Thermodynamic Properties of Ultracold Fermi Gases Across the BCS-BEC Crossover and the Bertsch Problem

Abstract

In this chapter we review the thermodynamic properties of ultracold Fermi gases in which the strength of the interaction is continuously varied. The system features a crossover between a state described by the BCS theory of superconductivity and a Bose-Einstein condensate. A discussion of the Bertsch problem is presented.

Appendix

In this appendix we review the definition of the partition function for a fermionic system with the use of Grassmann variables. Consider a fermionic system described by fermionic operators which are associated with Grassmann variables [27]:

$$\displaystyle \begin{aligned}\varPsi\left|\psi\right>=\psi\left|\psi\right> \quad \text{and}\quad \left<\bar{\psi}\right|\varPsi^\dagger=\left<\bar{\psi}\right|\bar{\psi},\end{aligned}$$

where $\left |\psi \right >$ and $\left <\bar {\psi }\right |$ indicate coherent states, which can be decomposed as:

$$\displaystyle \begin{aligned}\left|\psi\right>=\left|0\right>-\psi\left|1\right>.\end{aligned}$$

The states $\left |0\right >$ and $\left |1\right >$ are the only ones possible and indicate the states with zero and one particle respectively. The fermion operators anti-commutate and therefore we have:

$$\displaystyle \begin{aligned}\left\{\varPsi,\varPsi\right\}=\left\{\varPsi^\dagger,\varPsi^\dagger\right\}=0, \qquad \qquad \left\{\varPsi,\varPsi^\dagger\right\}=1.\end{aligned}$$

The following properties of the Grassmann variables are obtained:

$$\displaystyle \begin{aligned}\psi^2=\bar{\psi}^2=0;\end{aligned}$$

$$\displaystyle \begin{aligned}\left<\bar{\psi}|\psi\right>=\left<0|0\right>+\left<1|\bar{\psi}\,\psi|1\right>=e^{\bar{\psi}\psi};\end{aligned}$$

$$\displaystyle \begin{aligned}\int \psi\, d\psi=-\int d\psi\,\psi=1;\end{aligned}$$

$$\displaystyle \begin{aligned}\int 1\, d\psi=0.\end{aligned}$$

In general

$$\displaystyle \begin{aligned}\int x_{j_n}\dots x_{j_1}\, dx_{i_1}\dots dx_{i_n}=\epsilon \begin{pmatrix} j_1 & \dots & j_n \\ i_1 & \dots & i_n \end{pmatrix} \end{aligned}$$

and is different from zero only if $\left (j_1\dots j_n\right )$ coincide with $\left (i_1\dots i_n\right )$.

$$\displaystyle \begin{aligned}\epsilon\begin{pmatrix} j_1 & \dots & j_n \\ i_1 & \dots & i_n \end{pmatrix} = \left \{ \begin{array}{rl} 1 \quad \text{with an even number of permutations} \\ -1 \quad \text{with an odd number of permutations}\\ \end{array} \right . \end{aligned}$$

We then have the following relations:

• Gaussian integrals:

  $$\displaystyle \begin{aligned}\int e^{-\bar{\psi}M\psi}\mathscr{D}\left[\bar{\psi},\psi\right]=det\, M;\end{aligned}$$

• completeness relation:

  $$\displaystyle \begin{aligned}I=\int\left|\psi\right>\left<\bar{\psi}\right|e^{-\bar{\psi}\psi}\mathscr{D}\left[\bar{\psi},\psi\right];\end{aligned}$$

• trace of an operator:

  $$\displaystyle \begin{aligned}\text{Tr}\varOmega=\int \left<-\bar{\psi}\right|\varOmega\left|\psi\right>e^{-\bar{\psi}\psi}\mathscr{D}\left[\bar{\psi},\psi\right];\end{aligned}$$

where with $\mathscr {D}\left [\bar {\psi },\psi \right ]$ we mean the functional integral.

