Many Body Interactions–Introduction

Abstract

The Hamiltonian of a many particle system is usually of the form

$$H = \sum _i H_0(i) + \frac{1}{2} \sum _{i\ne j}V_{ij}$$

.

Appendices

Problems

11.1

Let us consider the paramagnetic state of a degenerate electron gas, in which $\overline{n}_{\varvec{k}\sigma }=1$ for $\varepsilon _{\varvec{k}\sigma } < \varepsilon _\mathrm{F}$ and zero otherwise.

• (a) Show that the exchange contribution to the energy of wave vector $\varvec{k}$ and spin $\sigma$ is

  $$ \varSigma _\mathrm{X\sigma }(\varvec{k}) = -\frac{1}{\varOmega }\sum _{\varvec{k}'} \overline{n}_{\varvec{k}'\sigma }\frac{4\pi e^2}{|\varvec{k}-\varvec{k}'|^2}. $$

• (b) Convert the sum over $\varvec{k}'$ to an integral and perform the integral to obtain

  $$ \varSigma _\mathrm{X\sigma }(\varvec{k}) = -\frac{e^2k_\mathrm{F}}{\pi } \left[ 1+\frac{1-x^2}{2x}\ln \mid \frac{1+x}{1-x}\mid \right] , $$

  where $x=k/k_\mathrm{F}$.

• (c) Plot $\varSigma _\mathrm{X\sigma }(k)$ as a function of $\frac{k}{k_\mathrm{F}}$.

• (d) Show that the total energy (kinetic plus exchange) for the N particle paramagnetic state in the Hartree–Fock approximation is

  $$ \begin{array}{ll} E_\mathrm{P}&{}=\sum _{\varvec{k}\sigma }\overline{n}_{\varvec{k}\sigma }\left[ \frac{\hbar ^2 k^2}{2m}+\varSigma _\mathrm{X\sigma }(\varvec{k})\right] \\ &{}=N\left( \frac{3}{5}\frac{\hbar ^2 k_\mathrm{F}^2}{2m}-\frac{3e^2k_\mathrm{F}}{4\pi }\right) . \end{array} $$

11.2

Consider the ferromagnetic state of a degenerate electron gas, in which $\overline{n}_{\varvec{k}\uparrow }=1$ for $k<k_\mathrm{F\uparrow }$ and $\overline{n}_{\varvec{k}\downarrow }=0$ for all k.

• (a) Determine the Hartree–Fock energy $E_{\varvec{k}\sigma }=\frac{\hbar ^2 k^2}{2m}+\varSigma _\mathrm{X\sigma }(\varvec{k}).$

• (b) Determine the value of $k_\mathrm{F}$ (Fermi wave vector of the nonmagnetic state) for which the ferromagnetic state is a valid Hartree–Fock solution.

• (c) Determine the value of $k_\mathrm{F}$ for which $E_\mathrm{F} = \sum _\mathrm{k} E_{\varvec{k}\uparrow }$ has lower energy than $E_\mathrm{P}$ obtained in Problem 11.1.

11.3

Evaluate $I_{xx}(q,\omega )$ in the same way as we evaluated $I_{zz}(q,\omega )$, which is given by (11.166).

11.4

The longitudinal dielectric function is written as

$$ \begin{array}{l} \epsilon ^{(l)}(q,\omega )=1-\frac{\omega _p^2}{\omega ^2}\left[ 1+I_{zz}(q,\omega \right) ]. \end{array} $$

Use $\ln (x+iy) = \frac{1}{2}\ln (x^2+y^2) +i \arctan \frac{y}{x}$ to evaluate $\epsilon _2^{(l)}(z, u)$, the imaginary part of $\epsilon ^{(l)}(q,\omega )$, where $z=q/2k_\mathrm{F}$ and $u=\omega /qv_\mathrm{F}$.

11.5

Let us consider the static dielectric function written as

$$ \epsilon ^{(l)} (q, 0)= 1+\frac{3\omega _\mathrm{p}^2}{q^2v_\mathrm{F}^2}F(z), $$

where $z=q/2k_\mathrm{F}$ and $F(z)=\frac{1}{2} +\frac{1}{4}\left( \frac{1}{z}-z\right) \ln \left( \frac{1+z}{1-z}\right)$.

