In quantum mechanics, the phase of the wavefunction for a physical system is unobservable. Therefore, both ψ = ψ(x) and ψ' = ψ(x)e(adsbygoogle = window.adsbygoogle || []).push({}); ^{iθ}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)e^{iθ}, 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)e^{iθ}than it is for ψ = ψ(x).

Am I correct?

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# Implications of an arbitrary phase for momentum operator

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