The completeness relation is obtained developing coherent states and the exponential:

$$\displaystyle \begin{aligned}\int\left(\left|0\right>+\left|1\right>\psi\right) & \left(\left<0\right|-\bar{\psi}\left<1\right|\right)\left(1-\bar{\psi}\psi\right)\mathscr{D}\left[\bar{\psi},\psi\right] = \\ & \int \left(-\left|0\right>\left<0\right|\bar{\psi}\psi+\left|1\right>\left<1\right|\right)\psi\bar{\psi})\mathscr{D}\left[\bar{\psi},\psi\right]= \left|0\right>\left<0\right|+\left|1\right>\left<1\right|=1\!\!1, \end{aligned}$$

while an operator trace is obtained from:

$$\displaystyle \begin{aligned}\text{Tr}\varOmega & =\int\left(\left<0\right|+\left<1\right|\bar{\psi}\right)\varOmega\left(\left|0\right>-\psi\left|1\right>\right)\left(1-\bar{\psi}\psi\right)\mathscr{D}\left[\bar{\psi},\psi\right]= \\ & \int\left(-\left<0\right|\varOmega\left|0\right>\bar{\psi}\psi -\left<1\right|\varOmega\left|1\right>\bar{\psi}\psi\right)\mathscr{D}\left[\bar{\psi},\psi\right]=\left<0\right|\varOmega\left|0\right>+\left<1\right|\varOmega\left|1\right>= \text{Tr} \varOmega. \end{aligned}$$

The partition function of a canonical ensemble is defined as:

$$\displaystyle \begin{aligned}Z=\sum_n e^{-\beta E_n}= \text{Tr} e^{-\beta E_n}.\end{aligned}$$

It can be written using the path integral approach, developing the exponential trace, using

$$\displaystyle \begin{aligned}e^{-\beta H}=\lim_{N\rightarrow\infty}\left(exp\left(-\frac{\beta H}{N}\right)\right)^N=\underbrace{\left(1-\frac{\beta H}{N}\right)\dots\left(1-\frac{\beta H}{N}\right)}_{\text{N times}},\end{aligned}$$

for large N. Then

$$\displaystyle \begin{aligned} Z &=\int\left<-\bar{\psi}_0\right|\left(1-\frac{\beta H}{N}\right)\dots\left(1-\frac{\beta H}{N}\right)\left|\psi_0\right>e^{-\bar{\psi}_0\psi_0}\mathscr{D}\left[\bar{\psi}_0,\psi_0\right]= \\ &= \int\left<-\bar{\psi}_0\right|\left(1-\frac{\beta H}{N}\right)\left|\psi_{N-1}\right> \left<\bar{\psi}_{N-1}\right|\left(1-\frac{\beta H}{N}\right)\left|\psi_{N-2}\right>\dots \\& \qquad \qquad \qquad \qquad \dots\left<\bar{\psi}_{1}\right|\left(1-\frac{\beta H}{N}\right)\left|\psi_{0}\right>\prod_{i=0}^{N-1}e^{-\bar{\psi}_i\psi_i}d\bar{\psi}_i\psi_i,\end{aligned}$$

where N − 1 completeness relations were inserted.

Since we are dealing with fermionic systems, it is necessary to impose anti-periodic boundary conditions

$$\displaystyle \begin{aligned}-\bar{\psi}_0=\bar{\psi}_N \quad \text{and}\quad -\psi_0=\psi_N.\end{aligned}$$

Then

$$\displaystyle \begin{aligned}Z=\int\prod_{i=0}^{N-1}\left(\left<\bar{\psi}_{i+1}\right|\left(1-\frac{\beta H}{N}\right)\left|\psi_i\right>e^{-\bar{\psi}_i\psi_i}d\bar{\psi}_i d\psi_i\right),\end{aligned}$$

from which

$$\displaystyle \begin{aligned}Z=\int\prod_{i=0}^{N-1}exp\bigg\{\frac{\beta}{N}\left[\frac{\bar{\psi}_{i+1}-\bar{\psi}_i}{\beta/N}\psi_i-H\left(\bar{\psi}_{i+1},\psi_i\right)\right]\bigg\}d\bar{\psi}_i d\psi_i.\end{aligned}$$

Define now τ as the imaginary time (the inverse of the temperature) that will vary between 0 and β. We can then (non-rigorously) see $\frac {\bar {\psi }_{i+1}-\bar {\psi }_i}{\beta /N}$ as to the derivative with respect to this time.

