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I’m taking this for granted and using it to show that the wave function in momentum space is the Fourier transform of $x$ space. But suddenly I don’t why this stands.

$$\langle p |x \rangle =e^{-ipx}/\sqrt{2\pi}.$$

Qmechanic
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RicknJerry
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  • Momentum eigenstates are plane waves. – Superfast Jellyfish Feb 23 '20 at 18:42
  • @FellowTraveller i know that, $\hat{p} \psi(x)=p \psi(x)$ and the solution is plane wave. But and then...? Do you mean that here \psi(x) =\langle x|p\rangle$ ? – RicknJerry Feb 23 '20 at 18:49
  • Possible duplicates: https://physics.stackexchange.com/q/41880/2451 and links therein. – Qmechanic Feb 23 '20 at 18:51
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    If $|p\rangle$ is an eigenstate of momentum then the wavefunction which is defined as $\langle x|p\rangle$ is nothing but plane waves. – Superfast Jellyfish Feb 23 '20 at 18:55
  • It;s just how Dirac notation works: For any state $|\psi \rangle$ the function $\psi(x)=\langle x|\psi\rangle $ is the wavefunction in the $x$ basis. Similarly $\langle p|\psi\rangle$ is the wavefunction in the $p$ basis. We have that $\langle x|\hat p|\psi\rangle = -i\partial_x \langle x|\psi \rangle$ and $\langle p|\hat p|\psi\rangle = p\langle p|\psi\rangle$ . – mike stone Feb 23 '20 at 19:42

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