• (a) Expand F(z) in power of z for $z \ll 1$. Repeat it in power of 1 / z for $z \gg 1$.

• (b) Determine the expressions of the static dielectric function $\epsilon ^{(l)} (q, 0)$ in the corresponding limits.

11.6

In the absence of a d.c. magnetic field, we see that $|\nu> = |k_x, k_y, k_z> \equiv |\varvec{k}>$, the free electron states.

• (a) Show that $<\varvec{k}'|\varvec{V}_q |\varvec{k}> = \frac{\hbar }{m} (\varvec{k}+\frac{\varvec{q}}{2})\delta _{\varvec{k}',\varvec{k}+\varvec{q}},$ where $\varvec{V}_q = \frac{1}{2}[\varvec{v}_0\mathrm{{e}}^{i\varvec{q}\cdot \varvec{r}}+\mathrm{{e}}^{i\varvec{q}\cdot \varvec{r}}\varvec{v}_0]$.

• (b) Derive the Lindhard form of the conductivity tensor given by

  $$ \underline{\sigma }(\varvec{q},\omega ) = \frac{\omega _\mathrm{p}^2}{4\pi i\omega } \left[ \underline{1}+\frac{m}{N}\sum _{\varvec{k}} \frac{f_0(E_{\varvec{k}+\varvec{q}})-f_0(E_{\varvec{k}})}{E_{\varvec{k}+\varvec{q}}-E_{\varvec{k}}-\hbar \omega }(\frac{\hbar }{m})^2(\varvec{k}+\frac{\varvec{q}}{2})(\varvec{k}+\frac{\varvec{q}}{2})\right] . $$

• (c) Show that the Lindhard form of the dielectric tensor is written as

  $$ \underline{\epsilon }(\varvec{q},\omega ) = (1-\frac{\omega _\mathrm{p}^2}{\omega ^2})\underline{1} -\frac{m\omega _\mathrm{p}^2}{N\omega ^2}\sum _{\varvec{k}} \frac{f_0(E_{\varvec{k}+\varvec{q}})-f_0(E_{\varvec{k}})}{E_{\varvec{k}+\varvec{q}}-E_{\varvec{k}}-\hbar \omega }(\frac{\hbar }{m})^2(\varvec{k}+\frac{\varvec{q}}{2})(\varvec{k}+\frac{\varvec{q}}{2}). $$

11.7

Suppose that a system has a strong and sharp absorption line at a frequency $\omega _\mathrm{A}$ and that $\epsilon _2(\omega )$ can be approximated by

$$ \epsilon _2(\omega ) = A\delta (\omega -\omega _\mathrm{A}) \text{ for } \omega >0. $$

• (a) Evaluate $\epsilon _1(\omega )$ by using the Kronig–Kramers relation.

• (b) Sketch $\epsilon _1(\omega )$ as a function of $\omega$.

11.8

The equation of motion of a charge ($-e$) of mass m harmonically bound to a lattice point $R_n$ is given, with $\varvec{x}= \varvec{r}-\varvec{R}$, by

$$ m(\ddot{\varvec{x}} +\gamma \dot{\varvec{x}} +\omega _0^2 \varvec{x}) = -e\varvec{E} \mathrm{{e}}^{i\omega t}. $$

Here $\omega _0$ is the oscillator frequency and the electric field $\varvec{E}= E \hat{x}$.

• (a) Solve the equation of motion for $\varvec{x} (t) = \varvec{X}(\omega ) \mathrm{{e}}^{i\omega t}$.

• (b) Let us consider the polarization $P(\omega )=-en_0 X(\omega )$, where $n_0$ is the number of oscillators per unit volume. Write $P(\omega )=\alpha (\omega )E$ and determine $\alpha (\omega )$.

• (c) Plot $\alpha _1(\omega )$ and $\alpha _2(\omega )$ vs. $\omega$, where $\alpha =\alpha _1+\alpha _2$.

• (d) Show that $\alpha (\omega )$ satisfies the Kronig-Kramers relation.

11.9

Take $H=\frac{1}{2m}\left( \varvec{p} +\frac{e}{c}\varvec{A}\right) ^2 -e\phi$ and $H^\prime = \frac{1}{2m}\left( \varvec{p} +\frac{e}{c}\varvec{A}^\prime \right) ^2 -e\phi ^\prime$ where $\varvec{A}^\prime = \varvec{A} + \varvec{\nabla }\chi$ and $\phi ^\prime = \phi -\frac{1}{c}\dot{\chi }$.

