A short note on the Bloch theorem

Bloch theorem

The Bloch theorem assumes the translational invariance of the potential $V(\mathbf{r})$. Here, the translational symmetry of $V(\mathbf{r})$ means that the potential at $\mathbf{r}$ is the same as the potential image at $\mathbf{r}-\mathbf{R}$ which is translated by an operator $T_{\mathbf{R}}$:

\[\begin{equation} V(\mathbf{r}) = T_{\mathbf{R}}V(\mathbf{r}) = V(\mathbf{r}-\mathbf{R}) \end{equation}\]

where $\mathbf{R}$ is the Bravais lattice vector, which is expressed as $\mathbf{R} = n_{1}\mathbf{a}_{1} + n_{2}\mathbf{a}_{2} + n_{3}\mathbf{a}_{3}$. In addition, the action of $T_{\mathbf{R}}$ on the $n$th derivatives of a function $f(\mathbf{r})$ yields:

\[\begin{equation} T_{\mathbf{R}}\left[\frac{\partial^{n}}{\partial \mathbf{r}^{n}}f(\mathbf{r})\right] = \frac{\partial^{n}}{\partial \mathbf{r}^{n}}f(\mathbf{r}-\mathbf{R}) = \frac{\partial^{n}}{\partial \mathbf{r}^{n}}T_{\mathbf{R}}f(\mathbf{r}) \end{equation}\]

This implies that $T_{\mathbf{R}}$ commutes with the momentum operator. Now for the Hamiltonian $H$ with the potential $V(\mathbf{r})$, the following relationships hold:

\[\begin{align} (T_{\mathbf{R}}H)f(\mathbf{r}) &= -\frac{\hbar^{2}}{2m}\nabla^{2}f(\mathbf{r})\Big\vert_{\mathbf{r}-\mathbf{R}} + V(\mathbf{r}-\mathbf{R})f(\mathbf{r}-\mathbf{R}), \nonumber\\ (HT_{\mathbf{R}})f(\mathbf{r}) &= -\frac{\hbar^{2}}{2m}\nabla^{2}f(\mathbf{r})\Big\vert_{\mathbf{r}-\mathbf{R}} + V(\mathbf{r})f(\mathbf{r}-\mathbf{R}) \end{align}\]

If and only if the potential is periodic; $V(\mathbf{r}) = V(\mathbf{r}-\mathbf{R})$, those relationships are equal to each other. This indicates that the Hamiltonian commutes with the translation operator:

\[\begin{equation} \boxed{[T_{\mathbf{R}},H] = 0.} \end{equation}\]

In other words, $H(\mathbf{r}) = H(\mathbf{r}-\mathbf{R})$. To grasp its physical meaning, let $\ket{\mathbf{r}}$ be the eigenstate of $H(\mathbf{r})$ such that $H(\mathbf{r})\ket{\mathbf{r}} = E\ket{\mathbf{r}}$. The translated image of $\ket{\mathbf{r}}$ becomes $T_{\mathbf{R}}\ket{\mathbf{r}} = \ket{\mathbf{r}-\mathbf{R}}$. Due to the commutation relation above, this becomes the eigenstate of $H(\mathbf{r}-\mathbf{R})$ with $H(\mathbf{r})\ket{\mathbf{r}-\mathbf{R}} = E\ket{\mathbf{r}-\mathbf{R}}$. Note that the eigenvalue $E$ is the same for two states.

Apparently, it seems one can obtain an infinite degeneracy in the energy only by changing $\mathbf{R}$. It is because one can think of the translated images of $\ket{\mathbf{r}}$ are independent of each other. However, this intuition is false.

This can be resolved by the following example: First, assume that $\ket{\mathbf{0}} = \ket{\mathbf{r}=\mathbf{0}}$ is two-fold degenerate with two states $\ket{\mathbf{0}^{(1)}}$ and $\ket{\mathbf{0}^{(2)}}$ as in the case of Ziman (p. 17). The translation $T$ of $\ket{\mathbf{0}}$ by $\mathbf{a}_{1}$ gives:

\[\begin{equation} \begin{bmatrix} \ket{\mathbf{a}^{(1)}_{1}} \\ \ket{\mathbf{a}^{(2)}_{1}} \end{bmatrix} = \begin{bmatrix} t_{11} & t_{12} \\ t_{21} & t_{22} \end{bmatrix} \begin{bmatrix} \ket{\mathbf{0}^{(1)}} \\ \ket{\mathbf{0}^{(2)}} \end{bmatrix} \end{equation}\]

