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Non-uniqueness of the k-vector in Bloch state

  1. Oct 8, 2015 #1
    How to understand that Bloch wave solutions can be completely characterized
    by their behaviour in a single Brillouin zone? Given Bloch wave:
    \begin{equation*}
    \psi_{\mathbf{k}}(\mathbf{r}) = u_{\mathbf{k}}(\mathbf{r}) \exp (i\mathbf{k}\mathbf{r})
    \end{equation*}
    I can write wavefunction for momentum ##\mathbf{k}' = \mathbf{k} + \mathbf{G}##.
    \begin{equation*}
    \psi_{\mathbf{k}+\mathbf{G}}(\mathbf{r}) = u_{\mathbf{k}'}(\mathbf{r})\exp(i\mathbf{G}\mathbf{r})\exp (i\mathbf{k}\mathbf{r})
    \end{equation*}
    As I understand ##u_{\mathbf{k}'}(\mathbf{r})\exp(i\mathbf{G}\mathbf{r}) \neq u_{\mathbf{k}}(\mathbf{r})##, so:
    \begin{equation*}
    \psi_{\mathbf{k}+\mathbf{G}}(\mathbf{r}) \neq \psi_{\mathbf{k}}(\mathbf{r}).
    \end{equation*}
    Or I am wrong? I know that another way to analyze this problem is to notice that
    ##u_k## and ##u_k'## are periodic and do the Fourier expansion:
    \begin{equation*}
    u_{\mathbf{k}'}(\mathbf{r})\exp(i\mathbf{G})=\sum_{\mathbf{G}_{i}}C_{\mathbf{k}',\mathbf{G}_{i}}\exp\left(i\mathbf{G}\mathbf{r}\right)\exp\left(i\mathbf{G}_{i}\mathbf{r}\right)
    \end{equation*}
    However I still don't see why [tex] C_{\mathbf{k}',\mathbf{G}_{i}}\exp\left(i\mathbf{G}\mathbf{r}\right) = C_{\mathbf{k},\mathbf{G}_{i}}[/tex]
     
  2. jcsd
  3. Oct 9, 2015 #2

    DrDu

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    You are right. The functions u have another index n, which specifies to which band they belong. If k is greater than K, so that k' is in the BZ, then u_n(k') would be different from that for a wavefunction for which k=k' in the first BZ with u_n'(k').
     
  4. Oct 9, 2015 #3
    Thanks for reply.
    So how to understand proposition, that every ##\mathbf{k}## can be mapped to first BZ. Can anyone give me practical example? There are many sources where relation ## \psi_{\mathbf{k}} = \psi_{\mathbf{k} + \mathbf{G}} ## is given.
     
  5. Oct 9, 2015 #4

    DrDu

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  6. Oct 10, 2015 #5

    cgk

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    The Bloch theorem is also not really that mysterious if you consider it's core meaning: If you have a bunch of symmetry operations which commute with the Hamiltonian, then the eigenstates of the Hamiltonian can be chosen in such a way that both the Hamiltonian and the symmetry generators are diagonalized simultaneously. For symmetry generators, this effectively means that all eigenstates of the Hamiltonian can be labelled by the irreducible representations of the symmetry groups. This is commonly done in molecular spectroscopy, where you attach labels like [itex]\Sigma_u[/itex] etc to states, indicating their irrep of the molecule's point group symmetry.

    For the "translation by integer multiples of a certain vector" group (i.e., lattice symmetry), the irreps happen to be labelled by k-vectors in the Brillouin zone. They correspond to eigenvectors of the form [itex]T_{\vec L}\vert\psi\rangle = e^{i\vec k\vec L}\vert\psi\rangle[/itex] regarding spatial translations [itex]T_{\vec L}[/itex] by multiples of the lattice vector [itex]\vec L[/itex], with eigenvalue [itex]e^{i\vec k\vec L}[/itex]. This also means that only eigenvalues up to 2π multiplies of [itex]i\vec k\vec L[/itex] are unique (this corresponds to the first BZ). Higher k-vectors thus do not need to be considered.
     
  7. Oct 14, 2015 #6
    Again, thanks for replies. However I still need to clear things up. Are theese points realy correct?
    1. Single bloch wave is not periodic in ##\mathbf{k}## space
    2. However we can map solutions to 1st Briullin zone ##\psi_{\mathbf{k}'} = \psi_{\mathbf{k} + \mathbf{G}} = \psi_{\mathbf{k}, j}## by assigning different ##j## for different Briullin zones -- we call it a band number
    3. Free particle is an example of Bloch function -- we still have periodicity ##V(\mathbf{r}) = V(\mathbf{r} + \mathbf{G})##
    If 2'nd point is correct, substitution of bloch function into Schrodinger equation gives us
    \begin{equation*}
    \left[
    \frac{1}{2m_e}
    \left(
    \frac{\hbar \nabla_{\mathbf{r}}}{i} + \hbar \mathbf{k}
    \right)
    + V(\mathbf{r})^2
    \right]u_{\mathbf{k}}(\mathbf{r}) = \varepsilon_{\mathbf{k}} u_{\mathbf{k}}(\mathbf{r})
    \end{equation*}
    so how to interpret different eigensolutions of this equation if we already assigned band numbers for single ##\mathbf{k}##.

