Fermi Gas

  • We’ll study Fermi gas with weak interactions (perturbation theory)

  • Illustrates Landau’s Fermi liquid theory, a ‘standard model’ of condensed matter

  • Landau’s theory applies more generally, even when interactions not weak

Weakly Interacting Fermi Gas

  • Fermi gas with short-ranged interactions

$$ H = \int d\br\left[ \sum_{s=\uparrow,\downarrow}\frac{1}{2m}\nabla\pdop_s\cdot\nabla\pop_s + U_0 \pdop_\uparrow\pdop_\downarrow\pop_\downarrow\pop_\uparrow\right] $$

  • Work in momentum space $$ H =\sum_{\bk,s} \epsilon(\bk)\adop_{\bk,s}\aop_{\bk,s} + \overbrace{\frac{U_0}{V}\sum_{\bk_1+\bk_2=\bk_3+\bk_4} \adop_{\bk_1,\uparrow}\adop_{\bk_2,\downarrow}\aop_{\bk_3,\downarrow}\aop_{\bk_4,\uparrow}}^{\equiv H_\text{int}} $$ with $\epsilon(\bk)=\bk^2/2m$, and $V$ the volume.

  • $U_0=0$: eigenstates are product states $\ket{\mathbf{N}}$ of single particle momentum states specified by the occupancies $N_{s}(\bk) = 0,1$

  • Ground state is Fermi sphere of radius $k_\text{F}$ in momentum space with $N_{s}(\bk) = \theta(k_F-\abs{\bk})$.

  • Low energy excited states will have $N_{s}(\bk)=1$ for $\abs{\bk}\ll k_\text{F}$ and $N_{s}(\bk)=0$ for $\abs{\bk}\gg k_\text{F}$.

  • In perturbation theory we can still label eigenstates by these occupation numbers even though eigenstates $\neq\ket{\mathbf{N}}$

  • Without interaction energy of a state $\ket{\mathbf{N}}$ is

$$ E^{(0)}(\mathbf{N}) = \sum_{\bk,s} \epsilon(\bk)N_{s}(\bk) $$

  • For $U_0\neq 0$ energy $E(\mathbf{N})$ is function of labels, but no longer linear

  • Second order expansion of energy in terms of deviation of occupancies from ground state values is key ingredient of Landau’s theory

Perturbation Theory to Second Order

  • Second order perturbation theory for the energies

$$ \begin{align*} E^{(1)}(\mathbf{N}) &= \braket{\mathbf{N}|H_\text{int}|\mathbf{N}}\\ E^{(2)}(\mathbf{N}) &= \sum_{\mathbf{N}'\neq \mathbf N}\frac{\abs{\braket{\mathbf{N'}|H_\text{int}|\mathbf{N}}}^2}{E^{(0)}(\mathbf{N})-E^{(0)}(\mathbf{N}')} \end{align*} $$

  • First order correction is easy $$ E^{(1)}(\mathbf{N}) = \frac{U_0}{V} \sum_{\bk,\bk'} N_{\uparrow}(\bk)N_{\downarrow}(\bk') = \frac{U_0}{V}N_\uparrow N_\downarrow. $$ (energy used in Stoner criterion in Lecture 6)
  • For second order we need $\braket{\mathbf{N}'|H_\text{int}|\mathbf{N}}$, nonzero if $$ \begin{align*} N'_{\bk_1,\uparrow} = N_{\uparrow}(\bk_1) + 1, \quad N'_{\downarrow}(\bk_2) = N_{\downarrow}(\bk_2) + 1\\ N'_{\downarrow}(\bk_3) = N_{\downarrow}(\bk_3) - 1, \quad N'_{\uparrow}(\bk_4) = N_{\uparrow}(\bk_4) - 1, \end{align*} $$ for $\bk_i$ satisfying $\bk_1+\bk_2=\bk_3+\bk_4$ $$ \braket{\mathbf{N}'|H_\text{int}|\mathbf{N}} = \frac{U_0}{V} \left(1-N_{\uparrow}(\bk_1)\right)\left(1-N_{\downarrow}(\bk_2)\right)N_{\downarrow}(\bk_3)N_{\uparrow}(\bk_4) $$ (ignoring any coinciding momenta; occupancies 0 or 1)

$$ E^{(2)}(\mathbf{N}) = \left(\frac{U_0}{V}\right)^2 \sum_{\bk_1+\bk_2=\bk_3+\bk_4}\frac{\left(1-N_{\uparrow}(\bk_1)\right)\left(1-N_{\downarrow}(\bk_2)\right)N_{\downarrow}(\bk_3)N_{\uparrow}(\bk_4)}{\epsilon(\bk_3)+\epsilon(\bk_4)-\epsilon(\bk_1)-\epsilon(\bk_2)} $$

