# What does the comma mean in this commutation rule between quantum operators?

The Theorem about quantum operators commutation relation says:

Consider pairs $$(U, V )$$ of unitary representations on a Hilbert space $$H$$, satisfying the commutation rule: $$U(x) V(y)=\exp (i \omega(x, y)) V(y) U(x).$$ Such pairs are all equivalent to multiples of the standard Schrödinger representation on $$L^2(\Re^n)$$.

The comma between $$x$$ and $$y$$, does it mean inner product or it means something else?

References:

1. J. Rosenberg A Selective History of the Stone-von Neumann Theorem, https://www.math.umd.edu/~jmr/StoneVNart.pdf page 6.
• This is just a guess, but isn't it just a function $\omega: G\times G\rightarrow \mathbb R$, so $x$ and $y$ are just elements of the group and the function $\omega$ has two arguments? Commented May 24, 2020 at 10:36

## 1 Answer

$$\omega$$ is presumably a bilinear form on $$\mathbb{R}^n$$, cf. the Heisenberg group. In other words, the comma separates the two arguments $$x,y\in\mathbb{R}^n$$ of $$\omega$$.

• Thank you for your quick answer. I tried to derive it by defining U(x)=exp(ixP) and V(y)=exp(iyQ) where P and Q are canonical operators but I did not know how to get the final result as in the original question. Can I derive it by defining U and V like this? If yes can you give me a small hint? Commented May 24, 2020 at 11:41