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I Berry connection

  1. Nov 30, 2016 #1
    The Berry connection ##\mathcal{A}_{k}(\lambda)## of a quantum system is given by

    $$\mathcal{A}_{k}(\lambda) \equiv -i \langle n|\frac{\partial}{\partial \lambda^{k}}|n\rangle,$$

    where the ket ##|n(\lambda)\rangle## depends on the parameters ##\lambda^{k}, k=1,2,\dots## in the system.

    The field strength ##\mathcal{F}_{kl}## of the Berry connection ##\mathcal{A}_{k}(\lambda)## is defined by

    $$\mathcal{F}_{kl} = \frac{\partial\mathcal{k}}{\partial\lambda_{l}}-\frac{\partial\mathcal{l}}{\partial\lambda_{k}}.$$

    Therefore, we can define an analog of Maxwell's theory with the Berry connection ##\mathcal{A}_{k}(\lambda)##. As such, we expect the Berry connection ##\mathcal{A}_{k}(\lambda)## to be gauge invariant. In other words, there must be a gauge redundancy in the definition of the Berry connection ##\mathcal{A}_{k}(\lambda)##


    I was wondering if you guys have any idea about the physical meaning of this gauge redundancy for some state ##|n(\lambda)\rangle##.
     
  2. jcsd
  3. Dec 1, 2016 #2
    This basically means that the Berry phase remains invariant under the gauge transformation ##|n(\lambda)\rangle \rightarrow e^{i\phi(\lambda)}|n(\lambda)\rangle## (i.e. addition of a phase factor to the instantaneous eigenstates). This transformation modifies the Berry connection ##\mathcal{A}(\lambda) \rightarrow \mathcal{A}(\lambda) + \nabla \phi##. However, if ##\phi(\lambda)## is single-valued, then the Berry phase does not change.
     
  4. Dec 1, 2016 #3
    Ok, but say you take a particle in a box and rotate the box in parameter space in a closed loop, like you usually do in Berry phase analysis.

    What is the physical source of the gauge transformation of the Berry connection in this case?
     
  5. Dec 1, 2016 #4
    As I understand it, there is no 'physical source' - simply put, it just means that
    1. There is a certain gauge freedom (similar to electrodynamics, in which different gauge choices do not affect the physical quantities observed), and this is nothing more than saying that ##|\psi\rangle## and ##e^{i \theta} |\psi\rangle ## correspond to the same physical state.
    2. The Berry phase fits the characteristics of being a true observable - it is not an artifact arising from our gauge choice.
     
  6. Dec 1, 2016 #5

    A. Neumaier

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    2016 Award

    Gauge invariance means precisely that there are degrees of freedom without any physical relevance, and corresponding gauge transformations that tell how the variables can be changed without changing the physics.
     
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