Landau $f$ function

  • $E^{(2)}(\mathbf{N})$ has three independent momentum sums! 🙀

  • We are after excitation energies, so expand in change in occupation

$$ N_{s}(\bk) = \theta(k_F-\abs{\bk}) + n_{s}(\bk) $$

  • $n_\bk$ is mean deviation from Fermi sphere in continuum limit:

drawing

  • Excitation energy expanded in $n_\bk$

$$ \Delta E = \sum_{\bk,s} \varepsilon_s(\bk)n_{s}(\bk) + \frac{1}{2V}\sum_{\bk, s,\bk’, s’} f_{s^{}s’}(\bk,\bk’)n_{s}(\bk)n_{s’}(\bk’) $$

  • This is Landau’s key idea, not restricted to perturbation theory

  • At first order $$ \begin{align*} \varepsilon_s(\bk) &= \epsilon(\bk) + \frac{U_0 N_{\bar s}}{V}+\cdots\\ f_{\uparrow\downarrow} &= f_{\downarrow\uparrow} = U_0+\cdots,\quad f_{\uparrow\uparrow}=f_{\downarrow\downarrow}=0+\cdots \end{align*} $$ $\bar s$ is $\bar\uparrow=\downarrow$, $\bar\downarrow=\uparrow$ 🥱

  • Second order contributions to $f$-function more interesting

$$ E^{(2)}(\mathbf{N}) = \left(\frac{U_0}{V}\right)^2 \sum_{\bk_1+\bk_2=\bk_3+\bk_4}\frac{\left(1-N_{\uparrow}(\bk_1)\right)\left(1-N_{\downarrow}(\bk_2)\right)N_{\downarrow}(\bk_3)N_{\uparrow}(\bk_4)}{\epsilon(\bk_3)+\epsilon(\bk_4)-\epsilon(\bk_1)-\epsilon(\bk_2)} $$

$$ N_{s}(\bk) = \theta(k_F-\abs{\bk}) + n_{s}(\bk) $$

$$ \begin{align*} f_{\uparrow\uparrow}(\bk,\bk') = -\frac{U_0^2}{V}\left[\sum_{\bk+\bk_3=\bk'+\bk_2} \frac{N_{\downarrow}(\bk_3)(1-N_{\downarrow}(\bk_2))}{\epsilon(\bk)+\epsilon(\bk_3)-\epsilon(\bk')-\epsilon(\bk_2)}\right.\nonumber\\ \left.+\sum_{\bk'+\bk_3=\bk+\bk_2}\frac{N_{\downarrow}(\bk_3)(1-N_{\downarrow}(\bk_2))}{\epsilon(\bk')+\epsilon(\bk_3)-\epsilon(\bk)-\epsilon(\bk_2)}\right] \end{align*} $$

  • $f_{\uparrow\uparrow}(\bk,\bk’)$ is more complicated $$ \begin{align*} f_{\uparrow\downarrow}(\bk,\bk') = U_0 + f_{\uparrow\uparrow}(\bk,\bk') +\frac{U_0^2}{V}\left[\sum_{\bk+\bk'=\bk_3+\bk_4}\frac{N(\bk_3)N(\bk_4)}{\epsilon(\bk_3)+\epsilon(\bk_4)-2E_\text{F}}\right.\nonumber\\ \left.\sum_{\bk+\bk'=\bk_1+\bk_2}\frac{(1-N(\bk_1))(1-N(\bk_2))}{2E_\text{F}-\epsilon(\bk_1)-\epsilon(\bk_2)}\right] \end{align*} $$ (just for reference)

  • $n_{s}(\bk)\neq 0$ in energy window of size $k_\text{B} T$ around Fermi surface

  • At low $T$ take $\abs{\bk}=\abs{\bk'}=k_\text{F}$. New feature at second order is nontrivial dependence of $f_{s^{}s’}(\bk,\bk’)$ on angle between $\bk$ and $\bk’$.

