# 第9章 Heisenberg 不确定关系

## 9.1 引言

## 9.2 不确定关系的推导

## 9.3 最小不确定度波包

## 9.4 不确定性原理的应用

**习题 9.4.1** 考虑处于状态 $|n=1\rangle$ 的谐振子，验证

$$
\left\langle\frac{1}{X^2}\right\rangle \simeq \frac{1}{\left\langle X^2\right\rangle} \simeq \frac{m \omega}{\hbar}
$$



**习题 9.4.2** (1) 参考附录 A.2 中的积分表，验证

$$
\psi=\frac{1}{\left(\pi a_0^3\right)^{1 / 2}} e^{-r / a_0}, \quad r=\left(x^2+y^2+z^2\right)^{1 / 2}
$$

是一个归一化波函数（氢原子的基态波函数）。注意，在三维中，对于只依赖于 $r$ 的函数，归一化条件为

$$
\begin{aligned}
\langle\psi \mid \psi\rangle & =\int \psi^*(r, \theta, \phi) \psi(r, \theta, \phi) r^2 \mathrm{d} r \mathrm{d}(\cos \theta) \mathrm{d} \phi \\
& =4 \pi \int \psi^*(r) \psi(r) r^2 \mathrm{d} r=1
\end{aligned}
$$

(2) 计算该状态中的 $(\Delta X)^2$ [说明 $(\Delta X)^2=\dfrac{1}{3}\left\langle r^2\right\rangle$]，并重新得到方程 (9.4.9) 中给出的结果。

(3) 证明在该状态中
$\langle 1 / r\rangle \simeq 1 /\langle r\rangle \simeq m e^2 / \hbar^2$。

**习题 9.4.3** 暂时忽略氢原子是一个三维系统这一事实，并假装

$$
H=\frac{P^2}{2 m}-\frac{e^2}{\left(R^2\right)^{1 / 2}} \quad\left(P^2=P_x^2+P_y^2+P_z^2, R^2=X^2+Y^2+Z^2\right)
$$

对应于一个一维问题。假设

$$
\Delta P \cdot \Delta R \geqslant \hbar / 2
$$

估算基态能量。

**习题 9.4.4** 计算 $\Delta T \cdot \Delta X$，其中 $T=P^2/2m$。为什么这一关系没有那么著名？

## 9.5 能量—时间不确定关系

# Chapter 9 The Heisenberg Uncertainty Relations
## 9.1 Introduction
## 9.2 Derivation of the Uncertainty Relations
## 9.3 The Minimum Uncertainty Packet
## 9.4 Applications of the Uncertainty Principle
**Exercise 9.4.1** Consider the oscillator in the state $|n=1\rangle$ and verify that
$$
\left\langle\frac{1}{X^2}\right\rangle \simeq \frac{1}{\left\langle X^2\right\rangle} \simeq \frac{m \omega}{\hbar}
$$



**Exercise 9.4.2** (1) By referring to the table of integrals in Appendix A.2, verify that
$$
\psi=\frac{1}{\left(\pi a_0^3\right)^{1 / 2}} e^{-r / a_0}, \quad r=\left(x^2+y^2+z^2\right)^{1 / 2}
$$
is a normalized wave function (of the ground state of hydrogen). Note that in three dimensions the normalization condition is
$$
\begin{aligned}
\langle\psi \mid \psi\rangle & =\int \psi^*(r, \theta, \phi) \psi(r, \theta, \phi) r^2 \mathrm{d} r \mathrm{d}(\cos \theta) \mathrm{d} \phi \\
& =4 \pi \int \psi^*(r) \psi(r) r^2 \mathrm{d} r=1
\end{aligned}
$$
for a function of just $r$.

(2) Calculate $(\Delta X)^2$ in this state [argue that $(\Delta X)^2=\dfrac{1}{3}\left\langle r^2\right\rangle$] and 
regain the result quoted in Eq. (9.4.9).

(3) Show that $\langle 1 / r\rangle \simeq 1 /\langle r\rangle \simeq m e^2 / \hbar^2$ in this state.

**Exercise 9.4.3** Ignore the fact that the hydrogen atom is a three-dimensional system and pretend that
$$
H=\frac{P^2}{2 m}-\frac{e^2}{\left(R^2\right)^{1 / 2}} \quad\left(P^2=P_x^2+P_y^2+P_z^2, R^2=X^2+Y^2+Z^2\right)
$$
corresponds to a one-dimensional problem. Assuming
$$
\Delta P \cdot \Delta R \geqslant \hbar / 2
$$
estimate the ground-state energy.

**Exercise 9.4.4** Compute $\Delta T \cdot \Delta X$, where $T=P^2/2m$. Why is this relation not so famous?

## 9.5 The Energy-Time Unvertainty Relation

§ QM-09 Heisenberg不确定性关系

§ QM-09 The Heisenberg Uncertainty Relations

# 第8章 量子理论的路径积分形式

## 8.1 路径积分处方

## 8.2 对该处方的分析

## 8.3 自由粒子 $U(t)$ 的一个近似

## 8.4 自由粒子传播子的路径积分计算

## 8.5 与 Schrödinger 方程的等价性

## 8.6 形如 $V=a+bx+cx^{2}+d\dot{x}+ex\dot{x}$ 的势

**习题 8.6.1** 验证

$$
U(x,t;x^{\prime},0)=A(t)\exp(\mathrm{i}S_{\text{cl}}/\hbar),~A(t)=\left(\dfrac{m}{2\pi\hbar\mathrm{i}t}\right)^{1/2}
$$

与 $V(x)=-fx$ 时的精确结果，即方程 (5.4.31)，一致。提示：从
$x_{\text{cl}}(t^{\prime\prime})=x_{0}+v_{0}t^{\prime\prime}+\dfrac{1}{2}(f/m)t^{\prime\prime 2}$
出发，并利用条件
$x_{\text{cl}}(0)=x^{\prime}$
和
$x_{\text{cl}}(t)=x$
求出常数 $x_{0}$ 和 $v_{0}$。



**习题 8.6.2** 证明对于 Lagrange量为

$$
\mathscr{L}=\dfrac{1}{2}m\dot{x}^{2}-\dfrac{1}{2}m\omega^{2}x^{2}
$$

的谐振子，

$$
U(x,t;x^{\prime})=A(t)\exp\left\{\dfrac{\mathrm{i}m\omega}{2\hbar\sin\omega t}[(x^{2}+x^{\prime 2})\cos\omega t-2xx^{\prime}]\right\}
$$

其中 $A(t)$ 是一个未知函数。（回忆习题 2.8.7。）

**习题 8.6.3** 我们知道，给定本征函数和本征值之后，可以构造传播子：
\begin{equation}\label{8.6.15}
U(x,t;x^{\prime},t^{\prime})=\sum_{n}\psi_{n}(x)\psi_{n}^{*}(x^{\prime})\mathrm{e}^{-\mathrm{i}E_{n}(t-t^{\prime})/\hbar}\tag{8.6.15}
\end{equation}
对于谐振子，考虑反过来的过程（因为路径积分方法能够直接给出 $U$）。

(1) 令 $x=x^{\prime}=t^{\prime}=0$。假设对于谐振子
$A(t)=(m\omega/2\pi\mathrm{i}\hbar\sin\omega t)^{1/2}$。
通过展开方程 (\ref{8.6.15}) 的两边，你应该得到
$E=\hbar\omega/2$、$5\hbar\omega/2$、$9\hbar\omega/2$、$\ldots$ 等。中间那些能级发生了什么？

(2) 现在考虑如何提取本征函数。令 $x=x^{\prime}$ 且 $t^{\prime}=0$。通过按
$\alpha=\exp(\mathrm{i}\omega t)$
的幂展开，求出
$E_{0}$、$E_{1}$、$|\psi_{0}(x)|^{2}$ 和 $|\psi_{1}(x)|^{2}$。

**习题 8.6.4** 回忆从方程 (8.5.4) 出发推导 Schrödinger 方程 (8.5.8) 的过程。注意，尽管我们把 $V$ 的宗量取为中点 $x+x^{\prime} / 2$，实际上这并没有太大影响：对于
$0\leqslant \alpha \leqslant 1$，
任何选择
$x+\alpha \eta$
（其中 $\eta=x^{\prime}-x$）
都会给出相同的结果，因为不同选择之间的差异是
$\eta \varepsilon \simeq \varepsilon^{3 / 2}$
阶的。这一切都得益于方程 (8.5.4) 中乘在 $V$ 前面的因子 $\varepsilon$，以及根据方程 (8.6.5) 有
$|\eta| \simeq \varepsilon^{1 / 2}$。

# Chapter 8 The Path Integral Formulation of Quantum Theory
## 8.1 The Path Integral Recipe
## 8.2 Analysis of the Recipe
## 8.3 An Approximation to $U(t)$ for the Free Particle
## 8.4 Path Integral Evaluation of the Free-Particle Propagator
## 8.5 Equivalence to the Schrödinger Equation
## 8.6 Potentials of the Form $V=a+bx+cx^{2}+d\dot{x}+ex\dot{x}$
**Exercise 8.6.1** Verify that
$$
U(x,t;x^{\prime},0)=A(t)\exp(\mathrm{i}S_{\text{cl}}/\hbar),~A(t)=\left(\dfrac{m}{2\pi\hbar\mathrm{i}t}\right)^{1/2}
$$
agrees with the exact result, Eq. (5.4.31), for $V(x)=-fx$. Hint: Start with $x_{\text{cl}}(t^{\prime\prime})=x_{0}+v_{0}t^{\prime\prime}+\dfrac{1}{2}(f/m)t^{\prime\prime 2}$ and find the constants $x_{0}$ and $v_{0}$ from the requirement that $x_{\text{cl}}(0)=x^{\prime}$ and $x_{\text{cl}}(t)=x$.



**Exercise 8.6.2** Show that for the harmonic oscillator with
$$
\mathscr{L}=\dfrac{1}{2}m\dot{x}^{2}-\dfrac{1}{2}m\omega^{2}x^{2}
$$
$$
U(x,t;x^{\prime})=A(t)\exp\left\{\dfrac{\mathrm{i}m\omega}{2\hbar\sin\omega t}[(x^{2}+x^{\prime 2})\cos\omega t-2xx^{\prime}]\right\}
$$
where $A(t)$ is an unknown function. (Recall Exercise 2.8.7.)

**Exercise 8.6.3** We know that given the eigenfunctions and the eigenvalues we can construct the propagator:
\begin{equation}\label{8.6.15}
U(x,t;x^{\prime},t^{\prime})=\sum_{n}\psi_{n}(x)\psi_{n}^{*}(x^{\prime})\mathrm{e}^{-\mathrm{i}E_{n}(t-t^{\prime})/\hbar}\tag{8.6.15}
\end{equation}
Consider the reverse process (since the path integral approach gives $U$ directly), for the case of the oscillator.

(1) Set $x=x^{\prime}=t^{\prime}=0$. Assume that $A(t)=(m\omega/2\pi\mathrm{i}\hbar\sin\omega t)^{1/2}$ for the oscillator. By expanding both sides of Eq. (\ref{8.6.15}), you should find that $E=\hbar\omega/2$, $5\hbar\omega/2$, $9\hbar\omega/2$, $\ldots$, etc. What happened to the levels in between?

(2) Now consider the extraction of the eigenfunctions. Let $x=x^{\prime}$ and $t^{\prime}=0$. Find $E_{0}$, $E_{1}$, $|\psi_{0}(x)|^{2}$, and $|\psi_{1}(x)|^{2}$ by expanding in powers of $\alpha=\exp(\mathrm{i}\omega t)$.

