Finding Momentum Mean & Variance from Wavefunction

[Kashmir]

I've a Gaussian momentum space wavefunction as $\phi(p)=\left(\frac{1}{2 \pi \beta^{2}}\right)^{1 / 4} e^{-\left(p-p_{0}\right)^{2} / 4 \beta^{2}}$

So that $|\phi(p)|^{2}=\frac{e^{-\left(p-p_{0}\right)^{2} / 2 \beta^{2}}}{\beta \sqrt{2 \pi}}$

Also then $\psi(x, t)=\frac{1}{\sqrt{2 \pi {\hbar}}} \int_{-\infty}^{} \phi(p) e^{i px/ {\hbar} } e^{-i p^{2} t / 2 m {\hbar} } d p$

hence $\phi(p, t)=\frac{1}{\sqrt{2 \pi h}} \int_{}^{} \psi(x, t) e^{-i px / {\hbar} } d x$I want to find $\langle p\rangle$ and $\Delta p$ at any time $t$. I can use the momentum or position wavefunction to do that, however I'm getting large number of integrals.

Is there any quicker way to find them in this case?

 

[PeroK]

Is there any quicker way to find them in this case?

Probably not. In any case, you have to learn to work with these integrals and Fourier transforms.

If you work in momentum space, then you need the momentum-space version of the SDE to get the time-dependent wave-function. Perhaps that is simplest?

 

[Kashmir]

Probably not. In any case, you have to learn to work with these integrals and Fourier transforms.

If you work in momentum space, then you need the momentum-space version of the SDE to get the time-dependent wave-function. Perhaps that is simplest?

Thank you. :)

 

[vanhees71]

But you have the solution in momentum space. Since $\hat{H}=\hat{p}^2/(2m)$ in momentum space the Schrödinger equation reads
$$\mathrm{i} \hbar \partial_t \phi(p,t)=\frac{p^2}{2m} \phi(p,t)$$
with the solution
$$\phi(p,t)=\exp \left (-\frac{\mathrm{i} p^2 t}{2m \hbar} \right) \phi_0(p).$$
Correspondingly the momentum distribution doesn't change,
$$|\phi(p,t)|^2=|\phi_0(p)|^2.$$

 

[Kashmir]

But you have the solution in momentum space. Since $\hat{H}=\hat{p}^2/(2m)$ in momentum space the Schrödinger equation reads
$$\mathrm{i} \hbar \partial_t \phi(p,t)=\frac{p^2}{2m} \phi(p,t)$$
with the solution
$$\phi(p,t)=\exp \left (-\frac{\mathrm{i} p^2 t}{2m \hbar} \right) \phi_0(p).$$
Correspondingly the momentum distribution doesn't change,
$$|\phi(p,t)|^2=|\phi_0(p)|^2.$$

I've not done Schrodinger equation in momentum space but this is my attempt. If it's wrong please tell me:

By operation of a momentum bra on first equation we've the second one

$\begin{aligned} i \hbar \frac{d}{d t}|\psi\rangle &=\frac{1}{2 m} \hat{p}^{2}|\psi\rangle \\ i{\hbar} \frac{d}{d t} \phi(p, t) &=\frac{1}{2 m}\left\langle p\left|\hat{p}^{2}\right| \psi\right\rangle \end{aligned}$Now $\langle p|\hat{p} \cdot \hat{p}| \psi\rangle$ can be evaluated by noting that $p<p \mid=\langle p| \hat{p}$

So We get $i \hbar \frac{d}{d t} \phi(p ,t)=(p^{2}/2m). \phi(p,t)$

 

[Kashmir]

Thank you @PeroK @vanhees71