  • Assume ground state unpolarized, i.e. $N_{s}(\bk)$ independent of $s$

  • It’s a bit fiddly to get at, but let’s work it out for the simpler case of $f_{\uparrow\uparrow}(\bk,\bk’)$! (only have one independent momentum 😀)

  • Continuum limit

$$ \begin{align*} f_{\uparrow\uparrow}(\bk,\bk') = \frac{U_0^2}{(2\pi)^3}\left[\int_{\substack{\abs{\bk_3}<k_\text{F},\abs{\bk_2}>k_\text{F}\\ \bk+\bk_3=\bk'+\bk_2 }} \frac{d\bk_3}{\epsilon(\bk_2)-\epsilon(\bk_3)} +\int_{\substack{\abs{\bk_3}<k_\text{F},\abs{\bk_2}>k_\text{F}\\ \bk'+\bk_3=\bk+\bk_2 }}\frac{d\bk_3}{\epsilon(\bk_2)-\epsilon(\bk_3)}\right] \end{align*} $$

  • Need integral

$$ \int_{\substack{\abs{\bk_3}<k_\text{F},\abs{\bk_2}>k_\text{F}\\ \bk+\bk_3=\bk'+\bk_2 }} \frac{d\bk_3}{\epsilon(\bk_2)-\epsilon(\bk_3)} $$

  • Denominator can be written ($\bk-\bk’$ is fixed) $$ \epsilon(\bk_2)-\epsilon(\bk_3)= \frac{1}{2m}\left(\bk_2+\bk_3\right)\cdot\left(\bk_2-\bk_3\right) = \frac{1}{2m}\left(\bk_2+\bk_3\right)\cdot\left(\bk-\bk’\right) $$

  • Notation $\mathbf{K} = \frac{1}{2}\left(\bk_2+\bk_3\right),\quad \bq = \frac{1}{2}\left(\bk_2-\bk_3\right)$

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  • Denominator becomes (for fixed $\bq$) $$ \epsilon(\bk_2)-\epsilon(\bk_3) = \frac{2}{m}\mathbf{K}\cdot\bq $$

  • Only angle $\theta$ between $\mathbf{K}$ and $\bq$ enters integral

  • Conditions $\abs{\bk_2}>k_\text{F}$ and $\abs{\bk_3}<k_\text{F}$ become $$ \begin{align*} \left(\mathbf{K}+\bq\right)^2>k_\text{F}^2,\quad \left(\mathbf{K}-\bq\right)^2<k_\text{F}^2 \end{align*} $$ which gives range of $K_-(\theta)<\abs{\mathbf{K}}<K_+(\theta)$ $$ K_{\pm}(\theta)=\pm q\abs{\cos\theta}+\sqrt{k_\text{F}^2-q^2\sin^2\theta},\qquad \theta<\pi/2 $$

  • In terms of these variables

$$ \begin{align*} \int_{\substack{\abs{\bk_3}<k_\text{F},\abs{\bk_2}>k_\text{F}\\ \bk+\bk_3=\bk'+\bk_2 }} \frac{d\bk_3}{\epsilon(\bk_2)-\epsilon(\bk_3)}&= \pi m\int_0^{\pi/2} d\theta \int_{K_-(\theta)}^{K_+(\theta)} \frac{K\sin\theta}{q\cos\theta} dK\nonumber\\ &=2\pi m\int_0^{\pi/2} d\theta \sin\theta \sqrt{k_\text{F}^2-q^2\sin^2\theta} \end{align*} $$

  • Other integral is same but in interval $(\pi/2,\pi)$. Finally… $$ \begin{align*} f_{\uparrow\uparrow}(\bk,\bk') &= \frac{U_0^2 m}{(2\pi)^2} \int_0^{\pi} d\theta \sin\theta \sqrt{k_\text{F}^2-q^2\sin^2\theta}\nonumber\\ &=\frac{U_0^2 m k_\text{F}}{(2\pi)^2}\left[1 - \frac{\cos^2\phi/2}{2\sin\phi/2}\log\left(\frac{1-\sin\phi/2}{1+\sin\phi/2}\right)\right] \end{align*} $$ $\phi$ is the angle between $\bk$ and $\bk'$ i.e. $\abs{\bk-\bk'}=2q=2k_\text{F}\sin\phi/2$

  • Our definition of $f$ implied quantization axis for spin

  • To write things in invariant way, think of occupation number $N(\bk)$ as $2\times 2$ matrix

$$ \mathsf{N}(\bk)=\begin{pmatrix} N_{\uparrow\uparrow}(\bk) & N_{\uparrow\downarrow}(\bk) \\ N_{\downarrow\uparrow}(\bk) & N_{\downarrow\downarrow}(\bk). \end{pmatrix} $$

