates as its argument approaches infinity is equal to its average value.]
## 19.6 Two-Particle Scattering
**Exercise 19.6.1** (1) Starting with Eqs. (19.6.16) and (19.6.17), show that the relation between $\mathrm{d}\sigma/\mathrm{d}\Omega$ and $\mathrm{d}\sigma/\mathrm{d}\Omega_L$ is
$$
\left.\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega_L}\right|_{\theta_0}=\left.\frac{\mathrm{d} \sigma}{\mathrm{d} \Omega}\right|_{2 \theta_0} 4 \cos \theta_0
$$

(2) Show that $\theta_L \leqslant \pi / 2$ by using just energy and momentum conservation.

(3) For unequal mass scattering, show that
$$
\tan \theta_L=\frac{\sin \theta}{\cos \theta+\left(m_1 / m_2\right)}
$$
where $m_2$ is the target mass.

**Exercise 19.6.2** Derive Eq. (19.6.21) using Eq. (19.3.16) for $f_c(\theta)$.

**Exercise 19.6.3** Assuming $f=f(\theta)$ show that $(\mathrm{d}\sigma/\mathrm{d}\Omega)_{\pi/2}=0$ for fermions in the triplet state.

§ QM-19 散射理论

§ QM-19 Scattering Theory

# 第18章 含时微扰论
## 18.1 问题
## 18.2 一阶微扰论
**习题 18.2.1** 证明：如果
$H^1(t)=-e \mathscr{E} X /\left[1+(t / \tau)^2\right]$，
那么在一阶近似下，
$$
P_{0 \rightarrow 1}=\frac{e^2 \mathscr{E}^2 \pi^2 \tau^2}{2 m \omega \hbar} \mathrm{e}^{-2 \omega \tau}
$$



**习题 18.2.2** 一个氢原子在 $t=-\infty$ 时处于基态。施加电场
$\mathbf{E}(t)=(\mathbf{k} \mathscr{E}) e^{-t^2 / \tau^2}$
直到 $t=\infty$。证明：在一阶近似下，原子最终处于任意一个 $n=2$ 状态的概率为
$$
P(n=2)=\left(\frac{e \mathscr{E}}{\hbar}\right)^2\left(\frac{2^{15} a_0^2}{3^{10}}\right) \pi \tau^2 e^{-\omega^2 \tau^2 / 2}
$$
其中
$\omega=\left(E_{2 l m}-E_{100}\right) / \hbar$。
答案是否取决于是否把自旋包括在描述中？

**习题 18.2.3** 考虑一个处于长度为 $L$ 的势箱基态中的粒子。根据半经典论证说明，与它相关的自然时间尺度为
$T \simeq m L^2 / \hbar \pi$。
如果势箱在时间
$\tau\ll T$
内关于中心对称地扩展为原来的两倍，那么在新势箱的基态中发现粒子的概率是多少？（见习题 5.2.1。）

**习题 18.2.4** 在 $\beta$ 衰变
$\mathrm{H}^3$
（核内两个中子和一个质子）
$\rightarrow\left(\mathrm{He}^3\right)^{+}$
（核内两个质子和一个中子）中，发射电子的动能为
$16~\mathrm{keV}$。
说明可以使用突然近似来描述一个最初处于 $\mathrm{H}^3$ 的 $1s$ 状态中的电子的响应。证明它处于
$\left(\mathrm{He}^3\right)^{+}$
基态的振幅为
$16(2)^{1 / 2} / 27$。
那么它处于
$$
|n=16, l=3, m=0\rangle~\text{of}~\left(\mathrm{He}^3\right)^{+}
$$
状态的概率是多少？

**习题 18.2.5** 一个谐振子处于
$H=H^0+H^1$
的基态，其中与时间无关的微扰
$H^1$
是线性势
$(-fx)$。
如果在 $t=0$ 时突然关闭 $H^1$，证明系统处于 $H^0$ 的第 $n$ 个本征态的概率由 Poisson 分布给出：
$$
P(n)=\frac{\mathrm{e}^{-\lambda} \lambda^n}{n!}, \quad \text { where } \quad \lambda=\frac{f^2}{2 m \omega^3 \hbar}
$$
[提示：使用公式
$$
\exp [A+B]=\exp [A] \exp [B] \exp \left[-\frac{1}{2}[A, B]\right]
$$
其中 $[A,B]$ 是一个 $c$ 数。]

**习题 18.2.6** 考虑一个受到微扰
$H^1(t)=H^1 \delta(t)$
的系统。证明：如果在
$t=0^{-}$
时系统处于状态
$\left|i^0\right\rangle$，
那么在
$t=0^{+}$
时处于状态
$\left|f^0\right\rangle$
的振幅在一阶近似下为
$$
d_f=\frac{-\mathrm{i}}{\hbar}\left\langle f^0\right| H^1\left|i^0\right\rangle \quad(f \neq i)
$$
注意：
(1) 系统的状态确实会瞬时发生变化；
(2) 即使微扰在 $t=0$ 时是“无穷大”的，只要它“曲线下面积”足够小，我们仍然可以使用一阶微扰论。

## 18.3 微扰论的高阶修正
**习题 18.3.1** 推导方程 (18.3.30)。

**习题 18.3.2** 在习题 14.4.3 的顺磁共振问题中，我们变换到了真实空间中的旋转参考系。证明这也等价于 Hilbert 空间中的一个旋转，但除了共振情形以外，它既不会把我们带到相互作用表象，也不会带到 Heisenberg 表象。在共振时，它对应于哪一种表象？
（如果
$\mathbf{B}=B_0 \mathbf{k}+B \cos \omega t \mathbf{i}-B \sin \omega t \mathbf{j}$，
将 $B_0$ 与 $H_S^0$ 对应，而将 $B$ 与 $H_S^1$ 对应。）

## 18.4 电磁相互作用的一般讨论
**习题 18.4.1** 对方程 (18.4.5) 取散度，证明如果 Maxwell 方程彼此相容，就必须满足连续性方程。

**习题 18.4.2** 利用方程 (18.4.7) 和 (18.4.9)，分别计算与
$(\mathbf{A}, \phi)$
和
$(\mathbf{A}^{\prime}, \phi^{\prime})$
对应的 $\mathbf{E}$ 和 $\mathbf{B}$，并验证上面的说法。

**习题 18.4.3** 假设给定某些不满足 Coulomb 规范条件的 $\mathbf{A}$ 和 $\phi$。让我们看看如何把它们变换到 Coulomb 规范。

(1) 证明如果选择
$$
\Lambda(\mathbf{r}, t)=-c \int_{-\infty}^t \phi\left(\mathbf{r}, t^{\prime}\right) \mathrm{d} t^{\prime}
$$
并变换到
$(\mathbf{A}^{\prime}, \phi^{\prime})$，
那么
$\phi^{\prime}=0$。
此时
$\mathbf{A}^{\prime}$
就是
$\mathbf{A}-\boldsymbol{\nabla} \Lambda$，
而
$\boldsymbol{\nabla} \cdot \mathbf{A}^{\prime}$
不一定为零。

(2) 证明如果再通过
$$
\Lambda^{\prime}=-\frac{1}{4 \pi} \int \frac{\nabla \cdot \mathbf{A}^{\prime}\left(\mathbf{r}^{\prime}, t\right) \mathrm{d}^3 \mathbf{r}^{\prime}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}
$$
进行一次规范变换，得到
$(\mathbf{A}^{\prime \prime}, \phi^{\prime \prime})$，
那么
$\boldsymbol{\nabla} \cdot \mathbf{A}^{\prime \prime}=0$。
[提示：回忆
$\nabla^2\left(1 /\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\right)=-4 \pi \delta^3\left(\mathbf{r}-\mathbf{r}^{\prime}\right)$。]

(3) 利用
$\nabla \cdot \mathbf{E}=0$
验证
$\phi^{\prime \prime}$
也为零。

(4) 证明：如果希望在 Coulomb 规范内部继续作规范变换，那么 $\Lambda$ 必须与时间无关，并满足
$\nabla^2 \Lambda=0$。
如果再要求当空间趋于无穷远时
$|\mathbf{A}| \rightarrow 0$，
那么 $\mathbf{A}$ 就变得唯一。

**习题 18.4.4** （Schrödinger 方法中规范不变性的证明）

(1) 写出处于势
$(A,\phi)$
中的粒子的 Hamilton量 $H$。

(2) 写出对势作规范变换之后得到的 Hamilton量
$H_{\Lambda}$。

(3) 证明：如果
$\psi(\mathbf{r}, t)$
是 Hamilton量 $H$ 对应的 Schrödinger 方程的解，那么方程 (18.4.33) 中给出的
$\psi_{\Lambda}(\mathbf{r}, t)$
就是
$H\rightarrow H_{\Lambda}$
后相应的解。

