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\begin_body

\begin_layout Title
The Kohn Anomaly: Simple Theory
\end_layout

\begin_layout Author
Daniel Sank, Max Hofmann, James Wenner
\end_layout

\begin_layout Date
March 9, 2009
\end_layout

\begin_layout Section*
Introduction
\end_layout

\begin_layout Standard
Phonons are essentially (plane wave) displacements of the ion cores in a
 solid.
 Since the ion cores are charged a phonon can be thought of as a travelling
 perturbation in the charge density distribution in a solid.
 Since electrons are also charged it's clear that electrons and phonons
 should interact in some way, and that the presence of each should affect
 the dispersion relation of the other.
 The Kohn anomaly refers to singularities in phonon dispersion relations
 at certain wave-vectors q, arising from an abrupt change in the ability
 of the electrons to screen the charge displacements induced by phonons
 at those wave-vectors.
\end_layout

\begin_layout Standard
In order to demonstrate the presence of these singularities theoretically
 we need to find some way of calculating the response of the electrons to
 the charge perturbation caused by the phonons.
 We do this using first order perturbation theory.
 In particular we calculate the perturbed expectation value of the electron
 density 
\begin_inset Formula $\Psi^{\dagger}\Psi$
\end_inset

 where the perturbing Hamiltonian is the electrostatic energy caused by
 the phonon distortion of the ionic cores in the solid.
 In the lingo of quantum many body theory this quantity is called the 
\begin_inset Quotes eld
\end_inset

density-density correlation function.
\begin_inset Quotes erd
\end_inset


\end_layout

\begin_layout Standard
Note that we are treating the electrostatic perturbation caused by the phonon
 as time independent.
 This may seem strange given that in real processes phonons are really travellin
g wave packets, but the time independent picture is justified by the fact
 that electrons equilibrate much faster than phonon lifetimes, and so as
 far as the electrons are concerned the affect of the phonon is time independent.
 
\end_layout

\begin_layout Section*
Simple Calculation of Zero Frequency Correlation Function
\end_layout

\begin_layout Standard
We use first order (nondegenerate) perturbation theory to calculate the
 density response of the free electron gas to an externally applied charge
 distribution.
 Recall that in first order perturbation theory the shift in the states
 is given by the formula,
\begin_inset Formula \[
|n^{1}\rangle=\sum_{m\neq n}\frac{\langle m^{0}|\hat{V}|n^{0}\rangle}{E_{n}^{0}-E_{m}^{0}}|m^{0}\rangle\]

\end_inset

Therefore, the expectation value of an operator 
\begin_inset Formula $A$
\end_inset

 to first order is
\begin_inset Formula \begin{eqnarray*}
\langle A\rangle & = & \langle n|A|n\rangle=\langle n^{0}+n^{1}|A|n^{0}+n^{1}\rangle\\
\langle A\rangle & = & \langle n^{0}|A|n^{0}\rangle+\langle n^{0}|A|n^{1}\rangle+\langle n^{1}|A|n^{0}\rangle+O(V^{2})\\
\langle A\rangle-\langle A\rangle_{0} & = & 2\Re\langle n^{0}|A|n^{1}\rangle\\
\langle A\rangle-\langle A\rangle_{0}\equiv\delta\langle A\rangle & = & 2\Re\left[\sum_{m\neq n}\frac{\langle n^{0}|A|m^{0}\rangle\langle m^{0}|V|n^{0}\rangle}{E_{n}^{0}-E_{m}^{0}}\right]\qquad(*)\end{eqnarray*}

\end_inset

Our perturbing Hamiltonian will be the energy due to an externally applied
 charge distribution,
\begin_inset Formula \[
V=\int d^{3}r\left[\rho_{\textrm{el}}(r)\right]\phi(r)=\int d^{3}r\left[\Psi^{\dagger}(r)\Psi(r)\right]\int dr'\frac{\rho_{\textrm{ext}}(r')}{|r-r'|}=\int d^{3}r\left(\Psi(r)\Psi(r)\right)\left[\rho*\frac{1}{r}\right](r)\]

