0
$\begingroup$

Suppose one has the following diagonal Hamiltonian

$$ \hat{H}=\omega_1 \hat{n}_1 - \omega_2 \hat{n}_2 $$

where $\omega_1$ and $\omega_2$ are greater than zero. Number operators $\hat{n}_1$ and $\hat{n}_2$ have, of course, non-negative integer eigenvalues.

Apparently, the spectrum of this system is not bounded from below, so $\hat{H}$ does not correspond to a physical system. Is this right?

Now let's introduce an additional assumption, i.e. that a conserved quantity is present. The conserved quantity is $$ \hat{K}=\hat{n}_1-\hat{n}_2 $$

So the Hamiltonian can be rewritten as

$$ \hat{H}= \omega_1(\hat{K}+\hat{n}_2) -\omega_2\hat{n}_2= $$ $$ = \hat{n}_2(\omega_1-\omega_2) + \mathrm{const} $$ where an additive term has been evidenced.

At this point, the stability condition should read: $$ \omega_1-\omega_2>0 $$ Is this right?

If this is correct, one can conclude that an Hamiltonian of the type $\hat{H}=\omega_1 \hat{n}_1 - \omega_2 \hat{n}_2$ is not necessarily unphysical. In fact, if one can find a conserved quantity capable of eliminating number operators associated to negative eigenfrequencies, one obtains a lower-bounded Hamiltonian, which is perfectly physical.

Here comes my question: if one has the diagonal Hamiltonian $$ \hat{H}=\omega_1 \hat{n}_1 - \omega_2 \hat{n}_2 $$ how can he be sure that it does is UNphysical? In other words: how can I be sure that no further conserved quantities (capable of making it lower-bounded) are present?

On top of that there is the problem of setting the conserved quantity. In principle the eigenvalues of $\hat{K}$ are integer numbers ranging from $-\infty$ to $+\infty$, so the ground state of this system is the one where $\hat{K}$ tends to $-\infty$ which, again, sounds like not having a lower bound.

$\endgroup$
  • $\begingroup$ Are you assuming that $[\hat{n}_1, \hat{n}_2] = 0$? $\endgroup$ – Michael Seifert Mar 26 '18 at 16:47
  • $\begingroup$ Yes, they commute. $\endgroup$ – AndreaPaco Mar 26 '18 at 16:48
-1
$\begingroup$

The problem with your thinking is that you are treating the Hamiltonian and the conserved quantities as independent. In fact the Hamiltonian determines the conserved quantities. Explicitly a quantity is conserved if and only if it commutes with the Hamiltonian $$ [\hat{H},\hat{Q}] = 0\;. $$

This means you cannot impose that $\hat{K}$ is conserved. Assuming $\hat{n}_1$ and $\hat{n}_2$ commute then $\hat{K}$ is conserved. This means that all you can possibly have done in the rest of your argument is rewrite the exact same Hamiltonian, which leads to your last paragraph.

On top of that there is the problem of setting the conserved quantity. In principle the eigenvalues of $\hat{K}$ are integer numbers ranging from $−∞$ to $+∞$, so the ground state of this system is the one where $\hat{K}$ tends to $−∞$ which, again, sounds like not having a lower bound.

This is indeed a problem and indeed means that your Hamiltonian is still unbounded from below, and therefore unphysical.

$\endgroup$
  • $\begingroup$ Thanks for your reply. Actually I do know that the Hamiltonian determines conserved quantities. The problem is wheter one is able to find and recognize them or not. In fact, the initial Hamiltonian (before the diagonalization process) can be very complicated and conserved quantites could be difficult to find. At the end of the diagonalization process, one ends up with an Hamiltonian of the type H=w1 n1 - w2 n2 and one cannot say if what has been found is physical or not, simply because one has not been able to find all conserved quantites. The ambiguity (and the origin of my question) is --> $\endgroup$ – AndreaPaco Mar 26 '18 at 17:33
  • $\begingroup$ if one does not recognize the existence of a conserved quantity, then the spectrum appears lower UNbounded thus making the problem UNphysical. On the contrary, if one has been able to identify all the conserved quantites, then the spectrum correctly is lower bounded and everything works well. Isn't this strange? $\endgroup$ – AndreaPaco Mar 26 '18 at 17:37
  • 1
    $\begingroup$ As is written in the answer, the spectrum is still not bounded from below, as $K$ can take any value. $\endgroup$ – Noiralef Mar 27 '18 at 13:38

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.