**Exercise 8.6.4** Recall the derivation of the Schrödinger equation (8.5.8) starting from Eq. (8.5.4). Note that although we chose the argument of $V$ to be the midpoint $x+x^{\prime} / 2$, it did not matter very much: any choice $x+\alpha \eta$, (where $\eta=x^{\prime}-x$) for $0\leqslant \alpha \leqslant 1$ would have given the same result since the difference between the choices is of order $\eta \varepsilon \simeq \varepsilon^{3 / 2}$. All this was thanks to the factor $\varepsilon$ multiplying $V$ in Eq. (8.5.4) and the fact that $|\eta| \simeq \varepsilon^{1 / 2}$, as per Eq. (8.6.5).

§ QM-08 量子理论的路径积分形式

§ QM-08 The Path Integral Formulation of Quantum Theory

# 第7章 谐振子
## 7.1 为什么要研究谐振子？
## 7.2 经典谐振子的回顾
## 7.3 谐振子的量子化（坐标基底）
**习题 7.3.1** 考虑这样一个问题：为什么对于方程 (7.3.11) 我们尝试了幂级数解，而对于方程 (7.3.8) 却没有这样做。把一个级数代入后一个方程，验证会得到 $C_{n+2}$、$C_n$ 和 $C_{n-2}$ 之间的三项递推关系，而从这个关系并不能很方便地得到解。问题在于，$\psi^{\prime\prime}$ 比 $2\varepsilon\psi$ 少两个 $y$ 的幂次，而 $-y^2$ 项则多两个 $y$ 的幂次。另一方面，在方程 (7.3.11) 中，三项 $u^{\prime\prime}$、$-2yu^{\prime}$ 和 $(2\varepsilon-1)u$ 中，后两项具有相同的 $y$ 幂次。



**习题 7.3.2** 验证 $H_3(y)$ 和 $H_4(y)$ 满足递推关系，即方程 (7.3.15)。

**习题 7.3.3** 如果 $\psi(x)$ 在 $x\rightarrow -x$ 下为偶函数，而 $\phi(x)$ 为奇函数，证明
$$
\int_{-\infty}^{\infty} \psi(x) \phi(x) \mathrm{d}x=0
$$
利用这一结果证明 $\psi_2(x)$ 和 $\psi_1(x)$ 正交。再利用附录 A.2 中 Gaussian 积分的结果，验证 $\psi_2(x)$ 和 $\psi_0(x)$ 正交。

**习题 7.3.4** 利用方程 (7.3.23)–(7.3.25)，证明
$$
\begin{aligned}
& \left\langle n^{\prime}\right| X|n\rangle=\left(\frac{\hbar}{2 m \omega}\right)^{1 / 2}\left[\delta_{n^{\prime}, n+1}(n+1)^{1 / 2}+\delta_{n^{\prime}, n-1} n^{1 / 2}\right] \\
& \left\langle n^{\prime}\right| P|n\rangle=\left(\frac{m \omega \hbar}{2}\right)^{1/2} \mathrm{i}\left[\delta_{n^{\prime}, n+1}(n+1)^{1 / 2}-\delta_{n^{\prime}, n-1} n^{1 / 2}\right]
\end{aligned}
$$

**习题 7.3.5** 利用习题 7.3.3 中的对称性论证，证明 $\langle n|X|n\rangle=\langle n|P|n\rangle=0$，从而在这些状态中
$\left\langle X^2\right\rangle=(\Delta X)^2$ 且 $\left\langle P^2\right\rangle=(\Delta P)^2$。
证明
$\langle 1| X^2|1\rangle=3 \hbar / 2 m \omega$
以及
$\langle 1| P^2|1\rangle=\dfrac{3}{2} m \omega \hbar$。
证明 $\psi_0(x)$ 恰好达到不确定关系下界
$\Delta X \cdot \Delta P \geqslant \hbar / 2$。

**习题 7.3.6** 考虑处于下列势中的粒子：
$$
\begin{aligned}
V(x) & =\frac{1}{2} m \omega^2 x^2, & & x>0 \\
& =\infty, & & x \leqslant 0
\end{aligned}
$$
此时波函数的边界条件是什么？求本征值和本征函数。

**习题 7.3.7** （动量空间中的谐振子）在 $P$ 基底中建立谐振子的本征值方程，并将其与方程 (7.3.2) 比较，证明动量空间中的本征函数可以由坐标空间中的本征函数通过代换
$x \rightarrow p$、$m \omega \rightarrow 1 / m \omega$
得到。因此，例如
$$
\psi_0(p)=\left(\frac{1}{m \pi \hbar \omega}\right)^{1 / 4} e^{-p^2 / 2 m \hbar \omega}
$$
还有若干其他量对，例如状态 $|n\rangle$ 中的 $\Delta X$ 和 $\Delta P$，也通过代换
$m \omega \rightarrow 1 / m \omega$
彼此联系。你可以留意这些关系。（回看习题 7.3.5。）

## 7.4 能量基底中的谐振子
**习题 7.4.1** 计算 $X$ 和 $P$ 在 $|n\rangle$ 基底中的矩阵元，并与习题 7.3.4 的结果比较。

**习题 7.4.2** 在状态 $|n\rangle$ 中求
$\langle X\rangle$、$\langle P\rangle$、$\left\langle X^2\right\rangle$、$\left\langle P^2\right\rangle$ 和 $\Delta X \cdot \Delta P$。

**习题 7.4.3** （Virial 定理）经典力学中的 Virial 定理指出，对于被势
$V(r)=a r^k$
束缚的粒子，其轨道平均动能与势能满足
$$
\overline{T}=c(k) \overline{V}
$$
其中 $c(k)$ 只依赖于 $k$。通过考虑圆轨道证明 $c(k)=k/2$。再利用上一习题的结果，证明对于谐振子 $(k=2)$，在量子态 $|n\rangle$ 中
$$
\langle T\rangle=\langle V\rangle
$$

**习题 7.4.4** 证明
$\langle n\mid X^{4}\mid n\rangle=(\hbar/2m\omega)^2[3+6n(n+1)]$。

**习题 7.4.5** 在 $t=0$ 时，粒子的初始状态为
$|\psi(0)\rangle=1/2^{1/2}(0\rangle+|1\rangle)$。

(1) 求 $|\psi(t)\rangle$；

(2) 求
$\langle X(0)\rangle=\langle\psi(0)| X|\psi(0)\rangle$、
$\langle P(0)\rangle$、
$\langle X(t)\rangle$、
$\langle P(t)\rangle$；

(3) 利用 Ehrenfest 定理求 $\langle\dot{X}(t)\rangle$ 和 $\langle\dot{P}(t)\rangle$，并由此解出 $\langle X(t)\rangle$ 和 $\langle P(t)\rangle$，再与第 (2) 问的结果比较。

**习题 7.4.6** 证明
$\langle a(t)\rangle=\mathrm{e}^{-\mathrm{i}\omega t}\langle a(0)\rangle$
以及
$\left\langle a^{\dagger}(t)\right\rangle=\mathrm{e}^{\mathrm{i} \omega t}\left\langle a^{\dagger}(0)\right\rangle$。

**习题 7.4.7** 对下列情形验证方程 (7.4.40)：

(1) $\Omega=X$，$\Lambda=X^{2}+P^{2}$。

(2) $\Omega=X^{2}$，$\Lambda=P^{2}$。

第二种情形说明了算符排序的不确定性。

**习题 7.4.8** 考虑经典力学中的三个角动量变量：
$$
\begin{aligned}
& l_{x}=yp_{z}-zp_{y} \\
& l_{y}=zp_{x}-xp_{z} \\
& l_{z}=xp_{y}-yp_{x}
\end{aligned}
$$

(1) 构造它们的量子对应物 $L_x$、$L_y$ 和 $L_z$，并注意这里不存在算符排序的不确定性。

(2) 验证
$\left\{l_{x}, l_{y}\right\}=l_{z}$
[Poisson 括号的定义见方程 (2.7.3)]。

(3) 验证
$\left[L_{x}, L_{y}\right]=\mathrm{i} \hbar L_{z}$。

**习题 7.4.9** 考虑如下非常规但完全可以接受的算符选取：
$$
\begin{aligned}
& X \rightarrow x \\
& P \rightarrow -\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} x}+f(x)
\end{aligned}
$$
在 $X$ 基底中。

(1) 验证正则对易关系得到满足。

(2) 可以把算符赋值的这一变化理解为 $X$ 基底发生酉变换的结果：
$$
|x\rangle \rightarrow|\tilde{x}\rangle=\mathrm{e}^{\mathrm{i} g(X) / \hbar}|x\rangle=\mathrm{e}^{\mathrm{i} g(x) / \hbar}|x\rangle
$$
其中
$$
g(x)=\int^x f\left(x^{\prime}\right) d x^{\prime}
$$

首先验证
$$
\langle\tilde{x}| X\left|\tilde{x}^{\prime}\right\rangle=x \delta\left(x-x^{\prime}\right)
$$
也就是说，
$$
X \xrightarrow[\text{新的}~X~\text{基底}]{} x
$$

接着验证
$$
\langle\tilde{x}| P\left|\tilde{x}^{\prime}\right\rangle=\left[-\mathrm{i} \hbar \frac{d}{d x}+f(x)\right] \delta\left(x-x^{\prime}\right)
$$
也就是说，
$$
P \xrightarrow[\text{新的}~x~\text{基底}]{}-\mathrm{i}\hbar \frac{\mathrm{d}}{\mathrm{d} x}+f(x)
$$

**习题 7.4.10** 回忆我们总是通过把 Cartesian 坐标 $x_1,\ldots,x_N$ 和动量 $p_1,\ldots,p_N$ 提升为满足正则对易关系的算符来量子化一个系统。如果在某些情况下非 Cartesian 坐标看起来更自然，例如具有球对称性的 Hamilton量本征值问题，我们首先在 Cartesian 坐标中建立微分方程，然后再变换到球坐标（第 4.2 节）。第 4.2 节指出，如果把 $\mathscr{H}$ 写成非 Cartesian 但正则的坐标 $q_1,\ldots,q_N$ 和 $p_1,\ldots,p_N$ 的函数，那么
$\mathscr{H}\left(q_i \rightarrow q_i, p_i \rightarrow-\mathrm{i}\hbar \partial / \partial q_i\right)$
并不能生成正确的 Hamilton量 $H$，尽管这种算符赋值满足正则对易关系。在本节中，我们重新讨论这个问题，以解释在不借助 Cartesian 坐标作为中间步骤而直接量子化非 Cartesian 坐标时出现的一些微妙之处。

(1) 考虑二维中的一个粒子，其
$$
\mathscr{H}=\frac{p_x^2+p_y^2}{2 m}+a\left(x^2+y^2\right)^{1 / 2}
$$
在坐标基底中对应于
$$
H \rightarrow \frac{-\hbar^2}{2 m}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)+a\left(x^2+y^2\right)^{1 / 2}
$$
由于该问题具有旋转对称性，我们采用极坐标
$$
\rho=\left(x^2+y^2\right)^{1 / 2}, \quad \phi=\tan ^{-1}(y / x)
$$
于是
$$
H \xrightarrow[\substack{\text { 坐标 } \\ \text { 基底 }}]{ } \frac{-\hbar^2}{2 m}\left(\frac{\partial^2}{\partial \rho^2}+\frac{1}{\rho} \frac{\partial}{\partial \rho}+\frac{1}{\rho^2} \frac{\partial^2}{\partial \phi^2}\right)+a \rho\tag{7.4.41}
$$
由于 $\rho$ 和 $\phi$ 不像 $x$ 和 $y$ 那样在 $\left(x^2+y^2\right)^{1 / 2}$ 项中混合在一起，因此极坐标形式更容易求解。

