- #1

- 1,170

- 3

## Main Question or Discussion Point

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.