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Why there is mod 2pi in Berry phase?

  1. Jul 15, 2015 #1
    Some books say that there is a gauge transform that we can put an extra phase e^{i \phi ( R(t))} to the wave function.
    Since R(t=0) = R(t=T), difference in \phi = 2 pi n, where n is any integers.
    As gauge transform would lead to 2 pi n difference, berry phase is determined up to 2pi n.

    However, when we calculate berry phase for bloch bands,
    it gives 2 pi n, and this is not determined up to 2 pi n (otherwise chern number is always zero).

    Would anybody solve my problem? Many thanks
     
  2. jcsd
  3. Jul 15, 2015 #2

    vanhees71

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    What system are you referring to? In the generality it's not clear, what the problem is. Quantum theory is invariant under arbitrary unitary transformations. If the unitary transformations are time dependent one says that you change from one picture of time evolution to another. The choice of the picture, by construction, doesn't matter for the physical observable quantities predicted by QT.
     
  4. Jul 15, 2015 #3
    I'm not sure to understand the question, but I think that the answers is that "berry phase" and "berry phase+2pi*n" are indistinguishable since the complex exponential is periodic.
     
  5. Jul 15, 2015 #4
    Thank you for the comments. I would like to clarify my question.

    Referring to
    https://en.wikipedia.org/wiki/Berry_connection_and_curvature,
    the geometrical phase, or the original berry phase, we can apply gauge transform
    [tex] |\psi'({\bf{R}})\rangle=e^{i\beta({\bf{R}})} |\psi({\bf{R}})\rangle,[/tex]
    with [tex]\beta({\bf{R}}(t=0))=\beta({\bf{R}}(t=T))+2n\pi,[/tex]
    where R(t=0)=R(t=T), is referring to the same point geometrically.
    This leads to the berry phase is determined up to 2pi.

    The quantum hall conductance or berry phase of bloch bands have the same integral form, but the parameter is k instead of R.
    For a 2 dimensional periodic system, berry phase for bloch band is in the form
    [tex] \gamma = \oint \langle u_{\bf{k}}| d u_{\bf{k}} \rangle.[/tex]
    The above gives multiple of 2 pi , according to first chern number.
    (ANNALS OF PHYSICS 160, 343-354 (1985))

    Corresponding gauge transform is
    [tex] | u'_{\bf{k}} \rangle= e^{i\beta(k_1,k_2)}| u_{\bf{k}} \rangle.[/tex]
    If I parametrize the path by λ, after a closed loop, it is
    [tex]\beta({\bf{k}}(\lambda=0))=\beta({\bf{k}}(\lambda=1))[/tex]
    but cannot
    [tex]\beta({\bf{k}}(\lambda=0))=\beta({\bf{k}}(\lambda=1))+ 2 n \pi[/tex]
    So in this case the phase is exact without determined up to 2 pi.

    May I know why the gauge is the same but cannot differ by 2 n pi for the bloch band berry phase?
     
    Last edited: Jul 16, 2015
  6. Jul 16, 2015 #5

    vanhees71

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    I hope in the beginning you meant the wave function, not a ket. In the latter case, I don't understand the notation ##|\psi(\mathbf{R}) \rangle##.
     
  7. Jul 18, 2015 #6

    DrDu

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    The meaning is that the ket psi depends parametrically on the classical external parameter R.
    For example, the wavefunctions for the electrons in molecules in the Born Oppenheimer approximation depends not only on the coordinates r of the electrons but also parametrically on the coordinates R of the nuclei.
     
    Last edited: Jul 18, 2015
  8. Jul 18, 2015 #7

    DrDu

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    A gauge transformation is always a unique function of the coordinates and also any phase factor is defined only up to 2n pi. However the real point is that only for special choices of n and gauge, the phase will be a ##\bf continuous## function of the parameter R. If you chose a discontinuous phase, physics doesn't change, but you will have to deal with delta functions in the vector potential where the phase has jumps.
     
  9. Jul 19, 2015 #8
    Thank you for the reply. Let us choose a fix n.
    For the case of hall conductance, it is an area integral with the gauge phase [tex]e^{if(k_1,k_2)}[/tex]
    If we use stroke's theorem, parametrize the path by [tex](k_1,k_2)=(k_1(\lambda),k_2(\lambda),[/tex] where [tex]\lambda=0[/tex] is the starting point while [tex]\lambda=1[/tex] is the ending point. As they are the same point, so
    [tex]f(k_1(\lambda=0),k_2(\lambda=0))=f((k_1(\lambda=1),k_2(\lambda=1))[/tex]
    But according to the berry phase in wiki page, the gauge transform allows
    [tex]f(k_1(\lambda=0),k_2(\lambda=0)=f((k_1(\lambda=1),k_2(\lambda=1))+2\pi[/tex]
    There is a violation, one are equal while one can be different by 2 pi. Of course continuous gauge is assumed.
     
  10. Jul 19, 2015 #9

    DrDu

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    That's the problem with wikipedia.
     
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