Proof for $\langle q| p \rangle = e^{iqp}$

Question

What would be the proof for $\langle q| p \rangle = e^{iqp}$?

Is it derived from canonical commutation relation?

($|q \rangle $ represents the position eigenstate, while $|p \rangle$ represents the momentum eigenstate.)

Answer 1

From the canonical commutation relation in 1 dimension

$$[q,p] = i$$

you can choose an eigenbasis for q. Then note that any function can be expanded in terms of these eigenstates:

$$ |\psi\rangle = \int f(q) |q\rangle $$

Then consider the linear operator D which differentiates f:

$$ D |\psi\rangle = \int f'(q) |q\rangle $$

and note that

$$ D q - q D = 1 $$

So that -iD obeys the canonical commutation relations. This means that

$$ [ p + iD , q ] = 0 $$

so that if $p+iD$ is nonzero, it is a function of q, say

$$ p + i D = G(q) $$

Then a position dependent phase rotation of the wavefunction gets rid of G. If you redefine the eigenstates of position as follows:

$$ |q\rangle = e^{iH(x)}|q\rangle $$

Then the derivative operator gets an additional term

$$ iD \rightarrow iD + H'$$

and by choosing H to be an indefinite integral of G, you get rid of G, and

$$ p = -iD$$

The eigenstates of D are found by solving the differential equation

$$ - iD |\psi\rangle = k |\psi\rangle $$

and in terms of the expansion in eigenstates

$$ - i f'(q) = k $$

or

$$ f(q) = e^{ikq}$$

This is the overlap $\langle p | q \rangle $. This argument is directly lifted from Dirac's quantum mechanics book.

The rigorous version of this argument is due to Stone and Von Neumann, and it proceeds by rewriting the canonical commutation relations in exponetiated form:

$$ e^{iPx} q e^{-iPx} = q + x $$

In this form, you can see that, up to the arbitary choice of phase, the p operator is the generator of translations. I find the formal argument unilluminating, since it is essentially a repeat of Dirac.

What is interesting is Feynman's observation that this was essentially a 1d argument. If you have canonical commutation relations in multiple dimensions

$$ [p_i,q_j] = \delta_{ij} $$

Then you cannot just use this relation to figure out that the p's can be turned into differential operators. You also need the relation

$$ [p_i , p_j] = 0 $$

Without imposing the second condition, you cannot guarantee that the phases you get when you translate around an infinitesimal box is zero. If you drop the latter condition, you end up introducing a magnetic field vector potential to describe the failure of p commutativity, and this allows you to "derive Maxwell's equations from quantum mechanics". Feynman phrased it this way, but I think this is a poor phrasing. I think Feynman knew it was a poor phrasing, and this is why he did not publish, but people wanted to know the argument, so Dyson wrote it up in the American Journal of Physics in the mid 90s.

Answer 2

$$ \langle{q}|\hat{p}|\alpha\rangle = \frac{\hbar}{i}\frac{\partial}{\partial q}\langle q|\alpha \rangle $$ So, $$\langle{q}|\hat{p}|p\rangle = \frac{\hbar}{i}\frac{\partial}{\partial q}\langle q|p\rangle $$ This is equivalent to $$p\langle{q}|p\rangle = \frac{\hbar}{i}\frac{\partial}{\partial q}\langle q|p\rangle $$ then its just a first order linear differential equation, so $$\langle q|p\rangle = Ne^{\frac{i p q}{\hbar}}$$ where $N$ is some normalisation factor depends on the dimension of $q$.

Ref:
J.J. Sakurai, 1994, 'Modern Quantum Mechanic'

Answer 3

You can solve that by a simple comparison, You will need the following fact: $$ \int dp \vert p \rangle \langle p \vert =1$$ which is always true since $\vert p \rangle$ are nice and orthonormal basis $\langle p_i\vert p_j \rangle=\delta(p_i-p_j)$, then we can use this basis to express any function we need. $$ \vert \psi \rangle = \int dp \vert p \rangle \langle p \vert \psi \rangle $$ where $$ \langle p \vert \psi \rangle = \psi(p)$$ Then we project with a bra describing a position vector basis $\langle q \vert $, we get: $$ \langle q \vert \psi \rangle = \int dp \langle q\vert p \rangle \psi(p) $$ Which means: $$ \psi(q) = \int dp \langle q\vert p \rangle \psi(p) $$ We compare this result with the Fourier transform of $\psi(q)$, which is written as follows: $$ \psi(q) = \frac{1}{(\sqrt{2})^3}\int dp e^{ipq} \psi(p) $$ Thus: $$\langle q\vert p \rangle=\frac{1}{(\sqrt{2})^3}e^{ipq}$$