Solution to Exercise 15

We consider the Hamiltonian,

$${\mathcal H}= \sum_i^N \frac{P_i^2}{2m} + \frac{1}{2} \sum_... ...\sum_i \int d {\bf r}^{\prime}V({\bf r}_i - {\bf r}^{\prime})$$ (1)

where $V(r) = e^2/r$ is the Coulomb potential with which the ions interact, and $\rho _0 = N/V$ is a uniform background charge for overall neutrality. To study small oscillations we will expand the potential term ($U$) in the Hamiltonian around the equilibrium position of each ion. Let ${\bf r}_i = {\bf R}_i + {\bf u}_i$, where ${\bf R}_i$ are the equilibrium positions and ${\bf u}_i$ are small deviations from equilibrium. Expanding upto second order we get,

$$U = \frac{1}{2} \sum_{i \neq j} V \left( {\bf R}_i - {\bf R}_j \right... ...o _0 \sum_i \int d {\bf r}^{\prime}V \left( {\bf R}_i - {\bf r}^{\prime}\right)$$

$$+ \frac{1}{2} \sum_{i \neq j} \left( \frac{\partial V} {\partial r^... ...l V }{\partial r^{\alpha }}({\bf R}_i - {\bf r}^{\prime}) \right) u_i^{\alpha }$$

$$+ \frac{1}{4} \sum_{i \neq j} \left( u_i^{\alpha } - u_j^{\alpha } ... ...al r^{\beta }}({\bf R}_i - {\bf r}^{\prime}) \right)u_i^{\alpha } u_i^{\beta }.$$

The first two terms in the above expansion are constants and they can be dropped out from the Hamiltonian. The third and fourth terms are zero because they represent the force on the ions at their equilibrium positions. Only the second order terms survive, and we can write the Hamiltonian for lattice vibrations as,

$${\mathcal H}= \sum_i^N \frac{P_i^2}{2m} + \frac{1}{2} \sum_... ...pha \beta} \left( {\bf R}_i - {\bf R}_j \right) u_j^{\beta }$$ (2)

where

$$C^{\alpha \beta}\left( {\bf R}_i - {\bf R}_j \right) = \lef... ... R}_i - {\bf r}^{\prime}) \right) & i=j \end{array} \right.$$ (3)

We expand $P_i^{\alpha }$ and $u_i^{\alpha }$ in Fourier series

$$u_i^{\alpha } = \frac{1}{N^{1/2}} \sum_{{\bf k}\lambda } e^{i {\bf k}\cdot {\bf R}_i} u_{{\bf k}\lambda } \varepsilon _{{\bf k}\lambda }^{\alpha } ,$$

$$P_i^{\alpha } = \frac{1}{N^{1/2}} \sum_{{\bf k}\lambda } e^{i {\bf k}\cdot {\bf R}_i} P_{{\bf k}\lambda } \varepsilon _{{\bf k}\lambda }^{\alpha }.$$

${\bf\varepsilon }_{{\bf k}\lambda }$ are the polarization vectors. For each wavevector ${\bf k}$ there is one longitudinal direction ($\lambda =1$) and two transverse directions $(\lambda = 2,3)$. They define an orthonormal basis for three dimensional vectors. Thus, $\varepsilon ^{\alpha }_{{\bf k}\lambda } \varepsilon ^{\alpha }_{{\bf k}\lambda ^{\prime }} = \delta _{\lambda \lambda ^{\prime }}$. Since $P_i^{\alpha }$ and $u_i^{\alpha }$ are real quantities, we impose the conditions $u_{{\bf k}\lambda }^{\ast} = u_{-{\bf k}\lambda }$, $P_{{\bf k}\lambda }^{\ast} = P_{-{\bf k}\lambda }$ and $\varepsilon _{- {\bf k}\lambda } = \varepsilon _{{\bf k}\lambda }$. It is important to note that in the above Fourier expansion, due to periodicity of the lattice, $u_{{\bf k}\lambda } = u_{{\bf k} + {\bf G}\lambda }$ where ${\bf G}$ is a reciprocal lattice vector. To avoid redundant Fourier components we restrict the wavevector ${\bf k}$ to the first Brillouin zone (BZ). The same consideration applies for $P_{{\bf k}\lambda }$. We also define the Fourier transform of the dynamical matrix $C^{\alpha \beta }({\bf R}_k)$ by

$$C^{\alpha \beta }({\bf k}) = \sum_k e^{- i {\bf k}\cdot {\bf R}_k } C^{\alpha \beta } ({\bf R}_k).$$ (4)