  • $f$-function then has four spin indices

$$ \frac{1}{2V}\sum_{\bk, s_1,s_2,\bk', s_3,s_4} f_{s_1s_2,s_3s_4}(\bk,\bk')n_{s_1s_2}(\bk)n_{s_3s_4}(\bk'), $$

$$ \begin{align*} f_{s_1s_2,s_3s_4}(\bk,\bk') = \frac{U_0}{2}\left[\left(1+ \frac{mU_0 k_\text{F}}{2\pi^2}\left[2+\frac{\cos\phi}{2\sin\phi/2}\log\frac{1+\sin\phi/2}{1-\sin\phi/2}\right]\right)\delta_{s_1s_3}\delta_{s_2s_4}\right.\nonumber\\ \left.\left(1+ \frac{mU_0 k_\text{F}}{2\pi^2}\left[1-\frac{1}{2}\sin\phi/2\log\frac{1+\sin\phi/2}{1-\sin\phi/2}\right]\right)\boldsymbol{\sigma}_{s_1s_3}\cdot\boldsymbol{\sigma}_{s_2s_4}\right] \end{align*} $$ (again, just for reference)

  • Key message: interaction between quasiparticles in an interacting Fermi gas is defined in terms of a pair of functions $$ \nu(E_F)f_{s_1s_2,s_3s_4}(\bk,\bk') = F(\phi) \delta_{s_1s_3}\delta_{s_2s_4} + G(\phi)\boldsymbol{\sigma}_{s_1s_3}\cdot\boldsymbol{\sigma}_{s_2s_4} $$
  • $F(\phi)$ and $G(\phi)$ made dimensionless by scaling by density of states at Fermi surface $\nu(E_F)\equiv k_{\text{F}}m/\pi^2$.

Quasiparticle energy $\varepsilon_s(\bk)$

$$ \Delta E = \sum_{\bk,s} \varepsilon_s(\bk)n_{s}(\bk) + \frac{1}{2V}\sum_{\bk, s,\bk’, s’} f_{s^{}s’}(\bk,\bk’)n_{s}(\bk)n_{s’}(\bk’). $$

  • $\varepsilon_s(\bk)$ will involve two momentum sums 😟

  • What can we say on general grounds? Near Fermi surface expect

$$ \varepsilon_s(\bk) - E_\text{F} \approx v_\text{F}(\abs{\bk}-k_\text{F}) $$

  • Fermi velocity $v_\text{F}$ altered by interactions. Define effective mass

$$ m_* = \frac{k_\text{F}}{v_\text{F}} $$

  • Can get $m_*$ using results we have alrady, using one simple trick

  • Shift momentum of each quasiparticle by $\delta\bk$, giving new $N_s(\bk)$

$$ \begin{align*} N_s(\bk-\delta\bk) &=\theta(k_F-\abs{\bk-\delta\bk}) + n_{s}(\bk-\delta\bk)+\cdots\nonumber\\ &=\theta(k_F-\abs{\bk}) + n_s(\bk) + \delta(k_F-\abs{\bk})\hat\bk\cdot\delta\bk - \delta\bk \nabla_\bk n_{s}(\bk)+\cdots. \end{align*} $$

drawing

  • Compute energy using Landau functional

$$ \begin{align*} N_s(\bk-\delta\bk) &=\theta(k_F-\abs{\bk}) + n_s(\bk) + \delta(k_F-\abs{\bk})\hat\bk\cdot\delta\bk - \delta\bk \nabla_\bk n_{s}(\bk)+\cdots. \end{align*} $$

  • Treat last three terms as $n_s(\bk)$. Excitation energy changes $$ \begin{align*} \Delta E(\text{after})-\Delta E(\text{before}) = \sum_{\bk,s} n_{s}(\bk)\delta\bk\cdot\nabla_\bk\varepsilon_s(\bk) \nonumber\\ +\frac{1}{V}\sum_{\bk, s,\bk', s'} f_{s^{}s'}(\bk,\bk')n_{s}(\bk)\left[\delta(k_F-\abs{\bk'})\hat\bk'\cdot\delta\bk - \nabla_{\bk'}n_{s'}(\bk')\cdot\delta\bk\right]. \end{align*} $$ (to first order in $\delta\bk$. Integrate by parts in first term)
  • On grounds of Galilean invariance, this change must be $$ \Delta E(\text{after})-\Delta E(\text{before}) = \frac{\mathbf{P}}{m}\cdot\delta\bk $$ where total momentum $\mathbf{P}$ is

$$ \mathbf{P} = \sum_{\bk,s} \bk n_{s}(\bk). $$

  • If two expressions for $\Delta E(\text{after})-\Delta E(\text{before})$ equal for all $n_s(\bk)$ and $\delta\bk$ to first order

$$ \frac{\bk}{m} = \nabla_\bk\varepsilon_s(\bk) + \sum_{s'}\int f_{s^{}s'}(\bk,\bk')\delta(k_F-\abs{\bk'})\hat\bk' \frac{d\bk'}{(2\pi)^3} $$

