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$B$ is an Hermitian operator in Hilbert space, and $|b\rangle$ is the eigenstate of $B$. We can have $[A, B] = 1$ where A is arbitary operator. Then we can calculate as below:

\begin{align} &\phantom{=}\langle b | [A,B] | b \rangle \notag\\ &= \langle b | b \rangle\notag\\ & = \langle b | AB | b \rangle - \langle b | BA | b \rangle \\ &= b \langle b | A | b \rangle - \langle b | A^\dagger B^\dagger | b \rangle ^* \notag\\ &= b \langle b | A | b \rangle - \langle b | A^\dagger B | b \rangle ^* \\ &= b \langle b | A | b \rangle - b^* \langle b | A^\dagger | b \rangle ^* \notag\\ &= b [ \langle b | A | b \rangle - \langle b | A | b \rangle ] \notag\\ &= 0. \end{align}

So it is shown that $1 = \langle b | [A,B] | b \rangle = 0$ which is clearly not right. But where does the problem in the process lie?

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  • $\begingroup$ What is $b$ where it occurs outside of a ket? Is that supposed to be the eigenvalue of $B$ associated with the state $\lvert b\rangle$? $\endgroup$ – David Z Jul 20 '18 at 14:18
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The answer is basically the same as the answer to this question. It's a very subtle point about infinite-dimensional Hilbert spaces. Note that the fact that your commutator is nonzero means that the operators must act on an infinite-dimensional Hilbert space (or else you could take the trace and would automatically get 0). So you need to be careful about the exact domain of definition of the operators, which will in general be smaller than the entire Hilbert space. It turns out that none of the eigenvectors of $B$ lie in the domain of $A$, so the ket $A|b \rangle$ is undefined, and your second equation is wrong.

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Simply $[A, B] $ cannot be equal to the identity.

Indeed, suppose such operators exists, and take the trace of the equation:

$Tr[A, B] =Tr[1]\Rightarrow Tr[AB] - Tr[BA] =Tr[1]$,

which should vanish via the cyclic property of the trace. Since the trace of the identity is not zero, the assumption of this operators exist, must be false.

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    $\begingroup$ Don’t forget about the canonical commutator though! $\endgroup$ – knzhou Jul 20 '18 at 12:47
  • $\begingroup$ @knzhou Reading the previous answer, I realize I forgot. $\endgroup$ – Alejandro Menaya Jul 20 '18 at 12:49
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    $\begingroup$ This is wrong. Operators can have a nontrivial commutator if the Hilbert space is infinite-dimensional and they aren't trace-class. $\endgroup$ – tparker Jul 20 '18 at 12:51
  • $\begingroup$ @tparker as pointed in my previous comment, I noticed. I assumed finite dimensional Hilbert space, when I should not. $\endgroup$ – Alejandro Menaya Jul 20 '18 at 12:56

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