For a free particle of mass $m$, with Hamiltonian

$$\hat{H} = \frac {\hat{P}^2} {2m},$$

where $$\hat{P} = -i \hbar \frac{\partial} {\partial x}.$$

The commutative relation is given by

$$[\hat{X}, \hat{H}] = \frac {i\hbar} {m} \hat{P}\tag{1}$$

In the common eigenstate of $\hat{H}$ and $\hat{P}$, $|e, p\rangle$, can we do the following?

$$\langle e, p| [\hat{X}, \,\hat{H}] |e, p\rangle = \langle e, p|\hat{X} (\hat{H}|e, p\rangle) - (\langle e, p|\hat{H}) \hat{X}|e, p\rangle \\ = \langle e, p|\hat{X} (e|e, p\rangle) - (\langle e, p|e) \hat{X}|e, p\rangle \\ = e( \langle e, p|\hat{X}|e, p\rangle - \langle e, p|\hat{X}|e, p\rangle ) \\ = 0 $$

Since the $\hat{H}$ is Hermitian, the above derivation doesn't seem to show any flaw. Given the commutative relation, Eq (1), we know the result is wrong. What's wrong with the above derivation?


Following the comment by Luboš Motl, I have worked out the solution and would like to share it here. The link provided by Qmechanic had the solution closely related to this question.

$$ \langle e', p'| [\hat{X}, \,\hat{H}] |e, p\rangle \\ = \langle e', p'|\hat{X} (\hat{H}|e, p\rangle) - (\langle e', p'|\hat{H}) \hat{X}|e, p\rangle \\ = (e - e') \langle e', p'|\hat{X}|e, p\rangle $$

Note that:

$$ e - e' = \frac{p^2}{2m} - \frac{p'^2}{2m} = \frac{(p+p')(p-p')}{2m} $$

$$ \langle e', p'|\hat{X}|e, p\rangle = -i\hbar \delta'(p - p') $$

where $\delta'(\cdot)$ is the derivative of the Dirac function, with respect to $p$.

Then we get

$$ (e - e') \langle e', p'|\hat{X}|e, p\rangle \\ = -i\hbar \frac{(p+p')}{2m} \cdot (p - p')\delta'(p - p') \\ = - \frac{i\hbar (p+p')}{2m} \cdot (-\delta(p - p')) \\ = \frac{i\hbar (p+p')}{2m} \delta(p - p') $$

As we take the limit $p \rightarrow p'$:

$$ lim_{p \rightarrow p'} \frac{i\hbar(p+p')}{2m} \delta(p - p') \\ \rightarrow \frac{i\hbar}{m} p \delta(p - p') $$

  • 3
    $\begingroup$ It's a very subtle flaw but have you tried to sandwich the operator in between somewhat more general states $\langle e,p|$ and $|e',p'\rangle$? $\endgroup$ Jun 24 '14 at 17:54
  • 3
    $\begingroup$ Related: physics.stackexchange.com/q/14116/2451 $\endgroup$
    – Qmechanic
    Jun 24 '14 at 18:39
  • $\begingroup$ @LubošMotl actually I did try the general sandwich case, but I dismissed it without giving it a thought. Since you mentioned it, I realized it's really a good point to attack: $lim_{e' \rightarrow e}$ ⟨e,p|$\hat{X}(\hat{H}$|e′,p′⟩) - (⟨e,p|$\hat{H})\hat{X}$|e′,p′⟩ = $lim_{e' \rightarrow e}$ (e' - e) ⟨e,p|$\hat{X}$|e′,p′⟩. Since $lim_{e' \rightarrow e} (e' - e) \rightarrow 0$ and $lim_{e' \rightarrow e} ⟨e,p|\hat{X}|e′,p′⟩ \rightarrow \infty$, we cannot dismiss either of two factors. I will try work it out in more detail. I voted you up. $\endgroup$
    – user36125
    Jun 24 '14 at 19:44
  • 2
    $\begingroup$ Hi user, yup. Just to be sure, try to think about the value of the function $x\delta'(x)$ at $x=0$. $\endgroup$ Jun 25 '14 at 4:08

As is customary in such question, I will point to this paper, which excellently discusses the problems the Dirac formalism has.

