I Finding Momentum Mean & Variance from Wavefunction

Kashmir
Messages
466
Reaction score
74
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?
 
Physics news on Phys.org
Kashmir said:
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?
 
PeroK said:
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. :)
 
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,
$$|\phi(p,t)|^2=|\phi_0(p)|^2.$$
 
vanhees71 said:
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)##
 
  • Like
Likes vanhees71 and PeroK
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
According to recent podcast between Jacob Barandes and Sean Carroll, Barandes claims that putting a sensitive qubit near one of the slits of a double slit interference experiment is sufficient to break the interference pattern. Here are his words from the official transcript: Is that true? Caveats I see: The qubit is a quantum object, so if the particle was in a superposition of up and down, the qubit can be in a superposition too. Measuring the qubit in an orthogonal direction might...
Back
Top