Finding momentum amplitude of a wave packet when initial wave form is given

Question

At time $t = 0$, a one-dimensional free wave packet for a particle of mass $m$ takes the form:

$$ \Psi(x,0) = \begin{cases} \frac{1}{\sqrt{L}}e^{i\alpha x} & \text{for } -L/2 < x < +L/2 \\ 0 & \text{elsewhere} \end{cases} $$

where $\alpha$ is a real constant. I need to find momentum amplitude $A(k)$ for this wave packet. And write the expression of the time-dependent wavefunction.

I know we can write a free wave packet as a linear combination of plane waves

$$ \Psi(x,t) = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}dk\,\tilde{\varphi}(k)e^{-i\omega(k)t}e^{ikx} = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{+\infty}dk\,\tilde{\varphi}(k,t)e^{ikx} $$

And we chose the amplitude for the different plane waves as a Gaussian

$$ \tilde{\varphi}(k,t=0) = \frac{1}{\sqrt{\sqrt{2\pi}\sigma_k}}e^{-(k-k_0)^2/4\sigma_k^2} $$

and further, I can use the fact that $(e^{i\theta}-e^{-i\theta})/2i=\sin\theta$

But apart from these pieces, I have no idea how to proceed.

Answer 1

The plane waves must be momentum eigenstates $\langle x| k\rangle= e^{ikx}$ normalized so that $$ \langle k|k'\rangle=\int_{-\infty}^{\infty}\langle k| x\rangle \langle x| k'\rangle \,dx = \int_{-\infty}^{\infty} e^{-ikx}e^{ik'x} \,dx = 2\pi \delta(k-k') $$ and with completeness relation $$ {\mathbb I}= \int_{-\infty}^{\infty}\frac {dk}{2\pi} |k\rangle\langle k|. $$ Then $$ A(k)\equiv \langle{k}|\Psi\rangle= \int_{-\infty}^{\infty}\langle k| x\rangle \langle x| \Psi\rangle \,dx= \int_{-\infty}^{\infty} e^{-ikx}\Psi(x) dx, $$ and $$ \Psi(x) \equiv \langle x| \Psi\rangle=\int_{-\infty}^{\infty}\frac {dk}{2\pi} \langle x| k\rangle \langle{k}|\Psi\rangle. $$ I leave you to do the easy integrals and to add the time dependence.

Whe re did you get your Gaussian idea from?