A particle of mass $m$ moves in one dimension in a potential $V(x)$ which satisfies $V(x)=V(-x)$. Show that the eigenstates of the Hamiltonian $H$ can be chosen so that they are also eigenstates of the parity operator $P$. For eigenstates with odd parity $\psi^{\text {odd }}(x)$, show that $\psi^{o d d}(0)=0$.

A potential $V(x)$ is given by

$$V(x)= \begin{cases}\kappa \delta(x) & |x|<a \\ \infty & |x|>a\end{cases}$$

State the boundary conditions satisfied by $\psi(x)$ at $|x|=a$, and show also that

$$\frac{\hbar^{2}}{2 m} \lim _{\epsilon \rightarrow 0}\left[\left.\frac{d \psi}{d x}\right|_{\epsilon}-\left.\frac{d \psi}{d x}\right|_{-\epsilon}\right]=\kappa \psi(0)$$

Let the energy eigenstates of even parity be given by

$$\psi^{\text {even }}(x)=\left\{\begin{array}{lc} A \cos \lambda x+B \sin \lambda x & -a<x<0 \\ A \cos \lambda x-B \sin \lambda x & 0<x<a \\ 0 & \text { otherwise } \end{array}\right.$$

Verify that $\psi^{\text {even }}(x)$ satisfies

$$P \psi^{\text {even }}(x)=\psi^{\text {even }}(x)$$

By demanding that $\psi^{e v e n}(x)$ satisfy the relevant boundary conditions show that

$$\tan \lambda a=-\frac{\hbar^{2}}{m} \frac{\lambda}{\kappa}$$

For $\kappa>0$ show that the energy eigenvalues $E_{n}^{\text {even }}, n=0,1,2, \ldots$, with $E_{n}^{\text {even }}<E_{n+1}^{\text {even }}$, satisfy

$$\eta_{n}=E_{n}^{e v e n}-\frac{1}{2 m}\left[\frac{(2 n+1) \hbar \pi}{2 a}\right]^{2}>0$$

Show also that

$$\lim _{n \rightarrow \infty} \eta_{n}=0,$$

and give a physical explanation of this result.

Show that the energy eigenstates with odd parity and their energy eigenvalues do not depend on $\kappa$.