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cscott
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Question
Free particle in 1D where V(x) = 0. There is a general boundary condition \psi(x+L)=e^{i\theta}\psi(x) used for box normalization which has arbitrary phase theta. E=k^2\hbar/(2m) is true for free particle energies.
Attempt
Comparing with the condition \psi(x+L)=\psi(x) I don't see how I will get different energies E since L is still the maximum wavelength, therefore \lambda = L/n = 2\pi/k or k = 2n\pi/L for n = 1, 2, ...; and then energies E_n can be computed.
How do I get theta dependence into the energies for the case \psi(x+L)=e^{i\theta}\psi(x)? 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 \psi_n(x) = L^{-1/2} \exp(i(2\pi n/L)x+i\theta)? ...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 \psi(x+L)=e^{i\theta}\psi(x) used for box normalization which has arbitrary phase theta. E=k^2\hbar/(2m) is true for free particle energies.
Attempt
Comparing with the condition \psi(x+L)=\psi(x) I don't see how I will get different energies E since L is still the maximum wavelength, therefore \lambda = L/n = 2\pi/k or k = 2n\pi/L for n = 1, 2, ...; and then energies E_n can be computed.
How do I get theta dependence into the energies for the case \psi(x+L)=e^{i\theta}\psi(x)? 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 \psi_n(x) = L^{-1/2} \exp(i(2\pi n/L)x+i\theta)? ...But I'm not sure that k is correct.
Can anyone clear this up for me? Much thanks.
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