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

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.

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.