Menu Outline Exercise Questions Appendices Lectures Student Projects Books Sections Introduction Atoms Molecules Crystal Structure Crystal Diffraction Crystal Binding Photons Phonons Electrons Band Model Crystal Physics Semiconductors

PHY.K02UF Molecular and Solid State Physics

## The translation operator

The translation operator of a periodic system $\textbf{T}$ translates the crystal by a lattice constant $a$. Consider a crystal with just 6 unit cells with periodic boundary conditions. The state of the crystal can be represented by a vector with six elements $u_1\cdots u_6$. The matrix representation of the translation operator in this case is,

$$\textbf{T}\vec{u}=\left[ \begin{matrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 \end{matrix} \right]\left[ \begin{matrix} u_1 \\ u_2 \\ u_3 \\ u_4 \\ u_5 \\ u_6 \end{matrix} \right]=\left[ \begin{matrix} u_2 \\ u_3 \\ u_4 \\ u_5 \\ u_6 \\ u_1 \end{matrix} \right].$$

When $\textbf{T}$ acts on a vector which specifies some quantity at every unit cell in the crystal, all of the elements of the vector get shifted by one position. Applying the translation operator twice shifts the elements two positions and applying it $N$ times leaves the vector unchanged because all of the elements get shifted back to their original positions. $\textbf{T}^N$ is the identity matrix $\textbf{I}$.

$$\textbf{T}^2\vec{u}=\left[ \begin{matrix} 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \end{matrix} \right]\left[ \begin{matrix} u_1 \\ u_2 \\ u_3 \\ u_4 \\ u_5 \\ u_6 \end{matrix} \right]=\left[ \begin{matrix} u_3 \\ u_4 \\ u_5 \\ u_6 \\ u_1 \\ u_2 \end{matrix} \right],\hspace{1.5cm}\textbf{T}^N\vec{u}=\left[ \begin{matrix} 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \end{matrix} \right]\left[ \begin{matrix} u_1 \\ u_2 \\ u_3 \\ u_4 \\ u_5 \\ u_6 \end{matrix} \right]=\left[ \begin{matrix} u_1 \\ u_2 \\ u_3 \\ u_4 \\ u_5 \\ u_6 \end{matrix} \right].$$

The inverse of the translation operator is $\textbf{T}^{-1}=\textbf{T}^{N-1}$.

$$\textbf{T}^{-1}\vec{u}=\textbf{T}^{N-1}\vec{u}=\left[ \begin{matrix} 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \end{matrix} \right]\left[ \begin{matrix} u_1 \\ u_2 \\ u_3 \\ u_4 \\ u_5 \\ u_6 \end{matrix} \right]=\left[ \begin{matrix} u_6 \\ u_1 \\ u_2 \\ u_3 \\ u_4 \\ u_5 \end{matrix} \right].$$

Since $\textbf{T}^N$ is the identity matrix and every vector is an eigenvector of the identity matrix with eigenvalue 1, the eigenvalues $\lambda$ of $\textbf{T}$ have the properties,

$$\textbf{T}\vec{u}=\lambda\vec{u}, \\ \textbf{T}^N\vec{u}=\lambda^N \vec{u}=\vec{u}, \\ \lambda^N = 1.$$

Therefore the $N$ eigenvalues of $\textbf{T}$ lie on the unit circle.

Once the eigenvalues are known, the eigenvectors can be found by solving the linear set of equations $(\textbf{T}-\lambda\textbf{I})\vec{u}=0$.

The eigenvectors of $\textbf{T}$ are,

$$\left[ \begin{matrix} e^{i2\pi j/N} \\ e^{i4\pi j/N}\\ e^{i6\pi j/N} \\ \vdots \\ e^{i2\pi (N-1)j/N} \\ 1 \end{matrix} \right]\hspace{1.5cm}j=1,2,\cdots,N.$$

This can be checked by applying the translation operator to the eigenvectors. For the case of $N=6$,

$$\textbf{T}\vec{u}=\left[ \begin{matrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 \end{matrix} \right]\left[ \begin{matrix} e^{i\pi j/3} \\ e^{i2\pi j/3}\\ e^{i\pi j/N} \\ e^{i4\pi j/3} \\ e^{i5\pi j/3} \\ 1 \end{matrix} \right]=\left[ \begin{matrix} e^{i2\pi j/3}\\ e^{i\pi j/N} \\ e^{i4\pi j/3} \\ e^{i5\pi j/3} \\ 1 \\ e^{i\pi j/3}\end{matrix} \right] = e^{i\pi j/3}\left[ \begin{matrix} e^{i\pi j/3} \\ e^{i2\pi j/3}\\ e^{i\pi j/N} \\ e^{i4\pi j/3} \\ e^{i5\pi j/3} \\ 1 \end{matrix} \right]\hspace{1.5cm}j=1,2,3,4,5,6.$$

where the identity $e^{i2\pi j}=1$ was used to get the final form of the vector. The eigenvectors of $\textbf{T}^2,\textbf{T}^3,\cdots ,\textbf{T}^N$ are the same as the eigenvectors of $\textbf{T}$.

The eigenvectors of the translation operator can be put in one-to-one correspondence with the $N$ allowed $k$-vectors in the first Brillouin zone of the crystal.

$$k=\frac{2\pi (j-N/2)}{L}=\frac{2\pi (j-N/2)}{Na}.$$

The using this correspondence, the eigenvalues and eigenvectors of the translation operator can be expressed in terms of the allowed values of $k$ in the first Brillouin zone.

$$\left[ \begin{matrix} 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 & 0 & 0 \end{matrix} \right]\left[ \begin{matrix} 1 \\ e^{ika}\\ e^{i2ka} \\ e^{i3ka} \\ e^{-i2ka} \\ e^{-ika} \end{matrix} \right]=\left[ \begin{matrix} e^{ika}\\ e^{i2ka} \\ e^{i3ka} \\ e^{-i2ka} \\ e^{-ika} \\ 1 \end{matrix} \right] = e^{ika}\left[ \begin{matrix} 1 \\ e^{ika}\\ e^{i2ka} \\ e^{i3ka} \\ e^{-i2ka} \\ e^{-ika} \end{matrix} \right]\hspace{1.5cm}k=-\frac{4\pi}{L},-\frac{2\pi}{L},0,\frac{2\pi}{L},\frac{4\pi}{L},\frac{6\pi}{L}.$$