# How can I take the Wigner transform of an operator with an absolute value?

I want to be able to find the Wigner transforms of operators of the form $\Theta(\hat{O})$, where $\Theta$ is the Heaviside function and $\hat{O}$ in general depends on both $x$ and $p$. For the operators of interest, $\tilde{O}$ is known. Since the Wigner transform is linear, I can use the expression $$\Theta(x)=\frac{1}{2} \frac{\left( |x| - x \right)}{x}$$

Unfortunately, I am no better off than I was when I started, as I can't figure out how get the Wigner transform of the absolute value of an arbitrary operator in terms of the Wigner transform of that operator.

Is there an expression for the Wigner transform of the absolute value of an operator in terms of the Wigner transform of that same operator?

If so, what is it?

• Comment to the question (v1): The right-hand side of the equation is not the Heavyside step function. – Qmechanic Jul 4 '14 at 7:35
• The most simple would be $W(|\hat O|) = |W(\hat O) |$... It should work at least for simple operators $\hat O$ like $Q^n$ or $P^n$. – Trimok Jul 4 '14 at 11:09
• @Qmechanic: Sorry! It was a typo. – Dan Jul 4 '14 at 17:16

The absolute value of the operator $\hat{O}$ is presumably $\sqrt{\hat{O}^\dagger \hat{O}}\equiv \hat{B}$, so the crucial question is what type of algorithm one would choose to evaluate the square root $\hat{B}$ of the operator $\hat{A}= \hat{O}^\dagger \hat{O}=\hat{B} \hat{B}$.
If, for example, a suitable satisfactory series in $\hat{A}$ were found for $\hat{B}$, The Weyl symbol/Wigner transform of $\hat{B}$ would be $\tilde{B}=$the same star-function of $\tilde{A}$, where, of course, $\tilde{A}=\tilde{O}^* \star\tilde{O}$, in the abstract. Beyond the above math stack sequence, for a bounded normalized $\hat{A}$, one might consider $\hat{B}= \sqrt{I+(\hat{A}-I)}$ $= \sum_{n=0}^{\infty} {{1/2}\choose{n}} (\hat{A} -I )^n$, etc. Transcription to Wigner transforms would merely replace operator powers by star-powers of Weyl symbols.
Unfortunately, this is merely formally correct, but an impractical setup. Even for (unbounded) $\hat{O}=\hat{x} +i\hat{p}$, finding a bona-fide $\tilde{B}$ is daunting.
One wonders if the $\tanh(k\hat{O})$ representation of the step function for large k were more practical, instead, so $|\hat{O}|=\lim_{k\to\infty} \tanh(k\hat{O}) ~\hat{O}$. See, e.g. this application 266886.