I'm now working on my Phd thesis on the area of deformation quantization and field theory. After doing all the "ground work" (definitions, motivations, basics of the theory etc) I have now to do some "dirty work" with many and many Moyal Products. So I have searched for some tools to make life easier. In particular I found this MO question with a nice comment by user @Comaszachos saying that we can "Bopp shift away". I look at the pdf linked in the comment and on page 27 we have a lemma stating:

$$f(x,p)*g(x,p)=f(\hat x,\hat p)g(x,p)=f(x+\frac{ih}{2} \partial_p , p-\frac{ih}{2} \partial_x)g(x,p)$$

So I tried to apply this to some simple examples and that is what I found: take $f(x,p)=g(x,p)=e^x e^p$. Than

$$f(x,p)*g(x,p)=e^{x+\frac{ih}{2} \partial_p}e^{p-\frac{ih}{2} \partial_x}e^xe^p$$

as $[x,\frac{ih}{2} \partial_p]=[p,\frac{ih}{2} \partial_x]=0$, we have

$$ f(x,p)*g(x,p)= (e^x e^{ \frac{ih}{2} \partial_p} e^p e^{-\frac{ih}{2} \partial_x}) (e^x e^p) $$

using Lagrange's formula for the shift operator

$$f(x,p)*g(x,p)=e^xe^{\frac{ih}{2} \partial_p}(e^pe^{x-\frac{ih}{2}}e^p)
=e^xe^{p+\frac{ih}{2}}e^{x-\frac{ih}{2}}e^{p+\frac{ih}{2}}
=e^{2x+2p+\frac{ih}{2}}$$

and this is wrong! We can see it by the fact that $f*g=fg+\frac{ih}{2}\{f,g\}+O(h^2)$
and $\{e^xe^p,e^xe^p\}=0$, but
$$e^{2x+2p+\frac{ih}{2}}=e^{2x+2p}+e^{2x+2p}\frac{ih}{2}+O(h^2)$$

As a comment, the exact correct answer is
$$f(x,p)*g(x,p)=e^{2x+2p}$$
where I used
$$f(x,p)*g(x,p)=\lim_{(x',p')\to (x,p)}e^{\frac{ih}{2}(\partial_x \partial_{p'}-\partial_{x'} \partial_p)}f(x,p)g(x',p')$$

with the fact $e^{\frac{ih}{2}(\partial_x \partial_{p'}-\partial_{x'} \partial_p)}=e^{\frac{ih}{2}\partial_x \partial_{p'}}e^{-\frac{ih}{2}\partial_{x'} \partial_p}$ and expanding and collecting terms.

**My question is:** Where is the error? How do I properly apply the Bop shift? Is there any restriction to use it?

This post imported from StackExchange MathOverflow at 2023-11-30 17:41 (UTC), posted by SE-user Diego Santos