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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?

Joshuah Heath
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Yuexin Zhang
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2 Answers2

16

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.

tparker
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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.

Alejandro Menaya
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