5
$\begingroup$

Let \begin{equation} \int\mathrm{d}z~ A(x,z) B(z,y) = \delta(x - y). \end{equation} Taking Wigner transform of both sides we readily obtain \begin{equation} A^W(X,p) \star B^W(X,p) = 1, \end{equation} where $\star$ is the Weyl–Groenewold (or Moyal) product. Now, to leading order in the gradient expansion (or in $\hbar$ if you prefer), $$A^W(X,p) \star B^W(X,p) = A^W(X,p) B^W(X,p) + O(\hbar),$$ which suggests that poles of $A^W$ coincide with zeros of $B^W$ and vice versa.

My question is: can this statement be generalized to the full case, i.e., without approximating the $\star$-product? My guess would be the answer is 'no' and that there is a simple counterexample showing this already on a Poisson-bracket level. But perhaps I am wrong?

$\endgroup$
5
  • $\begingroup$ Are you stipulating that the individual Weyl symbols A^W, etc have no explicit hbar dependence? This is untypical In lots of physical systems…. $\endgroup$ Commented Nov 8, 2022 at 14:20
  • $\begingroup$ @CosmasZachos I am not sure what made such an impression, but in any case, I don't stipulate anything. To be honest, in my community, we don't even attach much importance to hbar dependence. For example, the Weyl–Groenewold (or Moyal) product is understood more as an expansion in derivatives w.r.t. X and p, rather than in hbar. And yes, I am aware that's not how one usually thinks in the context of phase-space formulation of QM. In any case, I believe my question does not depend on any of that. It was "given that A^W \star B^W = 1, is it true that poles of A^W (B^W) are zeros of B^W (A^W)?" $\endgroup$
    – Gickle
    Commented Nov 8, 2022 at 14:41
  • 1
    $\begingroup$ Ok, if you appreciate the Laurent series in hbar, star inverses need not share in the singularity structure of multiplicative inverses. There is a bevy of wrong papers on this in the literature…. $\endgroup$ Commented Nov 8, 2022 at 19:41
  • $\begingroup$ @CosmasZachos if that's not too much trouble, could you give an example or two of such papers (the wrong ones)? And is there a simple way of seeing why the signularity structure does not follow that of a multiplicative inverse? I mean, I do understand that $\star$ involves derivatvies, so that even if, e.g., $A^W(X,p_0) = 0$, its derivatives may not vanish at $p_0$ spoiling the whole thing, but is there more to it? Has the question of the singularity structure of a star inverse been studied, by the way? I couldn't find anything on it in your "Concise Treatise", but maybe I simply missed it. $\endgroup$
    – Gickle
    Commented Nov 8, 2022 at 21:17
  • $\begingroup$ We did not cover the issue in the booklet because it is subtle and not settled. Takahashi, in his pioneering paper, reached wrong conclusions, and was followed by uncomfortably many others; most WFs for the oscillator are all non-analytic in ℏ... In any case, I have a modest (untypical ℏ-analytic) example in my answer, which illustrates unshifted poles in the star-inverse B, but of higher order than the simple zero of A. They cancel each other at each order in ℏ, and not by the measly simple zero of the (classical!) A... $\endgroup$ Commented Nov 11, 2022 at 17:49

1 Answer 1

2
$\begingroup$

I don't quite know, but I'd agree with your bet that the general answer is "no". I'll just jot down a few remarks and an unsatisfactory example that is easier to compute with; however, it is based on the "crypto-semiclassical" oscillator that most seasoned professionals appreciate is quite unrepresentative of generic QM situations, despite its clarity and usefulness.

