First some background information:
Eigenstates of the momentum operator with eigenvalue $p$ are given by $$\phi(p,x)=\frac{1}{\sqrt{2\pi \hbar}}\exp\left(\frac{i \, p x}{\hbar}\right)\tag{1}$$
Consider a square-wave wavepacket at time $t = 0$:
$$\psi_S(x)=\begin{cases}B\exp\left(\frac{i\, p_m \,x}{\hbar}\right) & -a/2 \lt x \lt a/2\\ 0 & \qquad\text{otherwise}\end{cases}\tag{2}$$
The momentum representation amplitude $a_S(p)$ is: $$\begin{align}a_S(p) &=\int_{-a/2}^{a/2}\phi^*(p,x)\psi_S(x)\,\mathrm{d}x \\&= \frac{B}{\sqrt{2 \pi \hbar}}\int_{-a/2}^{a/2}\exp\left(\frac{-i\, p\, x}{\hbar}\right)\exp\left(\frac{i\,p_m \,x}{\hbar}\right)\,\mathrm{d}x \\&=\frac{B}{\sqrt{2 \pi \hbar}}\int_{-a/2}^{a/2}\exp\left(\frac{-i\,(p-p_m) \,x}{\hbar}\right)\,\mathrm{d}x \\&= \frac{B}{\sqrt{2 \pi \hbar}}\frac{\sin\left[(p-p_m)a/(2 \hbar)\right]}{(p-p_m)/(2 \hbar)} \end{align}\tag{3}$$
Now here is my question:
If there is no potential acting, write down an expression for the wavefunction $\psi_S(x, t)$ as an integral involving $a_S(p)$. Is the square-wave shape preserved after $t = 0$ ? (You need not perform the integral.)
Here is my attempt at the question:
The question is asking me to write the following integral incorporating time dependence $e^{-i\,E \,t /\hbar}$ and using $E=p^2/(2m)$ along with $(1)$ and $(3)$:
$$\begin{align}\psi(x,t) &=\int \color{red}{a_S(p)}\color{#085}{\phi(p,x)}\color{blue}{\exp\left(\frac{-i E\, t}{\hbar}\right)}\,\mathrm{d}p \\ &= \int \color{red}{\frac{B}{\sqrt{2 \pi \hbar}}\frac{\sin\left[(p-p_m)a/(2 \hbar)\right]}{(p-p_m)/(2 \hbar)}}\color{#085}{\frac{1}{\sqrt{2 \pi \hbar}}{\exp\left(\frac{i \, p x}{\hbar}\right)}\color{blue}{\exp\left(\frac{-i\,p^2\,t}{2 m \hbar}\right)}}\,\mathrm{d}p \\ \end{align}$$
I have color coded the parts of the integrand corresponding to the factors in the the first integral above.
But the problem is that the correct answer is:
Answer: Unseen Momentum eigenvalues and energy eigenvalues are identical for the free particle, and each evolves with time-dependence $e^{-i E t / \hbar}$ with $E = p^2/2m$: \begin{align*} \psi(x,t) &= \frac{1}{\sqrt{2 \pi \hbar}} \int a_S(p) e^{+ ipx/\hbar} e^{-i p^2 t/(2 m \hbar)} \,\mathrm{d}p \\ &= 2 \hbar B \int \frac{\sin[(p-p_m)a/(2\hbar)]}{p-p_m} e^{-ipx/\hbar} e^{-i p^2 t / (2 m \hbar)} \,\mathrm{d}p \end{align*} This is not the inverse transform of the $\operatorname{sinc}$ function, due to the presence of the $e^{-i p^2 t / (2 m \hbar)}$ factor, so it’s not a square wave.
(Original image here.
This was a 2013-2014 past exam question for a quantum mechanics exam from Imperial College London.
Would I would like to know is how they managed to reach the final expression for $\psi(x,t)$. Firstly, I cannot figure out why there is a minus sign in front of the argument of the complex exponential for the momentum eigenstate. How did this happen?
Also, there were two factors of $\dfrac{1}{\sqrt{2\pi\hbar}}$ so why isn't there a factor $\pi$ in the denominator? Instead they put a $2\pi B$ in front of the integral which makes no sense to me.
Could someone please help me understand how they reached the final expression for $\psi(x,t)$?
Many thanks.