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A Why is brillouin zone a torus?

  1. Mar 5, 2016 #1
    Recently, topological concepts are popular in solid state physics, and berry connection and berry curvature are introduced in band theory. The integration of berry curvature, i.e. chern number, is quantized because Brillouin zone is a torus.

    However, I cannot justify the argument that Brillouin zone is a torus. Here is my analysis:

    1. I admit that ##\psi_k## and ##\psi_{k+G}## can be identified. However, since ##\psi_k(x+R) = e^{ikR}\psi_k(x)##, ##\psi_k##s do not live in the same Hilbert space because they fulfill different boundary conditions. (##R## is lattice vector; ##G## is reciprocal lattice vector)

    2. ##u_k##s, whose relation with ##\psi_k## is ##\psi_k(x)=e^{ikx} u_k(x)##, do live in the same Hilbert place because they are periodic with respect to lattice vector. However, since ##\psi_k = \psi_{k+G}##, ##u_k(x)=e^{ikG}u_{k+G}(x)##, which cannot be identified.

    So, how is Brillouin zone a torus?
     
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  3. Mar 7, 2016 #2

    DrDu

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    If ##\psi_k = \psi_{k+G}##, then ## \exp(ikx)u_k(x)=\exp(i(k+G)x)u_{k+G}(x)##, so ##u_k(x) =\exp(iGx)u_{k+G}(x)##, which is not what you wrote.
     
  4. Mar 7, 2016 #3
    What's the difference between your equation and mine? They look the same!
     
  5. Mar 7, 2016 #4

    DrDu

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    ##\exp(ikG) \neq \exp(iGx)##.
     
  6. Mar 7, 2016 #5
    Oh, I am sorry. You are right, but the question remains.
     
  7. Mar 7, 2016 #6

    DrDu

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    I fear I don't understand your problem, then.
     
  8. Mar 7, 2016 #7

    DrDu

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    Maybe an example is helpful: For an empty lattice, ## u_k(x)=1##. If you chose ##u_{k+G}=\exp(-iGx)## which has the periodicity of the lattice, then ## u_k(x)=\exp(iGx)u_{k+G}(x)##.
     
  9. Mar 7, 2016 #8

    radium

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    Bloch's theorem states that for a perfect crystal you can have a basis of wave functions so each is an energy eigenstate and can be written in the form of a plane wave times some function which has the same periodicity for whatever system you are talking about. If you restrict the range of ks then your wave functions will be unique. You get the torus by identifying the edges. If you have a 2d system with periodic BCs, you identify opposite edges corresponding to kx and ky to get a torus. Periodic BCs allow you to do this. This a theme that occurs in topology, you are looking at how systems react to smooth deformations which is why people say that a coffee cup and donut are topologically equivalent.

    The reason the torus picture is so important is that putting the system on a torus allows you to identify nontrivial topology in many systems. A way to diagnose topological order in a system is from a ground state degeneracy on a torus or a nontrivial topological invariant. I can give some examples below to motivate this conceptually. Some of this may not be very precise as these systems are often quite subtle (I myself have worked with some of these systems and I also have a lot of friends working on many of the more exotic phases) and terminology can get quite confusing.

    Topological invariants in many system arise due to quantization conditions, required by having a single valued wave function. In the quantum hall effect for example, you can find the hall conductance from adiabatically inserting a flux through a cylinder (you have periodicity in one direction) and measuring the charge pumped around in one cycle (the currents are only along the edge since the bulk is full, but that takes a lot more explanation. The reason you look at periodic boundary conditions is because the topological invariant comes from requiring the wave function to be single valued. Nontrivial topology in the QH makes it impossible to define a gauge globally, so you must define two and relate them, requiring a quantization condition. This is very similar to what happens in the Aharonov Bohm effect.

    Another example is a spin liquid state, which can occur in frustrated systems where there is no magnetic order even at T=0. It can be constructed from a "resonating valence bond" state where the state is in a superposition of all possible configurations. An SL is a topologically ordered system and has fractionalized excitations which usually cannot be created by themselves. In a Z2 spin liquid for example you have an excitation called the vison. If you create two visons and bring one all the way around the torus, that state is degenerate with the original state in the thermodynamic limit. Thus the ground state degeneracy is four.
     
  10. Mar 9, 2016 #9
    So, in the original question, I said wave functions (##\psi_k##)can be identified at opposite edges of brillouin zone. However, since ##\psi_k## in different points of brillouin zone do not live in the same Hilbert space, I don't know whether we can define berry phase using ##\psi_k##.
     
  11. Mar 9, 2016 #10

    DrDu

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    That's simply not true.
     
  12. Mar 9, 2016 #11
    Hey, they satisfy different boundary conditions: ##\psi_{k}(x+R)=e^{ikR}\psi(k)##
     
  13. Mar 9, 2016 #12

    DrDu

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    This does not mean that they live in different hilbert spaces. In fact, it is not even a boundary condition. E.g. for a free partikle ##\psi_k(x+R)=\exp(ikR)\psi_k(x)## for whatever choice of ##R##. You don't want to claim that the eigenfunctions of the free particle hamiltonian all live in different Hilbert spaces, do you?
     
    Last edited: Mar 9, 2016
  14. Mar 9, 2016 #13

    radium

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    PBCs on a spin chain for example state that \psi(x)=\psi(x+L) (you can also have APBCs where \psi(x)=-\psi(x+L). Why do you think that different \psi_k's live in different Hilbert spaces.

    The Berry phase is absolutely defined in k space. In fact, the hall conductance is the integral of the berry curvature over the Brillouin zone of the occupied bands.

    You should try looking at information about Bloch's theorem and I also think Ziman's Principles of the theory of solids would be good for this.
     
  15. Mar 9, 2016 #14
    Oh, yeah. I am sorry, ##\psi_{nk}##s do live in the same Hilbert space. However, in berry connection, we only use ##u_{nk}## and they do not satisfy PBC.
     
  16. Mar 9, 2016 #15

    radium

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    I don't really understand what your question is, but you can probably answer it if you read about the quantum Hall effect or the book by Bernevig about TIs. The book provides you with details on how to calculate the hall conductance.
     
  17. Mar 10, 2016 #16
    Thanks for the suggestion.
     
  18. Mar 10, 2016 #17

    DrDu

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    ##u_{nk}(x)## are lattice periodic.
     
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