Here again ${\bf k}$ is restricted to the first Brillouin zone. In terms of the Fourier components the Hamiltonian in equation (2) can be expressed as

$${\mathcal H}= \frac{1}{2m} \sum_{{\bf k}\lambda } P_{{\bf k... ...bf k} \lambda _1} \varepsilon ^{\beta }_{{\bf k}\lambda _2}.$$

To write the above equation we have used the orthonormality of the polarization vectors, and the relation $\sum_i e^{i ({\bf k}- {\bf k}^{\prime })\cdot {\bf R}_i} = N \delta _{{\bf k}, {\bf k}^{\prime }}$. We can choose the polarization vectors as the eigenvectors of the dynamical matrix , i.e.

$$C^{\alpha \beta }({\bf k}) \varepsilon ^{\beta }_{{\bf k}\l... ...2_{{\bf k}\lambda } \varepsilon ^{\alpha }_{{\bf k}\lambda }.$$ (5)

Then, the Hamiltonian for the phonons can be written as

$${\mathcal H}= \frac{1}{2m} \sum_{{\bf k}\lambda } P_{{\bf k... ...2_{{\bf k}\lambda } u_{{\bf k}\lambda } u_{-{\bf k}\lambda },$$ (6)

which describe oscillators whose frequency depend on wavevector. To find the phonon frequencies we have to solve the eigenvalue equation (5). For this we have to calculate $C^{\alpha \beta }({\bf k})$. We get, from equations (3) and (4)

$$C^{\alpha \beta }({\bf k}) = \sum_{{\bf R}_k \neq 0} e^{- i {\bf k}\cdot {\bf R}_k } C^{\alpha \beta } ({\bf R}_k) + C^{\alpha \beta }({\bf R}_k =0)$$

$$= \sum_{{\bf R}_k \neq 0} e^{- i {\bf k}\cdot {\bf R}_k } \left( - ... ... ^2 V}{\partial r^{\alpha } \partial r^{\beta }}({\bf R}_i - {\bf r}^{\prime })$$

$$= \sum_{{\bf R}_k} \left( 1 - e^{- i {\bf k}\cdot {\bf R}_k } \righ... ...2 V}{\partial r^{\alpha } \partial r^{\beta }}({\bf R}_i - {\bf r}^{\prime }) .$$

In the last line we could add the term ${\bf R}_k =0$ in the summation because it gives zero contribution to the sum. We express the Fourier transform of the Coulomb potential by $V(r) = \sum_{{\bf q}} e^{ i {\bf q}\cdot {\bf r}} V_{{\bf q}}$, where $V_{{\bf q}} = 4 \pi e^2/(V q^2)$. Note that, since the Coulomb potential does not have the periodicity of the lattice, wavevector ${\bf q}$ can take any value in the reciprocal space and is not restricted within the first Brillouin zone. Substituting in the above equation we get

$$C^{\alpha \beta }({\bf k}) = \sum_{{\bf R}_k} \left( 1 - e^... ...q}} \right) e^{i {\bf q}\cdot ({\bf R}- {\bf r}^{\prime })}.$$

Since we are interested in finding the phonon frequencies for small values of ${\bf k}$, it is tempting to take the limit ${\bf k}\rightarrow 0$ before evaluating the sums. Such a procedure would show that the dynamical matrix vanishes in this limit, implying that the phonon frequencies go to zero. However, this would be incorrect. This is because the Coulomb potential is long range, and so $V_{{\bf q}}$ is singular for small values of ${\bf q}$. We have to be careful while dealing with singular functions, and the correct procedure is to evaluate the sums and then take the limit ${\bf k}\rightarrow 0$. The integral over ${\bf r}^{\prime }$ gives $(2 \pi)^3 \delta ({\bf q})$. The Dirac delta function is converted into Kronecker delta by the replacement $(2 \pi)^3 \delta ({\bf q}) /V \rightarrow \delta _{{\bf q},0}$. This is the prescription for going from continuous ${\bf q}$ space to discrete ${\bf q}$ space. Since ${\bf q}$ is not restricted within the first Brillouin zone, we also get $\sum_{{\bf R}_k} e^{i {\bf q}\cdot {\bf R}_k} = N \sum_{{\bf G}} \delta _{{\bf q}, {\bf G}}$, where ${\bf G}$ are reciprocal lattice vectors. Using these in the above equation we get,

$$C^{\alpha \beta }({\bf k}) = -N \sum_{{\bf q}} q_{\alpha }q_{\beta } V_{{\bf q}} \left\{ \left... ... G}} \delta _{{\bf G}, {\bf q}- {\bf k}} \right) - \delta _{{\bf q},0} \right\}$$

$$= N \left[ k_{\alpha } k_{\beta } V_{{\bf k}} + \sum_{{\bf G}\neq 0... ... G}+ {\bf k}\right \vert} - G_{\alpha }G_{\beta } V_{{\bf G}} \right\} \right].$$ (7)

