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It is well known that the energy levels $$ E_n = \hbar \omega\left(n+\frac{1}{2}\right) $$ of the quantum harmonic oscillator verify the eigenvalue problem $$ \hat{H}|\psi_n\rangle = E_n |\psi_n \rangle $$ where $$ \hat{H} = \frac{\hat{p}^2}{2m} +\frac{1}{2}m\omega^2 \hat{x}^2. $$

Now let us assume that the harmonic potential is endowed with a uniform translation velocity, i.e.:

$$ \hat{H}_v = \frac{\hat{p}^2}{2m} +\frac{1}{2}m\omega^2 (\hat{x}-vt)^2. $$

For an observer moving with velocity $v$, the harmonic potential will look at rest, and hence I expect that the energy levels $E_n$ should be the usual ones (see above). Conversely, for an observer in the laboratory frame, I expect that the energy levels will be different (in Classical Mechanics, it is well known that the kinetic energy is different in different inertial frames), and, in general, they depend on $v$. My question is therefore: which is the expression of the energy levels in the lab frame?

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    $\begingroup$ What did you find by defining creation and annihilation operators and producing the spectrum algebraically? $\endgroup$ Commented Mar 1 at 1:12
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    $\begingroup$ Related: physics.stackexchange.com/q/384509 $\endgroup$
    – hft
    Commented Mar 1 at 2:43
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    $\begingroup$ I don't understand the votes to close. For me this seems like a perfectly reasonable question. $\endgroup$ Commented Mar 1 at 8:57
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    $\begingroup$ @Tobias Fünke. If the OP wishes to muse, given $\hat H_v= \exp (-ivt\hat p/\hbar) \hat H \exp (ivt\hat p/\hbar)$, so the two hamiltonians have the same spectrum, about the classical correspondence, he should state that explicitly. $\endgroup$ Commented Mar 1 at 14:24

2 Answers 2

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$\hat H_𝑣=\exp(−𝑖𝑣𝑡\hat p/ℏ)~\hat H\exp(𝑖𝑣𝑡\hat p/ℏ)$, so the two hamiltonians are canonically equivalent and have the same spectrum. You may also see this from $[\hat x -vt,\hat p]=i\hbar$, so $(\hat x,\hat p)\mapsto (\hat x -vt,\hat p)$ is a canonical transformation preserving the commutator, so defining $a$ and $a^\dagger$ with either pair will produce the same spectrum, $E_n$.

It is not clear to me what your classical limit argument is, since there is hardly an eigenvalue equation involved in your classical mechanics setup ("levels"?) expectations, which you may have to refine. In any case, classical physics is the same in all inertial frames.


Response to comments

Indeed, but you did not shift the momentum in concordance with the coordinate shift. What you wrote down in the OP is not the hamiltonian fo the TDSE of WP, $$ \breve H= 𝑈𝐻̂𝑈^\dagger + 𝑖ℏ\dot{U}𝑈^\dagger =\hat H_v + v\hat p= \frac{(\hat{p}+mv)^2}{2m}-\frac{mv^2}{2} +\frac{1}{2}m\omega^2 (\hat{x}-vt)^2. $$ The two hamiltonians have different spectra, differing by a constant shift term, amounting to a mere shift in the zero point energy, as you inferred. The level spacings are the same.

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  • $\begingroup$ Thank you for your kind explanation. My point is: are the two spectra exactly the same or is there an additive constant? According to Eq. (0) of this Reference [en.wikipedia.org/wiki/…, $H_v = UHU^\dagger + i\hbar \dot{U}U^\dagger$. I am interested precisely in the possible effect of the latter term on the system spectrum. $\endgroup$
    – AndreaPaco
    Commented Mar 1 at 23:46
  • $\begingroup$ Probably I was not clear, but my classical-limit argument is the following. A point particle which is at rest in the reference frame $S$ has energy $E_S=0$; On the other hand, when observed from a frame $S^\prime$ moving at velocity $v$ w.r.t. $S$, the same particle has energy $E_{S^\prime}=\frac{1}{2}mv^2$. I may be wrong, but I expected that, if the ground state energy of a quantum harmonic oscillator at rest is $E_S=\frac{1}{2}\hbar\omega$, the energy in reference frame $S^\prime$ should be something like $E_{S^\prime}=\frac{1}{2}\hbar\omega + \frac{1}{2}mv^2$. $\endgroup$
    – AndreaPaco
    Commented Mar 1 at 23:53
  • $\begingroup$ Thank you for your time. I have no doubts that the level spacings are the same. What I was curious about was indeed the "constant shift term" between the spectrum of $\hat{H}$ and the spectrum of $\hat{H}_v$. Do you agree that this shift is exactly $\Delta E = \frac{1}{2}mv^2$, as I anticipated in my last comment? $\endgroup$
    – AndreaPaco
    Commented Mar 2 at 11:34
  • $\begingroup$ Yes, didn’t I ? For the definitions of the OP, not your comment. You shouldn’t change them… $\endgroup$ Commented Mar 2 at 11:44
  • $\begingroup$ Seems to me you are asking about Galilean covariance and here, not your question title... $\endgroup$ Commented Mar 2 at 13:55
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OP seemingly wants to understand whether or not there is a contribution to the energy from the overall translation of the system with velocity $v$. Of course there should be, and we should expect something like we usually find when doing separation of variables to separate out the center of mass motion.