我们要讨论的问题如下：为什么不一开始就把 $\mathscr{H}$ 写成极坐标及其共轭动量的形式
$$
p_\rho=\mathbf{e}_\rho \cdot \mathbf{p}=\frac{x p_x+y p_y}{\left(x^2+y^2\right)^{1 / 2}}
$$
（其中 $\mathbf{e}_\rho$ 是径向单位矢量），以及
$$
p_\phi=x p_y-y p_x \quad\left(\text { 即角动量，也称为 } l_z\right)
$$
也就是说，
$$
\mathscr{H}=\frac{p_\rho^2}{2 m}+\frac{p_\phi^2}{2 m \rho^2}+a \rho \quad \text { （请验证） }
$$
然后直接把所有经典变量 $\rho$、$p_\rho$、$\phi$ 和 $p_\phi$ 提升为满足正则对易关系的量子算符呢？让我们这样做并看看会发生什么。如果选择满足对易关系的算符
$$
\begin{aligned}
& P_\rho \rightarrow-i \hbar \frac{\partial}{\partial \rho} \\
& P_\phi \rightarrow-i \hbar \frac{\partial}{\partial \phi}
\end{aligned}
$$
那么得到
$$
H \xrightarrow[\substack{\text { 坐标 } \\ \text { 基底 }}]{} \frac{-\hbar^2}{2 m}\left(\frac{\partial^2}{\partial \rho^2}+\frac{1}{\rho^2} \frac{\partial^2}{\partial \phi^2}\right)+a \rho\tag{7.4.42}
$$
这与方程 (7.4.41) 不一致。单从这一点本身来看还不算严重，因为正如上一习题所示，相同的物理内容可能隐藏在两个不同的方程中。但在当前情形中并非如此：正如我们将看到的，方程 (7.4.41) 和 (7.4.42) 中的 Hamilton量并不具有相同的本征值。（我们将看到，$P_{\rho}=-\mathrm{i}\hbar\mathrm{d}/\mathrm{d}\rho$，以及由它构造的 $H$，都不是 Hermitian 的。）我们知道方程 (7.4.41) 才是正确的，因为用 Cartesian 坐标进行量子化的程序具有实验支持。那么现在该怎么办？

(2) 一个解决办法来自如下事实：虽然选择
$P_\rho \rightarrow -\mathrm{i} \hbar \partial / \partial \rho$
可以得到正确的对易关系，但它并不是 Hermitian 的！验证
$$
\begin{aligned}
\left\langle\psi_1\right| P_\rho\left|\psi_2\right\rangle & =\int_0^{\infty} \int_0^{2 \pi} \psi_1^*\left(-\mathrm{i} \hbar \frac{\partial \psi_2}{\partial \rho}\right) \rho \mathrm{d} \rho \mathrm{d} \phi \\
& \neq \int_0^{\infty} \int_0^{2 \pi}\left(-\mathrm{i} \hbar \frac{\partial \psi_1}{\partial \rho}\right)^* \psi_2 \rho \mathrm{d} \rho \mathrm{d} \phi \\
& =\left\langle\mathbf{P}_\rho \psi_1 \mid \psi_2\right\rangle
\end{aligned}
$$
（可以假设当 $\rho\rightarrow 0$ 或 $\infty$ 时，$\rho \psi_1^* \psi_2 \rightarrow 0$。问题来自积分测度是 $\rho \mathrm{d}\rho \mathrm{d}\phi$，而不是 $\mathrm{d}\rho \mathrm{d}\phi$。）

但是证明
$$
P_\rho \rightarrow -\mathrm{i} \hbar\left(\frac{\partial}{\partial \rho}+\frac{1}{2 \rho}\right)
$$
确实是 Hermitian 的，同时也满足正则对易关系。角动量
$P_\phi \rightarrow -\mathrm{i} \hbar \partial / \partial \phi$
在单值函数上本身就是 Hermitian 的，即
$\psi(\rho, \phi)=\psi(\rho, \phi+2 \pi)$。

(3) 在 Cartesian 情形中我们看到，给
$-\mathrm{i} \hbar \partial / \partial x$
加上任意的 $f(x)$ 并不会产生任何物理影响；而在这里，给
$-\mathrm{i} \hbar \partial / \partial \rho$
加上一个 $\rho$ 的函数似乎很重要。为什么？[$f(x)$ 真的是完全任意的吗？它是不是必须为实函数？为什么？对于 $-\mathrm{i} \hbar / 2 \rho$ 这一项是否也是如此？]

(4) 代入新的动量算符 $P_\rho$，证明
$$
H\xrightarrow[\substack{\text{坐标}\\ \text{基底}}]{}\frac{-\hbar^2}{2 m}\left(\frac{\partial^2}{\partial \rho^2}+\frac{1}{\rho} \frac{\partial}{\partial \rho}-\frac{1}{4 \rho^2}+\frac{1}{\rho^2} \frac{\partial^2}{\partial \phi^2}\right)+a \rho
$$
它仍然与方程 (7.4.41) 不一致。我们已经满足了对易关系，也选择了 Hermitian 算符，却仍然没有得到正确的量子 Hamilton量。这个问题的关键在于，$\mathscr{H}$ 并不能唯一确定 $H$，因为 $H$ 中可能存在 $\hbar$ 一阶（或更高阶）的项，而这些项在 $\mathscr{H}$ 中并不存在。虽然这种不确定性在 Cartesian 情形中也存在，但在所有有实际意义的情形下，它都可以通过对称化来解决。对于非 Cartesian 坐标，这种不确定性更加严重。存在一些从 $\mathscr{H}$ 构造 $H$ 的方法（路径积分形式给出了一种提示），使得代换
$P_\rho \rightarrow-i \hbar(\partial / \partial \rho+1 / 2 \rho)$
能够导出方程 (7.4.41)。在当前情形中，与
$$
\mathscr{H}=\frac{p_\rho^2}{2 m}+\frac{p_\phi^2}{2 m \rho^2}+a \rho
$$
对应的量子 Hamilton量为
$$
H \xrightarrow[\substack{\text{坐标}\\ \text{基底}}]{} \mathscr{H}\left(\rho \rightarrow \rho, p_\rho \rightarrow -\mathrm{i} \hbar\left[\frac{\partial}{\partial \rho}+\frac{1}{2 \rho}\right] ; \phi \rightarrow \phi, p_\phi \rightarrow -\mathrm{i} \hbar \frac{\partial}{\partial \phi}\right)-\frac{\hbar^2}{8 m \rho^2}\tag{7.4.44}
$$
注意，附加项确实是 $\hbar$ 的非零阶项。

## 7.5 从能量基底到 $X$ 基底
**习题 7.5.1** 将方程 (7.5.1) 投影到 $P$ 基底上，得到 $\psi_0(p)$。

**习题 7.5.2** 将关系
$$
a\mid n\rangle=n^{1/2}\mid n-1\rangle
$$
投影到 $X$ 基底上，并利用方程 (7.3.22) 推导递推关系
$$
H_{n}^{\prime}(y)=2nH_{n-1}(y)
$$

**习题 7.5.3** 从
$$
a+a^{\dagger}=2^{1/2}y
$$
以及
$$
(a+a^{\dagger})\mid n\rangle=n^{1/2}\mid n-1\rangle+(n+1)^{1/2}\mid n+1\rangle
$$
和方程 (7.3.22) 出发，推导关系
$$
H_{n+1}(y)=2yH_{n}(y)-2nH_{n-1}(y)
$$

**习题 7.5.4** 谐振子的热力学。Boltzmann 公式
$$
P(i)=\mathrm{e}^{-\beta E(i)}/Z
$$
其中
$$
Z=\sum_{i}\mathrm{e}^{-\beta E(i)}
$$
给出了当一个系统与绝对温度为
$T=1/\beta k$
的热库处于热平衡时，在能量为 $E(i)$ 的状态 $i$ 中发现该系统的概率，其中
$k=1.4\times 10^{-16}~\mathrm{ergs}/^{\circ}\mathrm{K}$
是 Boltzmann 常数。（这里所说的“概率”是针对由许多相同经典系统组成的统计系综而言的，与量子力学中的概率无关。）

(1) 证明系统能量的热平均为
$$
\overline{E}=\sum_i E(i) P(i)=\frac{-\partial}{\partial \beta} \ln Z
$$

(2) 令系统为经典谐振子。此时指标 $i$ 是连续的，并对应于描述谐振子状态的变量 $x$ 和 $p$，即
$$
i \rightarrow x, p
$$
以及
$$
\sum_i \rightarrow \iint d x d p
$$
并且
$$
E(i) \rightarrow E(x, p)=\frac{p^2}{2 m}+\frac{1}{2} m \omega^2 x^2
$$
证明
$$
Z_{\mathrm{cl}}=\left(\frac{2 \pi}{\beta m \omega^2}\right)^{1 / 2}\left(\frac{2 \pi m}{\beta}\right)^{1 / 2}=\frac{2 \pi}{\omega \beta}
$$
以及
$$
\overline{E}_{\mathrm{cl}}=\frac{1}{\beta}=k T
$$
注意 $E_{\mathrm{cl}}$ 与 $m$ 和 $\omega$ 无关。

(3) 对于量子谐振子，量子数 $n$ 扮演指标 $i$ 的角色。证明
$$
Z_{\mathrm{qu}}=e^{-\beta \hbar \omega / 2}\left(1-e^{-\beta \hbar \omega}\right)^{-1}
$$
以及
$$
\overline{E}_{\mathrm{qu}}=\hbar \omega\left(\frac{1}{2}+\frac{1}{e^{\beta \hbar \omega}-1}\right)
$$

(4) 直观上很清楚，随着温度 $T$ 升高（于是 $\beta=1/kT$ 减小），谐振子会越来越容易被激发，并最终根据对应原理满足
$$
\overline{E}_{\mathrm{qu}} \xrightarrow[T \rightarrow \infty]{ } \overline{E}_{\mathrm{cl}}
$$
验证事实确实如此，并证明所谓“高温”意味着
$T \gg \hbar \omega / k$。

(5) 考虑一个含有 $N_0$ 个原子的晶体。对于小振动，它等价于 $3N_0$ 个彼此解耦的谐振子。晶体的平均热能 $\overline{E}_{\text{crystal}}$ 等于对所有简正模的 $\overline{E}_{\mathrm{cl}}$ 或 $\overline{E}_{\mathrm{qu}}$ 求和。证明如果把这些谐振子按经典方式处理，则每个原子的比热为
$$
C_{\mathrm{cl}}(T)=\frac{1}{N_0} \frac{\partial \overline{E}_{\text{crystal}}}{\partial T}=3 k
$$
它与 $T$ 以及谐振子的参数都无关，因此对所有晶体都相同。（更准确地说，是对那些原子可视为没有内部自由度的点粒子的晶体而言。）这一结果在高温下与实验相符，但在 $T\rightarrow 0$ 时不成立。实验上，
$$
\begin{aligned}
C(T) & \rightarrow 3 k & & (T~\text{较大}) \\
& \rightarrow 0 & & (T \rightarrow 0)
\end{aligned}
$$

按照 Einstein 的做法，将这些谐振子进行量子力学处理，并为简单起见假设它们都具有相同频率 $\omega$。证明
$$
C_{\mathrm{qu}}(T)=3 k\left(\frac{\theta_E}{T}\right)^2 \frac{\mathrm{e}^{\theta_E / T}}{\left(\mathrm{e}^{\theta_E / T}-1\right)^2}
$$
其中
$\theta_E=\hbar \omega / k$
称为 Einstein 温度，并且其数值随晶体而异。证明
$$
\begin{aligned}
& C_{\mathrm{qu}}(T) \xrightarrow[T \gg \theta_E]{ } 3k \\
& C_{\mathrm{qu}}(T) \xrightarrow[T \ll \theta_E]{ } 3 k\left(\frac{\theta_E}{T}\right)^2 \mathrm{e}^{-\theta_E / T}
\end{aligned}
$$

尽管当 $T\rightarrow 0$ 时
$C_{\mathrm{qu}}(T)\rightarrow 0$，
但这种指数衰减与实验观测到的
$C(T) \xrightarrow[T \rightarrow 0]{} T^3$
行为不一致。这一差异来自于假设所有简正模的频率都相同，而这显然一般并不成立。[回忆两个耦合质量块的情形，我们得到
$\omega_{\mathrm{I}}=(k / m)^{1 / 2}$
和
$\omega_{\mathrm{II}}=(3 k / m)^{1 / 2}$。]
这一差异后来由 Debye 消除了。

# Chapter 7 The Harmonic Oscillator
## 7.1 Why Study the Harmonic Oscillator?
## 7.2 Review of the Classical Oscillator
## 7.3 Quantization of the Oscillator (Coordinate Basis)
**Exercise 7.3.1** Consider the question why we tried a power-series solution for Eq. (7.3.11) but not Eq. (7.3.8). By feeding in a series into the latter, verify that a three-term recursion relation between $C_{n+2}$, $C_{n}$, and $C_{n-2}$ obtains, from which the solution does not follow so readily. The problem is that $\psi^{\prime\prime}$ has two powers of $y$ less than $2 \varepsilon \psi$, while the $-y^{2}$ piece has two more powers of $y$. In Eq. (7.3.11) on the other hand, of the three pieces $u^{\prime\prime}$, $-2yu^{\prime}$, and $(2\varepsilon-1)u$, the last two have the same powers of $y$.



