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Is a Bloch wave periodic in reciprocal space?

  1. Oct 12, 2015 #1
    A Bloch wave has the following form..

    ## \Psi_{nk}(r)=e^{ik\cdot r}u_{nk}(r)##

    The ##u_{nk}## part is said to be periodic in real space. But what about reciprocal space? I've had a hard time finding a direct answer to this question, but I seem to remember reading somewhere that the entire Bloch wave is periodic in k-space i.e. ##\Psi_{nk}(r)=\Psi_{n(k+G)}(r).## In that case, whatever additional exponential factor ##e^{iG\cdot r}## we gained from a k-space translation must occur as ##e^{-iG\cdot r}## in the ##u_{nk}## piece. How can we tell without knowing the exact form of the ##u_{nk}## piece though? I think that ##u_{nk}## comes from the fourier coefficients of the periodic potential. Maybe it has something to do with this?

    Thanks.
     
  2. jcsd
  3. Oct 13, 2015 #2

    DrDu

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    This is not a physical question but one of definition. Physics only fixes u and psi inside the first Brillouin zone. So you can simply impose periodicity in G as a definition.
     
  4. Nov 1, 2015 #3

    Henryk

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    The Bloch function can be written as ## \psi _{nk} (k) = \sum _G a_k (G) exp^{ i(k+G)r} ##. In reciprocal space, it will be non-zero at periodic points but the coefficients ## a_k (G) ## are, in general, different for different Gs. Some may be the same due to symmetry or other feature of the crystal potential.
    So, the answer to your question, is no, the Block function is not periodic in reciprocal space.
     
  5. Nov 2, 2015 #4

    DrDu

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    That's not correct. The Boch function can always be chosen to be periodic in reciprocal space (though alternative non-periodic choices are possible).
     
  6. Nov 2, 2015 #5

    DrDu

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    Specifically, if ##k'=k+K##, where K is some reciprocal lattice vector, you can write ## \psi _{nk} (k) = \sum _G a_k (G) \exp{ i(k+G)r}= \sum _G a_k (G)\exp(iKr) \exp{ i(k'+G-K)r}=\sum _G a_k (G+K)\exp(iKr) \exp{ i(k'+G)r}##. So you can simply define ##a_{k'}(G):=a_k(G+K)\exp(iKr)##.
     
  7. Nov 2, 2015 #6

    Henryk

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    DrDu,

    A function is periodic if ## f(r + T) = f(r)##, right?. Let's just write the Bloch function in reciprocal space. We picked a Bloch function with a specific value of ##k##, so let us label the reciprocal space variable as ##k'##. The function is reciprocal space is ## \psi(k') = \sum _G a_k(G) \delta(k' - (k+G)) ##. For the Bloch function to be periodic in reciprocal space, all the ##a_k(G)##'s would have to be equal. But they, in general, cannot. The Hamiltonian in reciprocal space is a matrix with off-diagonal coefficients equal to ##V(G)## and the ##a_k(G)##'s are defining eigenvectors of the matrix, that is, there is a well defined relationship between them. Re-labeling the coefficient does not change the relationship.
     
  8. Nov 3, 2015 #7

    DrDu

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    Ok, I see what you mean and this is obviously correct. However, I am not sure whether this is the answer to the OP's question. He is referring to the periodicity ##\Psi_{nk}(r)=\Psi_{n(k+G)}(r)## and what this does mean for the functions ##u_{kn}(r)##. The latter are given as
    ##u_{kn}=\sum_G a_{kn}(G)\exp(iGr)## for k in the first Brillouin zone. With the definition ##a_{k'n}=a_{kn}(G+K)exp(iKr)## for ##k'=k+K## the functions ##\Psi## become periodic in the sense of the OP.
     
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