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In Griffiths quantum mechanics, There is a problem that

"Find the momentum-space wave function $\varphi(p,t)$ for the $n$th stationary state of the infinite square well."

The $n$th stationary state has only one energy value: $E= n^2h^2/(8mL^2)$. First, I thought that the momentum $p$ can only be $\pm \sqrt{2mE}$ because of $E=\frac{p^2}{2m}$. Second, I use de Broglie wavelength. In $n$th stationary state, $\lambda=2L/n$ and $p=h/\lambda$. So I thought $p$ only can be $\pm nh/2L$, and those are same with $\pm\sqrt{2mE}$.

But, It was wrong, If I use Fourier transforms to find the momentum-space wave function, $p$ can be every real numbers because $n$th stationary state is not eigenfunction of momentum! Why is it possible? I couldn't find physical reason of that. What is my fault in my first and second opinion?

In summary : When we measure energy, the possible number of the momentum is determined to be one(or two because of plus-minus), and when we measure the momentum, the possible number of the energy is infinite? What is my error?

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The key point is the boundary condition.

  1. The momentum you got is the eigenvalue of momentum operator with the boundary condition that contains the information of potential that the particle experiences. This is a good quantum number.

  2. When you do Fourier transformation using $e^{ipx}$ as your basis which corresponding to a free particle, the $p$ you have in $\psi(p,t)$ is the eigenvalue of the momentum operator without any constraints. This is not a good quantum number. So, simply speaking, what you are doing is expanding an eigenstate in a 'bad' basis $e^{ipx}$, and of course, all real $p$ will come into your expression.

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A particle in a potential, whatever it is, does not have a certain value of momentum. Mathematically the momentum operator does not commute with the potential, and physically the potential gradient is a force that changes the momentum. A Fourier transform of the Hamiltonian eigenstate is possible and it gives you a wave function different from a plane wave, it is natural.

The mean value of momentum is zero, but the mean of the square value is not zero.

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  • $\begingroup$ Thank you for your answer. You mean that in x=0 or L, V(x) changes 0 to ∞ so the potential gradient can change the momentum? Then, is de Broglie wavelength method wrong? $\endgroup$ – sy youn Aug 25 '18 at 17:36
  • $\begingroup$ I have one more question. The standard deviation of momentum in n th stationary state is nh/2L, and it is same result that momentum is only ±nh/2L. Does it have some physical meanings? $\endgroup$ – sy youn Aug 25 '18 at 17:50
  • $\begingroup$ Yes, the physical meaning is simple: the stationary (standing) wave is a superposition of two running waves with the opposite propagation directions. $\endgroup$ – Vladimir Kalitvianski Aug 26 '18 at 4:49

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