Quantum Mechanics I, finding impuls wavefunction.

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
6 replies · 1K views
milkism
Messages
118
Reaction score
15
Homework Statement
Find the impuls wavefunction phi(p) from the position wavefunction.
Relevant Equations
Look solution.
I have this following Gaussian wavefunction.
1705233351985.png

I found the constant C to be $$\sqrt{\sqrt{\frac{2 \alpha}{\pi}}}$$.
Now they're asking me to find the normalized impuls wavefunction $$\phi(p)$$. I tried to use the fourier transform relation
$$\phi (p) = \int e^{-\frac{i ( p x)}{\hbar}} \Psi (x,t=0) dx$$
and i got a long answer
$$\sqrt{\sqrt{\frac{2 \alpha}{\pi}}} \sqrt{\frac{\pi}{\alpha}} e^{-\frac{q^2}{4\alpha} + \frac{pq}{2 \alpha \hbar} - \frac{p^2}{4 \alpha \hbar ^2}}$$
Is there an other way to solve this? Because next question is to find the expectation value of position from the normalized impuls wavefunction, which is going to be very hard.
 
Physics news on Phys.org
Is there a better way to find impuls wavefunctions from position wavefunctions?
 
BvU said:
How ?
Can you please post the complete problem statement ?

##\ ##
$$\int_{-\infty}^{\infty} \Psi ^{*} \Psi dx =1, \int_{-\infty}^{\infty} C^2 e^{-2 \alpha x^2} dx = C^2 \sqrt{\frac{\pi}{2\alpha}} = 1$$
 
Reply
  • Like
Likes   Reactions: BvU
Calculate the corresponding normalized wave function φ(p) in momentum space. Explicitly compute, based on the knowledge of φ(p), the expectation value 〈x〉.
 
Reply
  • Like
Likes   Reactions: BvU
milkism said:
and i got a long answer
##e^{-{1\over 4\alpha}\left(q-p\right )^2}\ ## doesn't look all that bad to me ...
a peak around p = q, so the exercise will probably end up at something moving to the right with momentum ##q##

(did you check there is no ##i## in there ?)

The exercise reminded me of the treatment in Merzbacher, QM 2nd ed (1970 !) chapter 2.2

Next step was $$\psi(x,t) = {1\over \sqrt{2\pi}}\int_{-\infty}^{+\infty} \phi(k) e^{i(kx-\omega t}) dk$$ but I'm too rusty to comfortably work that out ( i.e. ##\omega(k)## )

##\ ##
 
Reply
  • Like
Likes   Reactions: PhDeezNutz and milkism