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?


marked as duplicate by DanielSank, Qmechanic quantum-mechanics Jan 14 '16 at 19:05

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/178168/2451, physics.stackexchange.com/q/41880/2451 and links therein. $\endgroup$ – Qmechanic Jan 14 '16 at 18:52
  • $\begingroup$ I've always thought of them as two different notations for the same thing: the amplitude of finding the particle a $x$ given $p$, (or vice versa). $\endgroup$ – garyp Jan 14 '16 at 18:55
  • 1
    $\begingroup$ Possible duplicate of proof for $\langle q| p \rangle = e^{ipq}$ $\endgroup$ – DanielSank Jan 14 '16 at 19:00
  • $\begingroup$ Note that you do have broken logic. You have posited $\phi(x,p)=e^{ixp}/\sqrt{2\pi\hbar}$, and it's perfectly reasonable to expand $$|x⟩=\int \chi(x,p')|p'⟩\mathrm dp',$$ but then there is nothing to guarantee that $\phi=\chi$. For more details, see the linked duplicates. $\endgroup$ – Emilio Pisanty Jan 27 '16 at 10:48