2
$\begingroup$

The Hamiltonian operator $$H=\frac{{\bf p}^2}{2m} +\frac{m\omega^2}{2}{\bf r}^2-\Omega L_z$$ with $L_z=xp_y-yp_x$, can be written as $$H=\hbar\left(\omega+\Omega\right)\alpha^\dagger\alpha+\hbar\left(\omega-\Omega\right)\beta^\dagger\beta+\hbar\omega$$ where the ladder operators $\alpha$ and $\beta$ are given by $$ \alpha=\frac{1}{\sqrt{2}}(a_x+ia_y)$$ $$ \beta=\frac{1}{\sqrt{2}}(a_x-ia_y). $$ and $a_x$, $a_y$ are the standard ladder operators for the harmonic oscillator. That's a nice way to see that the energy for this problem doesn't have a lower bound (when $\Omega>\omega$).

QUESTION: Does anyone have a clear picture of what kind of excitations do the operators $\alpha^\dagger\alpha$ and $\beta^\dagger\beta$ represent?

Attempt answer: would it be correct to say that they represent clockwise and anticlockwise rotations?

Issue with the attempt: it doesn't make sense to have the lower bound on one of them and not on the other.

Second attempt: anything to do with maxons and rotons?

$\endgroup$
  • $\begingroup$ What do you mean by "lower bound"? Keep in mind that there is nothing that forces $\Omega$ to be positive. The two modes $\alpha$ and $\beta$ are completely symmetric as you've written them. $\endgroup$ – Emilio Pisanty Feb 2 '18 at 10:46
  • 1
    $\begingroup$ Maybe this is known to you, in Cohen-Tennoudji's Vol 1 book, at the end of the complement BVII, they use this representation. This is for the isotropic 3D Harmonic Oscillator, but some results hold for your case. There they indeed write $L_z = \hbar \left(\beta^{\dagger} \beta-\alpha^{\dagger} \alpha\right)$ but also $L_+ = \hbar \sqrt{2} \left( a^{\dagger}_z \alpha - \beta^{\dagger} a_z\right)$ which does not have such an straightforward interpretation. Then they match these quantum numbers with the more standard angular momentum ones. It might help. $\endgroup$ – secavara Feb 2 '18 at 10:50
  • $\begingroup$ Also, is the constant term in your Hamiltonian $\frac{\hbar \omega}{2}$ or $\hbar \omega$? I assume yours is a 2D case, right? $\endgroup$ – secavara Feb 2 '18 at 11:08
  • $\begingroup$ @Emilio Pisanty More precisely then for $|\Omega|>\omega$ the energy can become arbitrarily negative by adding excitations $\beta^\dagger\beta$ -> no ground state exists $\endgroup$ – Semola Feb 2 '18 at 16:02
  • $\begingroup$ @secavara You are right about the 1/2: it's a typo. Thanks for the Tannoudji's reference, I was not aware of it, I'll have a look! $\endgroup$ – Semola Feb 2 '18 at 16:05
3
$\begingroup$

For the sake of simplicity lets assume $\Omega > 0$.

I believe that your initial attempt is correct and that $\alpha$ and $\beta$ can be thought of as states where the particle is "moving in a circle around the $z$ axis". Lets look at the term you added to the standard harmonic Hamiltonian $$ -\Omega L_z\;. $$ This term says that orbiting around the $z$ axis in a negative direction requires energy, but orbiting in the positive direction reduces your energy. If this effect is larger than the energies associated with the harmonic oscillator, then you can reduce your energy as much as you like by orbiting in the appropriate direction. In other words this term does exactly what your "Issue with the attempt" was struggling to explain.

This explains why, as @secavra states in the comments, $L_z \propto \beta^\dagger\beta - \alpha^\dagger\alpha$, i.e. the angular momentum is given by the number of excitations for circling one way minus the number for circling in the opposite direction. It can also be seen by looking at the representations in terms of $a_x$ and $a_y$; $\alpha$ and $\beta$ represent harmonic motion in the $x$ and $y$ directions phase shifted by $\pm \frac{\pi}{2}$, which classically would be circular motion.

$\endgroup$
  • $\begingroup$ Thanks, great answer, I agree with what you are saying. Do you know by any chance if any of this can be related to rotons (in the sense of Landau)? More specifically is there any way to build rotons excitations from the rotating GPE perspective? [any reference you might have would be of course welcome] $\endgroup$ – Semola Feb 2 '18 at 17:33

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.