Commutator expectation value in quantum mechanics

Question

Suppose $A$ and $B$ are operators, $A$ is Hermitian, $B$ anti-hermitian, and their commutator is the identity, i.e. $$[A, B] = I \, .$$ Denoting the eigenvectors of $A$ as $\lvert a \rangle$, so that $A \lvert a \rangle = a \lvert a \rangle $, we have $$\langle a| [A, B] |a \rangle = \langle a | a \rangle \tag{1}$$ and $$ \langle a| (AB-BA) |a \rangle = a \langle a | B | a \rangle - a \langle a | B | a \rangle = 0 \, . \tag{2}$$

Summarizing, if $|a\rangle$ is normalized, $\langle a|a\rangle = 1$, we obtain $0=1$. Where is the mistake?

Answer 1

This is a very nice problem in the theory of operators in separable Hilbert spaces. The trick to notice is that your $|a\rangle$ is not in the domain of the commutator, therefore your equation 1) is meaningless. More precisely, we have the following lemma:

Lemma Let C be the commutator C(A, B; D(C)), in the sense that: $$\forall \phi \in D(C) \subseteq D(AB) \cap D(BA), (AB-BA)\phi = C\phi$$ Suppose furthermore that A is selfadjoint with a non-empty point spectrum. A necessary condition for the eigenvectors of A to belong to D(C) is that C maps each of the eigenvectors of A to its orthogonal complement.

Proof . Let Aϕ = aϕ and ϕ ∈ D(C). Because the eigenvalue a is real and because $|\langle ϕ|Bϕ\rangle| < ∞$, we have the equality $\langle ϕ|Cϕ\rangle = \langle ϕ|(AB − BA)ϕ\rangle = 0$. That is $\langle ϕ|Cϕ\rangle = 0,$ or ϕ is orthogonal to Cϕ.

Assume now that $C = 1_{\mathcal{H}}$. Then, in order to have the eigenvector $\phi$ in the domain of the commutator, it follows by the previous lemma that $\phi = 0$.

Answer 2

@Qmechanic answered your question in the comments, but evidently the message did not automatically sink in. Let me try to illustrate it with the routine demonstration you probably were exposed to when learning the uses of the Dirac bra-ket notation.

The short answer is that the tracefulness of the identity in the r.h.s. of your commutator equation for an infinite dimensional Hilbert space leads to $\langle a | a\rangle \neq$1, because it is actually singular. So your equation 1) is fine, since the r.h.s. is infinity. But your equation 2) is flawed, since the relevant expression involves a 0 multiplying a stronger infinity, amounting to infinity, again, as in 1).

I will illustrate this with $A=\hat x$ and $B=\hat p /\hbar$, as in standard QM courses. Absorb $\hbar$ in $\hat p$ to make the formalism more familiar.

Starting from the standard operator equation $[\hat x ,\hat p]=i 1\!\!1 $, first take its non-diagonal matrix elements, before building up to your 2), $$ \langle x|\hat x \hat p - \hat p \hat x|y\rangle = (x-y)\langle x|\hat p|y\rangle=(x-y)\int dp ~ \langle x| p\rangle \langle p| \hat p|y\rangle \\ =(x-y)\int dp ~ \langle x| p\rangle p \langle p |y\rangle \\ =\frac{ (x-y)}{2\pi}\int dp ~ p ~e^{i(x-y)p} =-i (x-y)\partial_x \delta (x-y) \\ =i \delta(x-y). $$ As always, $\langle x| p\rangle=\exp(ixp) ~/\sqrt{2\pi}$.   Check the last equality by operating on a well-behaved test function. It trivially reflects homogeneity of degree -1, $\delta(\lambda x)=\delta(x)/\lambda$, so differentiate this by λ and set λ=1.

That is, the expression diverges for $x\to y$, just like 1). The crucial point is that as the prefactor (x-y) decreases, the matrix element multiplying it diverges and faster.

All matrix elements of a commutator being available, as above, you may reconstitute your original operator equations from these, by insertion of resolutions of the identity on either side.