Derivation of plane wave from inner product of position ket and momentum ket In textbooks it seems to be taken for granted that
$$\langle \mathbf{r}|\mathbf{k}\rangle ~=~ \frac{1}{\sqrt{\Omega}}\exp(i\mathbf{k}\cdot\mathbf{r}).$$
I'm sure it's obvious but is there a mathematical way of showing this to be the case?
 A: Observe that
$$ p \langle x \vert p \rangle = \langle x \vert \hat{p} \vert p \rangle  =-\mathrm{i} \partial_x \langle x \vert p \rangle$$
since, with $\hbar = 1$, $\hat p = -\mathrm{i}\partial_x$ when acting on wavefunctions $\langle x \vert \psi \rangle$. The solution of this differential equation is then the inner product
$$ \langle x \vert p \rangle = A\mathrm{e}^{\mathrm{i}xp}$$
where $A$ is a normalization constant we are free to choose.
A: This is actually a cool proof that doesn't show up in a lot of intro quantum books. It involves infinitesimal translation operators. It's kinda lengthy, but essentially, you can show that infinitesimal translation is a unitary operator, and so can be written as $\hat{T}_{\epsilon}=\hat{1}-i\epsilon\hat{p}$, up to first order in the small translation $\epsilon$ and where $\hat{p}$ is the generator of translation (i.e. momentum). The rest of the proof is just a bit of calculus, but the main assumption is that momentum is in fact the generator of translation, which isn't too big of a leap of faith, since it is the generator of translation in classical mechanics as well. You can also prove the canonical commutation relation $[\hat{x},\hat{p}]=i\hbar$ with this assumption.
