Fermi Gas

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\mathbf{r}\left[ \sum_{s=\uparrow,\downarrow}\frac{1}{2m}\nabla\psi^\dagger_s\cdot\nabla\psi^{\vphantom{\dagger}}_s + U_0 \psi^\dagger_\uparrow\psi^\dagger_\downarrow\psi^{\vphantom{\dagger}}_\downarrow\psi^{\vphantom{\dagger}}_\uparrow\right]

  • Work in momentum space H =\sum_{\mathbf{k},s} \epsilon(\mathbf{k})a^\dagger_{\mathbf{k},s}a^{\vphantom{\dagger}}_{\mathbf{k},s} + \overbrace{\frac{U_0}{V}\sum_{\mathbf{k}_1+\mathbf{k}_2=\mathbf{k}_3+\mathbf{k}_4} a^\dagger_{\mathbf{k}_1,\uparrow}a^\dagger_{\mathbf{k}_2,\downarrow}a^{\vphantom{\dagger}}_{\mathbf{k}_3,\downarrow}a^{\vphantom{\dagger}}_{\mathbf{k}_4,\uparrow}}^{\equiv H_\text{int}} with \epsilon(\mathbf{k})=\mathbf{k}^2/2m, and V the volume

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

  • Ground state is Fermi sphere of radius k_\text{F} in momentum space with N_{s}(\mathbf{k}) = \theta(k_F-\lvert{\mathbf{k}}\rvert).

  • Low energy excited states will have N_{s}(\mathbf{k})=1 for \lvert{\mathbf{k}}\rvert\ll k_\text{F} and N_{s}(\mathbf{k})=0 for \lvert{\mathbf{k}}\rvert\gg k_\text{F}.

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

  • Without interaction energy of a state \lvert{\mathbf{N}}\rangle is

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

  • 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{\lvert{\braket{\mathbf{N'}|H_\text{int}|\mathbf{N}}}\rvert^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_{\mathbf{k},\mathbf{k}'} N_{\uparrow}(\mathbf{k})N_{\downarrow}(\mathbf{k}') = \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'_{\mathbf{k}_1,\uparrow} = N_{\uparrow}(\mathbf{k}_1) + 1, \quad N'_{\downarrow}(\mathbf{k}_2) = N_{\downarrow}(\mathbf{k}_2) + 1\\ N'_{\downarrow}(\mathbf{k}_3) = N_{\downarrow}(\mathbf{k}_3) - 1, \quad N'_{\uparrow}(\mathbf{k}_4) = N_{\uparrow}(\mathbf{k}_4) - 1, \end{align*} for \mathbf{k}_i satisfying \mathbf{k}_1+\mathbf{k}_2=\mathbf{k}_3+\mathbf{k}_4 \Braket{\mathbf{N}'|H_\text{int}|\mathbf{N}} = \frac{U_0}{V} \left(1-N_{\uparrow}(\mathbf{k}_1)\right)\left(1-N_{\downarrow}(\mathbf{k}_2)\right)N_{\downarrow}(\mathbf{k}_3)N_{\uparrow}(\mathbf{k}_4) (ignoring any coinciding momenta; occupancies 0 or 1)

E^{(2)}(\mathbf{N}) = \left(\frac{U_0}{V}\right)^2 \sum_{\mathbf{k}_1+\mathbf{k}_2=\mathbf{k}_3+\mathbf{k}_4}\frac{\left(1-N_{\uparrow}(\mathbf{k}_1)\right)\left(1-N_{\downarrow}(\mathbf{k}_2)\right)N_{\downarrow}(\mathbf{k}_3)N_{\uparrow}(\mathbf{k}_4)}{\epsilon(\mathbf{k}_3)+\epsilon(\mathbf{k}_4)-\epsilon(\mathbf{k}_1)-\epsilon(\mathbf{k}_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}(\mathbf{k}) = \theta(k_F-\lvert{\mathbf{k}}\rvert) + n_{s}(\mathbf{k})

  • n_\mathbf{k} is mean deviation from Fermi sphere in continuum limit:

  • Excitation energy expanded in n_\mathbf{k}

\Delta E = \sum_{\mathbf{k},s} \varepsilon_s(\mathbf{k})n_{s}(\mathbf{k}) + \frac{1}{2V}\sum_{\mathbf{k}, s,\mathbf{k}', s'} f_{s^{}s'}(\mathbf{k},\mathbf{k}')n_{s}(\mathbf{k})n_{s'}(\mathbf{k}')