## 18.5 原子与电磁辐射的相互作用
**习题 18.5.1** (1) 重新考察推导过程，说明可以通过在方程 (18.5.19) 中将
$\mathbf{p}_f$
替换为
$\mathbf{p}_f-\hbar \mathbf{k}$，
从而精确地计入因子
$\mathrm{e}^{\mathrm{i}\mathbf{k} \cdot \mathbf{r}}$。

(2) 验证上面关于电子动量分布的说法。

**习题 18.5.2** (1) 当逸出电子的动能为
$10~\mathrm{Ry}$
时，估算光电截面。将其与原子的几何截面
$\simeq \pi a_0^2$
比较。

(2) 证明：如果考虑电荷数为 $Z$ 的原子，其 $1s$ 状态发生光电发射，那么在极限
$p_f a_0 / Z \hbar \gg 1$
下，
$\sigma \propto Z^5$。

# Chapter 18 Time-Dependent Perturbation Theory
## 18.1 The Problem
## 18.2 First-Order Perturbation Theory
**Exercise 18.2.1** Show that if $H^1(t)=-e \mathscr{E} X /\left[1+(t / \tau)^2\right]$, then, to first order,
$$
P_{0 \rightarrow 1}=\frac{e^2 \mathscr{E}^2 \pi^2 \tau^2}{2 m \omega \hbar} \mathrm{e}^{-2 \omega \tau}
$$



**Exercise 18.2.2** A hydrogen atom is in the ground state at $t=-\infty$. An electric field $\mathbf{E}(t)=(\mathbf{k} \mathscr{E}) e^{-t^2 / \tau^2}$ is applied until $t=\infty$. Show that the probability that the atom ends up in any of the $n=2$ states is, to first order,
$$
P(n=2)=\left(\frac{e \mathscr{E}}{\hbar}\right)^2\left(\frac{2^{15} a_0^2}{3^{10}}\right) \pi \tau^2 e^{-\omega^2 \tau^2 / 2}
$$
where $\omega=\left(E_{2 l m}-E_{100}\right) / \hbar$. Does the answer depend on whether or not we incorporate spin in the picture?

**Exercise 18.2.3** Consider a particle in the ground state of a box of length $L$. Argue on semiclassical grounds that the natural time period associated with it is $T \simeq m L^2 / \hbar \pi$. If the box expands symmetrically to double its size in time $\tau \ll T$ what is the probability of catching the particle in the ground state of the new box? (See Exercise (5.2.1).)

**Exercise 18.2.4** In the $\beta$ decay $\mathrm{H}^3$ (two neutrons + one proton in the nucleus) $\rightarrow\left(\mathrm{He}^3\right)^{+}$ (two protons + one neuron in the nucleus), the emitted electron has a kinetic energy of $16~\mathrm{keV}$. Argue that the sudden approximation may be used to describe the response of an electron that is initially in the $1 s$ state of $\mathrm{H}^3$. Show that the amplitude for it to be in the ground state of $\left(\mathrm{He}^3\right)^{+}$is $16(2)^{1 / 2} / 27$. What is the probability for it to be in the state
$$
|n=16, l=3, m=0\rangle~\text{of}~\left(\mathrm{He}^3\right)^{+} ?
$$

**Exercise 18.2.5** An oscillator is in the ground state of $H=H^0+H^1$, where the time-independent perturbation $H^1$ is the linear potential ($-f x$). If at $t=0$, $H^1$ is abruptly turned off, show that the probability that the system is in the $n$th eigenstate of $H^0$ is given by the Poisson distribution
$$
P(n)=\frac{\mathrm{e}^{-\lambda} \lambda^n}{n!}, \quad \text { where } \quad \lambda=\frac{f^2}{2 m \omega^3 \hbar}
$$
[Hint: Use the formula
$$
\exp [A+B]=\exp [A] \exp [B] \exp \left[-\frac{1}{2}[A, B]\right]
$$
where $[A, B]$ is a $c$ number.]

**Exercise 18.2.6** Consider a system subject to a perturbation $H^1(t)=H^1 \delta(t)$. Show that if at $t=0^{-}$the system is in the state $\left|i^0\right\rangle$, the amplitude to be in a state $\left|f^0\right\rangle$ at $t=0^{+}$is, to first order,
$$
d_f=\frac{-\mathrm{i}}{\hbar}\left\langle f^0\right| H^1\left|i^0\right\rangle \quad(f \neq i)
$$
Notice that (1) the state of the system does change instantaneously; (2) Even though the perturbation is "infinite" at $t=0$, we can still use first-order perturbation theory if the “area under it” is small enough.
## 18.3 Higher Orders in Perturbation Theory
**Exercise 18.3.1** Derive Eq. (18.3.30).

**Exercise 18.3.2** In the paramagnetic resonance problem Exercise 14.4.3 we moved to a frame rotating in real space. Show that this is also equivalent to a Hilbert space rotation, but that it takes us neither to the interaction nor the Heisenberg picture, except at resonance. What picture is it at resonance? (If $\mathbf{B}=B_0 \mathbf{k}+B \cos \omega t \mathbf{i}-B \sin \omega t \mathbf{j}$, associate $B_0$ with $H_S^0$ and $B$ with $H_S^1$.)
## 18.4 A General Discussion of Electromagnetric Interactions
**Exercise 18.4.1** By taking the divergence of Eq. (18.4.5) show that the continuity equation must be obeyed if Maxwell's equations are to be mutually consistent.

**Exercise 18.4.2** Calculate $\mathbf{E}$ and $\mathbf{B}$ corresponding to $(\mathbf{A}, \phi)$ and $(\mathbf{A}^{\prime}, \phi^{\prime})$ using Eqs. (18.4.7) and (18.4.9) and verify the above claim.

**Exercise 18.4.3** Suppose we are given some A and $\phi$ that do not obey the Coulomb gauge conditions. Let us see how they can be transformed to the Coulomb gauge.

(1) Show that if we choose
$$
\Lambda(\mathbf{r}, t)=-c \int_{-\infty}^t \phi\left(\mathbf{r}, t^{\prime}\right) \mathrm{d} t^{\prime}
$$
and transform to ($\mathbf{A}^{\prime}, \phi^{\prime}$) then $\phi^{\prime}=0$. $\mathbf{A}^{\prime}$ is just $\mathbf{A}-\boldsymbol{\nabla} \Lambda$, with $\boldsymbol{\nabla} \cdot \mathbf{A}^{\prime}$ not necessarily zero.

(2) Show that if we gauge transform once more to ($\mathbf{A}^{\prime \prime}, \phi^{\prime \prime}$) via
$$
\Lambda^{\prime}=-\frac{1}{4 \pi} \int \frac{\nabla \cdot \mathbf{A}^{\prime}\left(\mathbf{r}^{\prime}, t\right) \mathrm{d}^3 \mathbf{r}^{\prime}}{\left|\mathbf{r}-\mathbf{r}^{\prime}\right|}
$$
then $\boldsymbol{\nabla} \cdot \mathbf{A}^{\prime \prime}=0$. [Hint: Recall $\nabla^2\left(1 /\left|\mathbf{r}-\mathbf{r}^{\prime}\right|\right)=-4 \pi \delta^3\left(\mathbf{r}-\mathbf{r}^{\prime}\right)$.]

(3) Verify that $\phi^{\prime \prime}$ is also zero by using $\nabla \cdot \mathbf{E}=0$.

(4) Show that if we want to make any further gauge transformations within the Coulomb gauge, $\Lambda$ must be time independent and obey $\nabla^2 \Lambda=0$. If we demand that $|\mathbf{A}| \rightarrow 0$ at spatial infinity, $\mathbf{A}$ becomes unique.

**Exercise 18.4.4** (Proof of Gauge Invariance in the Schrödinger Approach)

(1) Write $H$ for a particle in the potentials $(A, \phi)$.

(2) Write down $H_{\Lambda}$, the Hamiltonian obtained by gauge transforming the potentials.

(3) Show that if $\psi(\mathbf{r}, t)$ is a solution to Schrödinger's equation with the Hamiltonian $H$, then $\psi_{\Lambda}(\mathbf{r}, t)$ given in Eq. (18.4.33) is the corresponding solution with $H \rightarrow H_{\Lambda}$.
## 18.5 Interaction of Atoms with Electromagnetic Radiation
**Exercise 18.5.1** (1) By going through the derivation, argue that we can take the $\mathrm{e}^{\mathrm{i}\mathbf{k} \cdot \mathbf{r}}$ factor into account exactly, by replacing $\mathbf{p}_f$ by $\mathbf{p}_f-\hbar \mathbf{k}$ in Eq. (18.5.19).

(2) Verify the claim made above about the electron momentum distribution.

**Exercise 18.5.2** (1) Estimate the photoelectric cross section when the ejected electron has a kinetic energy of $10~\mathrm{Ry}$. Compare it to the atom's geometric cross section $\simeq \pi a_0^2$.

(2) Show that if we consider photoemission from the $1 s$ state of a charge $Z$ atom, $\sigma \propto Z^5$, in the limit $p_f a_0 / Z \hbar \gg 1$.

