IFNT said:
If I have H=p^2/2m+V(x), |a'> are energy eigenkets with eigenvalue E_{a'}, isn't the expectation value of [H,x] wrt |a'> not always 0? Don't I have that
<a'|[H,x]|a'> = <a'|(Hx-xH)|a'> = <a'|Hx|a'> - <a'|xH|a'> = 0?
But if I calculate the commutator, I get:
<a'|[H,x]|a'> = <a'|-i p \hbar / m|a'> \neq 0
I suddenly can't see where I did wrong...
Adding to all the other answers in this thread...
This seems like a variation on Dirac's famous "0 = 1" paradox: if Q,P are the
usual canonically-conjugate operators satisfying [Q,P] = ih, we can (naively) write:
<br />
\langle q|\,[Q,P]\,|q\rangle ~=~ i\hbar \langla q|I|q\rangle ~=~ i\hbar \langle q|q\rangle<br />
<br />
\frac{1}{i \hbar} \Big( \langle q|QP|q\rangle - \langle q|PQ|q\rangle \Big)<br />
~=~ \langle q|q \rangle ~.<br />
In the first term, let Q act on the bra; in the second, let Q act on the ket:
<br />
\frac{1}{i\hbar}<br />
\Big( q \langle q|P|q\rangle - q \langle q|P|q\rangle \!\!\Big) ~=~ \langle q|q \rangle<br />
~~,~~\Longrightarrow 0 = 1 ~(!)<br />
The resolution of this paradox in standard quantum theory is that the
spectra of both operators is continuous, so the inner products in the above
are meaningless. You can't validly manipulate adjoints as above for this case.
This is a consequence of another theorem that for canonically conjugate
operators as above, at least one of the operators must be unbounded.
(Proof can be found in Reed & Simon.) Another theorem (Hellinger-Toeplitz)
then says this means that Q and P cannot both have the whole Hilbert space
as their domain.
This is (one of) the primary motivations why people move to rigged Hilbert spaces.
(Anyone using the Dirac bra-ket formalism in this context is secretly using
rigged Hilbert spaces whether they realize it or not.)
For a gentle introduction, try Ballentine section 1.4
Also see quant-ph/0502053 by Rafael de la Madrid. It also gives a gentle
introduction, and also covers the example of the 1D rectangular barrier
potential -- which might shed some light more directly relevant to the original
post in this thread.
HTH.
PS: I wish there was a sticky thread about this... :-)
