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量子力学基础概念：涵盖量子态、算符、测量等各种常用量子力学概念的定义、性质和分类。
薛定谔方程：研究和应用于量子力学中的薛定谔方程及其解析解、数值解等。
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量子力学与统计力学：量子力学在统计力学领域的应用，包括量子统计和玻尔兹曼分布等。
量子力学与相对论：介绍量子力学和相对论的关系和量子场论的基本原理。
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Problem 1.1 Consider a particle and two normalized energy eigenfunctions $\psi_1(\mathbf{x})$ and $\psi_2(\mathbf{x})$ corresponding to the eigenvalues $E_1 \neq E_2$. Assume that the eigenfunctions vanish outside the two non-overlapping regions $\Omega_1$ and $\Omega_2$ respectively.
(a) Show that, if the particle is initially in region $\Omega_1$ then it will stay there forever.
(b) If, initially, the particle is in the state with wave function
$$
\psi(\mathbf{x}, 0)=\frac{1}{\sqrt{2}}\left[\psi_1(\mathbf{x})+\psi_2(\mathbf{x})\right]
$$
show that the probability density $|\psi(\mathbf{x}, t)|^2$ is independent of time.
(c) Now assume that the two regions $\Omega_1$ and $\Omega_2$ overlap partially. Starting with the initial wave function of case (b), show that the probability density is a periodic function of time.
(d) Starting with the same initial wave function and assuming that the two eigenfunctions are real and isotropic, take the two partially overlapping regions $\Omega_1$ and $\Omega_2$ to be two concentric spheres of radii $R_1>R_2$. Compute the probability current that flows through $\Omega_1$.
Solution
(a) Clearly $\psi(\mathbf{x}, t)=e^{-i E t / \hbar} \psi_1(\mathbf{x})$ implies that $|\psi(\mathbf{x}, t)|^2=\left|\psi_1(\mathbf{x})\right|^2$, which vanishes outside $\Omega_1$ at all times.
(b) If the two regions do not overlap, we have
$$
\psi_1(\mathbf{x}) \psi_2^*(\mathbf{x})=0
$$
everywhere and, therefore,
$$
|\psi(\mathbf{x}, t)|^2=\frac{1}{2}\left[\left|\psi_1(\mathbf{x})\right|^2+\left|\psi_2(\mathbf{x})\right|^2\right]
$$
which is time independent.
(c) If the two regions overlap, the probability density will be
$$
\begin{aligned}
|\psi(\mathbf{x}, t)|^2= & \frac{1}{2}\left[\left|\psi_1(\mathbf{x})\right|^2+\left|\psi_2(\mathbf{x})\right|^2\right] \
& +\left|\psi_1(\mathbf{x})\right|\left|\psi_2(\mathbf{x})\right| \cos \left[\phi_1(\mathbf{x})-\phi_2(\mathbf{x})-\omega t\right]
\end{aligned}
$$
where we have set $\psi_{1,2}=\left|\psi_{1,2}\right| e^{i \phi_{1,2}}$ and $E_1-E_2=\hbar \omega$. This is clearly a periodic function of time with period $T=2 \pi / \omega$.
(d) The current density is easily computed to be
$$
\mathcal{J}=\hat{\mathbf{r}} \frac{\hbar}{2 m} \sin \omega t\left[\psi_2^{\prime}(r) \psi_1(r)-\psi_1^{\prime}(r) \psi_2(r)\right]
$$
and vanishes at $R_1$, since one or the other eigenfunction vanishes at that point. This can be seen through the continuity equation in the following alternative way:
$$
\begin{aligned}
I_{\Omega_1} & =-\frac{d}{d t} \mathcal{P}{\Omega_1}=\int{S\left(\Omega_1\right)} d \mathbf{S} \cdot \mathcal{J}=\int_{\Omega_1} d^3 x \nabla \cdot \mathcal{J}=-\int_{\Omega_1} d^3 x \frac{\partial}{\partial t}|\psi(\mathbf{x}, t)|^2 \
& =\omega \sin \omega t \int_{\Omega_1} d^3 x \psi_1(r) \psi_2(r)
\end{aligned}
$$
The last integral vanishes because of the orthogonality of the eigenfunctions.
Problem 1.2 Consider the one-dimensional normalized wave functions $\psi_0(x)$, $\psi_1(x)$ with the properties
$$
\psi_0(-x)=\psi_0(x)=\psi_0^*(x), \quad \psi_1(x)=N \frac{d \psi_0}{d x}
$$
Consider also the linear combination
$$
\psi(x)=c_1 \psi_0(x)+c_2 \psi_1(x)
$$
with $\left|c_1\right|^2+\left|c_2\right|^2=1$. The constants $N, c_1, c_2$ are considered as known.
(a) Show that $\psi_0$ and $\psi_1$ are orthogonal and that $\psi(x)$ is normalized.
(b) Compute the expectation values of $x$ and $p$ in the states $\psi_0, \psi_1$ and $\psi$.