§ QM-18 含时微扰论

§ QM-18 Time-Dependent Perturbation Theory

# 第17章 不含时微扰论

## 17.1 形式

## 17.2 一些例子

**习题 17.2.1** 对谐振子问题考虑 $H^1=\lambda x^4$。

(1) 证明

$$
E_n^1=\frac{3 \hbar^2 \lambda}{4 m^2 \omega^2}\left[1+2 n+2 n^2\right]
$$

(2) 说明无论 $\lambda$ 多么小，只要 $n$ 足够大，微扰展开总会失效。其物理原因是什么？



**习题 17.2.2** 考虑一个旋磁比为 $\gamma$ 的自旋 $1/2$ 粒子，处于磁场
$\mathbf{B}=B \mathbf{i}+B_0 \mathbf{k}$
中。把 $B$ 视为微扰，计算基态能量的一阶和二阶修正，以及波函数的一阶修正。然后将精确结果展开到相应阶数并进行比较。

**习题 17.2.3** 在研究氢原子时，我们假设质子是一个点电荷 $e$。这会得到熟悉的电子—质子 Coulomb 相互作用
$\left(-e^2/r\right)$。

(1) 证明：如果质子是一个半径为 $R$、电荷密度均匀的电荷分布，那么相互作用势为

$$
\begin{aligned}
V(r) & =-\frac{3 e^2}{2 R}+\frac{e^2 r^2}{2 R^3}, & & r \leqslant R \\
& =-\frac{e^2}{r}, & & r>R
\end{aligned}
$$

(2) 计算这种修正导致的氢原子基态能量的一阶修正。可以假设
$\mathrm{e}^{-R / a_0} \simeq 1$。
你应该得到
$E^1=2 e^2 R^2 / 5 a_0^3$。

**习题 17.2.4** (1) 证明 Thomas-Reiche-Kuhn 求和规则

$$
\sum_{n^{\prime}}\left(E_{n^{\prime}}-E_n\right)|\langle n^{\prime}\mid X\mid n\rangle|^ 2=\sum_{n^{\prime}}\left(E_{n^{\prime}}-E_n\right)\langle n\mid X\mid n^{\prime}\rangle\langle n^{\prime}\mid X\mid n\rangle=\frac{\hbar^2}{2 m}
$$

其中 $|n\rangle$ 和 $\left|n^{\prime}\right\rangle$ 是
$H=P^2 / 2 m+V(X)$
的本征态。
[提示：用 $H$ 消去因子
$E_{n^{\prime}}-E_n$。]

(2) 在谐振子的第 $n$ 个状态上检验该求和规则。

**习题 17.2.5** 我们已经看到，如果忽略 He 原子基态中两个电子之间的排斥作用
$e^2 / r_{12}$，
则能量为
$-8 \mathrm{Ry}=-108.8~\mathrm{eV}$。
把
$e^2 / r_{12}$
视为微扰，证明

$$
\langle 100,100| H^1|100,100\rangle=\frac{5}{2} \mathrm{Ry}
$$

从而
$E_0^0+E_0^1=-5.5 . \mathrm{Ry}=-74.8~\mathrm{eV}$。
回忆实验测得的数值为
$-78.6~\mathrm{eV}$，
而变分估计为
$-77.5~\mathrm{eV}$。

[提示：
$\left\langle H^1\right\rangle$
可以看成两个同心、球对称并按指数衰减的电荷分布之间的相互作用。先求其中一个电荷分布产生的势
$\phi(r)$，
再计算该势与另一个电荷分布之间的相互作用能。]

**习题 17.2.6** 验证方程 (17.2.34)。

**习题 17.2.7** 对谐振子考虑
$H^1=-qfX$。
求一个满足方程 (17.2.38) 的 $\Omega$。把它代入方程 (17.2.39) 中的
$E_n^2$
并与前面的计算比较。

**习题 17.2.8** 补全从方程 (17.2.41) 到方程 (17.2.43) 的推导步骤。尽量利用对称性论证来减少计算积分的工作量。

## 17.3 简并微扰论

**习题 17.3.1** 利用偶极跃迁选择定则证明 $H^1$ 具有上面的形式，并完成 $\Delta$ 的计算。

**习题 17.3.2** 考虑一个自旋 1 粒子（没有轨道自由度）。令
$H=A S_z^2+B\left(S_x^2-S_y^2\right)$，
其中 $S_i$ 是 $3\times3$ 自旋矩阵，并且
$A \gg B$。
把 $B$ 项视为微扰，求
$H^0=A S_z^2$
的那些在微扰下保持稳定的本征态。计算能量对 $B$ 的一阶修正。它们与精确结果之间是什么关系？

**习题 17.3.3** 考虑
$H^0$
包含 Coulomb 相互作用和自旋—轨道相互作用，而
$H^1$
是弱磁场
$\mathbf{B}=B \mathbf{k}$
的效应。选取适当的基底，证明一阶能级修正与 $j_z$ 的关系为

$$
E^1=\left(\frac{e B}{2 m c}\right)\left(1 \pm \frac{1}{2 l+1}\right) j_z, \quad j=l \pm 1 / 2
$$

假设
$E^1 \ll E_{\text{f.s.}}^1$，
画出 $n=2$ 能级的示意图。

**习题 17.3.4** 这里讨论一些计算氢原子本征态中某些算符期望值的技巧。

(1) 假设我们想求
$\langle 1 / r\rangle_{nlm}$。
先考虑
$\langle\lambda / r\rangle$。
可以把
$\langle\lambda / r\rangle$
理解为微扰
$\lambda / r$
导致的一阶修正。这个问题实际上可以精确求解：只需在所有地方把
$e^2$
替换为
$e^2-\lambda$。
（为什么？）

因此，根据方程 (13.1.16)，精确能量为
$E(\lambda)= -\left(e^2-\lambda\right)^2 m / 2 n^2 \hbar^2$。
一阶修正就是关于 $\lambda$ 的线性项，即
$E^1=m e^2 \lambda / n^2 \hbar^2=\langle\lambda / r\rangle$，
由此得到
$\langle 1 / r\rangle=1 / n^2 a_0$，
与方程 (13.1.36) 一致。

为了后面使用，注意
$E(\lambda)=E^0+E^1+\cdots=E(\lambda=0)+\lambda(\mathrm{d}E/\mathrm{d}\lambda)*{\lambda=0}+\cdots$，
因此从精确结果中提取 $E^1$ 的一种方法就是计算
$\lambda(\mathrm{d}E/\mathrm{d}\lambda)*{\lambda=0}$。

(2) 现在考虑
$\left\langle\lambda / r^2\right\rangle$。
在这种情况下仍然可以得到精确解，因为该微扰只是把离心项修改为

$$
\frac{\hbar^2 l(l+1)}{2 m r^2}+\frac{\lambda}{r^2}=\frac{\hbar^2 l^{\prime}\left(l^{\prime}+1\right)}{2 m r^2}\tag{17.3.23}
$$

其中
$l^{\prime}$
是 $\lambda$ 的函数。根据方程 (13.1.14)，$E$ 对
$l^{\prime}(\lambda)$
的依赖为

$$
E\left(l^{\prime}\right)=\frac{-m e^4}{2 \hbar^2\left(k+l^{\prime}+1\right)^2}=E(\lambda)=E^0+E^1+\cdots
$$

证明

$$
\left\langle\frac{\lambda}{r^3}\right\rangle=E^1=\left.\lambda \frac{\mathrm{d}E}{\mathrm{d}\lambda}\right|_{\lambda=0}=\left(\frac{\mathrm{d}E}{\mathrm{d}l^{\prime}}\right)_{l^{\prime}=l} \cdot\left(\frac{\mathrm{d}l^{\prime}}{\mathrm{d} \lambda}\right)_{l^{\prime}=l} \cdot \lambda=\frac{\lambda}{n^3 a_0^2\left(l+\frac{1}{2}\right)}
$$

两边约去 $\lambda$，便得到方程 (17.3.11)。

(3) 最后考虑
$\left\langle l / r^3\right\rangle$。
由于 Coulomb Hamilton量中没有这样的项，我们采用另一个技巧。考虑径向动量算符
$p_r=-i \hbar(\partial / \partial r+1 / r)$，
利用它可以把 Hamilton量的径向部分

$$
\left(\frac{-\hbar^2}{2 m}\right)\left(\frac{1}{r^2} \frac{\partial}{\partial r} r^2 \frac{\partial}{\partial r}\right)
$$

写成
$p_r^2 / 2 m$。
（请验证。）

利用能量本征态中
$\left\langle\left[H, p_r\right]\right\rangle=0$
这一事实，并显式计算该对易子，证明

$$
\left\langle\frac{1}{r^3}\right\rangle=\frac{1}{a_0(l)(l+1)}\left\langle\frac{1}{r^2}\right\rangle
$$

再与第 (2) 问的结果结合，便得到方程 (17.3.20)。

(4) 利用第 (1) 问中的技巧求平均动能，这一次对质量作重新标度。重新得到 Virial 定理。

# Chapter 17 Time-Independence Perturbation Theory
## 17.1 The Formalism
## 17.2 Some Examples
**Exercise 17.2.1** Consider $H^1=\lambda x^4$ for the oscillator problem.

