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Box Normalization

  • Thread starter cscott
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Question

Free particle in 1D where V(x) = 0. There is a general boundary condition [tex]\psi(x+L)=e^{i\theta}\psi(x)[/tex] used for box normalization which has arbitrary phase theta. [tex]E=k^2\hbar/(2m)[/tex] is true for free particle energies.

Attempt

Comparing with the condition [tex]\psi(x+L)=\psi(x)[/tex] I don't see how I will get different energies E since L is still the maximum wavelength, therefore [tex]\lambda = L/n = 2\pi/k[/tex] or [tex]k = 2n\pi/L[/tex] for n = 1, 2, ...; and then energies [tex]E_n[/tex] can be computed.

How do I get theta dependence into the energies for the case [tex]\psi(x+L)=e^{i\theta}\psi(x)[/tex]? Or maybe the better question is do I need theta dependence in the energies for a correct solution? Shouldn't the phase of a wave function have no physical significance?

Given the k above is true then my normalized eigenfunctions would be [tex]\psi_n(x) = L^{-1/2} \exp(i(2\pi n/L)x+i\theta)[/tex]? ...But I'm not sure that k is correct.

Can anyone clear this up for me? Much thanks.
 
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Answers and Replies

  • #2
Hao
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The phase is arbitrary, but that does not mean that you can set it to zero when applying the boundary condition.

It may help to consider the physical meaning of the boundary condition above, which occurs when you have translational symmetry in crystal lattices (Bloch waves).

http://www.iue.tuwien.ac.at/phd/smirnov/node41.html
 

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