# Translation invariance in crystals

Context: solid state physics - dynamics of atoms in crystals.

Coupling constants are defined as:

\begin{align*} \frac{\partial^2\Phi}{\partial r_{nai}\partial r_{mbj}} = \Phi_{nai}^{mbj} \end{align*}

where $\Phi$ is the total energy of the crystal expressed as a function of nuclei coordinates. $r_{nai}$ is the position vector of the $n$-th cell, $a$-th atom at $i$-th spatial direction.

My book says that coupling constants have some properties. One of them is the "translation invariance", which is stated as:

\begin{align} \Phi_{nai}^{mbj} = \Phi_{0ai}^{(m-n)bj} \end{align}

I don't understand the zero index in RHS. Since the $n$ or $m$ represent the cell, what does it mean to refer to zero-th cell ?

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I may be misinterpreting the situation (as you didn't mention which book you're getting the expression from), but assuming that $\mathbf{n}=(n_x,n_y,n_z)$ is actually the location of the center of the cell in question, by assuming an infinite spatial extent you naturally have the identity \begin{align} \Phi_{\mathbf{n}ai}^{\mathbf{m}bj} = \Phi_{\mathbf{0}ai}^{(\mathbf{m}-\mathbf{n})bj} \end{align} by translational invariance. Note that $\Phi_{\mathbf{n}ai}^{\mathbf{m}bj}$ is basically a matrix element of the Hessian matrix of the system potential.

The visual intuition of this is simple: the differential change in system energy when moving the $a$th atom of the cell located at $\mathbf{n}$ in the direction $\hat{\mathbf{r}}_1$ and moving the $b$th atom of the cell located at $\mathbf{m}$ in the direction $\hat{\mathbf{r}}_2$ must be the same as the differential change in system energy when moving the $a$th atom of the cell located at $\mathbf{0}$ in the direction $\hat{\mathbf{r}}_1$ and moving the $b$th atom of the cell located at $\mathbf{m-n}$ in the direction $\hat{\mathbf{r}}_2$.

Why? Just translate your visual frame of reference.

For a purely mechanical proof, let $\mathbf{r}$ be the generalized coordinates of the system (all the $\mathbf{n}$'s, all the $a$'s, and the vector displacements of each atom). Letting $T_\mathbf{n}$ be a particular translation operator, by the translation invariance of the system you have $$\Phi(\mathbf{r})=\Phi(T_\mathbf{n}\mathbf{r})$$ and so by taking the gradient twice you obtain the matrix expression $$\nabla\nabla\Phi(\mathbf{r})=T_\mathbf{n}^\mathsf{T}\nabla\nabla\Phi(T_\mathbf{n}\mathbf{r})T_\mathbf{n}$$ which can alternately be written in component form as \begin{align} \Phi_{nai}^{mbj} = \Phi_{0ai}^{(m-n)bj} \end{align} when you take the subscripts to represent the row index and the superscripts as the column index.

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Is there a property for calculating $\nabla \Phi(T_n \mathbf{r})$ ? – Zet Mar 21 '14 at 19:59