Berry phase and gauge dependence

In summary: I could gauge away the Berry phase completely and in particular there would be no Aharonov-Bohm effect! This effect has been measured, so somehow ##\Omega / \Omega_0## can be chosen to be a non-integer.In summary, the discussion revolves around the definition and interpretation of the Berry phase in quantum mechanics. It is possible to get rid of the integer part of the Berry phase by a gauge transformation, but there are cases where the single-valuedness of the wavefunction demands that the Berry phase equal 2π times an integer. This has implications for the Aharonov-Bohm effect and other topological effects. However, in superconducting rings, the Meissner effect expels magnetic
  • #1
Baisl
4
0
Hi,

I have a question that has been bugging me recently. It's about the berry phase and something that I find contradictory.

One can see that it is possible to get rid of 2π x (integer) part of the Berry phase by means of a gauge transformation. This in general applies to phases (gauge transformation) that one can add to quantum states ##|ψ> → e^{\beta(R)} |ψ> ##, being R a bunch of parameters. Under a closed loop in R-space, single-valuedness of |ψ> demands ##\Delta \beta (R) = 2πn##, with n ∈ ℤ. So the Berry phase is defined up to shifts of ##2πn##.
Now, as far as I understand, in several places in literature, the same single-valuedness argument is used to impose that the Berry phase equals ##2πm## , with m ∈ ℤ !

For instance, in the Aharonov-Bohm example, I can define ##\lambda = -\phi/q ## , with ##\phi## being the azimuthal angle. Such a function is not globally defined, but ##A \rightarrow A + 1/q \nabla \phi## , ##|\psi> →| \psi> e^{i n \phi} ## transform in a single-valued way and hence ##\lambda## is an allowed gauge transformation. The effect of such ##\lambda## is piercing one extra unit of flux ##\Omega_0## (I can do the same process with an integer number of units of ##\Omega_0##). So the Berry phase, which in this case equals ##2\pi \Omega / \Omega_0## ( ##\Omega## is the total flux through the solenoid), is defined up to 2π×(integer). Only the non-integer part of ##\Omega /(2 \pi \Omega_0)## is gauge invariant.
So forthe particle going around a loop enclosing the flux and you get that the wavefunction is now ## e^{2 \pi i \Omega/ \Omega_0} |\psi>## and you can again worry about the single-valuedness of ##| \psi>##. And indeed, in some places they argue that ##\Omega## must be quantized in units of ##\Omega_0## to ensure single-valuedness. But, if ##\Omega / \Omega_0 \in \mathbb{Z}## I could gauge-away the Berry phase completely and in particular there would be no Aharonov-Bohm effect! This effect has been measured, so somehow ##\Omega / \Omega_0## can be chosen to be a non-integer.

I have seen similar arguments (using again single-valuedness) to show that the Berry phase under a periodic-in-time perturbation [Thouless, 1983] should be 2π×(integer). But then the same applies: couldn't I just gauge it away by a gauge tranformation ##|ψ> → e^{\beta(R)} |ψ> ## for a suitable ##\beta(R)##?

What am I missing?

Thank you very much and sorry for the long post!
 
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  • #2
Thinking as a physicist as opposed to a mathematician, I think it is helpful to stress that the single valued wavefunctions will in general be discontinuous. This discontinuity is brought about by the vector potential containing a delta function contribution which stems from the angle jumping back from 2 pi to 0. This delta function is quite ill defined, but you can smear it out and then take the limit of vanishing smearing. For finite smearing, you will find some finite value for the magnetic field on the ring. In the case of the AB effect, this is ok, as an electron can move in a magnetic field. So the AB effect exists for noninteger phases. But there are other cases where flux quantization occurs. Namely for superconducting rings. By the Meissner effect, a superconductor expells magnetic fields completely, so we have no possibility to assume some weak magnetic field on the SC and let it tend to 0. What you can do is to start from a Josephson junction. A non-integer magnetic field inside the junction will lead to rapid oscillations of the current in the junction. I suppose as you close the junction, the changing currents will expell just enough flux to end with an integer amount of flux in the ring.
 
  • #3
Many thanks for your reply DrDu!

Let me first try to understand your point:

DrDu said:
Thinking as a physicist as opposed to a mathematician, I think it is helpful to stress that the single valued wavefunctions will in general be discontinuous. This discontinuity is brought about by the vector potential containing a delta function contribution which stems from the angle jumping back from 2 pi to 0.

I see precisely what you mean up to here. Could you write down a few formulas to show how the finite magnetic field arises from the smearing?
So you're saying that non-integer ##\Omega/\Omega_0## corresponds to having a finite smearing of the jump of the phase?
Is allowing for such discontinuities of the wavefunction completely equivalent to allowing for it to be not single-valued, i.e. is it ok to consider a jump ## \psi \rightarrow \psi e^{i 2 \pi \Omega/\Omega_0 \phi},\ \Omega/\Omega_0 \not\in \mathbb{Z}## after one loop is completed?