In order to see this very explicitly, consider first the ground state of the potential in its rest frame: $$ \psi_0(x) = Ae^{-\frac{1}{2}x^2}\;, $$ where $A$ is the usual normalization constant ($1/\pi^{1/4}$, or whatever), and where I am setting $m=\hbar=\omega=1$ to help keep my typing brief. (But it is straightforward to fill those constants back in, if desired.)

It is also helpful to have a reference to look at for a similar problem. Griffiths' Quantum Mechanics book has a section in Chapter 2 on the "delta function potential." Griffiths also provides a related practice problem in Chapter 2 regarding how the ground state of the boosted delta function potential (i.e., the delta function potential with the replacement $x\to x-vt$) compares to the ground state of the stationary delta function potential.

By analogy with Griffiths, it is straightforward to see that our boosted solution should be: $$ \chi_0(x,t) = Ae^{-\frac{1}{2}(x-vt)^2}e^{-i(\frac{1}{2}\omega + \frac{1}{2}mv^2)t}e^{imvx}\;,\tag{C} $$ where I've now put back in some of the $m,$ $\hbar$, and $\omega$ variables to help orient the reader, but there are still some missing (the reader can fill them in on their own).

The reader can show that the function $\chi_0(x,t)$ satisfies the time-dependent Schrodinger equation: $$ i\frac{\partial \chi_0(x,t)}{\partial t} = -\frac{1}{2}\chi_0''(x,t)+\frac{1}{2}(x-vt)^2\chi_0(x,t)\;. $$

Now, a few words about the pieces in Eq. (C): $$ \chi_0(x,t) = \underbrace{Ae^{-\frac{1}{2}(x-vt)^2}}_{1} \underbrace{e^{-i(\frac{1}{2}\omega + \frac{1}{2}mv^2)t}}_{2}\underbrace{e^{imvx}}_{3}\;,\tag{C}\;. $$

  • Piece 1 is just the ground state of the stationary harmonic oscillator potential, but now evaluated at the moving position $(x-vt)$ instead of evaluated at $x$.
  • Piece 2 has the usual energy factor $e^{-iEt}$, but the energy $E$ is the energy of the ground state of the stationary potential plus the center of mass kinetic energy $\frac{1}{2}mv^2$. (I think this is the piece OP is mainly interested in.)
  • Piece 3 is a plane wave factor for a plane wave of momentum $mv$, which is just the momentum of the center of mass due to the overall translation of the system at velocity $v$.

The reader can calculate the expectation value of $\hat H_v$ on $\chi_0(x,t)$ and will find a result that they are probably expecting. (Namely, $\langle \chi_0|\hat H_v|\chi_0\rangle = \frac{1}{2}mv^2 + \frac{1}{2}\hbar\omega$.)

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  • $\begingroup$ Thank you for your kind explanation. My point is: are the two spectra exactly the same or is there an additive constant? According to Eq. (0) of this Reference [en.wikipedia.org/wiki/…, $H_v = UHU^\dagger + i\hbar \dot{U}U^\dagger$. I am interested precisely in the possible effect of the latter term on the system spectrum. $\endgroup$
    – AndreaPaco
    Commented Mar 1 at 23:47
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    $\begingroup$ You can put in an additive constant if you want. The Hamiltonian is only well-defined up to an additive constant. This is because the potential energy is only defined up to an additive constant. So, even if you are not considering a time-dependent potential or whatever there is always the freedom to add a constant to the energy if you would like to do so. $\endgroup$
    – hft
    Commented Mar 1 at 23:55
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    $\begingroup$ OK, yeah, I see what you are asking about there. I will try to update my answer later $\endgroup$
    – hft
    Commented Mar 1 at 23:57
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    $\begingroup$ OK, I made an update. Hope it helps. $\endgroup$
    – hft
    Commented Mar 2 at 0:41
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    $\begingroup$ Thank you, that's perfect! $\endgroup$
    – AndreaPaco
    Commented Mar 2 at 18:56

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