4
$\begingroup$

We have that a property of the transition operator defining c-equivalence (or star equivalence from equation 1 in Bertelson) is

\begin{align*} T(f\star_Mg)=T(f)\star'T(g)\,, \end{align*}

where $\star_M$ is the Moyal star product and $\star'$ is a star product for another ordering. Therefore, the stargenvalue equation, $H\star_MW=EW$, becomes, upon operation under the transition operator,

\begin{align*} T(H\star_MW)=T(EW)\implies T(H)\star'T(W)=ET(W)\,, \end{align*} where $H$ is the Hamiltonian, $W$ is the Wigner function, and $E$ is the energy spectrum. We note that $E$ does not change under the transition operator.

We can derive $H\star_MW=EW$ using the star exponential (defined in Section 4 in Bayen),

\begin{align*} Exp_{\star M}\left(\frac{tH}{i\hbar}\right)=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{t}{i\hbar}\right)^nH^{\star_Mn}\,, \end{align*}

where $t$ is the time and

\begin{align*} H^{\star_Mn}=\underbrace{H\star_MH\star_M\cdots\star_MH}_{\text{$n$ times}} \end{align*}

When switching to a different ordering using different star products, I've noticed that several papers (for example, Section 3.3 in Dito and Section 5 in Hirshfeld) write the star exponential as

\begin{align*} Exp_{\star'}\left(\frac{tH}{i\hbar}\right)=\sum_{n=0}^{\infty}\frac{1}{n!}\left(\frac{t}{i\hbar}\right)^nH^{\star'n}\,, \end{align*}

without using the transition operator. Why is this allowed?

I'm really confused about this because a different energy spectrum can then be calculated, which contradicts $T(H)\star'T(W)=ET(W)$.

$\Large{\text{Update}}$

Looking at the published version by D&T, I was able to keep track of the $\exp(\gamma t/2)$ term. So, I see how (pg. 313)

\begin{align*} T\left(Exp_{\star M}\left(\frac{tH}{i\hbar}\right)\right)=Exp_{\star\gamma}\left(\frac{tT(H)}{i\hbar}\right) \end{align*}

compares to

\begin{align*} Exp_{\star\gamma}\left(\frac{tH}{i\hbar}\right)(q,p)=\exp(\gamma t/2)\sum_{n=0}^{\infty}\big(-i(n+1/2)\omega t\big)T(\pi_n) \end{align*}

According to equation (10),

\begin{align*} Exp_{\star}\left(\frac{tH}{i\hbar}\right)=\sum_{n=0}^{\infty}e^{\frac{tE_n}{i\hbar}}\pi_n\,. \end{align*}

So, if $\star=\star_M$,

\begin{align*} T\left(Exp_{\star M}\left(\frac{tH}{i\hbar}\right)\right)&=T\left(\sum_{n=0}^{\infty}e^{\frac{tE_n}{i\hbar}}\pi_n\right)\\ &=\sum_{n=0}^{\infty}e^{\frac{tE_n}{i\hbar}}T(\pi_n)\,, \end{align*}

As

\begin{align*} Exp_{\star\gamma}\left(\frac{tT(H)}{i\hbar}\right)=Exp_{\star\gamma}\left(\frac{tH}{i\hbar}\right)\exp(-\gamma t/2)\,, \end{align*}

by using

\begin{align*} T\left(Exp_{\star M}\left(\frac{tH}{i\hbar}\right)\right)=Exp_{\star\gamma}\left(\frac{tT(H)}{i\hbar}\right)\,, \end{align*}

wouldn't that mean,

\begin{align*} T\left(Exp_{\star M}\left(\frac{tH}{i\hbar}\right)\right)=\sum_{n=0}^{\infty}\big(-i(n+1/2)\omega t\big)T(\pi_n)\,, \end{align*}

so that $E_n=(n+1/2)\hbar\omega$? However, if we use equation (10) with $\star=\star_{\gamma}$, then I immediately see how to get $E_n=\hbar\omega(n+1/2+i\gamma/2\omega$. Both ways seem valid to me, so I must be missing a crucial detail.

$\endgroup$
  • 2
    $\begingroup$ Hirschfeld and Henselder are magnificently sloppy. They start with the Husimi antinormal ordering, whose nonclassical Hamiltonian is actually well known to be the classical (5.3) +1/2, so the spectrum is (5.12)+1/2, as they mumble right thereafter. Since a constant like that does not alter the crucial step in (5.7), and modifies expressions trivially, they shrug it off. Compare with (5.23). The conversion is done correctly, consistently with the ruleof Bayen et al, in our book pp 59-60. $\endgroup$ – Cosmas Zachos Jul 6 '16 at 18:55
  • $\begingroup$ Okay. Now I see why we can keep using $H$, instead of $T(H)$. However, I am still little confused regarding why Dito is able to get a different energy spectrum. $T(H)\star_{\gamma}T(W)=ET(W)$ would imply that $E=(n+1/2)\hbar\omega$, rather than Dito's $E=(n+1/2+i\gamma/2\omega)$. Is there a subtlety that I am missing that would prevent $T(H)\star_{\gamma}T(W)=ET(W)$ from being used? $\endgroup$ – user85503 Jul 6 '16 at 20:46
4
$\begingroup$

It turns out that with your two examples you identified a commonplace ritual complacent sloppiness in the field. Your general treatment of equivalent star products is sound, and best summarized in Francois Bayen's superior and tasteful 1978 talk writeup in the GTMP conference proceedings (Austin, Texas, 1978; Lecture Notes in Physics, 94 , Springer 1978) pp (261-267), where the basic example is treated, as it is also treated in our book, Concise Treatise in Quantum Mechanics in Phase space, WS (2014) eqns (122-131) and especially Exercise 0.21 in that online version.

