Bloch's Theorem

In a perfect crystal, ions are arranged in a regular periodic array so an electron in the periodic potential \(U(\vec{r})\) experiences the periodicity of the lattice:

\[ U(\vec{r}+\vec{R}) = U(\vec{r}) \]

where

\[ \vec{R}\in \text{Bravais lattice (B.L.)} \]

Eigenstates \(\psi_{n\vec{k}}\) Hamiltonian is

\[ H = -\frac{\hbar^2}{2m}\nabla^2 + U(\vec{r}) \]

can be chosen to have the form

\[ \psi_{n\vec{k}}(\vec{r}) = e^{i\vec{k}\cdot \vec{r}} u_{n\vec{k}}(\vec{r}) \]

where

\[ u_{n\vec{k}}(\vec{r}+\vec{R}) = u_{n\vec{k}}(\vec{r}) \qquad \forall \vec{R}\in \text{B.L.} \]

Here \(n\) is the band index since for a given \(\vec{k}\), there can be many independent eigenstates, so we label them by \(n\).

Using the Bloch form,

\[ \psi_{n\vec{k}}(\vec{r}) = e^{i\vec{k}\cdot \vec{r}} u_{n\vec{k}}(\vec{r}) \]

we evaluate the wavefunction at \(\vec{r}+\vec{R}\):

\[ \psi_{n\vec{k}}(\vec{r}+\vec{R}) = e^{i\vec{k}\cdot(\vec{r}+\vec{R})} u_{n\vec{k}}(\vec{r}+\vec{R}) \]

Since

\[ u_{n\vec{k}}(\vec{r}+\vec{R}) = u_{n\vec{k}}(\vec{r}), \]

we get

\[ \psi_{n\vec{k}}(\vec{r}+\vec{R}) = e^{i\vec{k}\cdot \vec{R}} e^{i\vec{k}\cdot \vec{r}} u_{n\vec{k}}(\vec{r}) \]

Therefore,

\[ \boxed{ \psi_{n\vec{k}}(\vec{r}+\vec{R}) = e^{i\vec{k}\cdot \vec{R}} \psi_{n\vec{k}}(\vec{r}) } \]

This is another common statement of Bloch's theorem.

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Proof of Bloch's Theorem

Define a translation operator \(\hat{T}_{\vec{R}}\) such that it translates a function by a lattice vector \(\vec{R}\):

\[ \hat{T}_{\vec{R}} f(\vec{r}) = f(\vec{r}+\vec{R}) \]

Since the Hamiltonian is periodic,

\[ U(\vec{r}+\vec{R}) = U(\vec{r}), \]

the Hamiltonian commutes with the translation operator:

\[ \hat{T}_{\vec{R}}H\psi = H(\vec{r}+\vec{R})\psi(\vec{r}+\vec{R}) \]

But because

\[ H(\vec{r}+\vec{R}) = H(\vec{r}), \]

we have

\[ \hat{T}_{\vec{R}}H\psi = H\hat{T}_{\vec{R}}\psi. \]

Therefore,

\[ \boxed{ \hat{T}_{\vec{R}}H = H\hat{T}_{\vec{R}} } \]

or equivalently,

\[ [H,\hat{T}_{\vec{R}}]=0. \]

For two lattice translations \(\vec{R}\) and \(\vec{R}'\),

\[ \hat{T}_{\vec{R}'}\hat{T}_{\vec{R}}\psi(\vec{r}) = \hat{T}_{\vec{R}'}\psi(\vec{r}+\vec{R}) \]

This gives

\[ \hat{T}_{\vec{R}'}\hat{T}_{\vec{R}}\psi(\vec{r}) = \psi(\vec{r}+\vec{R}+\vec{R}') \]

Similarly,

\[ \hat{T}_{\vec{R}}\hat{T}_{\vec{R}'}\psi(\vec{r}) = \psi(\vec{r}+\vec{R}'+\vec{R}) \]

Since vector addition commutes,

\[ \vec{R}+\vec{R}'=\vec{R}'+\vec{R}, \]

we get

\[ \boxed{ \hat{T}_{\vec{R}'}\hat{T}_{\vec{R}} = \hat{T}_{\vec{R}}\hat{T}_{\vec{R}'} = \hat{T}_{\vec{R}+\vec{R}'} } \]

So the translation operators commute.

Since

\[ [H,\hat{T}_{\vec{R}}]=0, \]

the Hamiltonian and the translation operators can have simultaneous eigenstates.

Therefore, we can choose \(\psi\) such that

\[ H\psi = E\psi \]

and

\[ \hat{T}_{\vec{R}}\psi = C(\vec{R})\psi. \]

Here \(C(\vec{R})\) is the eigenvalue of the translation operator.