\end_inset

where 
\begin_inset Formula $\rho_{\textrm{el}}$
\end_inset

 is the charge density of the electrons, 
\begin_inset Formula $\rho$
\end_inset

 is the externally applied (perturbing) charge distribution, and the * indicates
 convolution.
 By the convolution theorem 
\begin_inset Formula ${\cal F}\left[f*g\right]={\cal F}\left[f\right]{\cal F}\left[g\right]$
\end_inset

, therefore
\begin_inset Formula \[
\left[\rho*\frac{1}{r}\right](r)=\int\frac{d^{3}k}{(2\pi)^{3}}e^{ikr}\rho(k)\frac{4\pi}{k^{2}}\]

\end_inset

where we've used the fact that that the Fourier transform of 
\begin_inset Formula $\frac{1}{r}$
\end_inset

 is 
\begin_inset Formula $\frac{4\pi}{k^{2}}$
\end_inset

.
 Plugging this into our expression for V we get 
\begin_inset Formula \begin{eqnarray*}
V & = & \int d^{3}r\left(\Psi^{\dagger}(r)\Psi(r)\right)\left[\rho*\frac{1}{r}\right](r)=\int d^{3}r\left(\Psi^{\dagger}(r)\Psi(r)\right)\int\frac{d^{3}k}{(2\pi)^{3}}e^{ikr}\rho(k)\frac{4\pi}{k^{2}}\\
V & = & \int d^{3}r\left(\Psi^{\dagger}(r)\Psi(r)\right)\int\frac{d^{3}k}{(2\pi)^{3}}e^{ikr}\rho(k)\frac{4\pi}{k^{2}}\end{eqnarray*}

\end_inset

Consider the case where 
\begin_inset Formula $\rho(r)=\rho_{0}e^{iqr}$
\end_inset

 so that 
\begin_inset Formula $\rho(k)=\rho_{0}(2\pi)^{3}\delta(k-q)$
\end_inset

.
 Then we can do the k integral easily, leaving,
\begin_inset Formula \begin{eqnarray*}
V & = & 4\pi\frac{\rho_{0}}{q^{2}}\int d^{3}r\left(\Psi^{\dagger}(r)\Psi(r)\right)e^{iqr}\end{eqnarray*}

\end_inset

Now Fourier transform the 
\begin_inset Formula $\Psi$
\end_inset

's, using 
\begin_inset Formula $\Psi(r)=\int\frac{d^{3}k}{(2\pi)^{3}}e^{-ikr}a_{k}$
\end_inset

 and 
\begin_inset Formula $\Psi^{\dagger}(r)=\int\frac{d^{3}k'}{(2\pi)^{3}}e^{ik'r}a_{k'}^{\dagger}$
\end_inset

,
\begin_inset Formula \[
V=4\pi\frac{\rho_{0}}{q^{2}}\int d^{3}r\int\frac{d^{3}k}{(2\pi)^{3}}\int\frac{d^{3}k'}{(2\pi)^{3}}e^{i(k-k'+q)r}a_{k}^{\dagger}a_{k'}\]

\end_inset

The integral over 
\begin_inset Formula $r$
\end_inset

 can be done easily and gives 
\begin_inset Formula $(2\pi)^{3}\delta(k-k'+q)$
\end_inset

.
 We can then do the 
\begin_inset Formula $k'$
\end_inset

 integral, and what's left is
\begin_inset Formula \[
V=4\pi\frac{\rho_{0}}{q^{2}}\int\frac{d^{3}k}{(2\pi)^{3}}a_{k}^{\dagger}a_{k+q}\]

