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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.
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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