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Consider the position and momentum operators $\hat q$ and $\hat p$, defined respectively in terms of creation and destruction operators in the usual way: $$ \hat q = c (\hat a + \hat a^\dagger), \qquad \hat p = \frac{i\hbar}{2c} ( \hat a^\dagger - \hat a), \tag 1 $$ for some constant $c$, where $\hat a$ and $\hat a^\dagger$ satisfy the canonical commutation relations: $$ [\hat a, \hat a^\dagger] = 1, \qquad [\hat a,\hat a] = [\hat a^\dagger, \hat a^\dagger] = 0. \tag 2$$ If follows immediately that, as expected, $[\hat q, \hat p] = i \hbar$.

My question is now: starting from these definitions of $\hat p$ and $\hat q$, how can I derive the common relation $\hat p = -i \hbar \frac{\partial}{\partial q}$, or some equivalent version of this relation? In other words, how can I compute the value of $\langle q | \hat p| \psi \rangle$, or equivalently of $\langle q| \hat p | q' \rangle$?

If I try to simply compute this using the definition of $\hat p$ given in (1) I get nonsensical results, I'm gessing because of $|q \rangle$ being not normalized eigenstates. What method should then be used in this context?

Qmechanic
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glS
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  • ...I might need to get used to this gold badge. If the marked question doesn't answer your question, ping me and I will reopen the question. – ACuriousMind Jun 13 '16 at 20:16
  • Once you arrive at the commutation relations for q and p, the question is a duplicate of http://physics.stackexchange.com/q/45248/. Maybe there is a more direct way. – jinawee Jun 13 '16 at 20:18
  • @ACuriousMind why do you say it's a duplicate of that question? There they basically start from $\hat p =-i \hbar \partial_q$, while my question is how to get there starting from the definition of $p, q$ in the context of a quantum harmonic oscillator – glS Jun 13 '16 at 20:18
  • @jinawee I see. Then we are basically saying that given the CCR the only possibility is that $p=-i\hbar \partial_q$ (up to a function added to this). But then I wonder, is there no way to reach this result that does not involve knowing the solution in advance or employing highly non-trivial math? – glS Jun 13 '16 at 20:24
  • Hm, yeah, you're right, the one jinawee linked is a better fit. The rigorous statement of "the momentum operator must be the derivative in the position representations" is the Stone-von Neumann theorem, I don't think you'll get to that without some sort of "highly non-trivial math". Quantum mechanics is mathematically non-trivial ;) – ACuriousMind Jun 13 '16 at 20:27
  • @ACuriousMind true. But still there are often hand-wavy, more physical arguments that can be made to more simply get results. For example, I wonder if in the present case one could consider an harmonic oscillator constrained in a finite space, which should make the eigenvalues of $q$ discrete and hence maybe remove some difficulties? – glS Jun 13 '16 at 20:32

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