\end_inset

Change variables so that 
\begin_inset Formula $k+q\rightarrow k$
\end_inset

 to get
\begin_inset Formula \[
V=4\pi\frac{\rho_{0}}{q^{2}}\int\frac{d^{3}k}{(2\pi)^{3}}a_{k-q}^{\dagger}a_{k}\]

\end_inset

What we really want to calculate is 
\begin_inset Formula $\langle m^{0}|V|n^{0}\rangle$
\end_inset

 where 
\begin_inset Formula $|m^{0}\rangle$
\end_inset

 is an arbitrary state of the system and 
\begin_inset Formula $|n^{0}\rangle$
\end_inset

 is the thermal state away from which we're perturbing.
 Let's consider the case 
\begin_inset Formula $T=0$
\end_inset

 so that 
\begin_inset Formula $|n^{0}\rangle$
\end_inset

 is just the pure state with all states filled up to the Fermi level, and
 all other states empty.
 Denote this state by 
\begin_inset Formula $|\Omega\rangle$
\end_inset

 from now on.
 With this notation the matrix element we need is
\begin_inset Formula \[
\langle m^{0}|V|\Omega\rangle=4\pi\frac{\rho_{0}}{q^{2}}\int\frac{d^{3}k}{(2\pi)^{3}}\langle m^{0}|a_{k-q}^{\dagger}a_{k}|\Omega\rangle\]

\end_inset

The inner product sitting in the integral is only nonzero if 
\begin_inset Formula $|m^{0}\rangle=a_{k-q}^{\dagger}a_{k}|\Omega\rangle$
\end_inset

, and then only if 
\begin_inset Formula $k<k_{f}$
\end_inset

 and 
\begin_inset Formula $|k-q|>k_{F}$
\end_inset

 (we exclude the case 
\begin_inset Formula $q=0$
\end_inset

 which is justified by charge neutrality of the system).
 Therefore we can write
\begin_inset Formula \[
\langle m^{0}|V|\Omega\rangle=4\pi\frac{\rho_{0}}{q^{2}}\int\frac{d^{3}k}{(2\pi)^{3}}\delta\left[|m^{0}\rangle,a_{k-q}^{\dagger}a_{k}|\Omega\rangle\right]\Theta(k_{F}-k)\Theta(|k-q|-k_{F})\]

\end_inset


\end_layout

\begin_layout Standard
Inserting our result for 
\begin_inset Formula $\langle m^{0}|V|\Omega\rangle$
\end_inset

 into our expression for the perturbed expectation value of 
\begin_inset Formula $A$
\end_inset

 we ge
\begin_inset Formula \begin{eqnarray*}
\delta\langle A\rangle & = & 2\Re\left[4\pi\frac{\rho_{0}}{q^{2}}\int\frac{d^{3}k}{(2\pi)^{3}}\sum_{m\neq\Omega}\frac{\delta\left[|m^{0}\rangle,a_{k-q}^{\dagger}a_{k}|\Omega\rangle\right]\Theta(k_{F}-k)\Theta(|k-q|-k_{F})\langle\Omega|A|m^{0}\rangle}{E_{\Omega}-E_{m}^{0}}\right]\\
\delta\langle A\rangle & = & 2\Re\left[4\pi\frac{\rho_{0}}{q^{2}}\int\frac{d^{3}k}{(2\pi)^{3}}\frac{\Theta\left(k_{F}-k\right)\Theta\left(|k-q|-k_{F}\right)\langle\Omega|A|a_{k-q}^{\dagger}a_{k}\Omega\rangle}{E_{k}^{0}-E_{k-q}^{0}}\right]\end{eqnarray*}

\end_inset

where 
\begin_inset Formula $E_{k}^{0}\equiv\hbar^{2}k^{2}/2m$
\end_inset

.
\end_layout

\begin_layout Standard
Remember that the operator who's perturbed expectation value we want to
 know is the electron density 
\begin_inset Formula $\Psi^{\dagger}(x)\Psi(x)$
\end_inset

