The Hamiltonian for a particle in a finite box is $$H = \frac{p^2}{2m} + V(x)$$ which will give time evolution as $$ i\hbar d/dt|{\psi(t)}\rangle = H|{\psi(t)}\rangle \, .$$ However, if I do a Galilean transformation into a moving frame at some velocity $v$, then one expects a similar equation $$ i\hbar d/dt|{\psi''(t)}\rangle = H''|{\psi''(t)}\rangle $$ where $$ H'' = \frac{p^2}{2m} + V(x-vt) \, .$$ Indeed, this is true, because $$ H'' = H' - i\hbar U \frac{dU^\dagger}{dt} = \frac{p^2}{2m} + V(x-vt) + \frac{1}{2}mv^2 $$ where $H'$ is the "transformed" operator given by $UHU^\dagger$ and $$ U= \exp\left(\frac{i}{\hbar}(pvt + xmv)\right) \, .$$ Therefore, the only difference between our expectations is a constant factor which contributes nothing to the eigenvectors except a shift of their respective eigenvalues.

Although the Hamiltonian operator for the stationary case is all well and good, and it clearly represents the observable of energy, what does $H''$ actually represent? It is time dependent so surely it can't represent the energy. Do we simply ignore any interpretation and say energy can only be defined well in the rest frame of the box where the Hamiltonian is $H$?

I've taken a year of undergraduate quantum, so that is about my level of expertise, although I have done quite a bit of independent learning beyond that.

It is clear that a $\frac{1}{2}mv^2$ makes no difference if the overall Hamiltonian is time independent, but I'm not sure if the two $H''$ I wrote out are indeed equivalent in the case of time dependence.

  • $\begingroup$ System energy plus a kinetic energy part? $\endgroup$ – Matt Jun 28 at 17:28
  • $\begingroup$ Not really sure what means. On some hand waving level possibly, but I was hoping for something more concrete. $\endgroup$ – Electric to be Jun 28 at 19:11
  • $\begingroup$ If you have a fixed potential $V(x)$, you already have picked out a preferred frame, so you would expect that Galilean symmetry isn't very useful. $\endgroup$ – knzhou Jun 30 at 1:42
  • $\begingroup$ Sure knzhou, I agree. In terms of solving the problem, you might as well use the original frame. I'm more concerned with the interpretation of using a moving frame. $\endgroup$ – Electric to be Jun 30 at 23:43

Everything is the same in the new frame, inertial frames are equivalent for formulation of the equations. Hamiltonian is still sum of kinetic and potential energy. If in the original frame the potential energy is $V(x)$, and if the transformation to the new frame is $$ x'' = x + vt $$ then in the new frame the new Hamiltonian will depend on $x''$ and its conjugate $p''$: $$ H'' = \frac{p''^2}{2m} + V(x'' - vt) $$ which is explicitly time dependent.

It is time dependent so surely it can't represent the energy.

Where is this idea coming from? It is an operator, not energy value: it can depend on time and represent energy in the above sense, i.e. sum of kinetic and potential energy.

If you're worried that energy should be time-independent, then this is obeyed for expectation values of $H$ and $H''$:

$$ \frac{d}{dt}\langle H\rangle = 0. $$

$$ \frac{d}{dt}\langle H''\rangle = 0. $$

This is possible because besides $H''$, $\psi''$ depends on time in precisely such a way (it moves along with the potential energy function) that it cancels the dependence and the result is independent of time.

  • $\begingroup$ A few things, in your H'' expression, I believe p'' and x'' should just be p and x. Because H'' is the result I obtained after I transformed p'' -> p + mv and x'' -> x + vt, and also added the time dependent term of the Hamiltonian (iℏUdU†dt). But second, I just don't think that a time dependent potential can really be counted as "potential energy", right? I'm not sure. I feel like the whole concept of defining a potential energy breaks down in this case, if we try to think about it classically. After all in classical mechanics, isn't H only = E in case of H independent of t? $\endgroup$ – Electric to be Jun 30 at 23:28
  • $\begingroup$ What do you think $H''$ is? I think it is Hamiltonian in the new frame of reference, no? So it should be expressed as function of operators in that frame of reference. $\endgroup$ – Ján Lalinský Jul 1 at 11:35
  • $\begingroup$ Yes but if you perform the transformation you also have to include a time dependent term because U is explicitly dependent on time. As a result the transformed P’’ simply becomes P. And X’’ becomes x - vt. This discussion also depends on what you define the Hamiltonian in the new frame to be. Is it the operator which gives time evolution in the new frame, or simply UHU’? Because there is a difference for time dependent U(t), as in the case of a boost to a new reference frame. $\endgroup$ – Electric to be Jul 1 at 20:17

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