Of course, it can't hold if operators are over finite-dimensional spaces, as is evident from trace considerations. Can it be true for infinite-dimensional spaces? I think not, but I don't see how we can argue in this case.
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related: https://en.wikipedia.org/wiki/Stone%E2%80%93von_Neumann_theorem. The Heisenberg relations are an example of $[A,B]\subset iI$, where $\subset$ means extension (as in selfadjoint extension). – Phoenix87 Sep 01 '15 at 13:08
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Yes, there is even a typical example. If $V$ is the space of all polynomials on $\mathbb C$, and $D,x:V\to V$ are defined by: $$(Df)(x)=\frac{df}{dx}\\(xf)(x)=xf(x).$$ Then $Dx-xD=I$. You could also take $V$ to be all meromorphic functions over $\mathbb C$, or the space of power series over $\mathbb C$.
Such pairs of operators actually are the underlying explanation for the Heisenberg Uncertainty Principle.
Thomas Andrews
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Let $V = \mathbb{K}[x]$ be the space of polynomials in one variable with coefficients from your ground field $\mathbb{K}$, and consider the two operators $\frac{d}{dx}$ (formal differentiation of polynomials) and $x$ (multiplication of polynomials by $x$).
Michael Joyce
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The canonical commutation relations for position and momentum operators in quantum mechanics is $$[\hat x,\hat p]=i \hbar.$$ Here the vector space is the Hilbert space over $\mathbb C$.
krvolok
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