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How to prove Bloch function is periodic in reciprocal lattice?

I saw in some textbooks this formula: $$ \Psi_{\mathbf{k}} (\mathbf{r}) = \sum_{\mathbf{G}} c_{\mathbf{k}+\mathbf{G}}e^{i(\mathbf{k}+\mathbf{G})\cdot \mathbf{r}} $$ which makes the statement of this question obvious. ($\mathbf{G}$ is reciprocal lattice vectors)

But I don't understand this formula. I know $$ \Psi_{\mathbf{k}}(\mathbf{r}) = e^{i\mathbf{k}\cdot\mathbf{r}}u_{\mathbf{k}}(\mathbf{r}) $$ and $u_{\mathbf{k}}(\mathbf{r})$ is periodic function of lattice, therefore can be written in Fourier series: $$ u_{\mathbf{k}}(\mathbf{r}) = \sum_{\mathbf{G}} c_{\mathbf{k},\mathbf{G}}e^{i\mathbf{G}\cdot\mathbf{r}} $$ Now I don't understand why $c_{\mathbf{k},\mathbf{G}}$ can be written as $c_{\mathbf{k}+\mathbf{G}}$ ?

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4 Answers 4

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Bloch functions are not necessarily periodic in reciprocal space. By the translation symmetry of the lattice, the wave function $\psi_{nk}(r)$ must satisfy the Bloch condition:

$$ \psi_{nk}(r-R) = e^{-ik\cdot R}\psi_{nk}(r) $$ where $R$ is a lattice vector. Now this is generically satisfied by a function of the form $$\psi_{nk}(r) = e^{ik\cdot r}u_{nk}(r) $$ where $u_{nk}(r-R)=u_{nk}(r)$. But the choice of $u_{nk}(r)$ is not unique. There is a gauge freedom meaning that we can take $u_{nk}(r)\mapsto e^{-iG\cdot r}u_{nk}(r)$ and the new wavefunction will still satisfy the Bloch condition. So does it matter which one we choose?

Well the convention is to choose the so-called periodic gauge condition, i.e. we choose to have the wavefunction $\psi_{nk}$ be periodic in reciprocal space: $\psi_{n,k+G}(r)=\psi_{nk}(r)$. For this to be true, we must choose a $u_{nk}(r)$ which satisfies

$$ u_{n,k+G}(r) = e^{-iG\cdot r} u_{nk}(r) $$

So this is what makes $\psi_{nk}(r)$ periodic in reciprocal space. We do not have to satisfy this condition, but it is conventional and convenient.

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    $\begingroup$ Just as an extra comment, the periodic gauge condition is not always possible. At least not in a smooth way. When the band is topological, ie has non-trivial Chern number, there is a topological obstruction to this. This, among other things, imply that the corresponding Wannier states are not localized. So the question of the OP is actually very good and non-trivial, though this important subtlety is often glossed over in books. References: arxiv.org/abs/cond-mat/0608527, arxiv.org/abs/math-ph/0601034, arxiv.org/abs/cond-mat/0606726. $\endgroup$
    – Heidar
    Commented Oct 17, 2021 at 1:13
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because the index of summation only relates to G,you can forget about "k",and also k=G+k(that shows the transnational symmetry). and look here.

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  • $\begingroup$ G=G+k ? Do you mean k = k+G ? $\endgroup$
    – Tim
    Commented Jan 26, 2015 at 20:51
  • $\begingroup$ Also your argument is not true, a function of two arguments $c_{k,G}$ is not necessary to only depend on $k+G$ $\endgroup$
    – Tim
    Commented Jan 26, 2015 at 20:54
  • $\begingroup$ yes,corrected my statement! $\endgroup$
    – user71065
    Commented Jan 26, 2015 at 20:54
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Because the reciprocal lattice $G$-periodic, the state with a wave vector $k+G$ describes the same state as that of wave vector $k$. You can therefore reduce your study to the first Brillouin Zone ($-\pi<k\leq\pi$). This means that the coefficient in your Fourier expansion will only depend on where you are within this zone. You can add or subtract as many times $G$ as you like from your $k$ vector, and the result will stay the same. At least in simple descriptions where no further corrections make the Bloch theorem only an approximation.

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I'm also not happy with the exposition found in most (solid state physics) textbooks and think one cannot rigorously prove this without group theory. The argument would be the following in a setting with periodic boundary conditions (Born-von Karman) where $\Psi (x + Na) = \Psi (x)$ (for simplicity in 1d):

Using $$ [H, T] = 0~, $$ where the translation operator is defined as $$ T f(x) = f(x + a)~, $$ $k$ labels the $N$ unique solutions $\Psi_k$ that can be distinguished by $T$ and yield $$ T \Psi_k (x) = {\rm e}^{{\rm i} k a} \Psi_k (x) \quad \text{with}\quad k \in \left\{ \frac{2 \pi n}{N a} : n \in \mathbb N^{[0, N)}\right\}~. $$ Now, for any $\Psi_{k'}$ with $k' = k + G$, where $G = 2\pi / a \cdot m$ is an integer multiple of the reciprocal lattice vector $b = 2\pi / a$, we would find $$ T \Psi_{k+G}(x) = {\rm e}^{{\rm i} k a} \Psi_{k+G} (x)~, $$ i.e. $\Psi_{k'}$ yields the same eigenvalues of $T$ as $\Psi_k$ and is therefore not distinguishable from $\Psi_k$ as it belongs to the same irreducible representation. We can therefore define $$ \Psi_{k + G} (x) \equiv \Psi_k (x) \quad\text{for any}\quad G=2\pi/a \cdot m~. $$ All the properties of the Fourier representation of $\Psi_k$ are a consequence of this and not the other way round.

Literature:

  • Dresselhaus, Group Theory, Chp. 10.2
  • Zee, Group Theory in a Nutshell, Chp. III.1
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  • $\begingroup$ I have the same doubt. Thank you much for your help. I have a doubt. I hope you can answer. You wrote: "Ψk′ yields the same eigenvalues of T as Ψk and is therefore not distinguishable from Ψk" Is this a property of the eigenfunctions? or it is part of group theory? $\endgroup$
    – Who
    Commented Nov 6, 2020 at 11:21
  • $\begingroup$ It is a property of the operator $T$ which only has $N$ eigenvalues and therefore eigenfunctions labelled by the $N$ values of $k$. Does that help? $\endgroup$
    – Floyd4K
    Commented Nov 19, 2020 at 15:53
  • $\begingroup$ Thank you for your response. Sorry, it is not clear yet to me. So the argument is that the operator T has only one set of eigenvalues, and since both states have the same eigenvalues then they must be the same state. But how we know this must be the case? what let us conclude this? how are we sure they arent different states with the same eigenvalues? $\endgroup$
    – Who
    Commented Nov 21, 2020 at 1:49
  • $\begingroup$ I didn't and wouldn't say must, but can, in the following sense: The two functions will "look" transform the same way under translations, so they won't be able to represent states differing in the translation properties. Of course this doesn't mean you cannot have more than one physical state at a given $k$. On the contrary, that is why we have a band structure. But all wavefunctions at this $k$ will transform similarly under translation by $a$. $\endgroup$
    – Floyd4K
    Commented Dec 8, 2020 at 9:55

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