The point is that constants are easy. For quadratic hamiltoninas like that of the SH oscillator's, and when T is a trivial Weierstrass transform, so, then, the exponential of a bilinear in derivatives, its action on such hamiltonians is trivial: it just shifts them by a constant, $T(H)=H+\gamma$. Since constant terms, hence factors upon exponentiation, etc, slip through star products unseen and unscathed, it is very easy to get complacent, and "do everything in one's head", without adequate explanation. So the unwholesome temptation presents itself to merely shift usage from T(H) to H and the eigenvalues from E+γ to E, and move on, mumbling "yeah, yeah, you know what I mean", thereby confusing readers like you.

Both examples you adduce commit this, coming from different places. However, the referee did his job properly in the published version of D&T since that version explains the stunt in the insert eqns after (29), and before (30), missing from the preprint you cite. (The object of their formal exercise is immaterial here: they translate the Enz trick of modifying the PBs to describe the damped QHO in phase space without any work.)

By contrast, the published version of H&H replicates the fast talking sloppiness, especially virulent for a pedagogical journal like AmJPhys, purporting to educate hapless readers---the referee there clearly failed the process. They use H=(5.3)=T(H)-1/2, so their spectrum is not the correct one, (5.12)+1/2, mumbling right thereafter that it is missing the zero point energy, "yeah, yeah", and forfeiting a teaching moment that journal purports to encourage. Since a constant 1/2 like that does not alter the crucial step in (5.7), and modifies expressions trivially, they shrug it off. Compare with (5.23). This is the celebrated/standard Husimi picture antinormal ordering example that Bayen and we review precisely for its pedagogical value.

Your basic instinct is right: when in doubt, ensure the spectrum is the same, and don't be above including additive factors in the hamiltonian, explicitly. We mostly strive to do this in our book.

Edit to address your update. All formulas and statements in your update are impeccable. Your final formula then says that the $\star_\gamma$ exponential of $H-\gamma/2\omega$ is the rhside you have. So you multiply both sides by $e^{\gamma/2\omega}$ and get the unnumbered eqn of that paper following their (31). You incorporate that factor into the energies of all modes, shifting each by $i\gamma/2\omega$. Their (10) is peculiar to $\star_M$... consider how it is different than Husimi's (Exercise 0.21 in our book). Their (30-31) achieves just what you suggested.

$\endgroup$
  • $\begingroup$ Okay. I think I grasp most of it now. Thanks. Just to make sure that I understand: Using $T(H\star_MW)=T(H)\star'T(W)=ET(W)$ is perfectly valid. If I try to calculate the energies via the $\star$-exponential approach, I should still get the same energy (making sure that I've accounted for any constants that I could factor out). Thinking about it, there's one more follow-up I have. (I needed to insert it under the original question so I would not run out of characters). $\endgroup$ – user85503 Jul 7 '16 at 19:10
  • $\begingroup$ I apologize for hammering away at this point, but I'm still trying to wrap my head around getting an energy different from $(n+1/2)\hbar\omega$. If I use the $\star_M$ exponential, I remember seeing a derivation of $H\star W=EW$, which I can apply a transition operator to, without changing the energy from $(n+1/2)\hbar\omega$. If I use the $\star_{\gamma}$ exponential and a $T(H)$ so I don't have any factors of $\exp(\gamma t/2)$ hanging around, I still get $(n+1/2)\hbar\omega$. $\endgroup$ – user85503 Jul 8 '16 at 20:28
  • $\begingroup$ Though if I use the $\star_{\gamma}$ exponential with an $H$, I'll end up with an extra $\exp(\gamma t/2)$, which I can just absorb into the energy (as done by D&T and mentioned in your answer). Why is this also physically correct? Is it because the $\gamma/2\omega$ is an offset, so like H&H, it doesn't matter? Looking at (33), the commutation relations seem to depend explicitly on this offset, so at first glance, it appears to be important. It is this apparent contradiction between the energies arising from the two different methods that is most confusing to me. $\endgroup$ – user85503 Jul 8 '16 at 20:28
  • 1
    $\begingroup$ I'm not sure what is troubling you...Forget about time evolution: it's all in the stargenvalue equation, as in our book. $H\star_M W=E W$, hence $(H-i\gamma/2)\star_\gamma T(W)=ET(W)$, that is $H\star_\gamma T(W)=(E+i\gamma/2) T(W)$. You may now time evolve and exponentiate, as they do in (30), but they have done the calculation for you. $\endgroup$ – Cosmas Zachos Jul 8 '16 at 21:40
  • 1
    $\begingroup$ Helpful trick: Just do everything, explicitly, stargenvalue equation, *E*=1/2, time evolution, exponentiation, on the ground state, the simple Gaussian in phase space: Apply the D&T T on it, etc... It is straightforward. You might be overwhelmed by the plethora of symbols, and the irrelevant generating function of Laguerre polynomials' jazz, which is irrelevant to your focus.... $\endgroup$ – Cosmas Zachos Jul 8 '16 at 21:46

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.