**Exercise 7.3.2** Verify that $H_{3}(y)$ and $H_{4}(y)$ obey the recursion relation, Eq. (7.3.15).

**Exercise 7.3.3** If $\psi(x)$ is even and $\phi(x)$ is odd under $x \rightarrow-x$, show that
$$
\int_{-\infty}^{\infty} \psi(x) \phi(x) \mathrm{d}x=0
$$
Use this to show that $\psi_2(x)$ and $\psi_1(x)$ are orthogonal. Using the values of Gaussian integrals in Appendix A.2 verify that $\psi_2(x)$ and $\psi_0(x)$ are orthogonal.

**Exercise 7.3.4** Using Eqs. (7.3.23)-(7.3.25), show that
$$
\begin{aligned}
& \left\langle n^{\prime}\right| X|n\rangle=\left(\frac{\hbar}{2 m \omega}\right)^{1 / 2}\left[\delta_{n^{\prime}, n+1}(n+1)^{1 / 2}+\delta_{n^{\prime}, n-1} n^{1 / 2}\right] \\
& \left\langle n^{\prime}\right| P|n\rangle=\left(\frac{m \omega \hbar}{2}\right)^{1/2} \mathrm{i}\left[\delta_{n^{\prime}, n+1}(n+1)^{1 / 2}-\delta_{n^{\prime}, n-1} n^{1 / 2}\right]
\end{aligned}
$$

**Exercise 7.3.5** Using the symmetry arguments from Exercise 7.3.3 show that $\langle n|X|n\rangle=\langle n|P|n\rangle=0$ and thus that $\left\langle X^2\right\rangle=(\Delta X)^2$ and $\left\langle P^2\right\rangle=(\Delta P)^2$ in these states. Show that $\langle 1| X^2|1\rangle=3 \hbar / 2 m \omega$ and $\langle 1| P^2|1\rangle=\dfrac{3}{2} m \omega \hbar$. Show that $\psi_0(x)$ saturates the uncertainty bound $\Delta X \cdot \Delta P \geqslant \hbar / 2$.

**Exercise 7.3.6** Consider a particle in a potential
$$
\begin{aligned}
V(x) & =\frac{1}{2} m \omega^2 x^2, & & x>0 \\
& =\infty, & & x \leqslant 0
\end{aligned}
$$
What are the boundary conditions on the wave functions now? Find the eigenvalues and eigenfunctions.

**Exercise 7.3.7** (The Oscillator in Momentum Space) By setting up an eigenvalue equation for the oscillator in the $P$ basis and comparing it to Eq. (7.3.2), show that the momentum space eigenfunctions may be obtained from the ones in coordinate space through the substitution $x \rightarrow p$, $m \omega \rightarrow 1 / m \omega$. Thus, for example,
$$
\psi_0(p)=\left(\frac{1}{m \pi \hbar \omega}\right)^{1 / 4} e^{-p^2 / 2 m \hbar \omega}
$$
There are several other pairs, such as $\Delta X$ and $\Delta P$ in the state $|n\rangle$, which are related by the substitution $m \omega \rightarrow 1 / m \omega$. You may wish to watch out for them. (Refer back to Exercise 7.3.5.)

## 7.4 The Oscillator in the Energy Basis
**Exercise 7.4.1** Compute the matrix elements of $X$ and $P$ in the $|n\rangle$ basis and compare with the result from Exercise 7.3.4.

**Exercise 7.4.2** Find $\langle X\rangle$, $\langle P\rangle$, $\left\langle X^2\right\rangle$, $\left\langle P^2\right\rangle$, $\Delta X \cdot \Delta P$ in the state $|n\rangle$.

**Exercise 7.4.3** (Virial Theorem) The virial theorem in classical mechanics states that for a particle bound by a potential $V(r)=a r^k$, the average (over the orbit) kinetic and potential energies are related by
$$
\overline{T}=c(k) \overline{V}
$$
when $c(k)$ depends only on $k$. Show that $c(k)=k / 2$ by considering a circular orbit. Using the results from the previous exercise show that for the oscillator $(k=2)$
$$
\langle T\rangle=\langle V\rangle
$$
in the quantum state $|n\rangle$.

**Exercise 7.4.4** Show that $\langle n\mid X^{4}\mid n\rangle=(\hbar/2m\omega)^2[3+6n(n+1)]$.

**Exercise 7.4.5** At $t=0$ a particle starts out in $|\psi(0)\rangle=1/2^{1/2}(0\rangle+|1\rangle)$.

(1)Find $|\psi(t)\rangle$;

(2)Find $\langle X(0)\rangle=\langle\psi(0)| X|\psi(0)\rangle,\langle P(0)\rangle,\langle X(t)\rangle,\langle P(t)\rangle$;

(3)Find $\langle\dot{X}(t)\rangle$ and $\langle\dot{P}(t)\rangle$ using Ehrenfest's theorem and solve for $\langle X(t)\rangle$ and $\langle P(t)\rangle$ and compare with part (2).

**Exercise 7.4.6** Show that $\langle a(t)\rangle=\mathrm{e}^{-\mathrm{i}\omega t}\langle a(0)\rangle$ and that $\left\langle a^{\dagger}(t)\right\rangle=\mathrm{e}^{\mathrm{i} \omega t}\left\langle a^{\dagger}(0)\right\rangle$.

**Exercise 7.4.7** Verify Eq. (7.4.40) for the case

(1)$\Omega=X$, $\Lambda=X^{2}+P^{2}$.

(2)$\Omega=X^{2}$, $\Lambda=P^{2}$.

The second case illustrates the ordering ambiguity.

**Exercise 7.4.8** Consider the three angular momentum variables in classical mechanics:
$$
\begin{aligned}
& l_{x}=yp_{z}-zp_{y} \\
& l_{y}=zp_{x}-xp_{z} \\
& l_{z}=xp_{y}-yp_{x}
\end{aligned}
$$
(1)Construct $L_{x}$, $L_{y}$, and $L_{z}$, the quantum counterparts, and note that there are no ordering ambiguities.

(2)Verify that $\left\{l_{x}, l_{y}\right\}=l_{z}$ [see Eq. (2.7.3) for the definition of the Poisson Bracket].

(3)Verify that $\left[L_{x}, L_{y}\right]=\mathrm{i} \hbar L_{z}$.

**Exercise 7.4.9** Consider the unconventional (but fully acceptable) operator choice
$$
\begin{aligned}
& X \rightarrow x \\
& P \rightarrow -\mathrm{i} \hbar \frac{\mathrm{d}}{\mathrm{d} x}+f(x)
\end{aligned}
$$
in the $X$ basis.

(1)Verify that the canonical commutation relation is satisfied.

(2)It is possible to interpret the change in the operator assignment as a result of a unitary change of the $X$ basis:
$$
|x\rangle \rightarrow|\tilde{x}\rangle=\mathrm{e}^{\mathrm{i} g(X) / \hbar}|x\rangle=\mathrm{e}^{\mathrm{i} g(x) / \hbar}|x\rangle
$$
where
$$
g(x)=\int^x f\left(x^{\prime}\right) d x^{\prime}
$$
First verify that
$$
\langle\tilde{x}| X\left|\tilde{x}^{\prime}\right\rangle=x \delta\left(x-x^{\prime}\right)
$$
i.e.,
$$
X \xrightarrow[\text{new}~X~\text{basis}]{} x
$$
Next verify that
$$
\langle\tilde{x}| P\left|\tilde{x}^{\prime}\right\rangle=\left[-\mathrm{i} \hbar \frac{d}{d x}+f(x)\right] \delta\left(x-x^{\prime}\right)
$$
i.e.,
$$
P \xrightarrow[\text{new}~x~\text{basis}]{}-\mathrm{i}\hbar \frac{\mathrm{d}}{\mathrm{d} x}+f(x)
$$
**Exercise 7.4.10** Recall that we always quantize a system by promoting the Cartesian coordinates $x_1, \ldots, x_N$; and momenta $p_1, \ldots, p_N$ to operators obeying the canonical commutation rules. If non-Cartesian coordinates seem more natural in some cases, such as the eigenvalue problem of a Hamiltonian with spherical symmetry, we first set up the differential equation in Cartesian coordinates and then change to spherical coordinates (Section 4.2). In Section 4.2 it was pointed out that if $\mathscr{H}$ is written in terms of non-Cartesian but canonical coordinates $q_1 \ldots q_N$; $p_1 \ldots p_N$; $\mathscr{H}\left(q_i \rightarrow q_i, p_i \rightarrow-\mathrm{i}\hbar \partial / \partial q_i\right)$ does not generate the correct Hamiltonian $H$, even though the operator assignment satisfies the canonical commutation rules. In this section we revisit this problem in order to explain some of the subtleties arising in the direct quantization of non-Cartesian coordinates without the use of Cartesian coordinates in intermediate stages.

(1) Consider a particle in two dimensions with
$$
\mathscr{H}=\frac{p_x^2+p_y^2}{2 m}+a\left(x^2+y^2\right)^{1 / 2}
$$
which leads to
$$
H \rightarrow \frac{-\hbar^2}{2 m}\left(\frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}\right)+a\left(x^2+y^2\right)^{1 / 2}
$$
in the coordinate basis. Since the problem has rotational symmetry we use polar coordinates
$$
\rho=\left(x^2+y^2\right)^{1 / 2}, \quad \phi=\tan ^{-1}(y / x)
$$
in terms of which
$$
H \xrightarrow[\substack{\text { coordinate } \\ \text { basis }}]{ } \frac{-\hbar^2}{2 m}\left(\frac{\partial^2}{\partial \rho^2}+\frac{1}{\rho} \frac{\partial}{\partial \rho}+\frac{1}{\rho^2} \frac{\partial^2}{\partial \phi^2}\right)+a \rho\tag{7.4.41}
$$
Since $\rho$ and $\phi$ are not mixed up as $x$ and $y$ are [in the $\left(x^2+y^2\right)^{1 / 2}$ term] the polar version can be more readily solved.

