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A Berry phase and gauge dependence

  1. Oct 13, 2016 #1
    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!
     
    Last edited: Oct 13, 2016
  2. jcsd
  3. Oct 14, 2016 #2

    DrDu

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    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.
     
  4. Oct 14, 2016 #3
    Many thanks for your reply DrDu!

    Let me first try to understand your point:

    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?

    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.
     
  5. Oct 17, 2016 #4

    DrDu

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    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 ?
     
  6. Oct 18, 2016 #5
    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.
     
  7. Oct 18, 2016 #6

    DrDu

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    Isn't this periodicity in k space?
     
  8. Oct 18, 2016 #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##.
     
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