Is my expansion of the state $| x \rangle$ correct? [duplicate]

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 14 '16 at 19:05

• 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). – garyp Jan 14 '16 at 18:55
• Possible duplicate of proof for $\langle q| p \rangle = e^{ipq}$ – DanielSank Jan 14 '16 at 19:00
• 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. – Emilio Pisanty Jan 27 '16 at 10:48