• (a) Show that $H^\prime - \frac{e}{c}\dot{\chi } = \mathrm{{e}}^{-\frac{ie\chi }{\hbar c}} H \mathrm{{e}}^{\frac{ie\chi }{\hbar c}}.$

• (b) Show that $\rho ^\prime =\mathrm{{e}}^{-\frac{ie\chi }{\hbar c}} \rho \mathrm{{e}}^{\frac{ie\chi }{\hbar c}}$ satisfies the same equation of motion, viz. $\frac{\partial \rho ^\prime }{\partial t} +\frac{i}{\hbar }\left[ H^\prime ,\rho ^\prime \right] _-=0$ as $\rho$ does.

11.10

Let us take $\tilde{\rho }_0(H,\eta )=\left[ \mathrm{exp}(\frac{H-\eta }{\varTheta })+1\right] ^{-1}$ as the local thermal equilibrium distribution function (or local equilibrium density matrix). Here $\eta (\varvec{r}, t)=\zeta +\zeta _1(\varvec{r}, t)$ is the local value of the chemical potential at position $\varvec{r}$ and time t, while $\zeta$ is the overall equilibrium chemical potential. Remember that the total Hamiltonian H is written as $H=H_0+H_1$. Write $\tilde{\rho }_0(H,\eta )=\rho _0(H_0,\zeta )+\rho _2$ and show that the matrix element of $\rho _2$ in the representation where $H_0$ is diagonal is given, to terms linear in the self-consistent field, by

$$ \langle \varvec{k}|\rho _2|\varvec{k}'\rangle = \frac{f_0(\varepsilon _{\varvec{k}^\prime })-f_0(\varepsilon _{\varvec{k}})}{\varepsilon _{\varvec{k}^\prime }-\varepsilon _{\varvec{k}}} \langle \varvec{k}|H_1-\zeta _1|\varvec{k}^\prime \rangle . $$

11.11

Longitudinal sound waves in a simple metal like Na or K can be represented by the relation $\omega ^2=\frac{\varOmega _\mathrm{p}^2}{\epsilon ^{(l)}(q,\omega )}$, where $\epsilon ^{(l)}(q,\omega )$ is the Lindhard dielectric function. We know that, for finite $\omega$, $\epsilon ^{(l)}(q,\omega )$ can be written as $\epsilon ^{(l)}(q,\omega )=\epsilon _1(q,\omega ) +i \epsilon _2(q,\omega )$. This gives rise to $\omega =\omega _1 +i\omega _2$, and $\omega _2$ is proportional to the attenuation of the sound wave via excitation of conduction electrons. Estimate $\omega _2(q)$ for the case $\omega _1^2\simeq \frac{q^2\varOmega _\mathrm{p}^2}{k_\mathrm{s}^2} \gg \omega _2^2$.

Summary

In this chapter we briefly introduced method of second quantization and Hartree–Fock approximation to describe the ferromagnetism of a degenerate electron gas and spin density wave states in solids. Equation of motion method is considered for density matrix to describe gauge invariant theory of linear responses in the presence of the most general electromagnetic disturbance. Behavior of Lindhard dielectric functions and static screening effects are examined in detail. Oscillatory behavior of the induced electron density in the presence of point charge impurity and an anomaly in the phonon dispersion relation are also discussed.

In the second quantization or occupation number representation, the Hamiltonian of a many particle system with two body interactions can be written as

$$ H=\sum _{k} \varepsilon _k c_{k}^\dag c_k +\frac{1}{2}\sum _{kk^\prime ll^\prime } \langle k^\prime l^\prime |V|kl\rangle c_{k^\prime }^\dag c_{l^\prime }^\dag c_l c_k, $$

where $c_k$ and $c_{k^\prime }^\dag$ satisfy commutation (anticommutation) relation for Bosons (Fermions).