(1) Show that
$$
E_n^1=\frac{3 \hbar^2 \lambda}{4 m^2 \omega^2}\left[1+2 n+2 n^2\right]
$$

(2) Argue that no matter how small $\lambda$ is, the perturbation expansion will break down for some large enough $n$. What is the physical reason?



**Exercise 17.2.2** Consider a spin-1/2 particle with gyromagnetic ratio $\gamma$ in a magnetic field $\mathbf{B}=B \mathbf{i}+B_0 \mathbf{k}$. Treating $B$ as a perturbation, calculate the first- and second-order shifts in energy and first-order shift in wave function for the ground state. Then compare the exact answers expanded to the corresponding orders.

**Exercise 17.2.3** In our study of the H atom, we assumed that the proton is a point charge $e$. This leads to the familiar Coulomb interaction $\left(-e^2/r\right)$ with the electron.

(1) Show that if the proton is a uniformly dense charge distribution of radius $R$, the interaction is
$$
\begin{aligned}
V(r) & =-\frac{3 e^2}{2 R}+\frac{e^2 r^2}{2 R^3}, & & r \leqslant R \\
& =-\frac{e^2}{r}, & & r>R
\end{aligned}
$$

(2) Calculate the first-order shift in the ground-state energy of hydrogen due to this modification. You may assume $\mathrm{e}^{-R / a_0} \simeq 1$. You should find $E^1=2 e^2 R^2 / 5 a_0^3$.

**Exercise 17.2.4** (1) Prove the Thomas-Reiche-Kuhn sum rule
$$
\sum_{n^{\prime}}\left(E_{n^{\prime}}-E_n\right)|\langle n^{\prime}\mid X\mid n\rangle|^ 2=\sum_{n^{\prime}}\left(E_{n^{\prime}}-E_n\right)\langle n\mid X\mid n^{\prime}\rangle\langle n^{\prime}\mid X\mid n\rangle=\frac{\hbar^2}{2 m}
$$
where $|n\rangle$ and $\left|n^{\prime}\right\rangle$ are eigenstates of $H=P^2 / 2 m+V(X)$. [Hint: Eliminate the $E_{n^{\prime}}-E_n$ factor in favor of $H$.]

(2) Test the sum rule on the $n$th state of the oscillator.

**Exercise 17.2.5** We have seen that if we neglect the repulsion $e^2 / r_{12}$ between the two electrons in the ground state of He , the energy is $-8 \mathrm{Ry}=-108.8~\mathrm{eV}$. Treating $e^2 / r_{12}$ as a perturbation, show that
$$
\langle 100,100| H^1|100,100\rangle=\frac{5}{2} \mathrm{Ry}
$$
so that $E_0^0+E_0^1=-5.5 . \mathrm{Ry}=-74.8~\mathrm{eV}$. Recall that the measured value is $-78.6~\mathrm{eV}$ and the variational estimate is $-77.5~\mathrm{eV}$. [Hint: $\left\langle H^1\right\rangle$ can be viewed as the interaction between two concentric, spherically symmetric exponentially falling charge distributions. Find the potential $\phi(r)$ due to one distribution and calculate the interaction energy between this potential and the other charge distribution.]

**Exercise 17.2.6** Verify Eq. (17.2.34).

**Exercise 17.2.7** For the oscillator, consider $H^1=-qfX$. Find an $\Omega$ that satisfies Eq. (17.2.38). Feed it into Eq. (17.2.39) for $E_n^2$ and compare with the earlier calculation.

**Exercise 17.2.8** Fill in the steps connecting Eqs. (17.2.41) and (17.2.43). Try to use symmetry arguments to reduce the labor involved in evaluating the integrals.
## 17.3 Degenerate Pertubation Theory
**Exercise 17.3.1** Use the dipole selection rules to show that $H^1$ has the above form and carry out the evaluation of $\Delta$.

**Exercise 17.3.2** Consider a spin-1 particle (with no orbital degrees of freedom). Let $H=A S_z^2+B\left(S_x^2-S_y^2\right)$, where $S_i$ are $3 \times 3$ spin matrices, and $A \gg B$. Treating the $B$ term as a perturbation, find the eigenstates of $H^0=A S_z^2$ that are stable under the perturbation. Calculate the energy shifts to first order in $B$. How are these related to the exact answers?

**Exercise 17.3.3** Consider the case where $H^0$ includes the Coulomb plus spin-orbit interaction and $H^1$ is the effect of a weak magnetic field $\mathbf{B}=B \mathbf{k}$. Using the appropriate basis, show that the first-order level shift is related to $j_z$ by
$$
E^1=\left(\frac{e B}{2 m c}\right)\left(1 \pm \frac{1}{2 l+1}\right) j_z, \quad j=l \pm 1 / 2
$$
Sketch the levels for the $n=2$ level assuming that $E^1 \ll E_{\text{f.s.}}^1$.

**Exercise 17.3.4** We discuss here some tricks for evaluating the expectation values of certain operators in the eigenstates of hydrogen.

(1) Suppose we want $\langle 1 / r\rangle_{nlm}$. Consider first $\langle\lambda / r\rangle$. We can interpret $\langle\lambda / r\rangle$ as the first order correction due to a perturbation $\lambda / r$. Now this problem can be solved exactly; we just replace $e^2$ by $e^2-\lambda$ everywhere. (Why?) So the exact energy, from Eq. (13.1.16) is $E(\lambda)=$ $-\left(e^2-\lambda\right)^2 m / 2 n^2 \hbar^2$. The first-order correction is the term linear in $\lambda$, that is, $E^1=m e^2 \lambda / n^2 \hbar^2=$ $\langle\lambda / r\rangle$, from which we get $\langle 1 / r\rangle=1 / n^2 a_0$, in agreement with Eq. (13.1.36). For later use, let us observe that as $E(\lambda)=E^0+E^1+\cdots=E(\lambda=0)+\lambda(\mathrm{d}E/\mathrm{d}\lambda)_{\lambda=0}+\cdots$, one way to extract $E^1$ from the exact answer is to calculate $\lambda(\mathrm{d}E/\mathrm{d}\lambda)_{\lambda=0}$.

(2) Consider now $\left\langle\lambda / r^2\right\rangle$. In this case, an exact solution is possible since the perturbation just modifies the centrifugal term as follows:
$$
\frac{\hbar^2 l(l+1)}{2 m r^2}+\frac{\lambda}{r^2}=\frac{\hbar^2 l^{\prime}\left(l^{\prime}+1\right)}{2 m r^2}\tag{17.3.23}
$$
where $l^{\prime}$ is a function of $\lambda$. Now the dependence of $E$ on $l^{\prime}(\lambda)$ is, from Eq. (13.1.14),
$$
E\left(l^{\prime}\right)=\frac{-m e^4}{2 \hbar^2\left(k+l^{\prime}+1\right)^2}=E(\lambda)=E^0+E^1+\cdots
$$
Show that
$$
\left\langle\frac{\lambda}{r^3}\right\rangle=E^1=\left.\lambda \frac{\mathrm{d}E}{\mathrm{d}\lambda}\right|_{\lambda=0}=\left(\frac{\mathrm{d}E}{\mathrm{d}l^{\prime}}\right)_{l^{\prime}=l} \cdot\left(\frac{\mathrm{d}l^{\prime}}{\mathrm{d} \lambda}\right)_{l^{\prime}=l} \cdot \lambda=\frac{\lambda}{n^3 a_0^2\left(l+\frac{1}{2}\right)}
$$
Canceling $\lambda$ on both sides, we get Eq. (17.3.11).

(3) Consider finally $\left\langle l / r^3\right\rangle$. Since there is no such term in the Coulomb Hamiltonian, we resort to another trick. Consider the radial momentum operator, $p_r=-i \hbar(\partial / \partial r+1 / r)$, in terms of which we may write the radial part of the Hamiltonian
$$
\left(\frac{-\hbar^2}{2 m}\right)\left(\frac{1}{r^2} \frac{\partial}{\partial r} r^2 \frac{\partial}{\partial r}\right)
$$
as $p_r^2 / 2 m$. (Verify this.) Using the fact that $\left\langle\left[H, p_r\right]\right\rangle=0$ in the energy eigenstates, and by explictly evaluating the commutator, show that
$$
\left\langle\frac{1}{r^3}\right\rangle=\frac{1}{a_0(l)(l+1)}\left\langle\frac{1}{r^2}\right\rangle
$$
combining which with the result from part (2) we get Eq. (17.3.20).

(4) Find the mean kinetic energy using the trick from part (1), this time rescaling the mass. Regain the virial theorem.