    P.S.
    Sorry for my poor English.
     
  8. Oct 15, 2015 #7

    DrDu

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    As I said, whether 1 is correct or not is entirely a question of definition as
    ##\psi_{k+G}=\exp(i(k+G)x) u_{k+G,n}(x)##. So if you define ##u_{k+G,n}(x)\exp(iGx)=u_{k,n}(x)##, then everything is periodic.
     
  9. Oct 15, 2015 #8
    Sorry, I feel a little bit stupid, but I don't understand it. For me it sounds like if we define ##2=1## then ##3 = 6##.
     
  10. Oct 15, 2015 #9

    DrDu

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    Science Advisor

    This is kind of trivial to check:
    Solving the definition I gave for u_k+G this becomes just use this in your##u_{k+G,n}(x)=\exp(-iGx)u_{k,n}(x)## Now Schroedinger equation for u
    \begin{equation*}
    \left[
    \frac{1}{2m_e}
    \left(
    \frac{\hbar \nabla_{\mathbf{r}}}{i} + \hbar (\mathbf{k+G})
    \right)
    + V(\mathbf{r})^2
    \right]u_{\mathbf{k+G}}(\mathbf{r}) = \varepsilon_{\mathbf{k+G}} u_{\mathbf{k+G}}(\mathbf{r})
    \end{equation*}
    to get
    \begin{equation*}
    \left[
    \frac{1}{2m_e}
    \left(
    \frac{\hbar \nabla_{\mathbf{r}}}{i} + \hbar \mathbf{k}
    \right)
    + V(\mathbf{r})^2
    \right]u_{\mathbf{k}}(\mathbf{r}) = \varepsilon_{\mathbf{k+G}} u_{\mathbf{k}}(\mathbf{r})
    \end{equation*}
    This is identical to the Schroedinger equation for u_k identifying the eigenvalues ##\varepsilon_{\mathbf{k+G},n}=\varepsilon_{\mathbf{k},n}##. While u itself is not periodic in k, psi is as the exponentials exp(iGx) and exp(-iGx) cancel. I assumed here that the band indices are identical ##u_{k+G,n}=u_{k,n}exp (-iGx)## and this is where the arbitrariness of definition is hidden. I could also have chosen ##u_{k+G,n}=u_{k,n'}exp (-iGx)## where the n' are some permutation of the n. Especially when considering a free electron in an empty lattice, it may seem more natural to let n increase when stepping one G to the left or right.
     
    Last edited: Oct 15, 2015
  11. Oct 17, 2015 #10
    Thanks, DrDu, however I think that I found a way to explicitly see this periodicity. My problem was that I assumed
    that free particle solution ##\psi_{\mathbf{k}} = A\exp(i\mathbf{k}\mathbf{r}) + A\exp(i\mathbf{k}\mathbf{r})##
    is a special form of a Bloch wave and I couldn't see any periodicity. However if I undestand correctly wavefunction
    form ##\psi = \exp(i\mathbf{k}\mathbf{r})u(\mathbf{r})## can only be assumed for a systems with periodic
    boundary conditions (Born--Von Karman boundary conditions). And if I plug in ##\psi = \exp(ikx)u(x)## into 1D
    free particle Schrodinger equation and apply aditional boundary conditions ##u(0)=u(a),\ u'(0)=u'(a)##,
    I get a different sollution
    \begin{equation}
    \psi(x)=A\underbrace{\exp\left(-i\frac{2\pi}{a}nx\right)}_{u_{n,k}(x)}\exp(ikr)
    \end{equation}
    which now depends not only on continuous parameter ##\mathbf{k}## but also on discrete integer ##n##. With dispersion relation
    \begin{equation}
    E_n(k) = \frac{\hbar^2 (k - \frac{2\pi}{a}n)}{2m_e}
    \end{equation}
    For fixed ##n## I get parabolas which are separated in ##E## vs ##k## graphs by reciprocal latice vectors.
    tmp6C167_thumb_thumb.jpg
    However if I redefine indexes ##n## in such way that:
    \begin{equation*}
    u^\ast_{k, 1} =
    \begin{cases}
    u_{k, 1} & \text{if } -2\pi/a \leq k \leq 2\pi/a \\
    u_{k + 2\pi/a, 2} & \text{if } 2\pi/a \leq k \leq 4\pi/a \\
    \cdots
    \end{cases}
    \end{equation*}
    I can have a periodic solutions:
    \begin{equation*}
    \psi^{\ast}_{n, k} = \exp(ikx)u^\ast_{k, n}(x) = \psi^{\ast}_{n, k + 2\pi/a}
    \end{equation*}
     
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