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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.
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  • $\begingroup$ 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? $\endgroup$ May 24, 2020 at 10:36

1 Answer 1

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$\omega$ 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 $\omega$.

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  • $\begingroup$ 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? $\endgroup$
    – Quantally
    May 24, 2020 at 11:41

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