§ QM-17 不含时微扰论

§ QM-17 Time-Independence Perturbation Theory

# 第16章 变分法与 WKB 方法
## 16.1 变分法
**习题 16.1.1** 对
$V=\dfrac{1}{2} m \omega^2 x^2$
取试探函数
$\psi=\exp \left(-\alpha x^2\right)$，
求 $\alpha_0$ 和 $E\left(\alpha_0\right)$。



**习题 16.1.2** 对于一个从 $-a$ 延伸到 $a$ 的势箱中的粒子，在势箱内取
$\psi=(x-a)(x+a)$，
并计算 $E$。这里没有可供变分的参数，但仍然能够得到一个上界。将它与真实能量 $E_0$ 比较。（说明 $\psi^{\prime \prime}$ 在 $x=\pm a$ 处的奇异性不会对能量作出贡献。）

**习题 16.1.3** 对吸引型 delta 函数势
$V=-a V_0 \delta(x)$，
使用 Gaussian 试探函数。计算 $E_0$ 的上界，并与精确结果
$-m a^2 V_0^2/2\hbar^2$
比较（见习题 5.2.3）。

**习题 16.1.4** 对谐振子取
$$
\begin{aligned}
\psi & =(x-a)^2(x+a)^2, & & |x| \leqslant a \\
& =0, & & |x|>a
\end{aligned}
$$
计算 $E(a)$，将其极小化，并与 $\hbar \omega / 2$ 比较。

**习题 16.1.5** 对处于势
$V=-e^2/r$
中的电子的 $l=1$ 状态求解变分问题。在试探函数中应包含：
(i) 当 $r\rightarrow 0$ 时与 $l=1$ 相适应的正确行为；
(ii) 为使能量最小所需的正确节点数；
(iii) 在 Coulomb 势中当 $r\rightarrow \infty$ 时波函数的正确行为（即指数衰减，而不是 Gaussian 衰减）。
对于 $Y_l^m$，选择哪个 $m$ 是否重要？讨论你得到的能量上界与精确结果之间的关系。

## 16.2 Wentzel-Kramers-Brillouin 方法
**习题 16.2.1** 考虑方程 (16.2.21a) 中引入的函数 $W(E)$。由于 $-E$ 是 $S(t)$ 对 $t$ 的导数，因此 $W(E)$ 必须是 $S$ 的 Legendre 变换。在这种情况下，$t$ 应当作为 $W(E)$ 对 $E$ 的导数出现。验证对公式
$$
W(E)=\int_{x^{\prime}}^x \sqrt{2 m\left(E-V\left(x^{\prime \prime}\right)\right)} \mathrm{d} x^{\prime \prime}
$$
关于 $E$ 求导，得到的是在预先给定能量下从起点运动到终点所需的时间 $t$。

**习题 16.2.2** 利用上面给出的方法考虑自由粒子问题。现在你已经显式知道波函数，利用围道积分计算方程 (16.2.18b) 中的积分，得到
$$
U\left(x, x^{\prime}, t\right)=\theta\left(x-x^{\prime}\right) \mathrm{e}^{(\mathrm{i}/\hbar) \sqrt{2 m E}\left(x-x^{\prime}\right)}+\theta\left(x^{\prime}-x\right) \mathrm{e}^{(-\mathrm{i}/\hbar) \sqrt{2 m E}\left(x-x^{\prime}\right)}
$$
在做围道积分时，要考虑对于给定的 $x-x^{\prime}$ 符号，应当在哪个半平面闭合围道。将上述结果与半经典结果比较，并确认它们彼此一致。

注意，这里没有必要构造组合
$U\left(x, x^{\prime}, t\right)+U^*\left(x, x^{\prime}, t\right)$；
由于我们显式知道被积函数对 $p$ 的依赖关系，因此可以分别计算主值部分和 delta 函数的贡献。为了更清楚地看到这一点，可以不采用围道积分，而改用上面给出的
$(x \pm i \varepsilon)^{-1}$
公式。计算主值积分和 delta 函数的贡献，并观察它们如何相加得到围道积分的结果。

虽然在这里两种方法都可行，但对于 $V\neq0$ 的问题，围道积分无法使用，因为我们不知道波函数在复 $p$ 平面中的 $p$ 依赖关系。使用 $U+U^*$ 方法的优点在于，它只涉及实轴上的量，而且只涉及一个能量值。

**习题 16.2.3** 让我们重新审视一下前面的推导。为了分离出某一个能量的本征函数，我们付出了相当多的努力：构造
$U\left(x, x^{\prime}, z\right)+U^*\left(x, x^{\prime}, z\right)$
来消去主值部分并筛选出 delta 函数。

现在很清楚，如果在方程 (16.2.18a) 中能够把积分范围扩展到
$-\infty\leqslant t\leqslant\infty$，
那么就会直接得到所需的 $\delta$ 函数。此前阻碍我们这样做的是：$U(t)$ 原本是为 $t>0$ 构造的。然而，如果对负时间使用时间演化算符
$\mathrm{e}^{-(\mathrm{i}/\hbar)Ht}$，
那么在假定 Hamilton量保持不变的前提下，它只是在告诉我们系统在更早时刻的状态。因此，我们同样可以赋予负时间下的算符以意义，并定义一个覆盖所有时间的变换。

在经典力学中也是如此。例如，如果从一座高楼上竖直向上抛出一块石头，并问它什么时候到达地面，我们会得到两个答案，其中负的 $t$ 对应于把给定初始条件向更早时间外推。

通过考察
$\langle x\mid \mathrm{e}^{-(\mathrm{i}/\hbar)Hf}\mid x^{\prime}\rangle$
在复共轭以及交换
$x \leftrightarrow x^{\prime}$
下的变化，验证
$U\left(x, x^{\prime}, z\right)+U^*\left(x, x^{\prime}, z\right)$
确实是 $U(t)$ 在整个时间轴上的变换。同样，考察把 $U_{\mathrm{cl}}$ 在所有时间上作变换时会发生什么。

此时你会发现，不管 $x-x^{\prime}$ 的符号如何，一条向右运动的经典轨道都可能作出贡献——只是稳定点的时间 $t^*$ 会随着 $x-x^{\prime}$ 的符号改变而改变符号。（对于向左运动的轨道也是如此。）

**习题 16.2.4** 动能为
$4.2~\mathrm{MeV}$
的 α 粒子通过隧穿从一个核电荷数
$Z=90$
的原子核中逸出（这里的 $Z$ 指发射之后的核）。假设
$V_0=0$、
$x_0=10^{-12}~\mathrm{cm}$，
并且当
$x\geqslant x_0$
时，$V(x)$ 完全是 Coulomb 势。（见图 16.1。）估算平均寿命。

[提示：证明
$\gamma=\left(8 Z e^2 / \hbar v\right)\left[\cos ^{-1} y^{1 / 2}-y^{1 / 2}(1-y)^{1 / 2}\right]$，
其中
$y=x_0 \mid x_e$。
证明 $y\ll1$，并在进行数值计算前使用
$\cos^{-1} y^{1/2} \simeq \dfrac{1}{2}\pi -y^{1/2}$。]

**习题 16.2.5** 1974 年发现了两个新粒子，称为 $\psi$ 和 $\psi^{\prime}$，它们的静止能量分别为
$3.1$ 和 $3.7~\mathrm{GeV}$，
其中
$1~\mathrm{GeV}=10^9~\mathrm{eV}$。
人们认为它们是一个质量为
$m=1.5~\mathrm{GeV}/c^2$
（即
$mc^2=1.5~\mathrm{GeV}$）
的粲夸克与一个同质量反夸克组成的非相对论束缚态，二者处于线性势
$V(r)=V_0+kr$
中。

假设这两个粒子分别是轨道角动量为零时的
$n=0$
和
$n=1$
束缚态，利用 WKB 公式计算 $V_0$。对于
$\psi^{\prime\prime}$，
即 $n=2$ 状态，你预测它的静止质量是多少？（实验测得约为
$4.2~\mathrm{GeV}/c^2$。）

[提示：(1) 使用 $\mathrm{GeV}$ 而不是 $\mathrm{eV}$ 进行计算。
(2) 没有必要显式求出 $k$。]

**习题 16.2.6** 利用标度技巧，对 $\lambda x^4$ 势推导方程 (16.2.39)。

**习题 16.2.7** 用 WKB 方法求谐振子的允许能级。

**习题 16.2.8** 考虑 Coulomb 问题中 $l=0$ 的径向方程。由于
$V(r)$
在转向点
$r=0$
处是奇异的，因此不能使用 $(n+3/4)$。

(1) 加性常数应当大于还是小于 $3/4$？

(2) 通过分析 $r=0$ 附近的精确方程，可以证明该常数等于 $1$。利用这一常数，证明 WKB 能级与精确结果一致。

# Chapter 16 Variational and WKB Methods
## 16.1 The Variatonal Method
**Exercise 16.1.1** Try $\psi=\exp \left(-\alpha x^2\right)$ for $V=\dfrac{1}{2} m \omega^2 x^2$ and find $\alpha_0$ and $E\left(\alpha_0\right)$.



