# Now what is Aharonov-Bohm really?

1. May 21, 2012

### nonequilibrium

I can only seem to find vague sources explaining Aharonov-Bohm, usually saying things as "the different paths of an electron interfere". I presume this is language borrowed from a Feynman path integral formulation of QM, but I'm not familiar with that yet, so I'd rather see it explained in "ordinary" QM math.

Some sources seem to suggest that the basic consequence of a vector potential is that the phase $S$ of the wavefunction $\psi(\mathbf r,t) = R(\mathbf r,t) e^{iS(\mathbf r,t)}$ gets an extra term, namely the path integral $\int_{\mathbf r_0}^{\mathbf r} \mathbf A(\mathbf r') \cdot \mathrm d \mathbf r'$ (where $\mathbf r_0$ is some arbitrary reference point). Then again, this can't really be true, cause then phase wouldn't be well-defined (since a different path, but also going from r_0 to r, could give a different result). I realize that this last remark is also the key concept in the AB effect, but still, not in the aforementioned way, right? After all, phase shouldn't be ambiguous (except for a 2pi multiple of course). Actually, it's exactly this argument (that different path integrals should give the same phase mod 2pi) that is used to argue flux quantization in a superconducting ring... So why does it not apply more generally?

As you can see, I'm a bit confused. Note that I'm not looking for a vague explanation, I'm looking for something concrete (and well-defined) mathematically. Thank you!

2. May 21, 2012

### The_Duck

One concrete thing you can do easily in the regular Schrodinger wave equation formalism is solve for the energy eigenstates of a charged particle confined to a circular 1D ring through the center of which a solenoid passes carrying some amount of magnetic flux. There is no B-field at the ring's radius, but there is an A-field. Write down the Schrodinger equation including the A-field and solve it. Once you figure out what 1D Schrodinger equation to write down, you should find that the solutions are almost as trivial as free-particle solutions. You should find that the energy eigenstates of the particle are shifted by the presence of the magnetic field, even though the particle cannot enter the B-field. Furthermore (as I recall) the clockwise- and counter-clockwise-moving energy eigenstates cease to be degenerate. This is one manifestation of the Aharonov-Bohm effect.

Not sure if this is helpful, but it is concrete and mathematically well-defined.

3. May 21, 2012

### strangerep

As with most aspects of "ordinary QM", Ballentine is a good place to start.
Try section 11.4.

This doesn't mean the phase is ill-defined. It just means the phase can be position dependent -- which is also a key concept in gauge theories of interaction.

Take a look at Ballentine and then see if any points remain unclear...

4. May 23, 2012

### Staff: Mentor

The setup is similar to a double-slit with photons, where you can add phase shifts at the slits (e.g. with material with a different refractive index). Similar to this, you have to constrain the electron path in order to get a phase which is nearly the same for all electrons in one path.

5. May 24, 2012

### DrDu

I also back up the approach proposed by The Duck.
You will find that the energy states depend on the magnetic flux through the loop which is an observable in contrast to A.
Nevertheless this is astonishing given that the electrons don't enter the region where B is non-vanishing.