Let $\hat{x}, \hat{p}$ and $H(\hat{x}, \hat{p})=\hat{p}^{2} / 2 m+V(\hat{x})$ be the position operator, momentum operator and Hamiltonian for a particle moving in one dimension. Let $|\psi\rangle$ be the state vector for the particle. The position and momentum eigenstates have inner products

$$\langle x \mid p\rangle=\frac{1}{\sqrt{2 \pi \hbar}} \exp (i p x / \hbar), \quad\left\langle x \mid x^{\prime}\right\rangle=\delta\left(x-x^{\prime}\right) \quad \text { and } \quad\left\langle p \mid p^{\prime}\right\rangle=\delta\left(p-p^{\prime}\right) .$$

Show that

$$\langle x|\hat{p}| \psi\rangle=-i \hbar \frac{\partial}{\partial x} \psi(x) \quad \text { and } \quad\langle p|\hat{x}| \psi\rangle=i \hbar \frac{\partial}{\partial p} \tilde{\psi}(p)$$

where $\psi(x)$ and $\tilde{\psi}(p)$ are the wavefunctions in the position representation and momentum representation, respectively. Show how $\psi(x)$ and $\tilde{\psi}(p)$ may be expressed in terms of each other.

For general $V(\hat{x})$, express $\langle p|V(\hat{x})| \psi\rangle$ in terms of $\tilde{\psi}(p)$, and hence write down the time-independent Schrödinger equation in the momentum representation satisfied by $\tilde{\psi}(p)$.

Consider now the case $V(x)=-\left(\hbar^{2} \lambda / m\right) \delta(x), \lambda>0$. Show that there is a bound state with energy $E=-\varepsilon, \varepsilon>0$, with wavefunction $\tilde{\psi}(p)$ satisfying

$$\tilde{\psi}(p)=\frac{\hbar \lambda}{\pi} \frac{1}{2 m \varepsilon+p^{2}} \int_{-\infty}^{\infty} \tilde{\psi}\left(p^{\prime}\right) d p^{\prime}$$

Hence show that there is a unique value for $\varepsilon$ and determine this value.