In my quantum mechanics textbook it says that the relation between the basis $|x\rangle$ and $|p\rangle$ is given by:
$\langle p | x \rangle = \Large \frac{e^{-ip x/ \hbar}}{\sqrt{2\pi \hbar}} \, .$
However, i'm not sure how to go about proving this relation. My idea was that the eigenstates of momentum can be written as below:
$\phi(x,p) = \Large \frac{1}{\sqrt{2\pi \hbar}} e^{-ipx/\hbar},$
which form an orthogonal basis.
But, we can expand any state, $|\text{state}\rangle$, in terms of the othorgonal basis right? So is my notation below correct for the expansion of the state $| x \rangle$ ?
$|x\rangle = \int_{-\infty}^{+\infty} \phi(p', x) |p'\rangle dp'$
If it were correct then the following would also be correct?
Using the fact that $\langle p | p' \rangle = \delta(p - p')$, $\langle p | x \rangle$ can be computed:
\begin{align} \langle p | x \rangle =& \int_{-\infty}^{+\infty} \phi(p', x) \langle p | p' \rangle dp' \\ =& \int_{-\infty}^{+\infty} \phi(p', x) \delta(p - p') dp' \\ =& \phi(p, x) \\ =& \Large \frac{e^{-ip x/ \hbar}}{\sqrt{2\pi \hbar}} \end{align} as required.
Does the logic work here?