The Hartree–Fock Hamiltonian is given by $H=\sum _i E_i c_i^\dag c_i,$ where

$$ E_i = \varepsilon _i +\sum _{j} \overline{n_j}\left[ \langle ij|V|ij\rangle -\langle ij|V|ji\rangle \right] . $$

The Hartree–Fock ground state energy of a degenerate electron gas in the paramagnetic phase is given by $E_{\varvec{ks}} = \frac{\hbar ^2 k^2}{2m} -\frac{e^2 k_\mathrm{F}}{2\pi }\left[ 2+\frac{k_\mathrm{F}^2 -k^2}{kk_\mathrm{F}}\ln \left( \frac{k_\mathrm{F}+k}{k_\mathrm{F}-k}\right) \right] .$ The total energy of the paramagnetic state is

$$ E_\mathrm{P} = N\left[ \frac{3}{5}\frac{\hbar ^2 k_\mathrm{F}^2}{2m} -\frac{3}{4\pi }e^2 k_\mathrm{F}\right] . $$

If only states of spin $\uparrow$ are occupied, we have

$$ E_{k\uparrow } = \frac{\hbar ^2 k^2}{2m} -\frac{2^{1/3}e^2 k_\mathrm{F}}{2\pi }\left[ 2+\frac{2^{2/3}k_\mathrm{F}^2 -k^2}{2^{1/3}k_\mathrm{F}k}\ln \left( \frac{2^{1/3}k_\mathrm{F}+k}{2^{1/3}k_\mathrm{F}-k}\right) \right] ; E_{k\downarrow } = \frac{\hbar ^2 k^2}{2m}. $$

The total energy in the ferromagnetic phase is

$$ E_\mathrm{F} = \sum E_{k\uparrow } =N\left[ 2^{2/3}\frac{3 }{5}\frac{\hbar ^2 k_\mathrm{F}^2 }{2m} -2^{1/3}\frac{3}{4\pi }e^2 k_\mathrm{F}\right] . $$

The exchange energy prefers parallel spin orientation, but the cost in kinetic energy is high for a ferromagnetic spin arrangement. In a spin density wave state, the (negative) exchange energy is enhanced with no costing as much in kinetic energy. The Hartree-Fock ground state of a spiral spin density wave can be written as $|\phi _{\varvec{k}}\rangle {=} \cos \theta _k |\varvec{k}\uparrow \rangle + \sin \theta _k|\varvec{k}+\varvec{Q}\downarrow \rangle .$

In the presence of the self-consistent (Hartree) field $\{\phi , \varvec{A}\}$, the Hamiltonian is written as $\mathcal {H}=\mathcal {H}_0 + \mathcal {H}_1$, where $\mathcal {H}_0$ is the Hamiltonian in the absence of the self-consistent field and $\mathcal {H}_1 = \frac{e}{2c} \left( \varvec{v}_0 \cdot \varvec{A}+\varvec{A}\cdot \varvec{v}_0\right) -e\phi ,$ up to terms linear in $\{\phi , \varvec{A}\}$. Here $\varvec{v}_0 = \frac{\varvec{p}}{m}$ and the equation of motion of $\rho$ is $\frac{\partial \rho }{\partial t}+\frac{i}{\hbar }\left[ H,\rho \right] _-=0.$

The current and charge densities at $(\varvec{r}_0,t)$ are given, respectively, by

$$ \varvec{j}(\varvec{r}_0,t) = \text{ Tr } \left[ -e\left( \frac{1}{2}\varvec{v}\delta (\varvec{r}-\varvec{r}_0)+\frac{1}{2}\delta (\varvec{r}-\varvec{r}_0)\varvec{v}\right) \hat{\rho }\right] ; n(\varvec{r}_0,t) = \text{ Tr } \left[ -e\delta (\varvec{r}-\varvec{r}_0)\hat{\rho }\right] . $$