Now, in your concrete example, the problem lies in the energy/momentum states $| p_0 \rangle$ themselves, which are non-normalizable, since the wave function associated is the Fourier transform of $\delta(p-p_0)$, which means that $\psi_{|p_0\rangle}(x) = \mathrm{e}^{-\frac{ixp_0}{\hbar}}$. If you now try to calculate the inner product, you find: $$\langle p_0 | p_0 \rangle = \int_{-\infty}^{\infty}\psi_{|p_0\rangle}(x)\bar{\psi}_{|p_0\rangle}(x) \mathrm{d}x = \int_{-\infty}^{\infty} \mathrm{e}^{-\frac{ixp_0}{\hbar}} \mathrm{e}^{\frac{ixp_0}{\hbar}}\mathrm{d}x = \int_{-\infty}^\infty 1 \mathrm{d}x $$

Thus, momentum eigenstates are non-normalizable, and writing things like $\langle p_0 |X |p_0\rangle - \langle p_0 |X |p_0\rangle$ is really non-sensical, because you are subtracting two infinities. In particular, it is not $0$.

  • $\begingroup$ that seems to be a nice paper - I voted you up because of that. I will look into the paper in more details. But I am not satisfied with your answer, which I heard somewhere else before. Let me explain why. I don't think anyone could actually work out a particular result, for subtracting the two infinities you mentioned: $<p_0|\hat{X}|p_0> - <p_0|\hat{X}|p_0>$. They are literally two identical entities, which was a puzzle from the beginning. I believe Luboš Motl pointed out the real solution - the expectation was not properly defined, except for a limiting manner. $\endgroup$
    – user36125
    Jun 24 '14 at 20:07
  • $\begingroup$ I don't think we disagree - no one can work out a result for the difference, because it is non-sensical, since its constituents don't exist. I certainly won't begrudge you being happier with Luboš' angle of attacking this problem, since it admittedly is a bit more to the point :) $\endgroup$
    – ACuriousMind
    Jun 24 '14 at 20:33
  • $\begingroup$ I do appreciate your response very much. Please allow me explaining my point a bit further. The motivation of my question was to seek the consensus between the two approaches of calculating $<p_0|[\hat{X}, \hat{H}]|p_0>$. The second way of applying the Hermitian property of $\hat{H}$ should come up with the same result, instead of merely stating its "no-go" consequence. For example, if we could claim the Hermitian-ness of $\hat{H}$ is invalid in this case, that would be considered a logical consensus. $\endgroup$
    – user36125
    Jun 25 '14 at 16:33
  • $\begingroup$ By stating the example above, I don't really mean $\hat{H}$ is non-Hermitian here. $\endgroup$
    – user36125
    Jun 25 '14 at 16:42
  • 1
    $\begingroup$ $\hat H$ is Hermitian (by the way, the "right" word for "Hermitian-ness" is Hermiticity), no doubt about it. It is not always self-adjoint, but that is not the problem here. After having reread the paper I linked, I think that the problem could rather be that $|p_0\rangle$ is not even in the domain of definition of the operator $\hat X$, but I need some time to try and work that out. $\endgroup$
    – ACuriousMind
    Jun 25 '14 at 22:16

A tricky question, really. Apart from the fact that your $\lvert e,p\rangle$ vector does not belong to $L^2$ (hence you cannot take scalar products of it), I don't see any other flaw. That, in my opinion, means you have a nice argument to prove the following mathematical statement:

Let $\mathscr{H}$ be a separable Hilbert space, $0\neq z\in\mathbb{C}$. There are no self-adjoint operators $A$ and $B$ with non-empty discrete spectrum different from zero such that $[A,B]=z$.