Having evaluated the dynamical matrix, we will now solve the eigenvalue equation (5) for phonons of long wavelength (i.e. ${\bf k}\rightarrow 0$ modes). Suppose ${\bf k}= (k, 0, 0)$. Then, the dynamical matrix has the form,

$${C^{\alpha \beta }({\bf k}) \! = \!\!\!\!}$$

$$\left( \!\!\!\! \begin{array}{ccc} k^2 V_{{\bf k}} \!\! + \!\! {\... ...bf G}+ {\bf k}\right \vert } - V_{{\bf G}} \right) \end{array} \!\!\!\! \right)$$

The non-diagonal elements of the matrix are zero due to the cubic symmetry of the lattice. The longitudinal mode ($\lambda =1$), which is along the wavevector ${\bf k}$, should have eigenvector ${\bf\varepsilon }_{{\bf k}1} = (\varepsilon , 0, 0)$. Putting this form in equation (5) and noting that the dynamical matrix is diagonal, we get,

$$m \omega ^2_{{\bf k}1} = N \left\{ k^2 V_{{\bf k}} + \sum_{... ...+ {\bf k}\right \vert } - G^2_x V_{{\bf G}} \right) \right\}.$$ (8)

In the limit $k \rightarrow 0$, only the first term on the right hand side contribute, and we get

$$\omega = \sqrt{\frac{4 \pi e^2 \rho _0}{m}},$$ (9)

which is the plasma frequency for the ions. If we keep the second term of equation (8), we will get well behaved ${\bf k}$ dependent corrections to the plasma frequency for finite ${\bf k}$. It is important to note that even in the long wavelength limit the longitudinal phonons have a finite frequency. This is contrary to what we expect from Goldstone's theorem. The breakdown of Goldstone's theorem is due to the long range nature of the Coulomb potential. We can see that clearly from the first term of equation (8) which gave us the finite plasma frequency. If the potential was short range, $V_{{\bf k}}$ would be finite for vanishing ${\bf k}$. Then, in the limit ${\bf k}\rightarrow 0$, $k^2 V_{{\bf k}}$ would vanish, and we would not have got any finite frequency. It is also important to note that to see this effect it is crucial to keep the contribution from the uniform background charge. The background gives a ${\bf k}$ independent contribution to the longitudinal part of $C^{\alpha \beta }({\bf k})$. Without this contribution, the spectrum of the longitudinal phonons will reduce by a constant amount equal to the plasma frequency. The two transverse modes have eigenvectors ${\bf\varepsilon }_{{\bf k}2} = (0, \varepsilon , 0)$ and ${\bf\varepsilon }_{{\bf k}3} = (0, 0, \varepsilon )$ with eigenvalue equations

$$m\omega ^2_{{\bf k}2} = N \sum_{{\bf G}\neq 0} G_y^2 \left( V_{\left \vert {\bf G}+ {\bf k} \right \vert } - V_{{\bf G}} \right),$$ (10)

$$m\omega ^2_{{\bf k}3} = N \sum_{{\bf G}\neq 0} G_z^2 \left( V_{\left \vert {\bf G}+ {\bf k} \right \vert } - V_{{\bf G}} \right).$$ (11)

The two equations above are equivalent, and they give the same eigenvalue. In the limit of $k \rightarrow 0$,

$$V_{\left \vert {\bf G}+ {\bf k}\right \vert } - V_{{\bf G}} = \frac{1}{2} \left( \frac{\partial ^2 V}{\partial q_{\alpha } \partial q_{\beta }} \right)_{\left\vert {\bf G} \right\vert} k_{\alpha }k_{\beta }$$

$$= \frac{4 \pi e^2}{V} \left( \frac{4 G_x^2}{\left\vert {\bf G}\right\vert^6} - \frac{1}{\left\vert {\bf G}\right\vert^4} \right) k^2$$ (12)

Using equations (10) and (12), we find that the frequency of the long wavelength transverse phonons is proportional to the wavevector, and is given by

$$\omega _{{\bf k}} = \left( \frac{ 4 \pi e^2 \rho _0}{m} \su... ...}{\left\vert {\bf G} \right\vert^4} \right) \right)^{1/2} k.$$ (13)