- #1
spaghetti3451
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- 33
In quantum mechanics, the phase of the wavefunction for a physical system is unobservable. Therefore, both ψ = ψ(x) and ψ' = ψ(x)eiθ are valid wavefunctions.
For ψ = ψ(x), we have the following:
[itex]\widehat{x}ψ = xψ[/itex]
[itex]\widehat{p}ψ = λψ[/itex]
For ψ' = ψ(x)eiθ, we have the following:
[itex]\widehat{x}ψ' = xψ'[/itex]
[itex]\widehat{p}ψ' = \left(λ+ħ\frac{∂θ}{∂x}\right)ψ'[/itex]
I think this is a problem because the same physical system cannot have two different eigenvalues upon a single measurement. This suggests that the momentum operator is different for ψ' = ψ(x)eiθ than it is for ψ = ψ(x).
Am I correct?
For ψ = ψ(x), we have the following:
[itex]\widehat{x}ψ = xψ[/itex]
[itex]\widehat{p}ψ = λψ[/itex]
For ψ' = ψ(x)eiθ, we have the following:
[itex]\widehat{x}ψ' = xψ'[/itex]
[itex]\widehat{p}ψ' = \left(λ+ħ\frac{∂θ}{∂x}\right)ψ'[/itex]
I think this is a problem because the same physical system cannot have two different eigenvalues upon a single measurement. This suggests that the momentum operator is different for ψ' = ψ(x)eiθ than it is for ψ = ψ(x).
Am I correct?