**Exercise 16.1.2** For a particle in a box that extends from $-a$ to $a$, try (within the box) $\psi=(x-a)(x+a)$ and calculate $E$. There is no parameter to vary, but you still get an upper bound. Compare it to the true energy, $E_0$. (Convince yourself that the singularities in $\psi^{\prime \prime}$ at $x= \pm a$ do not contribute to the energy.)

**Exercise 16.1.3** For the attractive delta function potential $V=-a V_0 \delta(x)$ use a Gaussian trial function. Calculate the upper bound on $E_0$ and compare it to the exact answer ($-m a^2 V_0^2/2\hbar^2$) (Exercise 5.2.3).

**Exercise 16.1.4** For the oscillator choose
$$
\begin{aligned}
\psi & =(x-a)^2(x+a)^2, & & |x| \leqslant a \\
& =0, & & |x|>a
\end{aligned}
$$
Calculate $E(a)$, minimize it and compare to $\hbar \omega / 2$.

**Exercise 16.1.5** Solve the variational problem for the $l=1$ states of the electron in a potential $V=-e^2/r$. In your trial function incorporate (i) correct behavior as $r \rightarrow 0$, appropriate to $l=1$, (ii) correct number of nodes to minimize energy, (iii) correct behavior of wave function as $r \rightarrow \infty$ in a Coulomb potential (i.e., exponential instead of Gaussian damping). Does it matter what $m$ you choose for $Y_l^m$ ? Comment on the relation of the energy bound you obtain to the exact answer.
## 16.2 The Wentzel-Kramers-Brillouin Method
**Exercise 16.2.1** Consider the function $W(E)$ introduced in Eq. (16.2.21a). Since $-E$ is the $t$ derivative of $S(t)$, it follows that $W(E)$, must be the Legendre transform of $S$. In this case $t$ must emerge as the derivative of $W(E)$ with respect to $E$. Verify that differentiation of the formula
$$
W(E)=\int_{x^{\prime}}^x \sqrt{2 m\left(E-V\left(x^{\prime \prime}\right)\right)} \mathrm{d} x^{\prime \prime}
$$
gives the time $t$ taken to go from start to finish at this preassigned energy.

**Exercise 16.2.2** Consider the free particle problem using the approach given above. Now that you know the wave functions explicitly, evaluate the integral in Eq. (16.2.18b) by contour integration to obtain
$$
U\left(x, x^{\prime}, t\right)=\theta\left(x-x^{\prime}\right) \mathrm{e}^{(\mathrm{i}/\hbar) \sqrt{2 m E}\left(x-x^{\prime}\right)}+\theta\left(x^{\prime}-x\right) \mathrm{e}^{(-\mathrm{i}/\hbar) \sqrt{2 m E}\left(x-x^{\prime}\right)}
$$
In doing the contour integrals, ask in which half-plane you can close the contour for a given sign of $x-x^{\prime}$. Compare the above result to the semiclassical result and make sure it all works out. Note that there is no need to form the combination $U\left(x, x^{\prime}, t\right)+U^*\left(x, x^{\prime}, t\right)$; we can calculate both the principal part and the delta function contributions explicitly because we know the $p$-dependence of the integrand explicitly. To see this more clearly, avoid the contour integral approach and use instead the formula for $(x \pm i \varepsilon)^{-1}$ given above. Evaluate the principal value integral and the contribution from the delta function and see how they add up to the result of contour integration. Although both routes are possible here, in the problem with $V \neq 0$ contour integration is not possible since the $p$-dependence of the wave function in the complex $p$ plane is not known. The advantage of the $U+U^*$ approach is that it only refers to quantities on the real axis and at just one energy.

**Exercise 16.2.3** Let us take a second look at our derivation. We worked quite hard to isolate the eigenfunctions at one energy: we formed the combination $U\left(x, x^{\prime}, z\right)+U^*\left(x, x^{\prime}, z\right)$ to get rid of the principal part and filter out the delta function. Now it is clear that if in Eq. (16.2.18a) we could integrate in the range $-\infty \leqslant t \leqslant \infty$, we would get the $\delta$ function we want. What kept us from doing that was the fact the $U(t)$ was constructed to be used for $t>0$. However, if we use the time evolution operator $\mathrm{e}^{-(\mathrm{i}/\hbar)Ht}$ for negative times, it will simply tell us what the system was doing at earlier times, assuming the same Hamiltonian. Thus we can make sense of the operator for negative times as well and define a transform that extends over all times. This is true in classical mechanics as well. For instance, if a stone is thrown straight up from a tall building and we ask when it will be at ground level, we get two answers, one with negative $t$ corresponding to the extrapolation of the given initial conditions to earlier times. Verify that $U\left(x, x^{\prime}, z\right)+U^*\left(x, x^{\prime}, z\right)$ is indeed the transform of $U(t)$ for all times by seeing what happens to $\langle x\mid \mathrm{e}^{-(\mathrm{i}/\hbar)Hf}\mid x^{\prime}\rangle$ under complex conjugation and the exchange $x \leftrightarrow x^{\prime}$. Likewise, ask what happens to $U_{\mathrm{cl}}$ when we transform it for all times. Now you will find that no matter what the sign of $x-x^{\prime}$ a single right moving trajectory can contribute-it is just that the time of the stationary point, $t^*$, will change sign with the sign of $x-x^{\prime}$. (The same goes for a left moving trajectory.)

**Exercise 16.2.4** Alpha particles of kinetic energy $4.2~\mathrm{MeV}$ come tunneling out of a nucleus of charge $Z=90$ (after emission). Assume that $V_0=0$, $x_0=10^{-12}~\mathrm{cm}$, and $V(x)$ is just Coulombic for $x \geqslant x_0$. (See Fig. 16.1) Estimate the mean lifetime. [Hint: Show $\gamma=\left(8 Z e^2 / \hbar v\right)\left[\cos ^{-1} y^{1 / 2}-y^{1 / 2}(1-y)^{1 / 2}\right]$, where $y=x_0 \mid x_e$. Show that $y \ll 1$ and use $\cos^{-1} y^{1/2} \simeq \dfrac{1}{2}\pi -y^{1/2}$ before calculating numbers.]

**Exercise 16.2.5** In 1974 two new particles called the $\psi$ and $\psi^{\prime}$ were discovered, with rest energies $3.1$ and $3.7~\mathrm{GeV}$, respectively ($1~\mathrm{GeV}=10^9~\mathrm{eV}$). These are believed to be nonrelativistic bound states of a “charmed” quark of mass $m=1.5~\mathrm{GeV}/c^2$ (i.e., $mc^2=1.5~\mathrm{GeV}$) and an antiquark of the same mass, in a linear potential $V(r)=V_0+kr$. By assuming that these are the $n=0$ and $n=1$ bound states of zero orbital angular momentum, calculate $V_0$ using the WKB formula. What do you predict for the rest mass of $\psi^{\prime \prime}$, the $n=2$ state? (The measured value is $\simeq 4.2~\mathrm{GeV}/c^2$.) [Hints: (1) Work with $\mathrm{GeV}$ instead of $\mathrm{eV}$. (2) There is no need to determine $k$ explicitly.]

**Exercise 16.2.6** Obtain Eq. (16.2.39) for the $\lambda x^4$ potential by the scaling trick.

**Exercise 16.2.7** Find the allowed levels of the harmonic oscillator by the WKB method.

**Exercise 16.2.8** Consider the $l=0$ radial equation for the Coulomb problem. Since $V(r)$ is singular at the turning point $r=0$, we can't use ($n+3/4$).

(1) Will the additive constant be more or less than $3/4$?

(2) By analyzing the exact equation near $r=0$, it can be shown that the constant equals $1$. Using this constant show that the WKB energy levels agree with the exact results.

