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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?

[EDIT]

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 $p \rightarrow p'$:

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

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3  
It's a very subtle flaw but have you tried to sandwich the operator in between somewhat more general states $\langle e,p|$ and $|e',p'\rangle$? –  Luboš Motl Jun 24 at 17:54
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Related: physics.stackexchange.com/q/14116/2451 –  Qmechanic Jun 24 at 18:39
    
@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. –  user36125 Jun 24 at 19:44
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Hi user, yup. Just to be sure, try to think about the value of the function $x\delta'(x)$ at $x=0$. –  Luboš Motl Jun 25 at 4:08

2 Answers 2

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 $0$.

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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. –  user36125 Jun 24 at 20:07
    
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 :) –  ACuriousMind Jun 24 at 20:33
    
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. –  user36125 Jun 25 at 16:33
    
By stating the example above, I don't really mean $\hat{H}$ is non-Hermitian here. –  user36125 Jun 25 at 16:42
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$\hat H$ 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. –  ACuriousMind Jun 25 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).

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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! –  user36125 Jun 25 at 16:16
    
I have edited the answer, my reply would have been too long for a comment. –  yuggib Jun 25 at 17:17
    
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)? –  user36125 Jun 25 at 20:50
    
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 $L^2(\mathbb{R}^d)$. –  yuggib Jun 25 at 22:07

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