Perturbation proportional to momentum

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Homework Statement


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]?

Homework Equations


[/B]

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
 

Answers and Replies

  • #2
TSny
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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].
Try playing around with ## \psi_n=e^{i\varphi x} \phi_n ##, where [itex] \varphi=\pm \frac{\lambda}{\hbar} [/itex].
 
  • #3
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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...
 
  • #4
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Try playing around with ## \psi_n=e^{i\varphi x} \phi_n ##, where [itex] \varphi=\pm \frac{\lambda}{\hbar} [/itex].
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.

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...
Yeah, it seems so. I wonder why I get ## E_n=E_n^{(0)} ## in position representation.
 
  • #5
TSny
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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##?
 
  • #6
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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.
 
  • #7
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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##?
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.
 
  • #8
TSny
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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##.
 
  • #9
2,792
594
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.
Yeah, that's what I recalled in my last post.
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##.
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?
 
  • #10
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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
 

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