The question we address is the following: why not start with $\mathscr{H}$ expressed in terms of polar coordinates and the conjugate momenta
$$
p_\rho=\mathbf{e}_\rho \cdot \mathbf{p}=\frac{x p_x+y p_y}{\left(x^2+y^2\right)^{1 / 2}}
$$
(where $\mathbf{e}_\rho$ is the unit vector in the radial direction), and
$$
p_\phi=x p_y-y p_x \quad\left(\text { the angular momentum, also called } l_z\right)
$$
i.e.,
$$
\mathscr{H}=\frac{p_\rho^2}{2 m}+\frac{p_\phi^2}{2 m \rho^2}+a \rho \quad \text { (verify this) }
$$
and directly promote all classical variables $\rho$, $p_\rho$, $\phi$, and $p_\phi$ to quantum operators obeying the canonical commutations rules? Let's do it and see what happens. If we choose operators
$$
\begin{aligned}
& P_\rho \rightarrow-i \hbar \frac{\partial}{\partial \rho} \\
& P_\phi \rightarrow-i \hbar \frac{\partial}{\partial \phi}
\end{aligned}
$$
that obey the commutation rules, we end up with
$$
H \xrightarrow[\substack{\text { coordinate } \\ \text { basis }}]{} \frac{-\hbar^2}{2 m}\left(\frac{\partial^2}{\partial \rho^2}+\frac{1}{\rho^2} \frac{\partial^2}{\partial \phi^2}\right)+a \rho\tag{7.4.42}
$$
which disagrees with Eq. (7.4.41). Now this in itself is not serious, for as seen in the last exercise the same physics may be hidden in two different equations. In the present case this isn't true: as we will see, the Hamiltonians in Eqs. (7.4.41) and (7.4.42) do not have the same eigenvalues. (What we will see is that $P_{\rho}=-\mathrm{i}\hbar\mathrm{d}/\mathrm{d}\rho$, and hence the $H$ constructed with it, are non-Hermitian.) We know Eq. (7.4.41) is the correct one, since the quantization procedure in terms of Cartesian coordinates has empirical support. What do we do now?

(2) A way out is suggested by the fact that although the choice $P_\rho \rightarrow -\mathrm{i} \hbar \partial / \partial \rho$ leads to the correct commutation rule, it is not Hermitian! Verify that
$$
\begin{aligned}
\left\langle\psi_1\right| P_\rho\left|\psi_2\right\rangle & =\int_0^{\infty} \int_0^{2 \pi} \psi_1^*\left(-\mathrm{i} \hbar \frac{\partial \psi_2}{\partial \rho}\right) \rho \mathrm{d} \rho \mathrm{d} \phi \\
& \neq \int_0^{\infty} \int_0^{2 \pi}\left(-\mathrm{i} \hbar \frac{\partial \psi_1}{\partial \rho}\right)^* \psi_2 \rho \mathrm{d} \rho \mathrm{d} \phi \\
& =\left\langle\mathbf{P}_\rho \psi_1 \mid \psi_2\right\rangle
\end{aligned}
$$
(You may assume $\rho \psi_1^* \psi_2 \rightarrow 0$ as $\rho \rightarrow 0$ or $\infty$. The problem comes from the fact that $\rho \mathrm{d} \rho \mathrm{d} \phi$ and not $\mathrm{d} \rho \mathrm{d} \phi$ is the measure for integration.)

Show, however, that
$$
P_\rho \rightarrow -\mathrm{i} \hbar\left(\frac{\partial}{\partial \rho}+\frac{1}{2 \rho}\right)
$$
is indeed Hermitian and also satisfies the canonical commutation rule. The angular momentum $P_\phi \rightarrow -\mathrm{i} \hbar \partial / \partial \phi$ is Hermitian, as it stands, on single-valued functions: $\psi(\rho, \phi)=\psi(\rho, \phi+2 \pi)$.

(3) In the Cartesian case we saw that adding an arbitrary $f(x)$ to $-\mathrm{i} \hbar \partial / \partial x$ didn't have any physical effect, whereas here the addition of a function of $\rho$ to $-\mathrm{i} \hbar \partial / \partial \rho$ seems important. Why? [Is $f(x)$ completely arbitrary? Mustn't it be real? Why? Is the same true for the $-\mathrm{i} \hbar / 2 \rho$ piece?]

(4) Feed in the new momentum operator $P_\rho$ and show that
$$
H\xrightarrow[\substack{\text{coordinate}\\ \text{basis}}]{}\frac{-\hbar^2}{2 m}\left(\frac{\partial^2}{\partial \rho^2}+\frac{1}{\rho} \frac{\partial}{\partial \rho}-\frac{1}{4 \rho^2}+\frac{1}{\rho^2} \frac{\partial^2}{\partial \phi^2}\right)+a \rho
$$
which still disagrees with Eq. (7.4.41). We have satisfied the commutation rules, chosen Hermitian operators, and yet do not get the right quantum Hamiltonian. The key to the mystery lies in the fact that $\mathscr{H}$ doesn't determine $H$ uniquely since terms of order $\hbar$ (or higher) may be present in $H$ but absent in $\mathscr{H}$. While this ambiguity is present even in the Cartesian case, it is resolved by symmetrization in all interesting cases. With non-Cartesian coordinates the ambiguity is more severe. There are ways of constructing $H$ given $\mathscr{H}$ (the path integral formulation suggests one) such that the substitution $P_\rho \rightarrow-i \hbar(\partial / \partial \rho+1 / 2 \rho)$ leads to Eq. (7.4.41). In the present case the quantum Hamiltonian corresponding to
$$
\mathscr{H}=\frac{p_\rho^2}{2 m}+\frac{p_\phi^2}{2 m \rho^2}+a \rho
$$
is given by
$$
H \xrightarrow[\substack{\text{coordinate}\\ \text{basis}}]{} \mathscr{H}\left(\rho \rightarrow \rho, p_\rho \rightarrow -\mathrm{i} \hbar\left[\frac{\partial}{\partial \rho}+\frac{1}{2 \rho}\right] ; \phi \rightarrow \phi, p_\phi \rightarrow -\mathrm{i} \hbar \frac{\partial}{\partial \phi}\right)-\frac{\hbar^2}{8 m \rho^2}\tag{7.4.44}
$$
Notice that the additional term is indeed of nonzero order in $\hbar$.
## 7.5 Passege from the Energy Basis to the $X$ Basis
**Exercise 7.5.1** Project Eq. (7.5.1) on the $P$ basis and obtain $\psi_{0}(p)$.

**Exercise 7.5.2** Project the relation
$$
a\mid n\rangle=n^{1/2}\mid n-1\rangle
$$
on the $X$ basis and derive the recursion relation
$$
H_{n}^{\prime}(y)=2nH_{n-1}(y)
$$
using Eq. (7.3.22).

**Exercise 7.5.3** Starting with
$$
a+a^{\dagger}=2^{1/2}y
$$
and
$$
(a+a^{\dagger})\mid n\rangle=n^{1/2}\mid n-1\rangle+(n+1)^{1/2}\mid n+1\rangle
$$
and Eq. (7.3.22), derive the relation
$$
H_{n+1}(y)=2yH_{n}(y)-2nH_{n-1}(y)
$$

**Exercise 7.5.4** Thermodynamics of Oscillators. The Boltzmann formula
$$
P(i)=\mathrm{e}^{-\beta E(i)}/Z
$$
where
$$
Z=\sum_{i}\mathrm{e}^{-\beta E(i)}
$$
gives the probability of finding a system in a state $i$ with energy $E(i)$, when it is in thermal equilibrium with a reservoir of absolute temperature $T=1/\beta k$, $k=1.4\times 10^{-16}~\mathrm{ergs}/^{\circ}\mathrm{K}$; being Boltzmann's constant. (The “probability” referred to above is in relation to a classical ensemble of similar systems and has nothing to do with quantum mechanics.)

(1) Show that the thermal average of the system's energy is
$$
\overline{E}=\sum_i E(i) P(i)=\frac{-\partial}{\partial \beta} \ln Z
$$

(2) Let the system be a classical oscillator. The index $i$ is now continuous and corresponds to the variables $x$ and $p$ describing the state of the oscillator, i.e.,
$$
i \rightarrow x, p
$$
and
$$
\sum_i \rightarrow \iint d x d p
$$
and
$$
E(i) \rightarrow E(x, p)=\frac{p^2}{2 m}+\frac{1}{2} m \omega^2 x^2
$$
Show that
$$
Z_{\mathrm{cl}}=\left(\frac{2 \pi}{\beta m \omega^2}\right)^{1 / 2}\left(\frac{2 \pi m}{\beta}\right)^{1 / 2}=\frac{2 \pi}{\omega \beta}
$$
and that
$$
\overline{E}_{\mathrm{cl}}=\frac{1}{\beta}=k T
$$
Note that $E_{\mathrm{cl}}$ is independent of $m$ and $\omega$.

(3) For the quantum oscillator the quantum number $n$ plays the role of the index $i$. Show that
$$
Z_{\mathrm{qu}}=e^{-\beta \hbar \omega / 2}\left(1-e^{-\beta \hbar \omega}\right)^{-1}
$$
and
$$
\overline{E}_{\mathrm{qu}}=\hbar \omega\left(\frac{1}{2}+\frac{1}{e^{\beta \hbar \omega}-1}\right)
$$

(4) It is intuitively clear that as the temperature $T$ increases (and $\beta=1 / k T$ decreases) the oscillator will get more and more excited and eventually (from the correspondence principle)
$$
\overline{E}_{\mathrm{qu}} \xrightarrow[T \rightarrow \infty]{ } \overline{E}_{\mathrm{cl}}
$$
Verify that this is indeed true and show that “large $T$” means $T \gg \hbar \omega / k$.

(5) Consider a crystal with $N_0$ atoms, which, for small oscillations, is equivalent to $3 N_0$ decoupled oscillators. The mean thermal energy of the crystal $\overline{E}_{\text{crystal}}$ is $\overline{E}_{\text{cl}}$ or $\overline{E}_{\text{qu}}$ summed over all the normal modes. Show that if the oscillators are treated classically, the specific heat per atom is
$$
C_{\mathrm{cl}}(T)=\frac{1}{N_0} \frac{\partial \overline{E}_{\text{crystal}}}{\partial T}=3 k
$$
which is independent of $T$ and the parameters of the oscillators and hence the same for all crystals. (More precisely, for crystals whose atoms behave as point particles with no internal degrees of freedom.) This agrees with experiment at high temperatures but not as $T \rightarrow 0$. Empirically,
$$
\begin{aligned}
C(T) & \rightarrow 3 k & & (T~\text{large}) \\
& \rightarrow 0 & & (T \rightarrow 0)
\end{aligned}
$$
Following Einstein, treat the oscillators quantum mechanically, assuming for simplicity that they all have the same frequency $\omega$. Show that
$$
C_{\mathrm{qu}}(T)=3 k\left(\frac{\theta_E}{T}\right)^2 \frac{\mathrm{e}^{\theta_E / T}}{\left(\mathrm{e}^{\theta_E / T}-1\right)^2}
$$
where $\theta_E=\hbar \omega / k$ is called the Einstein temperature and varies from crystal to crystal. Show that
$$
\begin{aligned}
& C_{\mathrm{qu}}(T) \xrightarrow[T \gg \theta_E]{ } 3k \\
& C_{\mathrm{qu}}(T) \xrightarrow[T \ll \theta_E]{ } 3 k\left(\frac{\theta_E}{T}\right)^2 \mathrm{e}^{-\theta_E / T}
\end{aligned}
$$
Although $C_{\mathrm{qu}}(T) \rightarrow 0$ as $T \rightarrow 0$, the exponential falloff disagrees with the observed $C(T) \xrightarrow[T \rightarrow 0]{} T^3$ behavior. This discrepancy arises from assuming that the frequencies of all normal modes are equal, which is of course not generally true. [Recall that in the case of two coupled masses we get $\omega_{\mathrm{I}}=(k / m)^{1 / 2}$ and $\omega_{\mathrm{II}}=(3 k / m)^{1 / 2}$.] This discrepancy was eliminated by Debye.

