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Symmetry of band structure

  1. Sep 19, 2015 #1
    Do any band structure (in absence of any external field) in general, is symmetric with respect to k? In other words, do we always have E(k)=E(-k).
     
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  3. Sep 19, 2015 #2

    Henryk

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    Energy is a quadratic function of k, so it should be symmetric with respect to inversion.
    In general, the properties of the crystal reflect the symmetry of the crystal.
     
  4. Sep 20, 2015 #3
    Your statement is not true in general.
     
  5. Sep 20, 2015 #4

    radium

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    Graphene and a lot of other materials are definitely not quadratic, they are linear and have relativistic dispersion. Strongly interacting Kondo and Mott insulators also have strange this going on.

    In regards to your statement, it is true if you have time reversal and inversion symmetries.
     
  6. Sep 20, 2015 #5

    DrDu

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    This symmetry is often broken when either external magnetic fields are present or there is an interaction with internal magnetic fields, e.g. spin orbit coupling in ferromagnetic materials. As radium mentioned, in these materials time reversal symmetry may be broken (inversion symmetry has to be broken, too).
     
  7. Sep 20, 2015 #6
    By this statement do you mean that to have E(k)=E(-k), both "time reversal and inversion symmetry" are necessary or one of them suffices?
     
  8. Sep 22, 2015 #7
    The band structure has the symmetry of the crystal's point group. So if the crystal has inversion symmetry (including all applied fields), then so does the band structure.

    The band structure is the Fourier transform of the real space energy landscape. If the real space energy is, well, real, then E(-k) should be the complex conjugate of E(k). So again, if E(k) is real, then E(k) should always be the same as E(-k), even in the absence of inversion symmetry.

    BTW, and counter intuitively, a magnetic field does not break inversion symmetry. L = r x p is even under inversion as both r and p are odd. An electric field, on the other hand, does break inversion symmetry.
     
  9. Sep 22, 2015 #8

    Henryk

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    hokhani, my apology,
    I meant the kinetic energy is a quadratic function of momentum operator. Of course energy is not a quadratic function of k, we wouldn't have band structure if it was.

    Now, a few comments.
    Time reversal symmetry is the property of Schrodinger equation but band structure is the result of solution of time-independent equation. Time reversal has nothing to do with it.

    The subject of E(k) = E(-k) kind of intrigued me. It has been a few years since I left grad school.
    Definitely, if there is inversion symmetry, the above is true. However, when I looked at my textbook it appeared that inversion symmetry is not necessary. I'm going to attempt to prove it.
    Suppose, we have a Bloch wavefunction ##\psi _k(r) = e^{ikr} \phi (r) ## corresponding to energy E. Then, this wavefunction satisfies Shrodinger equation
    $$ - \frac {\hbar^2}{2m} \nabla ^2 \psi _k (r) + U(r) \psi _k (r) = E \psi _k (r)$$
    Now, we can simply take a complex conjugate of the above equation !!
    The complex conjugate of ## \psi _k(r) ## is ## \psi _{-k}(r) ##, U(r) and E are real, then this gives us immediately
    $$ - \frac {\hbar^2}{2m} \nabla ^2 \psi _{-k} (r) + U(r) \psi _{-k} (r) = E \psi _{-k} (r)$$
    with the same energy and opposite momentum

    I guess you don't need space inversion symmetry neither time reversal symmetry
     
  10. Sep 23, 2015 #9

    radium

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    Time reversal has a LOT to do with band structure. Kramer's degeneracy says that if you have a time reversal symmetric system with an odd number of electrons you will always have at least a two fold degeneracy. These degeneracies at time reversal symmetric points in the Brillouin zone are called Kramer's pairs. In states like QSHE or 3D TIs time reversal symmetry is what is causing a very important feature in the band structure. If you have a finite system you will have gapless edge states which cannot be removed with perturbations that do not break time reversal.
     
  11. Sep 23, 2015 #10

    DrDu

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    ##(\psi _k(r) = e^{ikr} \phi_k (r))^*=e^{-ikr}\phi_k^*(r) ##. This only equals ##\psi_{-k}(r)## if ##\phi_k^*=\phi_{-k}##. The other point is that in the presence of spin orbit interaction or magnetic fields, U may not be real because the spin matrix ##\sigma_y## is imaginary.
     
  12. Sep 23, 2015 #11
    I think at systems with inversion symmetry, an electron moving towards one direction see the same environment as the electron moving in the opposite direction. Therefore it seems E(k)=E(-k) to be held regardless of whether the time reversal exists or not.
     
    Last edited: Sep 23, 2015
  13. Sep 24, 2015 #12

    DrDu

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    Of course, but we already said that E(k)=E(-k) may fail to hold if there is no inversion symmetry.
     
  14. Sep 27, 2015 #13

    Daz

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    E(k)=E(-k) irrespective of inversion symmetry. This follows directly from the Bloch wave solution to Schrodinger’s equation. (Which is implicitly time-reversible.)

    Most of the III-V and II-VI compound semiconductors have the face centred cubic zinc blende structure (bar-4 3 m space group) and are therefore not centrosymmetric but they most definitely do have symmetrical (with respect to k) band structure. Look up the band structure of GaAs or InP, for example.

    In fact, because of this symmetry, it is common practice to plot the band structure along two different crystallographic directions in the same diagram. (Commonly for FCC semiconductors one sees <100> and <111> plotted.) It is implicit in such plots that the -k half of the plot is of no interest because it is a mirror image of the +k half.

    Google “GaAs band structure” to see many examples.

    As a corollary, remember that E is periodic in k and when we plot the band structure, we are implicitly folding the (infinitely repeating) E-vs-k dispersion diagram down into just the first Brillouin zone. If E(k) did not equal E(-k) there would be discontinuities in the “unfolded” dispersion relation, which would be unphysical.
     
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