Weyl Quantization Integral

Question

I have some doubts when calculating the integral for Weyl Quantization symbol. If I understand correctly, quantization using the Weyl symbol takes a function in phase space and takes it to an operator in Hilbert space. The notation that the textbook I'm using is $Op(A) =\hat{A}$. As a first example, the textbook calculates the quantization for xp. And by definition we have:

$$Op[px] = \frac{1}{(2\pi\hbar)^{2}}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}pxe^{\frac{i\xi}{\hbar}(x-\hat{x})}e^{\frac{iy}{\hbar}(p-\hat{p})}d\xi dy dp dx.$$

The above integral can be written as:

$$Op[px] = \frac{1}{(2\pi\hbar)^{2}}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}pxe^{\frac{i}{\hbar}[\xi x+yp]}e^{\frac{-i}{\hbar}[\xi \hat{x}+y\hat{p}]}d\xi dy dp dx.$$

Note that when using the Baker-Campbell-Hausdorff identity in the second exponential, we obtain the following integral:

$$Op[px] = \frac{1}{(2\pi\hbar)^{2}}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}pxe^{\frac{i\xi}{\hbar}(x+\frac{y}{2}-\hat{x})}e^{\frac{iy}{\hbar}(p-\hat{p})}d\xi dy dp dx.$$

and through variable substitution $x'=x+\frac{y}{2}$, we have:

$$Op[px] = \frac{1}{(2\pi\hbar)^{2}}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}p(x'-\frac{y}{2})e^{\frac{i\xi}{\hbar}(x'-\hat{x})}e^{\frac{iy}{\hbar}(p-\hat{p})}d\xi dy dp dx'.$$

My doubts start here

1. If we are calculating the quantization $xp$, why are the calculations done with $Op[px]$ and not with $Op[xp]$?

The textbook continues the calculations and says that, when separating the last integral into two, the first part becomes:

$$Op[px] = \hat{x}\hat{p}-\frac{1}{(2\pi\hbar)^{2}}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}p\frac{y}{2}e^{\frac{i\xi}{\hbar}(x'-\hat{x})}e^{\frac{iy}{\hbar}(p-\hat{p})}d\xi dy dp dx'.$$

My second question is:

2. Why is the result of the first integral $\hat{x}\hat{p}$ and not $\hat{p}\hat{x}$, since we are calculating $Op[px]$? And how the integral is performed to obtain this result?

And my third, and final doubt, comes from the remaining integral. The book says that the result of integral is:

$$Op[px] = \hat{x}\hat{p}-\frac{1}{(2\pi\hbar)}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}p\frac{y}{2}e^{\frac{iy}{\hbar}(p-\hat{p})}dy dp .$$

My third question is:

3. How the integral in variables $\xi$ and $x'$ was calculated? It looks like the result was 1, and if it was 1, how was it calculated?

If anyone wants to consult the text I am studying to better understand my question. Here is the link:

https://www.math.uni-tuebingen.de/user/stke/teaching/wigner_weyl/wigner_weyl_color.pdf

Understanding this will help me a lot in developing my master's degree.

Answer 1

You have muffed your second expression, when you use the degenerate CBH formula routine in QM. The Weyl map, eq (134) here, dictates, instead, $$Op[px] = \frac{1}{(2\pi\hbar)^{2}}\iiiint_{-\infty}^{+\infty} \!\!px~e^{\frac{i}{\hbar}[\xi x+yp-\frac{xy}{2}]}e^{\frac{-i}{\hbar}[\xi \hat{x}+y\hat{p}]}d\xi dy dp dx \tag{2}~.$$

Alternatively, the second equation you wrote amounts to your third, and not your first!

You then, correctly, move on to your 4th, $$Op[px] = \frac{1}{(2\pi\hbar)^{2}} \iiiint_{-\infty}^{+\infty} \!p\left(x'-\frac{y}{2}\right)e^{\frac{i\xi}{\hbar}(x'-\hat{x})}e^{\frac{iy}{\hbar}(p-\hat{p})}d\xi dy dp dx'\tag{4}.$$ Clean up your act! Keppeler's (3.6) is plain wrong, and inconsistent with his (3.7)!

1. If we are calculating the quantization xp, why are the calculations done with Op[px] and not with Op[xp]?

The heart of the Weyl map is to convince you that $Op[xp]=Op[px]$. x and p are classical, commuting variables, corresponding to operators through the Weyl map: and so convince yourself you get the same Weyl map from either classical ordering, since there is no nontrivial meaning in ordering c-number variables.

• This is the very essence of the Weyl correspondence. This should be emphasized in your notes, and certainly is in our booklet.

Again, and as emphasized in our booklet, $$Op[xp]= Op[px]=\frac{\hat x \hat p+ \hat p \hat x}{2}\\ = \hat x \hat p +\frac{\hbar}{2i}= \hat p \hat x -\frac{\hbar}{2i}~~.\tag{Keppeler2.8}$$ It has to be symmetric in $\hat x$ and $\hat p$, since it came from a kernel symmetric in both, your Keppeler's (3.9).

2. Why is the result of the first integral $\hat{x}\hat{p}$ and not $\hat{p}\hat{x}$, since we are calculating Op[px]? And how is the integral performed to obtain this result?

Selectively: $$ \frac{1}{(2\pi\hbar)^{2}}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}px' ~e^{\frac{i\xi}{\hbar}(x'-\hat{x})}e^{\frac{iy}{\hbar}(p-\hat{p})}d\xi dy dp dx'\\ = \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} x' \delta(x'-\hat{x}) p \delta(p-\hat{p})~~ dp dx' = \hat{x}\hat{p}~.$$ Note that, from (4), all $\hat{x}$s are manifestly arranged before all $\hat{p}$s, preventing any possible ordering ambiguity. The order of the c-numbers x and p is immaterial/meaningless, as already stressed.

3. How was the integral in variables $\xi$ and x' calculated? It looks like the result was 1, and if it was 1, how was it calculated?

It is a routine Dirac δ, finally collapsed, $$ \frac{1}{(2\pi\hbar)}\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty} \large e^{\frac{i\xi}{\hbar}(x'-\hat{x})}\frac{dx'}{\hbar} d\xi\\ = \int_{-\infty}^{+\infty}\large e^{\frac{-i \xi}{\hbar}\hat{x}} \delta(\xi) d\xi =1.$$