  • The star product is just a (von Neumann-Baker) convolution in phase space, $$A^W(x,p)\star B^W(x,p)\\ = {1\over \hbar ^2 \pi^2}\int dp^{\prime} dp^{\prime\prime} dx' dx'' ~A^W(x',p')~B^W(x'',p'') \\ \times \exp \left({-2i\over \hbar} \left( p(x'-x'') + p'(x''-x)+p''(x-x') \right )\right) , $$ not qualitatively different than your Hilbert-space integral kernel convolution gambit expression. These convolutions, in sharp contrast to local multiplication composition of functions, smear them and unzip them and rezip them nonlocally, so their extended structure and their variations around a point add and cancel in very different ways, incur local imbalances "loans" of their value to be repaid elsewhere in their domain; so little can be said about the structure around their zeros and singularities, except by heavy-duty professional functional analysts, which I am not. Convolutions probe variation of functions, and not just local features at points.

  • The ℏ is not quite an irrelevant aside: By virtue of dimensionality, it amounts to the scale of relevant variation of the (Weyl-symbol) functions involved: $\sqrt \hbar$ must divide x and p, and therefore multiply their corresponding gradients. These variations are where all the "action" is in QM. For example, the Wigner functions of stationary pure states of the oscillator depend ferociously nonanalytically on it. I really don't want to send you off on a tangent chasing irrelevancies by pointing you to the "wall-of-shame" papers that misconstrue this fact. (However, lest you went on an unjustified wild-goose chase in Takahashi's paper, his section 2 misconstrues the classical limit.)

  • As a lark, you might consider a (poor, unrepresentative) example based on the rescaled and nondimensionalized oscillator, $H^W=p^2+ x^2$. In Hilbert space, $$ \hat A = 1\!\!1 -\hat H, \qquad \hat B= \sum_{k=0}^\infty \hat H ^k= 1\!\!1 +\hat H + \hat H^2+\hat H^3+ \ldots \\ \leadsto \qquad \hat A \hat B= 1\!\!1, $$ so $$ A(x,z)= \langle x|1\!\! 1-(\hat p^2+ \hat x^2) |z\rangle \\ = \delta (x-z) -x^2\delta (x-z) + \hbar^2 \delta '' (x-z), $$ etc. Before indulging in star products, see how these Hilbert-space nonlocal expressions address your question satisfactorily.

In phase space, this particular expression $A^W\star B^W$ is easy to see to be real and analytic in ℏ, a very unrepresentative circumstance. Given the Bopp shift $$ H^W \star H^W= \left ( (x+i\hbar{\partial_p\over 2})^2+(p-i\hbar{\partial_x\over 2})^2 \right )(x^2+p^2)\\ =(x^2+p^2) (x^2+p^2-\hbar^2), $$ etc, you see $$1= (1-H^W)\star \bigl ( 1+H^W+H^W\star H^W ~+H^W\star H^W\star H^W +\ldots \bigr ) \\ =\left (1-H^W +\frac{\hbar^2}{4}(\partial_x^2+\partial_p^2 )\right ) ~ \bigl ( 1+H^W+H^W\star H^W ~+H^W\star H^W\star H^W + \ldots \bigr ) $$ dictates that all $O(\hbar)$ terms cancel among themselves in the equation, so they cancel their HW (the effective variable$^\natural$) singularities and zeros order by order in ℏ. Note the recursion relation that follows from it! But, as indicated, this (analyticity in ℏ) is a scandalously smooth and untypical circumstance. A real condign example should transcend this...


$^\natural$ So, defining $z\equiv H^W$, you just have $$ (1-z+\hbar^2 \partial_z+ \hbar^2 z\partial_z^2) B^W(z, \hbar)=1, $$ if you like ODEs, whose singularities you may study better. So $$ (A^W+ \hbar^2 \partial_z ~ z~\partial_z)(a_0+ \sum_{n=1}^\infty \hbar^{2n} a_n )=1~~~~\leadsto\\ a_n= -\frac{1}{1-z} \partial_z ~ z~\partial_z ~a_{n-1} , ~~~a_0=\frac{1}{ A^W}=\frac{1}{1-z}. $$ Thus $B^W= {1\over 1-z}-\hbar^2 {1+z\over (1-z)^4}+O(\hbar^4)$, etc. Here, the poles are unshifted, but note the orders: The ℏ terms cancel among themselves, not with the classical part.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.