quantum mechanics - What does the Canonical Commutation Relation (CCR) tell me about the overlap between Position and Momentum bases?

I'm curious whether I can find the overlap $\langle q | p \rangle$ knowing only the following:

1. $|q\rangle$ is an eigenvector of an operator $Q$ with eigenvalue $q$.


2. $|p\rangle$ is an eigenvector of an operator $P$ with eigenvalue $p$.


3. $Q$ and $P$ are Hermitian.


4. $[Q,P] = i \hbar$.

I'm asking because books and references I've looked in tend to assume that $Q$ is a differential operator when viewed in the $P$-basis, then show that it satisfies the commutation relation. (I haven't read one yet that proves that the form given for $Q$ is the only possible one.) I've heard, though, that we can work purely from the hypothesis of the commutation relation and prove the properties of $Q$ and $P$ from it.

Answer

Those things are surely not enough to find the inner product $\langle q|p\rangle$ uniquely. For example, starting with the conventional $Q,P$, you may redefine them by a canonical transformation, for example by $$ Q\to Q'=Q, \quad P\to P'= P + Q^3 $$ Then $P', Q'$ obey all the four conditions in the same way as $P,Q$. They also have eigenstates and $|p'\rangle$ states are something else than $|p\rangle$. In fact, eigenstates with a large eigenvalue $P'$ are close to $Q$ eigenstates because $Q^3$ easily dominates. The inner product – which is nothing else than the wave function of the $P'$ eigenstate written in the $Q$ representation – will end up being different. It will be a complicated solution of the equation for $\psi$ saying that it's a $P+Q^3$ eigenstate.

The conditions you wrote can only tell you that $$\langle q | PQ | p \rangle = \langle q | QP | p \rangle - i\hbar \langle q | p \rangle = (qp - i\hbar )\langle q | p \rangle $$ so they're only enough to say how $Q$ acts on the $p$ eigenstates and vice versa, in this combination. But the inner product itself isn't related to itself in any sense, so it can't be determined.

Let me mention that even if you imposed additional conditions that would say that $P$ and $Q$ are physically what they should be – e.g. that they scale properly under the scaling of $Q,P$ – the inner product would still be undetermined because at least the phase of the $|p\rangle$ and $|q\rangle$ eigenstates is arbitrary. In fact, even the "absolute value part of the" normalization is a matter of conventions that could be modified.

More generally, and I would say that this point is often overlooked, many properties of "wave functions" similar to your inner products are intermediate, convention- and representation-dependent quantities that don't have a direct physical impact (i.e. direct link with observable quantities). It's really the properties of observables such as the four conditions that you described that may be considered objective facts.