quantum mechanics - Deriving the expectation of $[hat X,hat H]$

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 $p \rightarrow p'$:

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