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

  • At first order \begin{align*} \varepsilon_s(\mathbf{k}) &= \epsilon(\mathbf{k}) + \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_{\mathbf{k}_1+\mathbf{k}_2=\mathbf{k}_3+\mathbf{k}_4}\frac{\left(1-N_{\uparrow}(\mathbf{k}_1)\right)\left(1-N_{\downarrow}(\mathbf{k}_2)\right)N_{\downarrow}(\mathbf{k}_3)N_{\uparrow}(\mathbf{k}_4)}{\epsilon(\mathbf{k}_3)+\epsilon(\mathbf{k}_4)-\epsilon(\mathbf{k}_1)-\epsilon(\mathbf{k}_2)}

N_{s}(\mathbf{k}) = \theta(k_F-\lvert{\mathbf{k}}\rvert) + n_{s}(\mathbf{k})

\begin{align*} f_{\uparrow\uparrow}(\mathbf{k},\mathbf{k}') = -\frac{U_0^2}{V}\left[\sum_{\mathbf{k}+\mathbf{k}_3=\mathbf{k}'+\mathbf{k}_2} \frac{N_{\downarrow}(\mathbf{k}_3)(1-N_{\downarrow}(\mathbf{k}_2))}{\epsilon(\mathbf{k})+\epsilon(\mathbf{k}_3)-\epsilon(\mathbf{k}')-\epsilon(\mathbf{k}_2)}\right.\nonumber\\ \left.+\sum_{\mathbf{k}'+\mathbf{k}_3=\mathbf{k}+\mathbf{k}_2}\frac{N_{\downarrow}(\mathbf{k}_3)(1-N_{\downarrow}(\mathbf{k}_2))}{\epsilon(\mathbf{k}')+\epsilon(\mathbf{k}_3)-\epsilon(\mathbf{k})-\epsilon(\mathbf{k}_2)}\right] \end{align*}

  • f_{\uparrow\downarrow}(\mathbf{k},\mathbf{k}') is more complicated \begin{align*} f_{\uparrow\downarrow}(\mathbf{k},\mathbf{k}') = U_0 + f_{\uparrow\uparrow}(\mathbf{k},\mathbf{k}') +\frac{U_0^2}{V}\left[\sum_{\mathbf{k}+\mathbf{k}'=\mathbf{k}_3+\mathbf{k}_4}\frac{N(\mathbf{k}_3)N(\mathbf{k}_4)}{\epsilon(\mathbf{k}_3)+\epsilon(\mathbf{k}_4)-2E_\text{F}}\right.\nonumber\\ \left.\sum_{\mathbf{k}+\mathbf{k}'=\mathbf{k}_1+\mathbf{k}_2}\frac{(1-N(\mathbf{k}_1))(1-N(\mathbf{k}_2))}{2E_\text{F}-\epsilon(\mathbf{k}_1)-\epsilon(\mathbf{k}_2)}\right] \end{align*} (just for reference)

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

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

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

  • Let’s work out angular dependence for f_{\uparrow\uparrow}(\mathbf{k},\mathbf{k}') (simpler, only have one independent momentum 😀)

  • Thermodynamic limit

\begin{align*} f_{\uparrow\uparrow}(\mathbf{k},\mathbf{k}') = \frac{U_0^2}{(2\pi)^3}\left[\int_{\substack{\lvert{\mathbf{k}_3}\rvert<k_\text{F},\lvert{\mathbf{k}_2}\rvert>k_\text{F}\\ \mathbf{k}+\mathbf{k}_3=\mathbf{k}'+\mathbf{k}_2 }} \frac{d\mathbf{k}_3}{\epsilon(\mathbf{k}_2)-\epsilon(\mathbf{k}_3)} +\int_{\substack{\lvert{\mathbf{k}_3}\rvert<k_\text{F},\lvert{\mathbf{k}_2}\rvert>k_\text{F}\\ \mathbf{k}'+\mathbf{k}_3=\mathbf{k}+\mathbf{k}_2 }}\frac{d\mathbf{k}_3}{\epsilon(\mathbf{k}_2)-\epsilon(\mathbf{k}_3)}\right] \end{align*}

  • Need integral

\int_{\substack{\lvert{\mathbf{k}_3}\rvert<k_\text{F},\lvert{\mathbf{k}_2}\rvert>k_\text{F}\\ \mathbf{k}+\mathbf{k}_3=\mathbf{k}'+\mathbf{k}_2 }} \frac{d\mathbf{k}_3}{\epsilon(\mathbf{k}_2)-\epsilon(\mathbf{k}_3)}

  • Denominator can be written (\mathbf{k}-\mathbf{k}' is fixed) \epsilon(\mathbf{k}_2)-\epsilon(\mathbf{k}_3)= \frac{1}{2m}\left(\mathbf{k}_2+\mathbf{k}_3\right)\cdot\left(\mathbf{k}_2-\mathbf{k}_3\right) = \frac{1}{2m}\left(\mathbf{k}_2+\mathbf{k}_3\right)\cdot\left(\mathbf{k}-\mathbf{k}'\right)