.
 Using Fourier transform we can write this operator as 
\begin_inset Formula $A=\Psi^{\dagger}(x)\Psi(x)=\int\frac{d^{3}l}{(2\pi)^{3}}\int\frac{d^{3}s}{(2\pi)^{3}}e^{i(l-s)x}a_{l}^{\dagger}a_{s}$
\end_inset

, and inserting this into the equation for 
\begin_inset Formula $\delta\langle A\rangle$
\end_inset

 gives, 
\begin_inset Formula \[
\delta\langle\Psi^{\dagger}(x)\Psi(x)\rangle=2\Re\left[4\pi\frac{\rho_{0}}{q^{2}}\int\frac{d^{3}k}{(2\pi)^{3}}\int\frac{d^{3}l}{(2\pi)^{3}}\int\frac{d^{3}s}{(2\pi)^{3}}e^{i(l-s)x}\frac{\Theta(k_{F}-k)\Theta(|k-q|-k_{F})\langle\Omega|a_{l}^{\dagger}a_{s}a_{k-q}^{\dagger}a_{k}|\Omega\rangle}{E_{k}^{0}-E_{k-q}^{0}}\right]\]

\end_inset

Let's process the matrix element 
\begin_inset Formula $\langle\Omega|a_{l}^{\dagger}a_{s}a_{k-q}^{\dagger}a_{k}|\Omega\rangle$
\end_inset

.
 Some thought in a peaceful moment reveals that the matrix element is nonzero
 only when certain relations between 
\begin_inset Formula $l,s,k,$
\end_inset

 and 
\begin_inset Formula $q$
\end_inset

 hold.
 These relations, along with the resulting factors they produce in the integrand
 are summarized here
\begin_inset Formula \[
\begin{cases}
k<k_{F} & \quad\Theta\left(k_{F}-k\right)\\
|k-q|>k_{F} & \quad\Theta\left(|k-q|-k_{F}\right)\\
s=k-q & \quad(2\pi)^{3}\delta\left(s-(k-q)\right)\\
l=k & \quad(2\pi)^{3}\delta(l-k)\end{cases}\]

\end_inset

When these conditions are met the matrix element equals unity.
 Stuffing all of this into the previous line for 
\begin_inset Formula $\delta\langle A\rangle$
\end_inset

 leaves us with
\begin_inset Formula \[
\delta\langle\Psi^{\dagger}(x)\Psi(x)\rangle=2\Re\left[4\pi\frac{\rho_{0}}{q^{2}}\int\frac{d^{3}k}{(2\pi)^{3}}e^{iqx}\frac{\Theta(k_{F}-k)\Theta(|k-q|-k_{F})}{E_{k}^{0}-E_{k-q}^{0}}\right]\]

\end_inset

This integral can be done by fairly elementary means.
 The result is
\begin_inset Formula \[
\delta\langle\Psi^{\dagger}(x)\Psi(x)\rangle=(\textrm{constants})*\frac{\rho_{0}e^{iqx}}{q^{2}}\left\{ 1+\frac{k_{F}}{q}\left[1-\frac{q^{2}}{4k_{F}^{2}}\right]\log\bigg|\frac{q+2k_{F}}{q-2k_{F}}\bigg|\right\} \]

\end_inset

The relevant features of this result are as follows
\end_layout

\begin_layout Itemize
The electron density response is linear in the perturbation
\end_layout

\begin_layout Itemize
The amplitude of the response has a sharp peak at 
\begin_inset Formula $q=2k_{F}$
\end_inset

 indicating that the response of the electron gas to the charge perturbation
 of a phonon changes abruptly at this wavevector.
\end_layout

\begin_layout Itemize
Since the electron-phonon coupling changes abruptly, we expect the properties
 of the phonons and electrons to show interesting features when 
\begin_inset Formula $q$
\end_inset

 is near 
\begin_inset Formula $2k_{F}$
\end_inset

.
\end_layout

\end_body
\end_document