DrDu said:
But there are other cases where flux quantization occurs. Namely for superconducting rings. By the Meissner effect, a superconductor expells magnetic fields completely, so we have no possibility to assume some weak magnetic field on the SC and let it tend to 0. What you can do is to start from a Josephson junction. A non-integer magnetic field inside the junction will lead to rapid oscillations of the current in the junction. I suppose as you close the junction, the changing currents will expell just enough flux to end with an integer amount of flux in the ring.

Ok. So if I now compute the berry phase in a superconductor I'll probably get a result of the form ##2 \pi n##, ##n \in \mathbb{Z}##. As shown in my previous post, this amounts to a gauge transformation of the wavefunction. So there will be no physical AB effect in a SC. Correct?
Now, in applications to adiabatic transport, why is it claimed that a Berry phase of the form ##2 \pi n## is physical? Not only that, ##n## is frequently a topological number and gives rise to all kinds of interesting effects. Why isn't that pure gauge?

Thanks again.
 
  • #4
No, the smearing doesn't produce a field in the SC, I was wrong here. The smearing is a good method to make sense out of the sudden jump of the phase (from any phi back to 0), but it does not necessarily lead to a magnetic field. (away from the center). It is easy to show that a vector potential of the form ##A= f(\phi)/\rho## in cylindrical coordinates has vanishing rotation (and hence magnetic field) for ##\rho \neq 0##.
But in fact you can have deviations from flux quantisations in superconductors when the sc is so thin that magnetic fields can penetrate the sc.
With Thouless 1983 you mean:
http://journals.aps.org/prb/abstract/10.1103/PhysRevB.27.6083 ?
 
  • #5
DrDu said:
No, the smearing doesn't produce a field in the SC, I was wrong here. The smearing is a good method to make sense out of the sudden jump of the phase (from any phi back to 0), but it does not necessarily lead to a magnetic field. (away from the center). It is easy to show that a vector potential of the form ##A= f(\phi)/\rho## in cylindrical coordinates has vanishing rotation (and hence magnetic field) for ##\rho \neq 0##.
But in fact you can have deviations from flux quantisations in superconductors when the sc is so thin that magnetic fields can penetrate the sc.
With Thouless 1983 you mean:
http://journals.aps.org/prb/abstract/10.1103/PhysRevB.27.6083 ?

Hi,

I actually took the example from the review https://arxiv.org/pdf/0907.2021v1.pdf (Section II. ADIABATIC TRANSPORT AND ELECTRIC POLARIZATION, pages 8 and 9)

There you can see that they use single-valuedness to show that the Berry phase must be ##2 \pi n, n \in \mathbb{Z} ##. What I find confusing is:

- Couldn't I repeat the argument to show that the Aharonov-Bohm phase must also be of the form ##2 \pi n, n \in \mathbb{Z} ##?
- Couldn't I gauge away the Berry phase, given that it is just ##2 \pi \times \text{(integer)}##?

I think that this quantization condition applies to many examples, such as integer Hall conductivity for instance. Obviously I'm missing something because these effects can be measured, so they cannot be just pure gauge.

Thanks.
 
  • #6
Isn't this periodicity in k space?
 
  • #7
It is not clear to me, but it is possible that indeed you can see the quantization as arising from the winding number of the phase over the cycles of the torus.
Note however that they make explicit use of single-valuedness in (2.12)

But what I don't understand is why is it not possible to gauge away the total Berry phase ##2 \pi c_n##.
 

Related to Berry phase and gauge dependence

1. What is the Berry phase and why is it important in physics?

The Berry phase, also known as the geometric phase, is a phenomenon in quantum mechanics where a quantum system undergoing adiabatic evolution accumulates a phase factor that depends on the path taken in parameter space. It is important in physics because it provides a way to characterize the non-trivial geometry of the parameter space and has applications in many areas of physics, including condensed matter physics and quantum computing.

2. How is the Berry phase related to gauge dependence?

The Berry phase is related to gauge dependence because it is a gauge-invariant quantity. This means that it does not depend on the specific choice of gauge used to describe the system. This is important because different choices of gauge can lead to different physical interpretations, but the Berry phase remains the same regardless.

3. Can the Berry phase be observed experimentally?

Yes, the Berry phase has been observed experimentally in a variety of systems, including solid state systems, optical systems, and superconductors. It can be measured using interferometry techniques or by directly measuring the energy spectrum of the system.

4. How does the presence of degeneracy affect the Berry phase?

The presence of degeneracy, where two or more energy levels have the same energy, can affect the Berry phase by introducing additional non-trivial geometric properties. For example, in the case of two degenerate energy levels, the Berry phase can take on any value between 0 and 2π, whereas for non-degenerate levels it is restricted to 0 or π.

5. Are there any practical applications of the Berry phase?

Yes, the Berry phase has many practical applications. In condensed matter physics, it can be used to study the behavior of electrons in materials and has been applied to the study of topological insulators. In quantum computing, it is used in the design of quantum gates and in error correction protocols. It also has applications in optics, where it can be used to manipulate the polarization of light.

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