§ QM-16 变分法与 WKB 方法

§ QM-16 Variational and WKB Methods

# 第15章 角动量的合成
## 15.1 一个简单的例子
**习题 15.1.1** 推导方程 (15.1.10) 和 (15.1.11)。使用
$$
\mathbf{S}_1 \cdot \mathbf{S}_2=S_{1 z} S_{2 z}+\frac{1}{2}\left(S_{1+} S_{2-}+S_{1-} S_{2+}\right)\tag{15.1.12}
$$
可能会有所帮助。



**习题 15.1.2** 除了 Coulomb 相互作用之外，氢原子中的电子和质子之间还存在另一种相互作用，称为超精细相互作用。描述这种相互作用的 Hamilton量来自两个粒子的磁矩，其形式为
$$
H_{h f}=A \mathbf{S}_1 \cdot \mathbf{S}_2 \quad(A>0)\tag{15.1.22}
$$
（该公式假设电子的轨道状态为 $|1,0,0\rangle$。）因此，总 Hamilton量就是 Coulomb Hamilton量加上 $H_{hf}$。

\noindent (1) 证明 $H_{hf}$ 将基态分裂为两个能级：
$$
\begin{aligned}
E_{+} & =-\mathrm{Ry}+\frac{\hbar^2 A}{4} \\
E_{-} & =-\mathrm{Ry}-\frac{3 \hbar^2 A}{4}
\end{aligned}\tag{15.1.23}
$$
并证明相应的状态分别为三重态和单重态。

(2) 尝试估算原子从三重态跃迁到单重态时所发射辐射的频率。为此，可以把电子和质子近似看作两个相距 $a_0$ 的偶极子 $\mu_e$ 和 $\mu_p$，其相互作用能量的量级为（这里的描述经过了过度简化；实际上 $\mathscr{H}_{hf}$ 和 $H_{hf}$ 的推导都相当微妙。我们的目的只是估算 $|A|$，而不是讨论其精确来源。）
$$
\mathscr{H}_{h f} \cong \frac{\mu_e \cdot \mu_p}{a_0^3}
$$
证明这意味着方程 (15.1.22) 中的常数为
$$
A \sim \frac{2 e}{2 m c} \frac{(5.6) e}{2 M c} \frac{1}{a_0^3}
$$
（其中 5.6 是质子的 $g$ 因子），并且
$$
\Delta E=E_{+}-E_{-}=A \hbar^2
$$
相对于基态能量而言，是一个 $(m/M)\alpha^2$ 阶的修正。利用第 13 章中的估算口诀，估算所发射辐射的波长为几十厘米。实验测得的数值为 $21.4~\mathrm{cm}$。这种辐射称为 $21~\mathrm{cm}$ 线，是探测宇宙其他区域中氢的一种方法。

(3) 估算室温热平衡下氢原子处于三重态和单重态的概率之比
$P(\text{三重态})/P(\text{单重态})$。

## 15.2 一般问题
**习题 15.2.1** (1) 令
$J^2=J_1^2+J_2^2+2 J_{1 z} J_{2 z}+J_{1+} J_{2-}+J_{1-} J_{2+}$
作用在
$\left|j_1 j_1, j_2 j_2\right\rangle$
上，验证它确实是
$j=j_1+j_2$
的状态。

(2) 验证方程 (15.2.8) 右边确实具有角动量
$j=j_1+j_2-1$。

**习题 15.2.2** 求下列角动量合成中的 CG 系数：

(1)
$\dfrac{1}{2}\otimes 1=\dfrac{3}{2}\oplus \dfrac{1}{2}$

(2)
$1\otimes 1=2\oplus 1\oplus 0$

**习题 15.2.3** 说明
$\dfrac{1}{2}\otimes\dfrac{1}{2}\otimes\dfrac{1}{2}
=\dfrac{3}{2}\oplus\dfrac{1}{2}\oplus\dfrac{1}{2}$。

**习题 15.2.4** 推导方程 (15.2.19) 和 (15.2.20)。

**习题 15.2.5** (1) 证明
$\mathbb{P}_1=\dfrac{3}{4} I+\left(\mathbf{S}_1 \cdot \mathbf{S}_2\right) / \hbar^2$
以及
$\mathbb{P}_0=\dfrac{1}{4} I-\left(\mathbf{S}_1 \cdot \mathbf{S}_2\right) / \hbar^2$
是投影算符，即满足
$\mathbb{P}_i \mathbb{P}_j=\delta_{i j} \mathbb{P}_j$
[使用方程 (14.3.39)]。

(2) 证明它们分别投影到
$\dfrac{1}{2} \otimes \dfrac{1}{2}=1 \oplus 0$
中的自旋 1 和自旋 0 子空间。

**习题 15.2.6** 对于角动量合成
$\mathbf{L}+\mathbf{S}=\mathbf{J}$，
构造对应于
$j=l\pm1/2$
子空间的投影算符
$\mathbb{P}_{\pm}$。

**习题 15.2.7** 证明：当把 $j_1$ 与 $j_1$ 合成时，$j=2j_1$ 的状态是对称的。证明
$j=2j_1=1$
的状态是反对称的。
（先说明最高权态的对称性，并证明施加降算符不会改变其对称性。）
随着 $j$ 继续减小，这种对称性与反对称性交替出现的规律仍然成立，不过证明会更加困难。

## 15.3 不可约张量算符
**习题 15.3.1** (1) 考虑无穷小旋转，证明方程 (15.3.11) 可由方程 (15.3.10) 推出。
[提示：
$D_{q^{\prime} q}^{(k)}=\left\langle kq^{\prime}\right| I-(\mathrm{i}\delta\boldsymbol{\theta} \cdot \mathbf{J}) / \hbar|k q\rangle$。
例如先取
$\delta\boldsymbol{\theta}$
沿 $x$ 方向，再将结果推广到其他方向。]

(2) 验证按照方程 (15.3.15) 由 $V$ 构造出的球张量
$V_1^q$
满足方程 (15.3.11)。

**习题 15.3.2** 声称
$\sum\limits_q (-1)^q S_k^q T_k^{(-q)}$
是一个标量算符。

(1) 对 $k=1$，验证它就是
$\mathbf{S}\cdot\mathbf{T}$。

(2) 通过考察它在旋转下的变换，一般性地证明这一结论。
[提示：
$D_{-m,-m^{\prime}}^{(j)}=(-1)^{m-m^{\prime}}\left(D_{m, m^{\prime}}^{(j)}\right)^*$。]

**习题 15.3.3** (1) 利用
$\langle j j \mid j j, 10\rangle=[j /(j+1)]^{1 / 2}$
证明
$$
\langle\alpha j|\left|J_1\right|\left|\alpha^{\prime} j^{\prime}\right\rangle=\delta_{\alpha \alpha^{\prime}} \delta_{j j} \hbar[j(j+1)]^{1 / 2}
$$

(2) 利用
$\boldsymbol{J} \cdot \mathbf{A}=J_z A_z+\dfrac{1}{2}\left(J_{-} A_{+}+J_{+} A_{-}\right)$
（其中
$A_{\pm}=A_x \pm i A_y$），说明
$$
\left\langle\alpha^{\prime} j m^{\prime}\right| \mathbf{J} \cdot \mathbf{A}|\alpha j m\rangle=c\left\langle\alpha^{\prime} j\mid |A| \mid \alpha j\right\rangle
$$
其中 $c$ 是一个与
$\alpha,\alpha^{\prime}$
以及 $\mathbf{A}$ 无关的常数。证明
$$
c=\hbar[j(j+1)]^{1 / 2} \delta_{m, m^{\prime}}.
$$

(3) 利用上述结果，证明
$$
\left\langle\alpha^{\prime} j m^{\prime}\right| A^q|\alpha j m\rangle=\frac{\left\langle\alpha^{\prime} j m\right| \mathbf{J} \cdot \mathbf{A}|\alpha j m\rangle}{\hbar^2 j(j+1)}\left\langle j m^{\prime}\right| J^q|j m\rangle
$$

**习题 15.3.4** (1) 考虑一个系统，其角动量由两部分
$\mathbf{J}_1$
和
$\mathbf{J}_2$
组成，而磁矩为
$$
\boldsymbol{\mu}=\gamma_1 \mathbf{J}_1+\gamma_2 \mathbf{J}_2
$$
在状态
$\left|j m, j_1 j_2\right\rangle$
中，利用方程 (15.3.19) 证明
$$
\begin{aligned}
& \left\langle\mu_x\right\rangle=\left\langle\mu_y\right\rangle=0 \\
& \left\langle\mu_z\right\rangle=m \hbar\left[\frac{\gamma_1+\gamma_2}{2}+\frac{\left(\gamma_1-\gamma_2\right)}{2} \frac{j_1\left(j_1+1\right)-j_2\left(j_2+1\right)}{j(j+1)}\right]
\end{aligned}
$$

(2) 将这一结果应用到处于
${ }^2 P_{1 / 2}$
状态的质子 $(g=5.6)$，证明
$\left\langle\mu_z\right\rangle=\pm0.26$
个核磁子。

(3) 对于处于
${ }^2 P_{1 / 2}$
状态的电子，证明
$\left\langle\mu_z\right\rangle=\pm\dfrac{1}{3}$
个 Bohr 磁子。

**习题 15.3.5** 证明当
$k>2j$
时，
$\langle j m\mid T_k^q\mid j m\rangle=0$。

## 15.4 一些“偶然”简并的解释

# Chapter 15 Addition of Angular Momenta
## 15.1 A Simple Example
**Exercise 15.1.1** Derive Eqs. (15.1.10) and (15.1.11). It might help to use
$$
\mathbf{S}_1 \cdot \mathbf{S}_2=S_{1 z} S_{2 z}+\frac{1}{2}\left(S_{1+} S_{2-}+S_{1-} S_{2+}\right)\tag{15.1.12}
$$



**Exercise 15.1.2** In addition to the Coulomb interaction, there exists another, called the hyperfine interaction, between the electron and proton in the hydrogen atom. The Hamiltonian describing this interaction, which is due to the magnetic moments of the two particles is,
$$
H_{h f}=A \mathbf{S}_1 \cdot \mathbf{S}_2 \quad(A>0)\tag{15.1.22}
$$
(This formula assumes the orbital state of the electron is $|1,0,0\rangle$.) The total Hamiltonian is thus the Coulomb Hamiltonian plus $H_{h f}$.