Here $-e\left[ \frac{1}{2}\varvec{v}\delta (\varvec{r}-\varvec{r}_0)+\frac{1}{2}\delta (\varvec{r}-\varvec{r}_0)\varvec{v}\right]$ is the operator for the current density at position $\varvec{r}_0$, while $-e\delta (\varvec{r}-\varvec{r}_0)$ is the charge density operator. Fourier transform of $\varvec{j}(\varvec{r}_0,t)$ gives

$$ \varvec{j}(\varvec{q},\omega )=\underline{\varvec{\sigma }}(\varvec{q},\omega )\cdot \varvec{E}(\varvec{q},\omega ) $$

where the conductivity tensor is given by $\underline{\varvec{\sigma }}(\varvec{q},\omega ) = \frac{\omega _p^2}{4\pi i \omega }\left[ \varvec{1}+\underline{\varvec{I}}(\varvec{q},\omega )\right] .$ Here

$$ {\varvec{\underline{I}}}(\varvec{q},\omega ) = \frac{m}{N}\sum _{\varvec{k},\varvec{k}^\prime } \frac{f_0(\varepsilon _{k^\prime })-f_0(\varepsilon _k)}{\varepsilon _{k^\prime }-\varepsilon _k-\hbar \omega } \langle k^\prime |\varvec{V}_q|k\rangle \langle k^\prime |\varvec{V}_q|k\rangle ^* $$

and the operator $\varvec{V}_q$ is defined by $\varvec{V}_q = \frac{1}{2}\varvec{v}_0\mathrm{{e}}^{i\varvec{q}\cdot \varvec{r}}+\frac{1}{2}\mathrm{{e}}^{i\varvec{q}\cdot \varvec{r}}\varvec{v}_0.$

The longitudinal and transverse dielectric functions are written as

$$ \epsilon ^{(l)}(q,\omega )=1-\frac{\omega _p^2}{\omega ^2}\left[ 1+I_{zz}(q,\omega \right) ] ; \epsilon ^{(\mathrm{{Tr}})}(q,\omega ) =1-\frac{\omega _p^2}{\omega ^2}\left[ 1+I_{xx}(q,\omega \right) ]. $$

Real part ($\epsilon _1$) and imaginary part ($\epsilon _2$) of the dielectric function satisfy the relation

The power dissipation per unit volume is then written $\mathcal {P}(\varvec{q}, \omega )=\frac{\omega }{2\pi }\epsilon _2(q,\omega )\mid E_0 \mid ^2.$

Due to collisions of electrons with lattice imperfections, the conductivity of a normal metal is not infinite at zero frequency. In the presence of collisions, the equation of motion of the density matrix becomes, in a relaxation time approximation,

$$ \frac{\partial \rho }{\partial t}+\frac{i}{\hbar }\left[ H,\rho \right] _-=-\frac{\rho -\tilde{\rho }_0}{\tau }. $$

Here $\tilde{\rho }_0$ is a local equilibrium density matrix. Including the effect of collisions, the induced current density becomes

$$ \varvec{j}(\varvec{q},\omega ) = \frac{\omega _\mathrm{p}^2}{4\pi i \omega } \left\{ \underline{\varvec{1}}+\underline{\varvec{I}} -\frac{i\omega \tau }{1+i\omega \tau }\frac{(\varvec{K}_1-\varvec{K}_2)(\varvec{K}_1^\prime -\varvec{K}_2^\prime )}{L_1+i\omega \tau L_2}\right\} \cdot \varvec{E}. $$

In the static limit, the dielectric function reduces to

$$ \epsilon ^{(l)} (q, 0)= 1+\frac{3\omega _\mathrm{p}^2}{q^2v_\mathrm{F}^2}F(z), $$

where $F(z)=\frac{1}{2} +\frac{1}{4}\left( \frac{1}{z}-z\right) \ln \left( \frac{1+z}{1-z}\right)$ and $z=q/2k_\mathrm{F}$. The self-consistent screened potential is written as

$$ \phi (q) = \frac{4\pi e}{q^2+ k_\mathrm{s}^2F(q/2k_\mathrm{F})}. $$

where $k_\mathrm{s} = \sqrt{\frac{4k_\mathrm{F}}{\pi a_0}}$. For high density limit $(\pi a_0 k_\mathrm{F} \gg 1)$ and large distances from the point charge impurity, the induced electron density is given by

$$ n_1(r)=\frac{6n_0}{a_0 k_\mathrm{F}}\, \frac{\cos 2k_\mathrm{F} r}{(2k_\mathrm{F}r)^3}. $$

Electronic screening of the charge fluctuations in the ion density modifies the dispersion relation of phonons, for example,

$$ \omega ^2 \simeq \frac{s_l^2 q^2}{F(z)+\frac{\pi a_0}{4k_\mathrm{F}}q^2} $$

showing a small anomaly at $q=2k_\mathrm{F}$.