# Implications of an arbitrary phase for momentum operator

1. Sep 16, 2014

### spaghetti3451

In quantum mechanics, the phase of the wavefunction for a physical system is unobservable. Therefore, both ψ = ψ(x) and ψ' = ψ(x)e are valid wavefunctions.

For ψ = ψ(x), we have the following:

$\widehat{x}ψ = xψ$
$\widehat{p}ψ = λψ$

For ψ' = ψ(x)e, we have the following:

$\widehat{x}ψ' = xψ'$
$\widehat{p}ψ' = \left(λ+ħ\frac{∂θ}{∂x}\right)ψ'$

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 than it is for ψ = ψ(x).

Am I correct?

2. Sep 16, 2014

### ShayanJ

You shouldn't say they're both valid, you should say they're equivalent.
Good observation and correct conclusion.
If you want the Schrodinger equation to be invariant under the transformation you mentioned, you should have $\hat p \rightarrow \hat p-q\hat A$ with a special transformation law for $\hat A$ so that the extra term it gives, cancels the term you derived in your post. In such a situation, its not $\hat p$ that has a physical meaning, its $\hat p-q\hat A$. It then turns out that $\hat A$ is just the magnetic vector potential and so for the interaction of charged particles with a magnetic field, you have the change $\hat p \rightarrow \hat p-q\hat A$ in the Schrodinger equation.

3. Sep 16, 2014

### Staff: Mentor

States are really positive operators of unit trace. In this view the states you are talking about, called pure, are naturally gauge invariant because by definition they are positive operators of the form |u><u|.

When viewed that way all these issues disappear.

Local guage invariance however has interesting consequences:
http://quantummechanics.ucsd.edu/ph130a/130_notes/node296.html

Thanks
Bill

Last edited: Sep 16, 2014
4. Sep 16, 2014

### spaghetti3451

Thanks for the help!

I have a couple of additional questions, if you don't mind.

1. Our lecturer uses $\hat p \rightarrow \hat p+q\hat A$. Does it make a difference?
2. He calls this new definition of the momentum operator as 'minimal coupling procedure.' What exactly is this supposed to mean?
3. Why should it be that the unobservable nature of the phase be so intimately connected to the dependence of the momentum operator on magnetic vector potential? In other words, why could the quantity $q\hat A$ have not had any physical interpretation other than the electromagnetic interpretation?

5. Sep 16, 2014

### Staff: Mentor

Not really - it just affects the interpretation of the sign of q.

It means the interaction only involves the charge distribution and not higher powers.

Because a U(1) guage invariant 4 vector is required to have local quage invariance.

And that more or less defines EM.

If you can get a hold of the following paper it explains it:
http://inspirehep.net/record/157843?ln=en

But exactly how much it more or less does it is debated:

Thanks
Bill

Last edited: Sep 16, 2014
6. Sep 16, 2014

### DrDu

I use this transformation to show that the wavefunction may also be discontinuous. Just set theta equal to a step function times pi.

7. Sep 16, 2014

### atyy

I think a U(1) gauge redundancy could represent something else. I can't think of a physical example now in the U(1) case, but the equation for massless Dirac fermions is realized in graphene. If there were a physical system in which a massless U(1) gauge field emerged, it would have the same formal behaviour as the EM field, but it would not be the EM field. A proposal to experimentally realise a U(1) gauge field in 2+1D is http://arxiv.org/abs/1208.4299.

8. Sep 16, 2014

### DrDu

An observable consequence is the Aharonov Bohm effect for a particle on a ring. The magnetic vector potential due to a flux through the center of the ring exactly induces this kind of change of the momentum operator (well, rather L instead of p, i.e. derivation with respect to phi instead of x).