# Perturbation proportional to momentum

1. Jan 27, 2015

### ShayanJ

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

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

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 $\psi_n=e^{i\varphi} \phi_n$ probably with $\varphi=\frac{\lambda}{\hbar}$. 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. Jan 27, 2015

### TSny

Try playing around with $\psi_n=e^{i\varphi x} \phi_n$, where $\varphi=\pm \frac{\lambda}{\hbar}$.

3. Jan 27, 2015

### Goddar

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. Jan 27, 2015

### ShayanJ

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.

5. Jan 27, 2015

### TSny

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. Jan 27, 2015

### Goddar

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. Jan 27, 2015

### ShayanJ

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. Jan 27, 2015

### TSny

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. Jan 27, 2015

### ShayanJ

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?

10. Jan 27, 2015

### Goddar

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