  • Notation \mathbf{K} = \frac{1}{2}\left(\mathbf{k}_2+\mathbf{k}_3\right),\quad \mathbf{q}= \frac{1}{2}\left(\mathbf{k}_2-\mathbf{k}_3\right)

  • Denominator becomes (for fixed \mathbf{q}) \epsilon(\mathbf{k}_2)-\epsilon(\mathbf{k}_3) = \frac{2}{m}\mathbf{K}\cdot\mathbf{q}

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

  • Conditions \lvert{\mathbf{k}_2}\rvert>k_\text{F} and \lvert{\mathbf{k}_3}\rvert<k_\text{F} become

\begin{align*} \left(\mathbf{K}+\mathbf{q}\right)^2>k_\text{F}^2,\quad \left(\mathbf{K}-\mathbf{q}\right)^2<k_\text{F}^2 \end{align*}

which gives range of K_-(\theta)<\lvert{\mathbf{K}}\rvert<K_+(\theta)

K_{\pm}(\theta)=\pm q\lvert{\cos\theta}\rvert+\sqrt{k_\text{F}^2-q^2\sin^2\theta},\qquad \theta<\pi/2

  • In terms of these variables

\begin{align*} \int_{\substack{\lvert{\mathbf{k}_3}\rvert<k_\text{F},\lvert{\mathbf{k}_2}\rvert>k_\text{F}\\ \mathbf{k}+\mathbf{k}_3=\mathbf{k}'+\mathbf{k}_2 }} \frac{d\mathbf{k}_3}{\epsilon(\mathbf{k}_2)-\epsilon(\mathbf{k}_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}(\mathbf{k},\mathbf{k}') &= \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 \mathbf{k} and \mathbf{k}' i.e. \lvert{\mathbf{k}-\mathbf{k}'}\rvert=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(\mathbf{k}) as 2\times 2 matrix

\mathsf{N}(\mathbf{k})=\begin{pmatrix} N_{\uparrow\uparrow}(\mathbf{k}) & N_{\uparrow\downarrow}(\mathbf{k}) \\ N_{\downarrow\uparrow}(\mathbf{k}) & N_{\downarrow\downarrow}(\mathbf{k}) \end{pmatrix}

  • f-function then has four spin indices

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

\begin{align*} f_{s_1s_2,s_3s_4}(\mathbf{k},\mathbf{k}') = \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}(\mathbf{k},\mathbf{k}') = 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(\mathbf{k})

\Delta E = \sum_{\mathbf{k},s} \varepsilon_s(\mathbf{k})n_{s}(\mathbf{k}) + \frac{1}{2V}\sum_{\mathbf{k}, s,\mathbf{k}', s'} f_{s^{}s'}(\mathbf{k},\mathbf{k}')n_{s}(\mathbf{k})n_{s'}(\mathbf{k}').

  • \varepsilon_s(\mathbf{k}) will involve two momentum sums 😟

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

\varepsilon_s(\mathbf{k}) - E_\text{F} \approx v_\text{F}(\lvert{\mathbf{k}}\rvert-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 already, using one simple trick

  • Shift momentum of each quasiparticle by \delta\mathbf{k}, giving new N_s(\mathbf{k})

\begin{align*} N_s(\mathbf{k}-\delta\mathbf{k}) &=\theta(k_F-\lvert{\mathbf{k}-\delta\mathbf{k}}\rvert) + n_{s}(\mathbf{k}-\delta\mathbf{k})+\cdots\nonumber\\ &=\theta(k_F-\lvert{\mathbf{k}}\rvert) + n_s(\mathbf{k}) + \delta(k_F-\lvert{\mathbf{k}}\rvert)\hat{\mathbf{k}}\cdot\delta\mathbf{k}- \delta\mathbf{k}\nabla_\mathbf{k}n_{s}(\mathbf{k})+\cdots. \end{align*}

  • Compute energy using Landau functional

\begin{align*} N_s(\mathbf{k}-\delta\mathbf{k}) &=\theta(k_F-\lvert{\mathbf{k}}\rvert) + n_s(\mathbf{k}) + \delta(k_F-\lvert{\mathbf{k}}\rvert)\hat{\mathbf{k}}\cdot\delta\mathbf{k}- \delta\mathbf{k}\nabla_\mathbf{k}n_{s}(\mathbf{k})+\cdots. \end{align*}