  • For momenta close to the Fermi surface

$$ \frac{\bk}{m} = \frac{\bk}{m_*} +\frac{1}{m} \int F(\phi) \bk' \frac{d\Omega_{\bk'}}{4\pi} $$

  • If $\bk'=\cos\phi \bk + \sin\phi \bk_\perp$, with $\bk_\perp\cdot\bk=0$, we get

$$ \frac{1}{m} = \frac{1}{m_*} +\frac{1}{m} \int F(\phi) \cos\phi \frac{\sin\phi d\phi}{2} $$

  • For the $F(\phi)$ that we found in $2^{\text{nd}}$ order perturbation theory $$ \frac{m_*}{m} = 1 + \frac{1}{30\pi^4}(7\log 2 - 1)\left(mU_0k_\text{F}\right)^2+\cdots. $$ (Use the substitution $u=\sin\phi/2$ to do the integral.)

  • In systems with strong interactions $m_*/m$ may be much larger!

  • In heavy fermion materials $m_*/m$ can approach 1000! Landau’s picture of fermionic quasiparticles still applies.

Eigenstates in Perturbation Theory: What is a Quasiparticle?

  • What do these quasiparticle states look like? Let’s see what perturbation theory says

  • At first order we have

$$ \ket{\mathbf{N}^{(1)}} = \sum_{\mathbf{N}'\neq \mathbf N}\frac{\braket{\mathbf{N'}|H_\text{int}|\mathbf{N}}}{E^{(0)}(\mathbf{N})-E^{(0)}(\mathbf{N}')}\ket{\mathbf{N}'} $$

  • Consider the Fermi sea ground state $\ket{\text{FS}}$. What states $\ket{\mathbf{N'}}$ can appear this case?

  • Only possibility is that the interaction creates two particle-hole pairs out of the Fermi sea, with total momentum zero.

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  • What about an excited state? If we consider the state $$ \adop_{\bk,s}\ket{\text{FS}} $$ two kinds of states can contribute

    1. Create a pair of particle-hole pairs

    2. Create a single particle-hole pair and move particle at $\bk$

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  • Consider $\adop_{\bk,s}\ket{0}$, creating a particle in exact ground state

  • Since $\ket{0}$ includes the first type of state (2 particle-hole pair states), $\adop_{\bk,s}\ket{0}$ is only missing second kind

  • Single quasiparticle state is $$ \begin{align*} \ket{\bk,s} &= \sqrt{\frac{z_k}{\braket{0|\aop_{\bk,s}\adop_{\bk,s}|0}}}\adop_{\bk,s}\ket{0} \\ &+ \frac{U_0}{V}\sum_{\substack{\bk_1+\bk_2=\bk_3+\bk\\ s,s'}}\frac{\adop_{\bk_1,s}\adop_{\bk_2,s'}\aop_{\bk_3,s'}\ket{\text{FS}}}{\epsilon(\bk_1)+\epsilon(\bk_2)-\epsilon(\bk_3)-\epsilon(\bk)} \end{align*} $$ $\sqrt{z_k}$ is a normalization factor

Normalizing, $1-z_\bk$ is

$$ \begin{align*} \left(\frac{U_0}{V}\right)^2\sum_{\substack{\bk_1+\bk_2=\bk_3+\bk\\\abs{\bk_3}<k_\text{F},\abs{\bk_2},\abs{\bk_1}>k_\text{F}\\ s,s'}}\frac{1}{\left[\epsilon(\bk_1)+\epsilon(\bk_2)-\epsilon(\bk_3)-\epsilon(\bk)\right]^2}+\cdots \end{align*} $$

  • This is overlap of single quasiparticle state $\ket{\bk,s}$ with $\adop_{s,\bk}\ket{0}$.

$$ z_\bk = \frac{\abs{\braket{\bk,s|\adop_{\bk,s}|0}}^2}{\braket{0|\aop_{\bk,s}\adop_{\bk,s}|0}} $$

  • A finite overlap – for quasiparticles at the Fermi surface – is requirement for Fermi liquid picture to hold

  • If it were to vanish, any resemblance of the quasiparticle to a free fermion would disappear!

  • For our example (hard!): $$ z_{\abs{\bk}=k_\text{F}} = 1 - \frac{(mUk_\text{F})^2}{8\pi^4}\left[\log 2 + \frac{1}{3}\right] $$

This is also the occupation number of the original fermions $\braket{0|\adop_{\bk,s}\aop_{\bk,s}|0}$ (not the quasiparticles!) just below the Fermi surface in the ground state (see Problem Set 3. There is a corresponding result just above. Even with interactions, there is a finite step in the distribution function at the Fermi surface.

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