§ QM-07 谐振子

§ QM-07 The Harmonic Oscillator

§ QM-06 经典极限

§ QM-06 The Classical Limit

# 第5章 一维中的简单问题
## 5.1 自由粒子
**习题 5.1.1** 证明方程 (5.1.9) 可以改写为对 $E$ 的积分以及对 $\pm$ 指标的求和：
$$
U(t)=\sum_{\alpha= \pm} \int_0^{\infty}\left[\frac{m}{(2 m E)^{1 / 2}}\right]|E, \alpha\rangle\langle E, \alpha| \mathrm{e}^{-\mathrm{i} E t / \hbar} \mathrm{d} E
$$


**解答:**
$$
\begin{aligned}
        E=\dfrac{p^{2}}{2m} \rightarrow p&=\alpha\sqrt{2mE}\\
        \mathrm{d}p&=\dfrac{\alpha m}{\sqrt{2mE}}\mathrm{d}E
    \end{aligned}
$$
其中 $\alpha=\pm 1$。因此
$$
\begin{aligned}
U(t)&=\int_{-\infty}^{+\infty} \mathrm{d}p\,|p\rangle\langle p| \mathrm{e}^{-\mathrm{i}Et/\hbar}\\&=\int_{-\infty}^{0}\mathrm{d}p\,|p\rangle\langle p| \mathrm{e}^{-\mathrm{i}Et/\hbar}+\int_{0}^{+\infty}\mathrm{d}p\,|p\rangle\langle p| \mathrm{e}^{-\mathrm{i}Et/\hbar}\\
        &=\int_{-\infty}^{0}\mathrm{d}E \dfrac{-m}{\sqrt{2mE}}\,|E, -\rangle\langle E, -| \mathrm{e}^{-\mathrm{i}Et/\hbar}+\int_{0}^{+\infty}\mathrm{d}E\dfrac{m}{\sqrt{2mE}}\,|E, +\rangle\langle E, +| \mathrm{e}^{-\mathrm{i}Et/\hbar}\\
        &=\int_{0}^{+\infty}\mathrm{d}E \dfrac{m}{\sqrt{2mE}}\,|E, -\rangle\langle E, -| \mathrm{e}^{-\mathrm{i}Et/\hbar}+\int_{0}^{+\infty}\mathrm{d}E\dfrac{m}{\sqrt{2mE}}\,|E, +\rangle\langle E, +| \mathrm{e}^{-\mathrm{i}Et/\hbar}\\
        &=\sum_{\alpha= \pm} \int_0^{\infty}\left[\frac{m}{(2 m E)^{1 / 2}}\right]|E, \alpha\rangle\langle E, \alpha| \mathrm{e}^{-\mathrm{i} E t / \hbar} \mathrm{d} E
\end{aligned}
$$
$$
~\tag*{$\blacksquare$}
$$

**习题 5.1.2** 通过在 $X$ 基底中求解本征值方程 (5.1.3)，重新得到方程 (5.1.8)，也就是说，证明能量为 $E$ 的一般解为
$$
\psi_E(x)=\beta \frac{\exp \left[i(2 m E)^{1 / 2} x / \hbar\right]}{(2 \pi \hbar)^{1 / 2}}+\gamma \frac{\exp \left[-i(2 m E)^{1 / 2} x / \hbar\right]}{(2 \pi \hbar)^{1 / 2}}
$$
[因子 $(2 \pi \hbar)^{-1 / 2}$ 是任意的，可以吸收到 $\beta$ 和 $\gamma$ 中。] 尽管当 $E<0$ 时 $\psi_E(x)$ 仍然满足该方程，但这些函数属于 Hilbert 空间吗？

**解答:**

在 $X$ 基底中，(5.1.3) 为
$$
\langle x\mid H\mid E\rangle=E\langle x\mid E\rangle
$$
它变为
$$
-\dfrac{\hbar^{2}}{2m}\psi_{E}^{\prime\prime}=E\psi_{E}
$$
一般解具有如下形式
$$
\psi_{E}(x)=A_{+}\mathrm{e}^{\frac{\mathrm{i}}{\hbar}\sqrt{2mE}x}+A_{-}\mathrm{e}^{-\frac{\mathrm{i}}{\hbar}\sqrt{2mE}x}
$$
因此，
$$
\langle x\mid E\rangle=\beta\langle x\mid E,+\rangle+\gamma\langle x\mid E,-\rangle
$$
如果记
$$
\begin{aligned}
\beta&=A_{+}\sqrt{2\pi k}\\
\gamma&=A_{-}\sqrt{2\pi k}
\end{aligned}
$$
最终得到能量为 $E$ 的一般解
$$
\psi_E(x)=\beta \frac{\exp \left[\mathrm{i}(2 m E)^{1 / 2} x / \hbar\right]}{(2 \pi \hbar)^{1 / 2}}+\gamma \frac{\exp \left[-\mathrm{i}(2 m E)^{1 / 2} x / \hbar\right]}{(2 \pi \hbar)^{1 / 2}}
$$
如果 $E<0$，这些解不属于 Hilbert 空间，因为此时当 $x\to+\infty$ 或 $x\to-\infty$ 时，其中一项会指数发散。

$$
~\tag*{$\blacksquare$}
$$

**习题 5.1.3** 我们已经看到，$U(t)$ 还有另一种表达式，即 $U(t)=e^{-i H t / \hbar}$。对于自由粒子，它变为
$$
U(t)=\exp \left[\frac{\mathrm{i}}{\hbar}\left(\frac{\hbar^2 t}{2 m} \frac{\mathrm{d}^2}{\mathrm{d} x^2}\right)\right]=\sum_{n=0}^{\infty} \frac{1}{n!}\left(\frac{\mathrm{i} \hbar t}{2 m}\right)^n \frac{\mathrm{d}^{2 n}}{\mathrm{d} x^{2 n}}\tag{5.1.18}
$$
考虑方程 (5.1.14) 中的初始状态，取 $p_0=0$，并令 $\Delta=1$、$t^{\prime}=0$：
$$
\psi(x, 0)=\frac{\mathrm{e}^{-x^2 / 2}}{(\pi)^{1 / 4}}
$$
利用上面的方程 (5.1.18) 求 $\psi(x, t)$，并与方程 (5.1.15) 比较。

[提示：(1) 将 $\psi(x, 0)$ 写成幂级数：
$$
\psi(x, 0)=(\pi)^{-1 / 4} \sum_{n=0}^{\infty} \frac{(-1)^n x^{2 n}}{n!(2)^n}
$$

(2) 求下列前几项
$$
1,\quad \left(\frac{\mathrm{i} \hbar t}{2 m}\right) \frac{\mathrm{d}^2}{\mathrm{d} x^2}, \quad \frac{1}{2!}\left(\frac{\mathrm{i} \hbar t}{2 m} \frac{\mathrm{d}^2}{\mathrm{d} x^2}\right)^2
$$
等等作用在该幂级数上的结果。

(3) 收集具有相同 $x$ 次幂的各项。

(4) 在 $x^{2 n}$ 的系数中寻找如下级数展开：
$$
\left(1+\frac{\mathrm{i} t \hbar}{m}\right)^{-n-1/2}=1-(n+1 / 2)\left(\frac{\mathrm{i} \hbar t}{m}\right)+\frac{(n+1 / 2)(n+3 / 2)}{2!}\left(\frac{\mathrm{i} \hbar t}{m}\right)^2+\cdots
$$

(5) 反复整理，直到得到答案。]

**解答:**

$$
~\tag*{$\blacksquare$}
$$

**习题 5.1.4** 考虑波函数
$$
\begin{aligned}
\psi(x, 0) & =\sin \left(\frac{\pi x}{L}\right), & & |x| \leqslant L / 2 \\
& =0, & & |x|>L / 2
\end{aligned}
$$
显然，对这个函数求任意阶导数都会得到另一个仍局限在区间 $|x| \leqslant L / 2$ 内的函数。因此，算符
$$
U(t)=\exp \left[\frac{\mathrm{i}}{\hbar}\left(\frac{\hbar^2 t}{2 m}\right) \frac{\mathrm{d}^2}{\mathrm{d} x^2}\right]
$$
作用在该函数上，似乎仍然得到一个局限于 $|x| \leqslant L / 2$ 的函数。那么波包的扩散又该如何理解？

**解答:**
答案在 Shankar 中已经给出：考虑边界处的导数。这里给出了一个算符幂级数（指数算符）不收敛的例子。注意，算符幂级数是否收敛，不仅取决于算符本身，也取决于它所作用的对象。因此并不存在矛盾：如果函数像上面那样突然截断，看起来似乎会产生矛盾，那么它在边界处的导数是奇异的；而如果它连续衰减，那么只要时间足够长，无论衰减得多么迅速，波函数最终一定会向外泄露。

此外，还应当意识到，这不可能是 Schrödinger 方程的解。Schrödinger 方程的解应满足下列条件：

(1) 波函数连续；

(2) 对于有限势，波函数的一阶导数通常连续。

因此，给定的波函数并不具有物理意义（它在 $x=\pm\dfrac{L}{2}$ 处不连续）。
$$
~\tag*{$\blacksquare$}
$$

## 5.2 箱中粒子
**习题 5.2.1** 一个粒子处于长度为 $L$ 的一维无限深势箱的基态。突然间，势箱关于中心对称地扩展为原来的两倍，而波函数来不及发生变化。证明在新势箱中发现粒子处于基态的概率为 $(8 / 3 \pi)^2$。

**习题 5.2.2** (a) 证明对于任意归一化的 $|\psi\rangle$，都有 $\langle\psi|H| \psi\rangle \geqslant E_0$，其中 $E_0$ 是最低能量本征值。（提示：将 $|\psi\rangle$ 在 $H$ 的本征基底中展开。）

(b) 证明下列定理：一维中的任何吸引势至少存在一个束缚态。提示：由于 $V$ 是吸引势，如果定义 $V(\infty)=0$，则对所有 $x$ 都有 $V(x)=-|V(x)|$。为了证明存在一个 $E<0$ 的束缚态，考虑
$$
\psi_\alpha(x)=\left(\frac{\alpha}{\pi}\right)^{1 / 4} e^{-\alpha x^2 / 2}
$$
并计算
$$
E(\alpha)=\left\langle\psi_\alpha|H| \psi_\alpha\right\rangle, \quad H=-\frac{\hbar^2}{2 m} \frac{d^2}{d x^2}-|V(x)|
$$
证明通过适当选择 $\alpha$，可以使 $E(\alpha)$ 为负。所需结论便由上面证明的定理得到。

**习题 5.2.3** 考虑 $V(x)=-a V_0 \delta(x)$。证明它存在一个能量为 $E=-m a^2 V_0^2 / 2 \hbar^2$ 的束缚态。是否还存在其他束缚态？提示：对 $E<0$，在势以外求解 Schrödinger 方程，只保留在无穷远处具有正确行为并且在 $x=0$ 处连续的解。画出波函数，可以看到在 $x=0$ 处有一个尖点，也就是斜率发生不连续变化。计算斜率的变化，并将其与
$$
\int_{-\varepsilon}^{+\varepsilon}\left(\frac{d^2 \psi}{d x^2}\right) d x
$$
相等起来，其中 $\varepsilon$ 为无穷小，而上式由 Schrödinger 方程确定。