  • Treat last three terms as n_s(\mathbf{k}). Excitation energy changes \begin{align*} \Delta E(\text{after})-\Delta E(\text{before}) = \sum_{\mathbf{k},s} n_{s}(\mathbf{k})\delta\mathbf{k}\cdot\nabla_\mathbf{k}\varepsilon_s(\mathbf{k}) \nonumber\\ +\frac{1}{V}\sum_{\mathbf{k}, s,\mathbf{k}', s'} f_{s^{}s'}(\mathbf{k},\mathbf{k}')n_{s}(\mathbf{k})\left[\delta(k_F-\lvert{\mathbf{k}'}\rvert)\hat{\mathbf{k}}'\cdot\delta\mathbf{k}- \nabla_{\mathbf{k}'}n_{s'}(\mathbf{k}')\cdot\delta\mathbf{k}\right]. \end{align*} (to first order in \delta\mathbf{k}. 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\mathbf{k} where total momentum \mathbf{P} is

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

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

\frac{\mathbf{k}}{m} = \nabla_\mathbf{k}\varepsilon_s(\mathbf{k}) + \sum_{s'}\int f_{s^{}s'}(\mathbf{k},\mathbf{k}')\delta(k_F-\lvert{\mathbf{k}'}\rvert)\hat{\mathbf{k}}' \frac{d\mathbf{k}'}{(2\pi)^3}

  • For momenta close to the Fermi surface

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

  • If \mathbf{k}'=\cos\phi \mathbf{k}+ \sin\phi \mathbf{k}_\perp, with \mathbf{k}_\perp\cdot\mathbf{k}=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

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

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

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

  • What about an excited state? If we consider the state a^\dagger_{\mathbf{k},s}\lvert{\text{FS}}\rangle 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 \mathbf{k}

  • Consider a^\dagger_{\mathbf{k},s}\lvert{0}\rangle, creating a particle in exact ground state

  • Since \lvert{0}\rangle includes the first type of state (2 particle-hole pair states), a^\dagger_{\mathbf{k},s}\lvert{0}\rangle is only missing second kind

  • Single quasiparticle state is \begin{align*} \lvert{\mathbf{k},s}\rangle &= \sqrt{\frac{z_k}{\braket{0|a^{\vphantom{\dagger}}_{\mathbf{k},s}a^\dagger_{\mathbf{k},s}|0}}}a^\dagger_{\mathbf{k},s}\lvert{0}\rangle \\ &+ \frac{U_0}{V}\sum_{\substack{\mathbf{k}_1+\mathbf{k}_2=\mathbf{k}_3+\mathbf{k}\\ s'}}\frac{a^\dagger_{\mathbf{k}_1,s}a^\dagger_{\mathbf{k}_2,s'}a^{\vphantom{\dagger}}_{\mathbf{k}_3,s'}\lvert{\text{FS}}\rangle}{\epsilon(\mathbf{k}_1)+\epsilon(\mathbf{k}_2)-\epsilon(\mathbf{k}_3)-\epsilon(\mathbf{k})} \end{align*} \sqrt{z_k} is a normalization factor

Normalizing, 1-z_\mathbf{k} is

\begin{align*} \left(\frac{U_0}{V}\right)^2\sum_{\substack{\mathbf{k}_1+\mathbf{k}_2=\mathbf{k}_3+\mathbf{k}\\\lvert{\mathbf{k}_3}\rvert<k_\text{F},\lvert{\mathbf{k}_2}\rvert,\lvert{\mathbf{k}_1}\rvert>k_\text{F}\\ s'}}\frac{1}{\left[\epsilon(\mathbf{k}_1)+\epsilon(\mathbf{k}_2)-\epsilon(\mathbf{k}_3)-\epsilon(\mathbf{k})\right]^2}+\cdots \end{align*}

  • This is overlap of single quasiparticle state \lvert{\mathbf{k},s}\rangle with a^\dagger_{s,\mathbf{k}}\lvert{0}\rangle.

z_\mathbf{k}= \frac{\lvert{\braket{\mathbf{k},s|a^\dagger_{\mathbf{k},s}|0}}\rvert^2}{\braket{0|a^{\vphantom{\dagger}}_{\mathbf{k},s}a^\dagger_{\mathbf{k},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_{\lvert{\mathbf{k}}\rvert=k_\text{F}} = 1 - \frac{(mUk_\text{F})^2}{8\pi^4}\left[\log 2 + \frac{1}{3}\right]

  • This is also occupation number of fermions \braket{0|a^\dagger_{\mathbf{k},s}a^{\vphantom{\dagger}}_{\mathbf{k},s}|0} (not quasiparticles!) just below k_\text{F} in ground state (see Problem Set 3

  • There is a corresponding result just above

  • Even with interactions, finite step in distribution function at k_\text{F}