Using the definition of the translation operator,

\[ \hat{T}_{\vec{R}}\psi(\vec{r}) = \psi(\vec{r}+\vec{R}) = C(\vec{R})\psi(\vec{r}) \]

Using the translation composition rule,

\[ \hat{T}_{\vec{R}'}\hat{T}_{\vec{R}} = \hat{T}_{\vec{R}+\vec{R}'}, \]

we apply it to \(\psi\):

\[ \hat{T}_{\vec{R}'}\hat{T}_{\vec{R}}\psi = C(\vec{R})\hat{T}_{\vec{R}'}\psi \]

Since

\[ \hat{T}_{\vec{R}'}\psi = C(\vec{R}')\psi, \]

we get

\[ \hat{T}_{\vec{R}'}\hat{T}_{\vec{R}}\psi = C(\vec{R})C(\vec{R}')\psi. \]

But also,

\[ \hat{T}_{\vec{R}'}\hat{T}_{\vec{R}}\psi = \hat{T}_{\vec{R}+\vec{R}'}\psi = C(\vec{R}+\vec{R}')\psi. \]

Therefore,

\[ \boxed{ C(\vec{R}+\vec{R}') = C(\vec{R})C(\vec{R}') } \]

For primitive lattice vectors \(\vec{a}_i\), we may write

\[ C(\vec{a}_i)=e^{2\pi i \alpha_i} \]

where \(\alpha_i\) are real numbers.

A general Bravais lattice vector is

\[ \vec{R}=n_1\vec{a}_1+n_2\vec{a}_2+n_3\vec{a}_3. \]

Using the multiplicative property,

\[ C(\vec{R}) = C(\vec{a}_1)^{n_1} C(\vec{a}_2)^{n_2} C(\vec{a}_3)^{n_3} \]

Therefore,

\[ C(\vec{R}) = e^{2\pi i n_1\alpha_1} e^{2\pi i n_2\alpha_2} e^{2\pi i n_3\alpha_3} \]

\[ C(\vec{R}) = e^{2\pi i(n_1\alpha_1+n_2\alpha_2+n_3\alpha_3)}. \]

We define a vector \(\vec{k}\) such that

\[ \vec{k}\cdot \vec{R} = 2\pi(n_1\alpha_1+n_2\alpha_2+n_3\alpha_3). \]

Then

\[ C(\vec{R})=e^{i\vec{k}\cdot \vec{R}}. \]

Thus,

\[ \boxed{ \hat{T}_{\vec{R}}\psi = \psi(\vec{r}+\vec{R}) = C(\vec{R})\psi = e^{i\vec{k}\cdot \vec{R}}\psi(\vec{r}) } \]

We have shown that

\[ \psi(\vec{r}+\vec{R}) = e^{i\vec{k}\cdot \vec{R}}\psi(\vec{r}). \]

Now define

\[ u_{\vec{k}}(\vec{r}) = e^{-i\vec{k}\cdot \vec{r}}\psi_{\vec{k}}(\vec{r}). \]

Then

\[ \psi_{\vec{k}}(\vec{r}) = e^{i\vec{k}\cdot \vec{r}} u_{\vec{k}}(\vec{r}). \]

Check the periodicity of \(u_{\vec{k}}\):

\[ u_{\vec{k}}(\vec{r}+\vec{R}) = e^{-i\vec{k}\cdot(\vec{r}+\vec{R})} \psi_{\vec{k}}(\vec{r}+\vec{R}) \]

Using

\[ \psi_{\vec{k}}(\vec{r}+\vec{R}) = e^{i\vec{k}\cdot \vec{R}} \psi_{\vec{k}}(\vec{r}), \]

we get

\[ u_{\vec{k}}(\vec{r}+\vec{R}) = e^{-i\vec{k}\cdot \vec{r}} e^{-i\vec{k}\cdot \vec{R}} e^{i\vec{k}\cdot \vec{R}} \psi_{\vec{k}}(\vec{r}) \]

\[ u_{\vec{k}}(\vec{r}+\vec{R}) = e^{-i\vec{k}\cdot \vec{r}} \psi_{\vec{k}}(\vec{r}) \]

Therefore,

\[ \boxed{ u_{\vec{k}}(\vec{r}+\vec{R}) = u_{\vec{k}}(\vec{r}) } \]

So the wavefunction can be written as

\[ \boxed{ \psi_{n\vec{k}}(\vec{r}) = e^{i\vec{k}\cdot \vec{r}} u_{n\vec{k}}(\vec{r}) } \]

with

\[ \boxed{ u_{n\vec{k}}(\vec{r}+\vec{R}) = u_{n\vec{k}}(\vec{r}) } \]

This is Bloch's theorem.