In many textbooks it is stated that, in position space, the wave function of a particle with definite momentum $p$ is given by $e^{ipx/\hbar}$. I know that the $\hbar$ comes from the de Broglie hypothesis, but where does the rest of this wave function come from? How do we know it is of this form?
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1$\begingroup$ Well, $\lambda = h/p= 2\pi\hbar/p$ is de-Broglie. After that, you're just writing down the general expression for a plane wave with that wavelength. So the reason that we write that wave function down is that it's essentially the same as de-Broglie's hypothesis, and in the end (after a tortuous path) it matches experimental realities. $\endgroup$– marchCommented Feb 2, 2023 at 6:23
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1$\begingroup$ Also, it is the solution to Schrodinger's equation and, actually, also other more general relativistic wave equations. $\endgroup$– RhinoCommented Feb 2, 2023 at 7:00
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$\begingroup$ In which context? In (non-relativistic) QM? $\endgroup$– Tobias FünkeCommented Feb 2, 2023 at 7:45
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1$\begingroup$ Well, if you accept that $[X,P]=i \hbar I$, then you can derive that $\langle x|p\rangle =: \psi_p(x) =\ldots$... It is an easy exercise $\endgroup$– Tobias FünkeCommented Feb 2, 2023 at 8:06
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1$\begingroup$ To expand my previous comment: With $X$ as a multiplication and $P$ as a derivative (up to constants) operator. Essentially, this boils down to the fact that states with definite momentum are eigenfunctions of the momentum operator (by definition). $\endgroup$– Tobias FünkeCommented Feb 2, 2023 at 8:16
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1 Answer
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The eigenfunctions of the Schrodinger equation for a free particle are the infinite plane waves:
$$ \psi(x,t) = \exp(i(kx + \omega t)) \tag{1} $$
where for a plane wave $k$ is the wave vector, $k = 2\pi/\lambda$, and $\omega$ is the angular frequency, $\omega = 2\pi f$. The de Broglie hypothesis is that $p = h/\lambda = \hbar k$, and we also have the energy relationship $E = hf = \hbar\omega$. We can use these to substitute for $k$ and $\omega$ to get:
$$ \psi(x,t) = \exp\left(i\left(\tfrac{p}{\hbar}x + \tfrac{E}{\hbar} t\right)\right) \tag{2} $$
The equation you cite is the spatial part of equation (2).
answered Feb 2, 2023 at 7:37
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$\begingroup$ This plane wave is an eigenfunction of the momentum operator. $\endgroup$ Commented Feb 2, 2023 at 7:58