1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Perturbation proportional to momentum

  1. Jan 27, 2015 #1

    ShayanJ

    User Avatar
    Gold Member

    1. The problem statement, all variables and given/known data
    I'm struggling with the problem below:
    If [itex] E_n^{(0)} \ \ (n\in \mathbb N)[/itex] are the energy eigenvalues of a system with Hamiltonian [itex] H_0=\frac{p^2}{2m}+V(x) [/itex], what are the exact energy eigenvalues of the system if the Hamiltonian is changed to [itex] H=H_0+\frac{\lambda}{m} p \ [/itex]?
    2. Relevant equations


    3. The attempt at a solution
    [itex] H=\frac{p^2}{2m}+V(x)+\frac{\lambda}{m} p+\frac{\lambda^2}{2m}-\frac{\lambda^2}{2m}=\frac{(p+\lambda)^2}{2m}+V(x)-\frac{\lambda^2}{2m}[/itex]
    So it seems that I should write:
    [itex]
    H \psi_n=E_n \psi_n \Rightarrow [\frac{(p+\lambda)^2}{2m}+V(x)]\psi_n=(E_n+\frac{\lambda^2}{2m})\psi_n
    [/itex]
    But from the statement of the problem, we know that:
    [itex]
    H_0 \phi_n=E_n^{(0)} \phi_n \Rightarrow [\frac{p^2}{2m}+V(x)]\phi_n=E_n^{(0)}\phi_n
    [/itex]

    My problem is, I can't find a connection between the two equations. One idea I have is that because the first equation has its momentum shifted by a constant, we may have [itex] \psi_n=e^{i\varphi} \phi_n [/itex] probably with [itex] \varphi=\frac{\lambda}{\hbar} [/itex]. But it seems to me that this can be true only for plane waves and so we should decompose the wavefunctions to plane waves and sum them again later. But this again has the problem that because of the presence of the potential, plane waves are not solutions to the TISE in general.
    Any ideas?
    Thanks
     
  2. jcsd
  3. Jan 27, 2015 #2

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Try playing around with ## \psi_n=e^{i\varphi x} \phi_n ##, where [itex] \varphi=\pm \frac{\lambda}{\hbar} [/itex].
     
  4. Jan 27, 2015 #3
    In momentum space, it seems like Ψn(p) = φn(p + λ) will automatically be an eigenfunction of the new Hamiltonian, with energy levels shifted to En(0)+ λ2/2m...
     
  5. Jan 27, 2015 #4

    ShayanJ

    User Avatar
    Gold Member

    With ## \varphi=-\frac{\lambda}{\hbar} ##, I get ## E_n=E_n^{(0)} ##!(## \varphi=\frac{\lambda}{\hbar} ## doesn't make sense!).
    Actually I'd have no problem with such a conclusion but the question is a multiple choice question and all four options show a change in energy.

    Yeah, it seems so. I wonder why I get ## E_n=E_n^{(0)} ## in position representation.
     
  6. Jan 27, 2015 #5

    TSny

    User Avatar
    Homework Helper
    Gold Member

    Suppose you let ##\psi_n = \phi_n e^{-i\lambda x/\hbar}##. What do you get if you apply the momentum operator to ##\psi_n##? Then, what do you get for the expression ##(\hat{p} + \lambda)\psi_n##?
     
  7. Jan 27, 2015 #6
    It reminds me of the more familiar problem of adding a term proportional to x to the harmonic oscillator potential: by completing the square, one gets a new H.O. equation with the energies shifted. The analogy works because the H.O. is quadratic in position, the way momentum is here...
    And with P = p + λ, we have dP = dp so in momentum representation, < V(ih∂p) > is not affected by the shift.
     
  8. Jan 27, 2015 #7

    ShayanJ

    User Avatar
    Gold Member

    Oh...It seems by the simultanuous transformations ## \hat p \rightarrow \hat p+\lambda ## and ## \phi_n \rightarrow e^{-i\frac \lambda \hbar x} \phi_n ##, nothing changes, so much like the transformations ## \hat H\rightarrow \hat H+\hbar \omega ## and ## e^{-i \frac E \hbar t}\rightarrow e^{-i \frac E \hbar t} e^{- i \omega t} ##(did I wrote the correct form?). Does this mean we're just doing a translation? But then its strange that we encounter only a phase multiplied by the wavefunction because potential depends on position and so there is no translation symmetry in general! The change in energy is still unclear to me but it seems a little natural because we're doing a translation and so the difference in the potential energy of the different positions should get into our equations. But I have problem thinking about the exact calculations.
     
  9. Jan 27, 2015 #8

    TSny

    User Avatar
    Homework Helper
    Gold Member

    You should be able to show that ##\psi_n = \phi_n e^{-i\frac{\lambda}{\hbar}x}## satisfies the eigenfunction equation ##[\frac{(p+\lambda)^2}{2m}+V(x)-\frac{\lambda^2}{2m}]\psi_n=E_n\psi_n## and discover the eigenvalues ##E_n##.
     
  10. Jan 27, 2015 #9

    ShayanJ

    User Avatar
    Gold Member

    Yeah, that's what I recalled in my last post.
    Oh..sorry man, I had a mistake in my calculations. Now I get ## E_n=E_n^{(0)}-\frac{\lambda^2}{2m} ##.
    Thanks

    Anyway, this ## \frac \lambda m p ## term doesn't seem to be a like other interactions. What is it actually? Are we giving the particle a push?
     
  11. Jan 27, 2015 #10
    Indeed, by taking Ψn(p) = φn(p + λ) you get:
    Ψn(x) = (2πh)-1/2∫dp φn(p + λ) eipx/h = (2πh)-1/2∫dP φn(P) ei(P–λ)x/h = φn(x) e-iλx/h
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Perturbation proportional to momentum
  1. Perturbation Theory (Replies: 2)

  2. Perturbation theory (Replies: 1)

  3. Perturbation theory (Replies: 2)

Loading...