A particle in one dimension has position and momentum operators $\hat{x}$ and $\hat{p}$ whose eigenstates obey

$$\left\langle x \mid x^{\prime}\right\rangle=\delta\left(x-x^{\prime}\right), \quad\left\langle p \mid p^{\prime}\right\rangle=\delta\left(p-p^{\prime}\right), \quad\langle x \mid p\rangle=(2 \pi \hbar)^{-1 / 2} e^{i x p / \hbar}$$

Given a state $|\psi\rangle$, define the corresponding position-space and momentum-space wavefunctions $\psi(x)$ and $\tilde{\psi}(p)$ and show how each of these can be expressed in terms of the other. Derive the form taken in momentum space by the time-independent Schrödinger equation

$$\left(\frac{\hat{p}^{2}}{2 m}+V(\hat{x})\right)|\psi\rangle=E|\psi\rangle$$

for a general potential $V$.

Now let $V(x)=-\left(\hbar^{2} \lambda / m\right) \delta(x)$ with $\lambda$ a positive constant. Show that the Schrödinger equation can be written

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

and verify that it has a solution $\tilde{\psi}(p)=N /\left(p^{2}+\alpha^{2}\right)$ for unique choices of $\alpha$ and $E$, to be determined (you need not find the normalisation constant, $N$ ). Check that this momentum space wavefunction can also be obtained from the position space solution $\psi(x)=\sqrt{\lambda} e^{-\lambda|x|}$.