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?

  • $\begingroup$ What are you asking for? It is not clear. $\endgroup$ Oct 16, 2018 at 21:55
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    $\begingroup$ @AlbertoNavarro The OP has work that essentially shows that $0=1$ and wants others to resolve the discrepancy $\endgroup$ Oct 16, 2018 at 21:56
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    $\begingroup$ $|a>$ is not normalizable. Related: physics.stackexchange.com/q/10230/2451 $\endgroup$
    – Qmechanic
    Oct 16, 2018 at 22:00
  • $\begingroup$ @Qmechanic. Not really. $\endgroup$
    – DanielC
    Oct 16, 2018 at 22:09
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    $\begingroup$ @DanielC but you show that it isn't normalizable don't you? $\endgroup$ Oct 16, 2018 at 23:00

2 Answers 2


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 $\phi = 0$.

  • $\begingroup$ Beautiful answer to a very important and popular question among theoretical physicists! The crucial point is " |a⟩ is not in the domain of the commutator". Thanks! $\endgroup$
    – Nikodem
    Nov 13, 2022 at 1:19

@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.

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    $\begingroup$ ? So you dismiss what every course following Dirac's book and his notation works out? Physicists were never stymied by unnormalizable states, or plane waves, or non-trace class operators. Fishermen on the pier hardly heed mathematical proofs of the non-existence of fish. Thanks to Dirac's distribution, physicists leapfrogged mathematicians by decades (... for a while). $\endgroup$ Oct 18, 2018 at 21:54
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    $\begingroup$ I am sorry, but my respect for mathematically sound physics requires dismissing sources not in agreement with mathematics. Yes, QM is a mathematical theory and has been so for 91 years, with the first involvement of von Neuman at the suggestion of David Hilbert. Nothing can change that. One can choose to ignore mathematical rigor in physics, I choose not to. $\endgroup$
    – DanielC
    Oct 18, 2018 at 22:05
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    $\begingroup$ @DanielC Are you saying anything using the Dirac delta function isn't valid? I just want to understand where you are coming from. $\endgroup$ Oct 19, 2018 at 0:35
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    $\begingroup$ @DanielC Cosmas recognizes this, but still discusses how to pull relevant conclusions. You don't need perfect mathematical rigor to understand the argument. Just like how in mathematics rigourous proofs are needed to make sure everything is valid, but we could still learn from just understanding the idea of the proof. In other words, mathematical rigor is very important, but I don't think it is always required when talking about or learning concepts, especially if the rigor would get in the way of a better learning experience. $\endgroup$ Oct 19, 2018 at 17:15
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    $\begingroup$ There is hardly anything in the above that is not in Dirac's book--including the homogeneity of the delta function. Weyl, who introduced the fake paradox, also virtually resolved it with QM around the clock. $\endgroup$ Oct 19, 2018 at 18:21

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