Closely related to that fact, the following result of Von Neumann: up to multiplicity and unitary equivalence, the relations $[A,B]=i$ (in their exponentiated form) are uniquely realized by $A=x$ (multiplication operator) and $B=-i\nabla_x$, that indeed have no discrete spectrum.

EDITED (in reply to the comment, also the statement above has been edited slightly, to be more precise):

A number $\lambda\in \mathbb{R}$ is in the discrete spectrum of $A$ ( called $\sigma_{disc}(A)$ ) if there exists at least one $\psi_{\lambda}\in \mathscr{H}$ ( the Hilbert space, usually $L^2(\mathbb{R}^d)$ ) such that $$A\psi_\lambda=\lambda\psi_\lambda\; .$$ Suppose there exist $A$ and $B$ self-adjoint such that $0\neq \lambda \in \sigma_{disc}(B)$ and $[A,B]=z$ (on a suitable dense domain). Now it follows that (on another suitable domain) $$[A,B^2]=2zB\; .$$ Let $\psi_\lambda\in \mathscr{H}$ be one of the eigenfunctions of $B$ associated to $\lambda$. On one hand, $$2z\langle\psi_\lambda, B\, \psi_\lambda\rangle_{\mathscr{H}}=2z\lambda \lVert \psi_\lambda \rVert^2_{\mathscr{H}}\; ;$$ on the other $$\langle\psi_\lambda, AB^2\, \psi_\lambda\rangle_{\mathscr{H}}- \langle\psi_\lambda, B^2A\, \psi_\lambda\rangle_{\mathscr{H}}=0$$ as you suggested. That is absurd, since $z$, $\lambda$ and $\lVert \psi_\lambda \rVert^2_{\mathscr{H}}$ are different from zero.

It follows that you cannot have two self-adjoint operators such that $[A,B]=z$ and $\sigma_{disc}(B)\neq \{0\},\emptyset$. The reasoning above does not work if there is no eigenfunction $\psi_\lambda\in \mathscr{H}$ (because with formal eigenfunctions you are not allowed to take scalar products or norms: they are not finite).

  • $\begingroup$ excuse me for my insufficient background in rigorous math analysis - I have a question on what you really mean. Why does the fact that "∣e,p⟩ vector does not belong to $L^2$" have anything to do with the statement: "no self-adjoint operators A and B with non-empty discrete spectrum such that [A,B]=z"? More particularly, what's the big deal about the non-empty discrete spectrum? Thanks! $\endgroup$
    – user36125
    Jun 25 '14 at 16:16
  • $\begingroup$ I have edited the answer, my reply would have been too long for a comment. $\endgroup$
    – yuggib
    Jun 25 '14 at 17:17
  • $\begingroup$ I see what you are saying - yet I'd like to mention a few points: 1) your spectrum (in the first equation) should be for operator B, instead of A; 2) from your edited answer, it doesn't seem to matter whether the spectrum is discrete or continuous; 3) for the specific example in the question, are you indicating $\hat{H}$ is non-Hermitian (my understanding self-adjoint means the same thing)? $\endgroup$
    – user36125
    Jun 25 '14 at 20:50
  • $\begingroup$ Well: 1) in the first equation I am defining the meaning of discrete spectrum, so $A$ or $B$ does not matter, but I agree it may be a little bit confusing with the notation below; 2) It matters, because I am defining the discrete spectrum in a way that there is at least one eigenfunction in the Hilbert space (it is not the case for all the values in the spectrum in general!); 3) self-adjoint means symmetric (hermitian) and something more (its domain is equal to the domain of the adjoint) you have to be careful! Your $H$, also known as Laplace operator, is self-adjoint on $L^2(\mathbb{R}^d)$. $\endgroup$
    – yuggib
    Jun 25 '14 at 22:07

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