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So the free particle wave functions are of the type:
ψ(x) = Aexp(ikx) + Bexp(-ikx) (1)
In a problem I am doing I am supposed to find the energy levels for a particle which is sliding on a frictionless ring and the exercise says that to do so I should use the fact that
ψ(x+L)=ψ(x) (2)
BUT! Since the phase of the wave-function in QM carries no physical significance, shouldn't the most general treatment add a phase to (1) such that:
ψ(x+L)=exp(iα)ψ(x) (3) , where α is an arbitrary real number.
Unfortunately when I do so and solve for the energy levels of the system I don't get the same result as when I use (1). α modifies the energy levels which of course shouldn't happen if the phase carries no physical significance. So is (3) a wrong assumption and if so, why? Here's what I did btw, maybe I made an error somewhere. Using (3) on (1):
Aexp(ikx)exp(ikL) + Bexp(-ikx)exp(-ikL) = exp(iα)(Aexp(ikx) + Bexp(-ikx))
This holds particularly for x=0 and x=k/2π yielding:
Aexp(ikL) + Bexp(-ikL) = exp(iα)(A+B)
Aexp(ikL) - Bexp(-ikL) = exp(iα)(A-B)
Which gives:
2Aexp(ikL) = 2Aexp(iα)
So either A=0 or kL-α = 2πn
But the appearance of α in the last equation alters the possible values of k and hence the possible energy values for the system.
ψ(x) = Aexp(ikx) + Bexp(-ikx) (1)
In a problem I am doing I am supposed to find the energy levels for a particle which is sliding on a frictionless ring and the exercise says that to do so I should use the fact that
ψ(x+L)=ψ(x) (2)
BUT! Since the phase of the wave-function in QM carries no physical significance, shouldn't the most general treatment add a phase to (1) such that:
ψ(x+L)=exp(iα)ψ(x) (3) , where α is an arbitrary real number.
Unfortunately when I do so and solve for the energy levels of the system I don't get the same result as when I use (1). α modifies the energy levels which of course shouldn't happen if the phase carries no physical significance. So is (3) a wrong assumption and if so, why? Here's what I did btw, maybe I made an error somewhere. Using (3) on (1):
Aexp(ikx)exp(ikL) + Bexp(-ikx)exp(-ikL) = exp(iα)(Aexp(ikx) + Bexp(-ikx))
This holds particularly for x=0 and x=k/2π yielding:
Aexp(ikL) + Bexp(-ikL) = exp(iα)(A+B)
Aexp(ikL) - Bexp(-ikL) = exp(iα)(A-B)
Which gives:
2Aexp(ikL) = 2Aexp(iα)
So either A=0 or kL-α = 2πn
But the appearance of α in the last equation alters the possible values of k and hence the possible energy values for the system.