Suppose I have Hamiltonian $H_0(\hat{p},\hat{r})$, it satisfies $H_0\psi(p,r)=E(p)\psi(p,r)$. If I make a change from $\hat{p}\to\hat{p}+p_0$, what is the form of the wave function of the Hamiltonian $H_1=H(\hat{p}+p_0, \hat{r})$ with the eigenvalue $E_p$?

Should it be $\psi(p+p_0, r)$ or $\psi(p-p_0, r)$? or neither?

For the simplest case, consider $H_0(\hat{p},\hat{r})=\hat{p}^2/2m$, if I choose the wave function to be $\psi(p,r)=e^{ipr/\hbar}$, it seems $H_1\psi(p-p_0, r)=E_p\psi(p-p_0,r)$;

if I if I choose the wave function to be $\psi(p,r)=e^{-ipr/\hbar}$, it seems $H_1\psi(p+p_0, r)=E_p\psi(p+p_0,r)$;

  • 5
    $\begingroup$ Wavefunctions are not functions of both $x$ and $p$. They are either functions of $x$ in position space or functions of $p$ in momentum space. It's not clear to me what you're trying to do here. $\endgroup$ – ACuriousMind Aug 17 at 14:01
  • $\begingroup$ Let's suppose in r space and p is some quantum number related to $\hat{p}$, for the plane wave example is in r space. $\endgroup$ – an offer can't refuse Aug 17 at 15:16
  • $\begingroup$ I think if you worked in p-space altogether, your question would be clear. It would take a few sign flips... $\endgroup$ – Cosmas Zachos Aug 18 at 13:00

You skipped the "either" option. Let's work in one dimension, and recall your unlocalized plane wave wavefunctions $\langle x| p\rangle\propto \exp(ixp/\hbar)$ and $\langle x|- p\rangle\propto \exp(-ixp/\hbar)$ are pointless for your free particle example: why don't you work in momentum space, instead? So, let's work in momentum space, now ultralocalized, and note your may always convert the final answer to coordinate space at the end. (In momentum space, the wavefunctions are δ-functions of momenta.) In representation-independent ket space,

$$ 2mH_0|p\rangle = p^2 |p\rangle \qquad \qquad 2mH_0|-p\rangle = p^2 |-p\rangle \\ 2mH_1|p\rangle = (p+p_0)^2 |p\rangle \qquad 2mH_1|-p\rangle = (p-p_0)^2 |-p\rangle \\ 2mH_1|p-p_0\rangle = p^2 |p-p_0\rangle \qquad 2mH_1|-(p+p_0)\rangle = p^2 |-(p+p_0)\rangle ~~. $$ If I understand your desideratum properly, you wish to shift the location of the eigenfunctions in momentum space so that their eigenvalues under $H_1$ coincide with the eigenvalues of $H_0$.

You can do this for parity doublets in p space, since the eigenvectors of $H_0$ are symmetric under parity; but those of $H_1$ are symmetric under inversion around $-p_0$ instead, $$ 2mH_1|p-p_0\rangle = p^2 |p-p_0\rangle \qquad 2mH_1|-p_0-p\rangle = p^2 |-p_0-p\rangle , $$ so your shifted momentum origin $-p_0$ will need to be accounted for, as well.

But that's all you are doing: you are redefining the origin of momenta (note you have not altered the commutation relations, so the x s are not impacted on substantially) and marveling at the notational complications of the shift.

Formally, you are translating momenta by a constant, $$ e^{-ip_0 \hat x/\hbar}~ \hat p ~ e^{ip_0 \hat x/\hbar} = \hat p +p_0, $$ so that $$ H_1= e^{-ip_0 \hat x/\hbar} H_0 e^{ip_0 \hat x/\hbar} , $$ and you may convert $$ e^{ip_0 \hat x/\hbar}H_1 |\psi\rangle= H_0 e^{ip_0 \hat x/\hbar} |\psi\rangle , $$ including the above free case.

So, the wave function to adopt, in any rep or language, is $e^{-ip_0 \hat x/\hbar} |\psi_0\rangle $ where now $|\psi_0\rangle$ is the corresponding eigenvector of $H_0$. So, in the momentum representation, $\psi_0 (p)$ goes to $$ \langle p| e^{-ip_0 \hat x/\hbar} |\psi_0\rangle = \langle p+p_0 | \psi_0\rangle= \psi_0( p+p_0). $$ Your $\psi_0 (-p)$ will then go to $\psi_0( -p+p_0)$, mutatis mutandis.

  • A note on your language. It became clear to me you were actually meaning $\psi_p (x)$ instead of $\psi (p,x)$ which appears like a phase-space function and is an oxymoron, qua standard QM. For a free particle, the eigenfunctions may indeed be indexed by the momentum, but this is basically the only place this happens. Look at the SHO, where the eigenfunctions and energies are indexed by a level integer, instead. Still, the above formal similarity transformation allows you to preserve the spectrum there, as well.
  • $\begingroup$ From the discussion, if the wave function is in momentum space, then $e^{-ip_0\hat{x}/\hbar}|\psi_0(p)>=|\psi_0(p+p_0)>$ for any Hamiltonian in general(for example Hamiltonian w/o parity symmetry)? $\endgroup$ – an offer can't refuse Aug 18 at 6:07
  • $\begingroup$ Indeed, $\langle p| e^{-ip_0 \hat x/\hbar}|\psi_0\rangle= \langle p+p_0|\psi_0\rangle=\psi_0(p+p_0)$. The p-dependence is provided by the bra. $\endgroup$ – Cosmas Zachos Aug 18 at 10:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.