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Apr 26, 2016 at 15:20 comment added user101311 @EmilioPisanty To confirm what you stated in your answer, which references finite dimensional linear algebra. If you want to express $\Psi$ in terms of the basis of Hamiltonian eigenfunctions i.e. $|n \rangle.$ Then $\Psi(x,t) = \sum_nc_ne^{\frac{-iE_n t}{\hbar}}\psi_n(x)$, does this correspond to $\langle x | n \rangle = c_n e^{\frac{-iE_n t}{\hbar}}$ and $\psi_n(x) = e_j \cdot A$?
Apr 17, 2016 at 20:01 comment added Emilio Pisanty Because it assumes the position-space wavefunction of momentum states, $⟨y|p⟩=e^{ipy/\hbar}/\sqrt{2\pi\hbar}$, for the inner product. (Note also that $⟨x|=\delta(x-y)$ and $|p⟩=e^{ipx/\hbar}/\sqrt{2\pi\hbar}$ are incorrect and inconsistent.) I'm not particularly interested in explaining further - if you still have a question, you can ask it separately.
Apr 17, 2016 at 16:43 comment added user114445 Okay but is the equation $\langle x| p \rangle = \frac{1}{\sqrt{2 \pi \hbar}}\int_{- \infty}^{\infty} \delta(x-y)e^{\frac{ipy}{\hbar}}dy = \frac{1}{\sqrt{2 \pi \hbar} }e^{\frac{ipx}{\hbar}}$ correct? Also, I'm not following why it is a circular argument, I'm simply taking the inner product ($\langle f , g \rangle := \int_{-\infty}^{\infty} f^* g dx$) of the eigenvector $\langle x| = \delta(x-y)$ and $| p \rangle = \frac{1}{\sqrt{2 \pi \hbar} }e^{\frac{ipx}{\hbar}}$.
Apr 15, 2016 at 14:01 comment added Emilio Pisanty @LucioD That's a circular argument. The formula for $⟨x|p⟩$ follows from a number of different things, depending on what you want to think of as fundamental. That's for a separate question, though.
Apr 15, 2016 at 13:13 comment added user114445 @EmilioPisanty Does $\langle x| p \rangle = \frac{1}{\sqrt{2 \pi \hbar}}e^{\frac{ipx}{\hbar}}$ follow from $\langle x|p \rangle = \int^{\infty}_{-\infty}\delta(x-y)e^{\frac{ipy}{\hbar}}dy = \frac{1}{\sqrt{2 \pi \hbar}}e^{\frac{ipx}{\hbar}}$?
Apr 15, 2016 at 12:50 history edited Emilio Pisanty CC BY-SA 3.0
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Apr 14, 2016 at 13:45 vote accept Alex
Apr 14, 2016 at 11:35 history edited AccidentalFourierTransform CC BY-SA 3.0
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Apr 14, 2016 at 11:22 history answered Emilio Pisanty CC BY-SA 3.0