where $t_{ij}$ is the translation coefficient and the matrix element of a unitary matrix $T$. However, $\ket{\mathbf{0}^{(1)}}$ and $\ket{\mathbf{0}^{(2)}}$ are not unique so that their linear combination can be constructed as the follows:

\[\begin{equation} \begin{bmatrix} \ket{\tilde{\mathbf{0}}^{(1)}} \\ \ket{\tilde{\mathbf{0}}^{(2)}} \end{bmatrix} = \begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{bmatrix} \begin{bmatrix} \ket{\mathbf{0}^{(1)}} \\ \ket{\mathbf{0}^{(2)}} \end{bmatrix} \end{equation}\]

where $c_{ij}$ is now the expansion coefficient and the matrix element of another unitary matrix $C$. One can find the matrix $C$ that diagonalizes $T$ so that $CTC^{-1}$ is diagonal:

\[\begin{equation} CTC^{-1} = \begin{bmatrix} d_{11} & 0 \\ 0 & d_{22} \end{bmatrix} \end{equation}\]

Feeding Eq. 5 to Eq. 4, one obtains the states that are like non-degenerate by the translation:

\[\begin{align} \begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{bmatrix} \begin{bmatrix} \ket{\mathbf{a}^{(1)}_{1}} \\ \ket{\mathbf{a}^{(2)}_{1}} \end{bmatrix} &= \begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{bmatrix} \begin{bmatrix} t_{11} & t_{12} \\ t_{21} & t_{22} \end{bmatrix} \begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{bmatrix}^{-1} \begin{bmatrix} \ket{\tilde{\mathbf{0}}^{(1)}} \\ \ket{\tilde{\mathbf{0}}^{(2)}} \end{bmatrix} \nonumber \\ \begin{bmatrix} \ket{\tilde{\mathbf{a}}^{(1)}_{1}} \\ \ket{\tilde{\mathbf{a}}^{(2)}_{1}} \end{bmatrix} &= \begin{bmatrix} d_{11} & 0 \\ 0 & d_{22} \end{bmatrix} \begin{bmatrix} \ket{\tilde{\mathbf{0}}^{(1)}} \\ \ket{\tilde{\mathbf{0}}^{(2)}} \end{bmatrix} \end{align}\]

using

\[\begin{equation} \begin{bmatrix} \ket{\mathbf{0}^{(1)}} \\ \ket{\mathbf{0}^{(2)}} \end{bmatrix} = \begin{bmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{bmatrix}^{-1} \begin{bmatrix} \ket{\tilde{\mathbf{0}}^{(1)}} \\ \ket{\tilde{\mathbf{0}}^{(2)}} \end{bmatrix} \end{equation}\]

Since the norm of states ${\ket{\tilde{\mathbf{a}}^{(i)}_{1}}}$ is equal to $1$, $d_{11}$ and $d_{22}$ are given by $e^{ik_{1}}$ and $e^{ik_{1}’}$, respectively. Note that $k_{1}$ and $k_{1}’$ are called the crystal momenta and have the same direction. However, their magnitudes are different. Now, think of the translation $T$ of $\ket{\mathbf{0}}$ by $\mathbf{a}_{2}$. This operator commutes with the translation by $\mathbf{a}_{1}$. This is because $T_{\mathbf{a}_{1}}T_{\mathbf{a}_{2}}$ and $T_{\mathbf{a}_{2}}T_{\mathbf{a}_{1}}$ to $\ket{\mathbf{0}}$ result in the same state $\ket{\mathbf{a}_{1}+\mathbf{a}_{2}}$ (identical with $\ket{\mathbf{a}_{2}+\mathbf{a}_{1}}$):

\[\begin{equation} \begin{rcases} T_{\mathbf{a}_{2}}T_{\mathbf{a}_{1}}\ket{\mathbf{0}} &= T_{\mathbf{a}_{2}}\ket{\mathbf{a}_{1}} \\ T_{\mathbf{a}_{1}}T_{\mathbf{a}_{2}}\ket{\mathbf{0}} &= T_{\mathbf{a}_{1}}\ket{\mathbf{a}_{2}} \end{rcases} \rightarrow \ket{\mathbf{a}_{1}+\mathbf{a}_{2}} \end{equation}\]

This indicates that there is a matrix that simultaneously diagonalizes $T_{\mathbf{a}_{1}}$ and $T_{\mathbf{a}_{2}}$.