**习题 5.2.4** 考虑质量为 $m$ 的粒子处于长度为 $L$ 的势箱中的状态 $|n\rangle$。假设随着势箱尺寸变化，粒子始终保持在该势箱的第 $n$ 个状态，求缓慢向内推动箱壁时所遇到的力 $F=-\partial E / \partial L$。再考虑该势箱中能量为 $E_n$ 的经典粒子。求它的速度、与某一指定箱壁碰撞的频率、每次碰撞的动量传递，并由此求出平均力。将其与上面计算得到的 $-\partial E / \partial L$ 比较。

**习题 5.2.5** 如果势箱的范围是从 $x=0$ 到 $L$（而不是从 $-L / 2$ 到 $L / 2$），证明
$\psi_n(x)=(2 / L)^{1 / 2} \sin (n \pi x / L)$，$n=1,2, \ldots, \infty$，并且
$E_n=\hbar^2 \pi^2 n^2 / 2 m L^2$。

**习题 5.2.6** 方势阱。考虑粒子处于如下方势阱中：
$$
V(x)= \begin{cases}0, & |x| \leqslant a \\ V_0, & |x| \geqslant a\end{cases}
$$

由于当 $V_0 \rightarrow \infty$ 时就得到无限深势箱，让我们先猜测降低势垒高度会对各个状态产生什么影响。首先，我们只关心束缚态，而所有束缚态都满足 $E \leq V_0$。其次，低能级的波函数看起来会类似于箱中粒子的波函数，明显的区别是 $\psi$ 在势阱边界处不再为零，而是以指数尾的形式延伸到阱外。本征函数仍然会依次具有偶、奇、偶等宇称。

(1) 证明偶解的能量满足超越方程
$$
k \tan k a=\kappa\tag{5.2.23}
$$
而奇解的能量满足
$$
k \cot k a=-\kappa\tag{5.2.24}
$$
其中 $k$ 和 $i \kappa$ 分别是势阱内部和外部的实波数与复波数。注意 $k$ 和 $\kappa$ 满足关系
$$
k^2+\kappa^2=2 m V_0 / \hbar^2\tag{5.2.25}
$$
验证当 $V_0$ 趋于 $\infty$ 时，可以重新得到无限深势箱中的能级。

(2) 方程 (5.2.23) 和 (5.2.24) 必须通过图解法求解。在 $(\alpha=k a, \beta=\kappa a)$ 平面上，设想一个满足方程 (5.2.25) 的圆。束缚态便由曲线 $\alpha \tan \alpha=\beta$ 或 $\alpha \cot \alpha=-\beta$ 与该圆的交点给出。（记住 $\alpha$ 和 $\beta$ 都为正。）

(3) 证明总是存在一个偶解，并且除非 $V_0 \geqslant \hbar^2 \pi^2 / 8 m a^2$，否则不存在奇解。当 $V_0$ 恰好达到这一要求时，$E$ 是多少？注意习题 5.2.2(b) 中的一般结论仍然成立。

## 5.3 概率的连续性方程
**习题 5.3.1** 考虑 $V=V_r-i V_i$ 的情形，其中虚部 $V_i$ 是常数。Hamilton量是 Hermitian 的吗？重新推导连续性方程，并证明发现粒子的总概率按 $e^{-2 V_i t / \hbar}$ 指数衰减。这类复势常用于描述粒子被汇吸收的过程。

**习题 5.3.2** 说明：如果 $\psi=c \tilde{\psi}$，其中 $c$ 为常数（实数或复数），而 $\tilde{\psi}$ 为实函数，那么相应的 $\mathbf{j}$ 为零。

**习题 5.3.3** 考虑
$$
\psi_{\mathbf{p}}=\left(\frac{1}{2 \pi \hbar}\right)^{3 / 2} e^{i(\mathbf{p}\cdot\mathbf{r}) / \hbar}
$$
求 $\mathbf{j}$ 和 $P$，并将它们之间的关系与电磁学方程 $\mathbf{j}=\rho \mathbf{v}$ 比较，其中 $\mathbf{v}$ 为速度。由于 $\rho$ 和 $\mathbf{j}$ 都是常数，注意连续性方程 (5.3.7) 显然自动满足。

**习题 5.3.4** 在一维中考虑
$\psi=A e^{i p x / \hbar}+B e^{-i p x / \hbar}$。
证明
$j=\left(|A|^2-|B|^2\right) p / m$。
由于 $\psi$ 中向右传播和向左传播的两部分之间不存在交叉项，我们可以把 $j$ 的两部分分别与 $\psi$ 的相应两部分联系起来。

## 5.4 单阶跃势：一个散射问题
**习题 5.4.1** 计算方程 (5.416) 中的第三部分，并将所得的 $T$ 与方程 (5.4.21) 比较。[提示：在 $k_1=k_0$ 附近展开因子 $\left(k_1^2-2 m V_0 / \hbar^2\right)^{1 / 2}$，Taylor 级数中只保留一阶导数项。]

**习题 5.4.2** (a) 对势
$V(x)=V_0 a \delta(x)$
的散射，计算 $R$ 和 $T$。

(b) 对下列情形做同样的计算：当 $|x|>a$ 时 $V=0$，当 $|x|<a$ 时 $V=V_0$。假设能量为正但小于 $V_0$。

**习题 5.4.3** 考虑一维中受到恒力 $f$ 作用的粒子。在动量空间中求传播子，并得到
$$
U\left(p, t ; p^{\prime}, 0\right)=\delta\left(p-p^{\prime}-f t\right) e^{i\left(p^{\prime 3}-p^3\right) / 6 m\hbar f}
$$
再变换回坐标空间，得到
$$
U\left(x, t ; x^{\prime}, 0\right)=\left(\frac{m}{2 \pi \hbar i t}\right)^{1 / 2} \exp \left\{\frac{i}{\hbar}\left[\frac{m\left(x-x^{\prime}\right)^2}{2 t}+\frac{1}{2} f t\left(x+x^{\prime}\right)-\frac{f^2 t^3}{24 m}\right]\right\}
$$
[提示：将 $\psi_E(p)$ 归一化，使得 $\left\langle E \mid E^{\prime}\right\rangle=\delta\left(E-E^{\prime}\right)$。注意 $E$ 不要求为正。]

## 5.5 双缝实验
## 5.6 一些定理

# Chapter 5 Simple Problems in One Dimension
## 5.1 The Free Particle
**Exercise 5.1.1** Show that Eq. (5.1.9) may be rewritten as an integral over $E$ and a sum over the $\pm$ index as
$$
U(t)=\sum_{\alpha= \pm} \int_0^{\infty}\left[\frac{m}{(2 m E)^{1 / 2}}\right]|E, \alpha\rangle\langle E, \alpha| \mathrm{e}^{-\mathrm{i} E t / \hbar} \mathrm{d} E
$$


**Solution.**
$$
\begin{aligned}
        E=\dfrac{p^{2}}{2m} \rightarrow p&=\alpha\sqrt{2mE}\\
        \mathrm{d}p&=\dfrac{\alpha m}{\sqrt{2mE}}\mathrm{d}E
    \end{aligned}
$$
where $\alpha=\pm 1$. Hence
$$
\begin{aligned}
U(t)&=\int_{-\infty}^{+\infty} \mathrm{d}p\,|p\rangle\langle p| \mathrm{e}^{-\mathrm{i}Et/\hbar}\\&=\int_{-\infty}^{0}\mathrm{d}p\,|p\rangle\langle p| \mathrm{e}^{-\mathrm{i}Et/\hbar}+\int_{0}^{+\infty}\mathrm{d}p\,|p\rangle\langle p| \mathrm{e}^{-\mathrm{i}Et/\hbar}\\
        &=\int_{-\infty}^{0}\mathrm{d}E \dfrac{-m}{\sqrt{2mE}}\,|E, -\rangle\langle E, -| \mathrm{e}^{-\mathrm{i}Et/\hbar}+\int_{0}^{+\infty}\mathrm{d}E\dfrac{m}{\sqrt{2mE}}\,|E, +\rangle\langle E, +| \mathrm{e}^{-\mathrm{i}Et/\hbar}\\
        &=\int_{0}^{+\infty}\mathrm{d}E \dfrac{m}{\sqrt{2mE}}\,|E, -\rangle\langle E, -| \mathrm{e}^{-\mathrm{i}Et/\hbar}+\int_{0}^{+\infty}\mathrm{d}E\dfrac{m}{\sqrt{2mE}}\,|E, +\rangle\langle E, +| \mathrm{e}^{-\mathrm{i}Et/\hbar}\\
        &=\sum_{\alpha= \pm} \int_0^{\infty}\left[\frac{m}{(2 m E)^{1 / 2}}\right]|E, \alpha\rangle\langle E, \alpha| \mathrm{e}^{-\mathrm{i} E t / \hbar} \mathrm{d} E
\end{aligned}
$$
$$
~\tag*{$\blacksquare$}
$$

**Exercise 5.1.2** By solving the eigenvalue equation (5.1.3) in the $X$ basis, regain Eq. (5.1.8), i.e., show that the general solution of energy $E$ is
$$
\psi_E(x)=\beta \frac{\exp \left[i(2 m E)^{1 / 2} x / \hbar\right]}{(2 \pi \hbar)^{1 / 2}}+\gamma \frac{\exp \left[-i(2 m E)^{1 / 2} x / \hbar\right]}{(2 \pi \hbar)^{1 / 2}}
$$
[The factor $(2 \pi \hbar)^{-1 / 2}$ is arbitrary and may be absorbed into $\beta$ and $\gamma$.] Though $\psi_E(x)$ will satisfy the equation even if $E<0$, are these functions in the Hilbert space?

**Solution.**

In $X$ basis, (5.1.3) is
$$
\langle x\mid H\mid E\rangle=E\langle x\mid E\rangle
$$
it becomes
$$
-\dfrac{\hbar^{2}}{2m}\psi_{E}^{\prime\prime}=E\psi_{E}
$$
The general solution has the following form
$$
\psi_{E}(x)=A_{+}\mathrm{e}^{\frac{\mathrm{i}}{\hbar}\sqrt{2mE}x}+A_{-}\mathrm{e}^{-\frac{\mathrm{i}}{\hbar}\sqrt{2mE}x}
$$
Therefore,
$$
\langle x\mid E\rangle=\beta\langle x\mid E,+\rangle+\gamma\langle x\mid E,-\rangle
$$
if we denote
$$
\begin{aligned}
\beta&=A_{+}\sqrt{2\pi k}\\
\gamma&=A_{-}\sqrt{2\pi k}
\end{aligned}
$$
Finally, we get the general solution of energy $E$
$$
\psi_E(x)=\beta \frac{\exp \left[\mathrm{i}(2 m E)^{1 / 2} x / \hbar\right]}{(2 \pi \hbar)^{1 / 2}}+\gamma \frac{\exp \left[-\mathrm{i}(2 m E)^{1 / 2} x / \hbar\right]}{(2 \pi \hbar)^{1 / 2}}
$$
If $E<0$, these solutions are not in the Hilbert space, since then the two terms grow exponentially either as $x\to+\infty$ or as $x\to -\infty$.

$$
~\tag*{$\blacksquare$}
$$

**Exercise 5.1.3** We have seen that there exists another formula for $U(t)$, namely, $U(t)=e^{-i H t / \hbar}$. For a free particle this becomes
$$
U(t)=\exp \left[\frac{\mathrm{i}}{\hbar}\left(\frac{\hbar^2 t}{2 m} \frac{\mathrm{d}^2}{\mathrm{d} x^2}\right)\right]=\sum_{n=0}^{\infty} \frac{1}{n!}\left(\frac{\mathrm{i} \hbar t}{2 m}\right)^n \frac{\mathrm{d}^{2 n}}{\mathrm{d} x^{2 n}}\tag{5.1.18}
$$
Consider the initial state in Eq. (5.1.14) with $p_0=0$, and set $\Delta=1$, $t^{\prime}=0$ :
$$
\psi(x, 0)=\frac{\mathrm{e}^{-x^2 / 2}}{(\pi)^{1 / 4}}
$$
Find $\psi(x, t)$ using Eq. (5.1.18) above and compare with Eq. (5.1.15).

