Does anyone know how to prove that for the product of two operators $\hat{A}\hat{B}$ the Weyl-Wigner correspondence reads $$ (AB)(x,p) = A\left (x-\frac{\hbar}{2i}\frac{\partial}{\partial p}, p+\frac{\hbar}{2i}\frac{\partial}{\partial x}\right )B(x,p), $$ without using the star product? Here $B(x,p)$ represents the usual Wigner map image $$ B(x,p) = \int_{-\infty}^\infty d\xi e^{-ip\xi/\hbar}\langle x + \xi/2|\hat{B}|x-\xi/2 \rangle. $$

I think one has to consider the operators as a function of $\hat{x}$ and $\hat{p}$ and then do a Taylor expansion, but I'm not sure.

  • $\begingroup$ But... your first equation is the star product in the Bopp-shift representation, no? $\endgroup$ Mar 18, 2019 at 21:10
  • $\begingroup$ Yes (?), however I'm thinking more in deriving the equation, rather than using a new definition. I.e. if I forget about the star product, how can I derive the equation using only the Weyl-Wigner representation of the operators? $\endgroup$ Mar 18, 2019 at 21:46
  • $\begingroup$ Yes, that's the notation as it is in Schelich (hoho again). $\endgroup$ Mar 18, 2019 at 22:08
  • 1
    $\begingroup$ Related. $\endgroup$ Mar 19, 2019 at 13:29

1 Answer 1


Your first line is the Bopp-shift definition of the star product, and it is proven directly as in Groenewold's original paper inventing it, always mindful of Lagrange's translation operator, $e^{\epsilon \partial_z} f(z)= f(z+\epsilon)$, of course. However, it is crucial to indicate the derivatives in A strictly only act on the arguments of B! That is the reason authors either define primed phase space coordinates for the trailing expression, and unprime them at the very end, or they use the far more sensible arrow notation (we use). To the extent you appreciate that, I will stick to your notation.

$$ A\left(x+i\frac{\hbar}{2}\frac{\partial}{\partial p}, p-i\frac{\hbar}{2}\frac{\partial}{\partial x}\right )B(x,p)\\ = \int_{-\infty}^\infty d\zeta \langle x + \zeta/2+\frac{i\hbar}{2}\partial_p '|\hat{A}|x-\zeta/2 +\frac{i\hbar}{2}\partial_p '\rangle e^{-ip\zeta/\hbar +\frac{\zeta}{2} \partial_x } ~~ \int_{-\infty}^\infty d\xi e^{-ip'\xi/\hbar}\langle x + \xi/2|\hat{B}|x-\xi/2 \rangle\\ =\int d\zeta d\xi ~~\langle x + \zeta/2+\xi /2 |\hat{A}|x-\zeta/2 +\xi/2 \rangle e^{-i(p /\hbar)(\zeta+\xi) } \langle x + \xi/2-\zeta/2|\hat{B}|x-\xi/2 -\zeta/2\rangle. $$ In the last line, we set $p'=p$, since there cannot be any ambiguity anymore. This is the only place with momentum dependence.

You then define light-cone coordinates $\rho\equiv \zeta+\xi$ and $\sigma\equiv x+(\xi -\zeta)/2$, whose Jacobian is unity, so that the above integral readily collapses to $$ \int d\rho d\sigma ~~\langle x + \rho/2 |\hat{A}|\sigma \rangle e^{-i(p /\hbar)(\rho) } \langle \sigma|\hat{B}|x-\rho/2 \rangle= \int d\rho ~e^{-ip \rho/\hbar }\langle x + \rho/2 |\hat{A} \hat{B}|x-\rho/2 \rangle=(AB)(x,p). $$

This is Groenewold's fundamental isomorphism theorem (1946), and, no matter what language you use, you find yourself doing the same symplectic manipulation again and again, and yet again. (In our booklet, this is the only theorem!)

  • $\begingroup$ Two questions. You have so partial derivatives with respect to $p'$, but there are no $p'$ in the expression to derive with. How did you get the third equality? I.e. why $\partial_{p'}\sim \xi/2$? $\endgroup$ Mar 18, 2019 at 22:54
  • $\begingroup$ It's ok, I see it now. Maybe you can just remove the primes. $\endgroup$ Mar 18, 2019 at 23:10
  • $\begingroup$ One last question. The star product in your notes, is there a derivation to find it, or is it just an observation (physical if one likes to see it that way), that this operator is the good one? $\endgroup$ Mar 18, 2019 at 23:13
  • 1
    $\begingroup$ Well, it is a brilliant discovery that Groenewold made. Once you have it, it is trivially provable as here, or (117-119) in our book.We used more variables there to make sure there is no ambiguity as here. $\endgroup$ Mar 18, 2019 at 23:16
  • 1
    $\begingroup$ Indeed, (4.27) of his thesis paper is one of the major achievement in this field. $\endgroup$ Mar 18, 2019 at 23:33

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.