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

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

In the common eigen-state of $\hat{H}$ and $\hat{P}$, $|e, p>$, can wo do the following?

$\lt e, p| [\hat{X}$, $\hat{H}] |e, p\gt$

$ = \lt e, p|\hat{X} (\hat{H}|e, p\gt) - (\lt e, p|\hat{H}) \hat{X}|e, p\gt $

$ = \lt e, p|\hat{X} (e|e, p\gt) - (\lt e, p|e) \hat{X}|e, p\gt $

$ = e( \lt e, p|\hat{X}|e, p\gt - \lt e, p|\hat{X}|e, p\gt ) $

$ = 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?

  • $\begingroup$ When you check your result physically: Your calculation implies that the expectation-value of the momentum operator P on a momentum-eigenstate always gives zero. This smells fishy :) $\endgroup$ – Mario Krenn Jun 24 '14 at 17:44
  • $\begingroup$ So when you take X operator hitting e|e,p> state where in the momentum representation X is a differential operator with respect to momentum. But e is a function of p. So you did not take the derivative. Just choose plane wave function as your basis and you will see X's action upon plane wave basis. $\endgroup$ – user45765 Jun 24 '14 at 17:53
  • $\begingroup$ @user45765 I well understood the point about "X's action upon plane wave basis", which is exactly what the commutative relation (1) means. What puzzled me is why the Hermitian property of $\hat{H}$, as a general conclusion, doesn't work in this case. Or the $\hat{H}$ might not be Hermitian in some states? $\endgroup$ – user36125 Jun 24 '14 at 18:32
  • $\begingroup$ P is always hermitian which you can check by its definition. Power of a hermitian operator is always hermitian, unless you have complex interacting potential which you do not in this case. It is hermitian in all spectrum. Xe=i$\frac{de}{dp}$+ eX has to be used in your derivation of $\langle\psi|[X,H]|\psi\rangle$ where e is the energy, $\frac{p^{2}}{2m}$ $\endgroup$ – user45765 Jun 24 '14 at 19:37
  • 1
    $\begingroup$ @user45765 I believe I got the solution in another thread: physics.stackexchange.com/questions/121653/…. In a nutshell, the expectation was not properly defined, except for a limiting manner. It's a diagonal element in $|e,p>$ basis, yet as a limit obtained from the non-diagonal elements. $\endgroup$ – user36125 Jun 24 '14 at 20:18

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