[Hints: (1) Write $\psi(x, 0)$ as a power series:
$$
\psi(x, 0)=(\pi)^{-1 / 4} \sum_{n=0}^{\infty} \frac{(-1)^n x^{2 n}}{n!(2)^n}
$$
(2) Find the action of a few terms
$$
1,\quad \left(\frac{\mathrm{i} \hbar t}{2 m}\right) \frac{\mathrm{d}^2}{\mathrm{d} x^2}, \quad \frac{1}{2!}\left(\frac{\mathrm{i} \hbar t}{2 m} \frac{\mathrm{d}^2}{\mathrm{d} x^2}\right)^2
$$
etc., on this power series.

(3) Collect terms with the same power of $x$.

(4) Look for the following series expansion in the coefficient of $x^{2 n}$ :
$$
\left(1+\frac{\mathrm{i} t \hbar}{m}\right)^{-n-1/2}=1-(n+1 / 2)\left(\frac{\mathrm{i} \hbar t}{m}\right)+\frac{(n+1 / 2)(n+3 / 2)}{2!}\left(\frac{\mathrm{i} \hbar t}{m}\right)^2+\cdots
$$

(5) Juggle around till you get the answer.]

**Solution.**

$$
~\tag*{$\blacksquare$}
$$

**Exercise 5.1.4** Consider the wave function
$$
\begin{aligned}
\psi(x, 0) & =\sin \left(\frac{\pi x}{L}\right), & & |x| \leqslant L / 2 \\
& =0, & & |x|>L / 2
\end{aligned}
$$
It is clear that when this function is differentiated any number of times we get another function confined to the interval $|x| \leqslant L / 2$. Consequently the action of
$$
U(t)=\exp \left[\frac{\mathrm{i}}{\hbar}\left(\frac{\hbar^2 t}{2 m}\right) \frac{\mathrm{d}^2}{\mathrm{d} x^2}\right]
$$
on this function is to give a function confined to $|x| \leqslant L / 2$. What about the spreading of the wave packet?

**Solution.**
The answer is already given in Shankar: Consider the derivatives at the boundary. We have here an example where the (exponential) operator power series doesn't converge. Notice that the convergence of an operator power series depends not just on the operator but also on the operand. So there is no paradox: if the function dies abruptly as above. so that there seems to be a paradox, the derivatives are singular at the boundary, while if it falls off continuously. the function will definitely leak out given enough time. no matter how rapid the falloff.

Moreover, it is important to realize that this cannot be a solution of Schrödinger equation. Solutions of Schrödinger equation should satisfy the following conditions:

(1) the wavefunction is continuous;

(2) the first order derivative of the wavefunction is usually continuous for finite potentials.

Therefore, the given wavefunction is not physical (not continuous at $x=\pm\dfrac{L}{2}$.)
$$
~\tag*{$\blacksquare$}
$$
## 5.2 The Particle in a Box
**Exercise 5.2.1** A particle is in the ground state of a box of length $L$. Suddenly the box expands (symmetrically) to twice its size, leaving the wave function undisturbed. Show that the probability of finding the particle in the ground state of the new box is $(8 / 3 \pi)^2$.

**Exercise 5.2.2** (a) Show that for any normalized $|\psi\rangle,\langle\psi|H| \psi\rangle \geqslant E_0$, where $E_0$ is the lowest-energy eigenvalue. (Hint : Expand $|\psi\rangle$ in the eigenbasis of $H$.)

(b) Prove the following theorem: Every attractive potential in one dimension has at least one bound state. Hint: Since $V$ is attractive, if we define $V(\infty)=0$, it follows that $V(x)=$ $-|V(x)|$ for all $x$. To show that there exists a bound state with $E<0$, consider
$$
\psi_\alpha(x)=\left(\frac{\alpha}{\pi}\right)^{1 / 4} e^{-\alpha x^2 / 2}
$$
and calculate
$$
E(\alpha)=\left\langle\psi_\alpha|H| \psi_\alpha\right\rangle, \quad H=-\frac{\hbar^2}{2 m} \frac{d^2}{d x^2}-|V(x)|
$$
Show that $E(\alpha)$ can be made negative by a suitable choice of $\alpha$. The desired result follows from the application of the theorem proved above.

**Exercise 5.2.3** Consider $V(x)=-a V_0 \delta(x)$. Show that it admits a bound state of energy $E=-m a^2 V_0^2 / 2 \hbar^2$. Are there any other bound states? Hint: Solve Schrödinger's equation outside the potential for $E<0$, and keep only the solution that has the right behavior at infinity and is continuous at $x=0$. Draw the wave function and see how there is a cusp, or a discontinuous change of slope at $x=0$. Calculate the change in slope and equate it to
$$
\int_{-\varepsilon}^{+\varepsilon}\left(\frac{d^2 \psi}{d x^2}\right) d x
$$
(where $\varepsilon$ is infinitesimal) determined from Schrödinger's equation.

**Exercise 5.2.4** Consider a particle of mass $m$ in the state $|n\rangle$ of a box of length $L$. Find the force $F=-\partial E / \partial L$ encountered when the walls are slowly pushed in, assuming the particle remains in the $n$th state of the box as its size changes. Consider a classical particle of energy $E_n$ in this box. Find its velocity, the frequency of collision on a given wall, the momentum transfer per collision, and hence the average force. Compare it to $-\partial E / \partial L$ computed above.

**Exercise 5.2.5** If the box extends from $x=0$ to $L$ (instead of $-L / 2$ to $L / 2$) show that $\psi_n(x)=(2 / L)^{1 / 2} \sin (n \pi x / L), n=1,2, \ldots, \infty$ and $E_n=\hbar^2 \pi^2 n^2 / 2 m L^2$.

**Exercise 5.2.6** Square Well Potential. Consider a particle in a square well potential:
$$
V(x)= \begin{cases}0, & |x| \leqslant a \\ V_0, & |x| \geqslant a\end{cases}
$$

Since when $V_0 \rightarrow \infty$, we have a box, let us guess what the lowering of the walls does to the states. First of all, all the bound states (which alone we are interested in), will have $E \leq V_0$. Second, the wave functions of the low-lying levels will look like those of the particle in a box, with the obvious difference that $\psi$ will not vanish at the walls but instead spill out with an exponential tail. The eigenfunctions will still be even, odd, even, etc.

(1) Show that the even solutions have energies that satisfy the transcendental equation
$$
k \tan k a=\kappa\tag{5.2.23}
$$
while the odd ones will have energies that satisfy
$$
k \cot k a=-\kappa\tag{5.2.24}
$$
where $k$ and $i \kappa$ are the real and complex wave numbers inside and outside the well, respectively. Note that $k$ and $\kappa$ are related by
$$
k^2+\kappa^2=2 m V_0 / \hbar^2\tag{5.2.25}
$$
Verify that as $V_0$ tends to $\infty$, we regain the levels in the box.

(2) Equations (5.2.23) and (5.2.24) must be solved graphically. In the $(\alpha=k a, \beta=\kappa a)$ plane, imagine a circle that obeys Eq. (5.2.25). The bound states are then given by the intersection of the curve $\alpha \tan \alpha=\beta$ or $\alpha \cot \alpha=-\beta$ with the circle. (Remember $\alpha$ and $\beta$ are positive.)

(3) Show that there is always one even solution and that there is no odd solution unless $V_0 \geqslant \hbar^2 \pi^2 / 8 m a^2$. What is $E$ when $V_0$ just meets this requirement? Note that the general result from Exercise 5.2.2b holds.

## 5.3 The Continuity Equation for Probability
**Exercise 5.3.1** Consider the case where $V=V_r-i V_i$, where the imaginary part $V_i$ is a constant. Is the Hamiltonian Hermitian? Go through the derivation of the continuity equation and show that the total probability for finding the particle decreases exponentially as $e^{-2 V_i t / \hbar}$. Such complex potentials are used to describe processes in which particles are absorbed by a sink.

**Exercise 5.3.2** Convince yourself that if $\psi=c \tilde{\psi}$, where $c$ is constant (real or complex) and $\tilde{\psi}$ is real, the corresponding $\mathbf{j}$ vanishes.

**Exercise 5.3.3** Consider
$$
\psi_{\mathbf{p}}=\left(\frac{1}{2 \pi \hbar}\right)^{3 / 2} e^{i(\mathbf{p}\cdot\mathbf{r}) / \hbar}
$$
Find $\mathbf{j}$ and $P$ and compare the relation between them to the electromagnetic equation $\mathbf{j}=\rho \mathbf{v}$, $\mathbf{v}$ being the velocity. Since $\rho$ and $\mathbf{j}$ are constant, note that the continuity Eq. (5.3.7) is trivially satisfied.

**Exercise 5.3.4** Consider $\psi=A e^{i p x / \hbar}+B e^{-i p x / \hbar}$ in one dimension. Show that $j=\left(|A|^2-|B|^2\right) p / m$. The absence of cross terms between the right- and left-moving pieces in $\psi$ allows us to associate the two parts of $j$ with corresponding parts of $\psi$.
## 5.4 The Single-Step Potential: A Problem in Scattering
**Exercise 5.4.1** Evaluate the third piece in Eq. (5.416) and compare the resulting $T$ with Eq. (5.4.21). [Hint: Expand the factor $\left(k_1^2-2 m V_0 / \hbar^2\right)^{1 / 2}$ near $k_1=k_0$, keeping just the first derivative in the Taylor series.]

**Exercise 5.4.2** (a) Calculate $R$ and $T$ for scattering of a potential $V(x)=V_0 a \delta(x)$. (b) Do the same for the case $V=0$ for $|x|>a$ and $V=V_0$ for $|x|<a$. Assume that the energy is positive but less than $V_0$.

**Exercise 5.4.3** Consider a particle subject to a constant force $f$ in one dimension. Solve for the propagator in momentum space and get
$$
U\left(p, t ; p^{\prime}, 0\right)=\delta\left(p-p^{\prime}-f t\right) e^{i\left(p^{\prime 3}-p^3\right) / 6 m\hbar f}
$$
Transform back to coordinate space and obtain
$$
U\left(x, t ; x^{\prime}, 0\right)=\left(\frac{m}{2 \pi \hbar i t}\right)^{1 / 2} \exp \left\{\frac{i}{\hbar}\left[\frac{m\left(x-x^{\prime}\right)^2}{2 t}+\frac{1}{2} f t\left(x+x^{\prime}\right)-\frac{f^2 t^3}{24 m}\right]\right\}
$$
[Hint: Normalize $\psi_E(p)$ such that $\left\langle E \mid E^{\prime}\right\rangle=\delta\left(E-E^{\prime}\right)$. Note that $E$ is not restricted to be positive.]
## 5.5 The Double-Slit Experiment
## 5.6 Some Theorems

第9章 Heisenberg 不确定关系

9.1 引言

9.2 不确定关系的推导

9.3 最小不确定度波包

9.4 不确定性原理的应用

第8章量子理论的路径积分形式

8.1 路径积分处方

8.2 对该处方的分析

8.3 自由粒子 $U(t)$ 的一个近似

8.4 自由粒子传播子的路径积分计算

8.5 与 Schrödinger 方程的等价性

8.6 形如 $V=a+bx+cx^{2}+d\dot{x}+ex\dot{x}$ 的势

第7章谐振子

7.1 为什么要研究谐振子？

7.2 经典谐振子的回顾

7.3 谐振子的量子化（坐标基底）

第5章一维中的简单问题

5.1 自由粒子