Analogously we can perform the approximation

$$\displaystyle \begin{aligned}H\left(\bar{\psi}_{i+1},\psi_i\right)=H\left(\bar{\psi}\left(\tau+\frac{\beta}{N}\right),\psi\left(\tau\right)\right).\end{aligned}$$

Introducing the action

$$\displaystyle \begin{aligned}S=\sum_{i=0}^{N}\left[\bar{\psi}_i\left(\psi_i-\psi_{i-1}\right)+\frac{\beta}{N}H\left(\bar{\psi}_i,\psi_{i-1}\right)\right],\end{aligned}$$

in the limit $\epsilon =\frac {\beta }{N}\rightarrow 0$ and then for N →∞ one has

$$\displaystyle \begin{aligned}S=\int_0^\beta d\tau\left[\bar{\psi}\left(\tau\right)\partial_\tau\psi\left(\tau\right)+H\left(\bar{\psi}\left(\tau\right),\psi\left(\tau\right)\right)\right]. \end{aligned}$$

In conclusion it is possible to rewrite the system partition function as

$$\displaystyle \begin{aligned}Z=\int\mathscr{D}\left[\bar{\psi},\psi\right]e^{-S\left[\bar{\psi},\psi\right]},\end{aligned}$$

or, more explicitly

$$\displaystyle \begin{aligned}Z=\int \mathscr{D}\left[\bar{\psi},\psi\right]exp\bigg\{\int_0^\beta \bar{\psi}\left(\tau\right)\left(-\frac{\partial}{\partial \tau}-H\right)\psi\left(\tau\right)d\tau\bigg\}.\end{aligned}$$

In the grand canonical ensemble, we can introduce the number operator $\hat {N}$

$$\displaystyle \begin{aligned}N=\sum_{k,\sigma=\uparrow\downarrow}c^\dagger_{k\sigma}c_{k\sigma},\end{aligned}$$

with $c^\dagger _{k\sigma }$ e c _kσ are defined from

$$\displaystyle \begin{aligned}\psi_\sigma\left(\mathbf{x}\right)=\sum_k\frac{e^{i\mathbf{rk}}}{\sqrt{V}}c_{k\sigma},\qquad \psi^\dagger_\sigma\left(\mathbf{x}\right)=\sum_k\frac{e^{-i\mathbf{rk}}}{\sqrt{V}}c^\dagger_{k\sigma}.\end{aligned}$$

Therefore, in a fermionic system, with anti-periodic boundary conditions in imaginary time:

$$\displaystyle \begin{aligned}\psi_\sigma\left(\beta,\mathbf{x}\right)=-\psi_\sigma\left(0,\mathbf{x}\right) \quad \text{and}\quad \bar{\psi}_\sigma\left(\beta,\mathbf{x}\right)=-\bar{\psi}_\sigma\left(0,\mathbf{x}\right),\end{aligned}$$

the action of the system becomes

$$\displaystyle \begin{aligned}S\left[\bar{\psi}_\sigma,\psi_\sigma\right]=\int_0^\beta d\tau\int d\mathbf{x}\sum_{\sigma}\bar{\psi}_\sigma \left(x\right)\left(\frac{\partial}{\partial \tau}-\mu\right)\psi_\sigma\left(x\right)+H\left(x\right),\end{aligned}$$

with $x=\left (\tau ,\mathbf {x}\right )$.