(c) Compute the expectation value of the kinetic energy $T$ in the state $\psi_0$ and demonstrate that
$$
\left\langle\psi_0\left|T^2\right| \psi_0\right\rangle=\left\langle\psi_0|T| \psi_0\right\rangle\left\langle\psi_1|T| \psi_1\right\rangle
$$
and that
$$
\left\langle\psi_1|T| \psi_1\right\rangle \geq\langle\psi|T| \psi\rangle \geq\left\langle\psi_0|T| \psi_0\right\rangle
$$
Solution
(a) We have
$$
\begin{aligned}
\left\langle\psi_0 \mid \psi_1\right\rangle & =N \int d x \psi_0^* \frac{d \psi_0}{d x}=N \int d x \psi_0 \frac{d \psi_0}{d x} \
& =\frac{N}{2} \int d x \frac{d \psi_0^2}{d x}=\frac{N}{2}\left{\psi_0^2(x)\right}_{-\infty}^{+\infty}=0
\end{aligned}
$$
The normalization of $\psi(x)$ follows immediately from this and from the fact that $\left|c_1\right|^2+\left|c_2\right|^2=1$
(b) On the one hand the expectation value $\left\langle\psi_0|x| \psi_0\right\rangle$ vanishes because the integrand $x \psi_0^2(x)$ is odd. On the other hand, the momentum expectation value in this state is
$$
\begin{aligned}
\left\langle\psi_0|p| \psi_0\right\rangle & =-i \hbar \int d x \psi_0(x) \psi_0^{\prime}(x) \
& =-\frac{i \hbar}{N} \int d x \psi_0(x) \psi_1(x)=-\frac{i \hbar}{N}\left\langle\psi_0 \mid \psi_1\right\rangle=0
\end{aligned}
$$
as we proved in the solution to (a). Similarly, owing to the oddness of the integrand $x \psi_1^2(x)$, the expectation value $\left\langle\psi_1|x| \psi_1\right\rangle$ vanishes. The momentum expectation value is
$$
\begin{aligned}
\left\langle\psi_1|p| \psi_1\right\rangle & =-i \hbar \int d x \psi_1^* \psi_1^{\prime}=-i \hbar \frac{N}{N^} \int d x \psi_1 \psi_1^{\prime} \ & =-i \hbar \frac{N}{2 N^} \int d x \frac{d \psi_1^2}{d x}=-i \hbar \frac{N}{2 N^*}\left{\psi_1^2\right}_{-\infty}^{+\infty}=0
\end{aligned}
$$
(c) The expectation value of the kinetic energy squared in the state $\psi_0$ is
$$
\begin{aligned}
\left\langle\psi_0\left|T^2\right| \psi_0\right\rangle & =\frac{\hbar^4}{4 m^2} \int d x \psi_0 \psi_0^{\prime \prime \prime \prime}=-\frac{\hbar^4}{4 m^2} \int d x \psi_0^{\prime} \psi_0^{\prime \prime \prime} \
& =\frac{\hbar^2}{2 m|N|^2}\left\langle\psi_1|T| \psi_1\right\rangle
\end{aligned}
$$
Note however that
$$
\begin{aligned}
\left\langle\psi_0|T| \psi_0\right\rangle & =-\frac{\hbar^2}{2 m} \int d x \psi_0 \psi_0^{\prime \prime}=\frac{\hbar^2}{2 m} \int d x \psi_0^{\prime} \psi_0^{\prime} \
& =\frac{\hbar^2}{2 m|N|^2}\left\langle\psi_1 \mid \psi_1\right\rangle=\frac{\hbar^2}{2 m|N|^2}
\end{aligned}
$$
Therefore, we have
$$
\left\langle\psi_0\left|T^2\right| \psi_0\right\rangle=\left\langle\psi_0|T| \psi_0\right\rangle\left\langle\psi_1|T| \psi_1\right\rangle
$$
Consider now the Schwartz inequality
$$
\left|\left\langle\psi_0 \mid \psi_2\right\rangle\right|^2 \leq\left\langle\psi_0 \mid \psi_0\right\rangle\left\langle\psi_2 \mid \psi_2\right\rangle=\left\langle\psi_2 \mid \psi_2\right\rangle
$$
where, by definition,
$$
\psi_2(x) \equiv-\frac{\hbar^2}{2 m} \psi_0^{\prime \prime}(x)
$$
The right-hand side can be written as
$$
\left\langle\psi_2 \mid \psi_2\right\rangle=\left\langle\psi_0\left|T^2\right| \psi_0\right\rangle=\left\langle\psi_0|T| \psi_0\right\rangle\left\langle\psi_1|T| \psi_1\right\rangle
$$
Thus, the above Schwartz inequality reduces to
$$
\left\langle\psi_0|T| \psi_0\right\rangle \leq\left\langle\psi_1|T| \psi_1\right\rangle
$$
In order to prove the desired inequality let us consider the expectation value of the kinetic energy in the state $\psi$. It is
$$
\langle\psi|T| \psi\rangle=\left|c_1\right|^2\left\langle\psi_0|T| \psi_0\right\rangle+\left|c_2\right|^2\left\langle\psi_1|T| \psi_1\right\rangle
$$
The off-diagonal terms have vanished due to oddness. The right-hand side of this expression, owing to the inequality proved above, will obviously be smaller than
$$
\left|c_1\right|^2\left\langle\psi_1|T| \psi_1\right\rangle+\left|c_2\right|^2\left\langle\psi_1|T| \psi_1\right\rangle=\left\langle\psi_1|T| \psi_1\right\rangle
$$
Analogously, the same right-hand side will be larger than
$$
\left|c_1\right|^2\left\langle\psi_0|T| \psi_0\right\rangle+\left|c_2\right|^2\left\langle\psi_0|T| \psi_0\right\rangle=\left\langle\psi_0|T| \psi_0\right\rangle
$$
Thus, finally, we end up with the double inequality
$$
\left\langle\psi_0|T| \psi_0\right\rangle \leq\langle\psi|T| \psi\rangle \leq\left\langle\psi_0|T| \psi_0\right\rangle
$$
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