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For any $n\times n$ matrices $A,B$, we have $\text{Tr}(AB)=\text{Tr}(BA)$. This can be easily proved by algebraic summation. Thus $\text{Tr}([A,B])=\text{Tr}(AB-BA)=0$. Also, $\text{Tr}(kI)=kn$. Therefore $[A,B]=kI$ is impossible for all $A,B$.

But if we look at infinite dimensional Hilbert spaces (let's use $L^2(\mathbb R)$ in quantum mechanics as an example), we can actually have the conical commutation relation $[x,p]=i\hbar$ for the position operator and momentum operator $x,p$. That means the above conclusion about finite square matrix does not apply here anymore. I struggle to see why this is the case.

I try to rewrite the proof of the result about $n\times n$ for infinite dimensions. The problem is that I can no longer prove $\text{Tr}(AB)=\text{Tr}(BA)$. $$ \text{Tr}(AB)=\sum_n \langle n|AB|n\rangle\\ \text{Tr}(BA)=\sum_n \langle n|BA|n\rangle $$ I can no longer write out the entries of $A,B$ explicitly.

So why $[A,B]=kI$ is impossible in finite dimensions but possible in infinite dimensions? What exactly makes this difference?

EDIT: a comment below reminds me that the trace of an operator in infinite dimension can be undefined. This is a problem. But can anyone give more details about this?

Ma Joad
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    Obligatory remark: it can happen that $[A,B]=I$ in a finite-dimensional vector space if the ground field is an appropriate one with a finite characteristic and the matrices have the right sizes. – user1551 Jun 14 '19 at 11:57
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    The trace is no longer defined on all operators on an infinite-dimensional space. For example, let $V$ have infinite dimension and consider the identity operator $I: V \to V$. Then the trace should be $\dim(V)$, which is infinite. – Joppy Jun 14 '19 at 12:47
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    You should change the title. You say "I struggle to see why this is the case" for the infinite-dimensional case, and not for the "finite matrix" case. There it is clear using the trace. – Dietrich Burde Jun 14 '19 at 14:08
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    There is a section dedicated exactly to this in the book "Methods of modern mathematical physics" of Reed and Simon. – Giuseppe Negro Jun 14 '19 at 14:18
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    Precisely, it is Example 2 on pg. 274 on the first volume. google books – Giuseppe Negro Jun 14 '19 at 14:23
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    An interesting example to play with might be $V = \operatorname{span}_{\mathbb{R}}{1, x, x^2, \ldots}$ the space of polynomials with real coefficients, and the operators $A: p(x) \mapsto xp(x)$ and $B: p(x) \mapsto p'(x)$ of multiplication by $x$ and differentiation. Then $[B, A]$ is the identity, and you can try writing out "matrices" for $A$ and $B$ to check what is going on, and why this cannot happen in finite dimensions. – Joppy Jun 14 '19 at 14:26

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For $n\times n$ matrices over a field $K$ of characteristic zero, it is exactly because of the trace, that $[A,B]=I$ is impossible. Actually, all matrices in $M_n(K)$ of trace zero are commutators of the form $[A,B]$. This is usually proved with Lie algebra theory, because the Lie algebra $\mathfrak{sl}_n(K)$ of trace zero matrices with Lie bracket $[A,B]=AB-BA$ is simple and every element is a commutator - see this reference:

Is every element of a complex semisimple Lie algebra a commutator?

The infinite-dimensional case usually is much more complicated and it is not surprising that many statements are no longer valid.

Dietrich Burde
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