According to Gardiner-Zoller (Quantum Noise), operators acting on the density matrix can be mapped via e.g. (I'm taking Wigner space as an example, but the same holds for P and Q)

$$a\rho\leftrightarrow\left(\alpha+\frac{1}{2}\frac{\partial}{\partial\alpha^*}\right)W( \alpha,\alpha^*)$$ $$\rho a^\dagger\leftrightarrow\left(\alpha^*+\frac{1}{2}\frac{\partial}{\partial \alpha}\right) W(\alpha,\alpha^*)$$

Below, an example is given using the P-function, frow which it is clear that if multiple operators are applied on the left or the right of the density matrix, the same correspondences hold, as long as the operators closest to $\rho$ are applied first (i.e. the phase-space representation most to the right, so closest to $W$).

Now my question is: what if there are operators acting on both sides of $\rho$? In the simplest case of $a\rho a^\dagger$ this does not seem to be an issue, because $\left(\alpha+\frac{1}{2}\frac{\partial}{\partial\alpha^*}\right)\left(\alpha^*+\frac{1}{2}\frac{\partial}{\partial \alpha}\right)$ =$\left(\alpha^*+\frac{1}{2}\frac{\partial}{\partial \alpha}\right)\left(\alpha+\frac{1}{2}\frac{\partial}{\partial\alpha^*}\right)$,

but it does lead to ambiguity for example for $aa\rho a^\dagger a^\dagger$.

I would expect that the proper way of doing this is still from the inside out, alternating operators from the left and the right; also because this way I obtain a result that is real. Is this correct? How to properly see this?

  • $\begingroup$ You have two different derivative operators, $D, D^*$, which commute. Where is the ambiguity in $DDD^*D^*$? $\endgroup$ Apr 30, 2019 at 21:52
  • $\begingroup$ @Zachos thanks, the wp seems fine. This seems a really strange situation where the derivative operators commute, but their squares not. Especially the product of the two cross-terms seems to mess up. $\endgroup$
    – Wouter
    Apr 30, 2019 at 22:10
  • $\begingroup$ Okay, I seem to have missed a term. Glad that commutation relations still behave as they should $\endgroup$
    – Wouter
    May 1, 2019 at 7:04

1 Answer 1


$$a\rho\leftrightarrow\left(\alpha+\frac{1}{2}\frac{\partial}{\partial\alpha^*}\right)W( \alpha,\alpha^*)\equiv D W( \alpha,\alpha^*),$$ $$\rho a^\dagger\leftrightarrow\left(\alpha^*+\frac{1}{2}\frac{\partial}{\partial \alpha}\right) W(\alpha,\alpha^*)\equiv D^* W( \alpha,\alpha^*).$$

You proved $[D,D^*]=0$, and, of course, both D s commute with themselves. Thus there cannot be an ambiguity in $$ DDD^*D^*= D^*D^*DD= D^*DDD^*=... $$ In chiral symmetry, that means the left and right action groups commute--they don't know about each other.

Check $$a^n\rho a^{\dagger ~n}\leftrightarrow D^n D^{*~n} W( \alpha,\alpha^*),$$ etc, if you must stick to hermitian/real objects.


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.