Here one can see the path independence:

\[\begin{equation} \boxed{T_{\mathbf{a}\_{1}}T_{\mathbf{a}\_{2}} = T_{\mathbf{a}\_{2}}T_{\mathbf{a}\_{1}}.} \end{equation}\]

This implies that the translation operator is abelian. However, if there is a uniform magnetic field, such a path independence does not hold anymore; $T_{\mathbf{a}_{1}}T_{\mathbf{a}_{2}} \neq T_{\mathbf{a}_{2}}T_{\mathbf{a}_{1}}$, instead $T_{\mathbf{a}_{1}}T_{\mathbf{a}_{2}} = e^{i\phi}T_{\mathbf{a}_{2}}T_{\mathbf{a}_{1}}$, where the additional phase factor $e^{i\phi}$ is called “Peierls phase.”

That is, ${\ket{\tilde{\mathbf{a}}^{(i)}_{2}}}$ will be expressed as:

\[\begin{equation} \ket{\tilde{\mathbf{a}}^{(1)}_{2}} = e^{ik_{2}}\ket{\tilde{\mathbf{0}}^{(1)}},~\text{and}~\ket{\tilde{\mathbf{a}}^{(2)}_{2}} = e^{ik'_{2}}\ket{\tilde{\mathbf{0}}^{(2)}} \end{equation}\]

This will hold for $\mathbf{a}_{3}$, and there exists a crystal momentum $\mathbf{k}$ such that:

\(\begin{equation} \boxed{\ket{\mathbf{R}} = e^{i\mathbf{k}\cdot\mathbf{R}}\ket{\mathbf{0}}} \end{equation}\)

[Not Safe] This is Bloch’s theorem in its careful form. The eigenstates of a lattice-periodic Hamiltonian can be chosen so that each is multiplied by a pure phase $e^{i\mathbf{k}\cdot\mathbf{R}}$ under any lattice translation. An arbitrary state drawn from a degenerate level need not itself be a Bloch state: it is a superposition of Bloch states of the same energy but, in general, of different $\mathbf{k}$.

This also disposes of the apparent paradox above. Translating a solution does not manufacture an endless supply of new ones: all the translated states $T_{\mathbf{R}}\ket{0}$ lie inside the same $g$-dimensional eigenspace, so at most $g$ of them are linearly independent. In the non-degenerate case $g=1$, and the translated state is the original one up to a phase.

Bloch function is expressed as a function of a phase factor $e^{i\mathbf{k}\cdot\mathbf{r}}$ multiplied by a function $\ket{u_{n\mathbf{k}}}$ that has the periodic symmetry of the lattice ($u_{n\mathbf{k}}(\mathbf{r}) = u_{n\mathbf{k}}(\mathbf{r}-\mathbf{R})$):

\[\begin{equation} \ket{\psi_{n\mathbf{k}}} = e^{i\mathbf{k}\cdot\mathbf{r}}\ket{u_{n\mathbf{k}}} \end{equation}\]

with the crystal momentum $\mathbf{k}$ that labels the state, and the band index $n$. If one applies the translation operator for both sides, one gets:

\[\begin{align} T_{\mathbf{R}}\ket{\psi_{n\mathbf{k}}} &= T_{\mathbf{R}}[e^{i\mathbf{k}\cdot\mathbf{r}}\ket{u_{n\mathbf{k}}}] \nonumber \\ &= e^{i\mathbf{k}\cdot(\mathbf{r}-\mathbf{R})}\ket{u_{n\mathbf{k}}} \end{align}\]

This is another statement that the eigenvalue of a Bloch state for the translation operator is given by the phase factor $e^{-i\mathbf{k}\cdot\mathbf{R}}$. Multiplying $\bra{\mathbf{r}}$, the relationship leads to:

\[\begin{gather} \braket{\mathbf{r}|T_{\mathbf{R}}|\psi_{n\mathbf{k}}} = \braket{\mathbf{r}-\mathbf{R}|\psi_{n\mathbf{k}}} = e^{i\mathbf{k}\cdot(\mathbf{r}-\mathbf{R})}\braket{\mathbf{r}|u_{n\mathbf{k}}} \rightarrow e^{-i\mathbf{k}\cdot\mathbf{R}}\braket{\mathbf{r}|\psi_{n\mathbf{k}}}, \nonumber \\\\ \boxed{\therefore \braket{\mathbf{r}-\mathbf{R}|\psi_{n\mathbf{k}}} = e^{-i\mathbf{k}\cdot\mathbf{R}}\braket{\mathbf{r}|\psi_{n\mathbf{k}}}} \end{gather}\]

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References

1. Ziman, J. M. Principles of the Theory of Solids. (Cambridge University Press, Cambridge, 1972).