\noindent (1) Show that $H_{h f}$ splits the ground state into two levels:
$$
\begin{aligned}
E_{+} & =-\mathrm{Ry}+\frac{\hbar^2 A}{4} \\
E_{-} & =-\mathrm{Ry}-\frac{3 \hbar^2 A}{4}
\end{aligned}\tag{15.1.23}
$$
and that corresponding states are triplets and singlet, respectively.

(2) Try to estimate the frequency of the emitted radiation as the atom jumps from the triplet to the singlet. To do so, you may assume that the electron and proton are two dipoles $\mu_e$ and $\mu_p$ separated by a distance $a_0$, with an interaction energy of the order (The description here is oversimplified; both $\mathscr{H}_{hf}$ and $H_{hf}$ are rather tricky to derive. Our aim is just to estimate $|A|$ and not to get into its precise origin.)
$$
\mathscr{H}_{h f} \cong \frac{\mu_e \cdot \mu_p}{a_0^3}
$$
Show that this implies that the constant in Eq. (15.1.22) is
$$
A \sim \frac{2 e}{2 m c} \frac{(5.6) e}{2 M c} \frac{1}{a_0^3}
$$
(where 5.6 is the $g$ factor for the proton), and that
$$
\Delta E=E_{+}-E_{-}=A \hbar^2
$$
is a correction of order $(m / M) \alpha^2$ relative to the ground-state energy. Estimate that the frequency of emitted radiation is a few tens of centimeters, using the mnemonics from Chapter 13. The measured value is $21.4~\mathrm{cm}$. This radiation, called the $21-\mathrm{cm}$ line, is a way to detect hydrogen in other parts of the universe.

(3) Estimate the probability ratio $P(\text{triplet})/P(\text{singlet})$ of hydrogen atoms in thermal equilibrium at room temperature.
## 15.2 The General Problem
**Exercise 15.2.1** (1) Verify that $\left|j_1 j_1, j_2 j_2\right\rangle$ is indeed a state of $j=j_1+j_2$ by letting $J^2=J_1^2+J_2^2+2 J_{1 z} J_{2 z}+J_{1+} J_{2-}+J_{1-} J_{2+}$ act on it.

(2) Verify that the right-hand side of Eq. (15.2.8) indeed has angular momentum $j=j_1+j_2-1$.

**Exercise 15.2.2** Find the CG coefficients of

(1) $\dfrac{1}{2}\otimes 1=\dfrac{3}{2}\oplus \dfrac{1}{2}$

(2) $1\otimes 1=2\oplus 1\oplus 0$

**Exercise 15.2.3** Argue that $\dfrac{1}{2}\otimes\dfrac{1}{2}\otimes\dfrac{1}{2}=\dfrac{3}{2}\oplus\dfrac{1}{2}\oplus\dfrac{1}{2}$.

**Exercise 15.2.4** Derive Eqs. (15.2.19) and (15.2.20).

**Exercise 15.2.5** (1) Show that $\mathbb{P}_1=\dfrac{3}{4} I+\left(\mathbf{S}_1 \cdot \mathbf{S}_2\right) / \hbar^2$ and $\mathbb{P}_0=\dfrac{1}{4} I-\left(\mathbf{S}_1 \cdot \mathbf{S}_2\right) / \hbar^2$ are projection operators, i.e., obey $\mathbb{P}_i \mathbb{P}_j=\delta_{i j} \mathbb{P}_j$ [use Eq. (14.3.39)].

(2) Show that these project into the spin-1 and spin-0 spaces in $\dfrac{1}{2} \otimes \dfrac{1}{2}=1 \oplus 0$.

**Exercise 15.2.6** Construct the project operators $\mathbb{P}_{\pm}$ for the $j=l \pm 1 / 2$ subspaces in the addition $\mathbf{L}+\mathbf{S}=\mathbf{J}$.

**Exercise 15.2.7** Show that when we add $j_1$ to $j_1$, the states with $j=2 j_1$ are symmetric. Show that the states with $j=2 j_1=1$ are antisymmetric. (Argue for the symmetry of the top states and show that lowering does not change symmetry.) This pattern of alternating symmetry continues as $j$ decreases, but is harder to prove.
## 15.3 Irreducible Tensor Operators
**Exercise 15.3.1** (1) Show that Eq. (15.3.11) follows from Eq. (15.3.10) when one considers infinitesimal rotations. (Hint: $D_{q^{\prime} q}^{(k)}=\left\langle kq^{\prime}\right| I-(\mathrm{i}\delta\boldsymbol{\theta} \cdot \mathbf{J}) / \hbar|k q\rangle$. Pick $\delta \boldsymbol{\theta}$ along, say, the $x$ direction and then generalize the result to the other directions.)

(2) Verify that the spherical tensor $V_1^q$ constructed out of $V$ as in Eq. (15.3.15) obeys Eq. (15.3.11).

**Exercise 15.3.2** It is claimed that $\sum\limits_q (-1)^q S_k^q T_k^{(-q)}$ is a scalar operator.

(1) For $k=1$, verify that this is just $\mathbf{S} \cdot \mathbf{T}$.

(2) Prove it in general by considering its response to a rotation. [Hint: $D_{-m,-m^{\prime}}^{(j)}=(-1)^{m-m^{\prime}}\left(D_{m, m^{\prime}}^{(j)}\right)^*$.]

**Exercise 15.3.3** (1) Using $\langle j j \mid j j, 10\rangle=[j /(j+1)]^{1 / 2}$ show that
$$
\langle\alpha j|\left|J_1\right|\left|\alpha^{\prime} j^{\prime}\right\rangle=\delta_{\alpha \alpha^{\prime}} \delta_{j j} \hbar[j(j+1)]^{1 / 2}
$$

(2) Using $\boldsymbol{J} \cdot \mathbf{A}=J_z A_z+\dfrac{1}{2}\left(J_{-} A_{+}+J_{+} A_{-}\right)$(where $A_{\pm}=A_x \pm i A_y$ ) argue that
$$
\left\langle\alpha^{\prime} j m^{\prime}\right| \mathbf{J} \cdot \mathbf{A}|\alpha j m\rangle=c\left\langle\alpha^{\prime} j\mid |A| \mid \alpha j\right\rangle
$$
where $c$ is a constant independent of $\alpha, \alpha^{\prime}$ and $\mathbf{A}$. Show that $c=\hbar[j(j+1)]^{1 / 2} \delta_{m, m^{\prime}}$.

(3) Using the above, show that
$$
\left\langle\alpha^{\prime} j m^{\prime}\right| A^q|\alpha j m\rangle=\frac{\left\langle\alpha^{\prime} j m\right| \mathbf{J} \cdot \mathbf{A}|\alpha j m\rangle}{\hbar^2 j(j+1)}\left\langle j m^{\prime}\right| J^q|j m\rangle
$$

**Exercise 15.3.4** (1) Consider a system whose angular momentum consists of two parts $\mathbf{J}_1$ and $\mathbf{J}_2$ and whose magnetic moment is
$$
\boldsymbol{\mu}=\gamma_1 \mathbf{J}_1+\gamma_2 \mathbf{J}_2
$$
In a state $\left|j m, j_1 j_2\right\rangle$ show, using Eq. (15.3.19), that
$$
\begin{aligned}
& \left\langle\mu_x\right\rangle=\left\langle\mu_y\right\rangle=0 \\
& \left\langle\mu_z\right\rangle=m \hbar\left[\frac{\gamma_1+\gamma_2}{2}+\frac{\left(\gamma_1-\gamma_2\right)}{2} \frac{j_1\left(j_1+1\right)-j_2\left(j_2+1\right)}{j(j+1)}\right]
\end{aligned}
$$

(2) Apply this to the problem of a proton $(g=5.6)$ in a ${ }^2 P_{1 / 2}$ state and show that $\left\langle\mu_z\right\rangle=$ $\pm 0.26$ nuclear magnetons.

(3) For an electron in a ${ }^2 P_{1 / 2}$ state show that $\left\langle\mu_z\right\rangle=\pm \dfrac{1}{3}$ Bohr magnetons.

**Exercise 15.3.5** Show that $\langle j m\mid T_k^q\mid j m\rangle=0$ if $k>2 j$.
## 15.4 Explanation of Some “Accidental” Degeneracies

第19章散射理论

19.1 引言

19.2 一维散射的回顾与概述

19.3 Born 近似（含时描述）

第18章含时微扰论

18.1 问题

18.2 一阶微扰论

第17章不含时微扰论

17.1 形式

17.2 一些例子

第16章变分法与 WKB 方法

16.1 变分法

第15章角动量的合成

15.1 一个简单的例子

第19章 散射理论