Quantum gases

First of all we assume that we are dealing with a weakly quantum gas, i.e. ${\frac {n}{n_{Q}}}<1$$\frac{n}{n_{Q}}<1$. The system is a gran-canonical ensemble. Hence the partition function is:

$$Z_{GC}=\sum _{N,E}e^{-\beta (E-N\mu )}g(E,N)=\sum _{n_{1}}\sum _{n_{2}}\dots \left(e^{-\beta \sum _{i=1}^{N}\left(n_{i}\epsilon _{i}-\mu n_{i}\right)}\right)$$

$$Z_{GC}=\sum _{N,E}e^{-\beta (E-N\mu )}g(E,N)=\sum _{n_{1}}\sum _{n_{2}}\dots \left(e^{-\beta \sum _{i=1}^{N}\left(n_{i}\epsilon _{i}-\mu n_{i}\right)}\right)$$

Which is equivalent to:

$$Z_{GC}=\prod _{i=1}^{N}\left(\sum _{n_{i}}e^{-\beta (\epsilon _{i}-\mu )n_{i}}\right)={\begin{cases}\prod _{i}\left(1+e^{-\beta (\epsilon _{i}-\mu )}\right)\\\prod _{i}\left({\frac {1}{1-e^{-\beta (\epsilon _{i}-\mu )}}}\right)\end{cases}}$$

$$Z_{GC}=\prod _{i=1}^{N}\left(\sum _{n_{i}}e^{-\beta (\epsilon _{i}-\mu )n_{i}}\right)=\begin{cases} \prod _{i}\left(1+e^{-\beta (\epsilon _{i}-\mu )}\right) \\ \prod _{i}\left(\frac{1}{1-e^{-\beta (\epsilon _{i}-\mu )}}\right) \end{cases}$$

Where the first expression holds for fermions, while the second one for bosons. In a gran canonical ensemble, the thermodynamic potential is given by the Landau thermodynamic potential:

$$\Omega (V,T,\mu )=F-{\frac {\partial F}{\partial N}}N=F-\mu N=-kT\log Z_{GC}$$

$$\Omega (V,T,\mu )=F-\frac{\partial F}{\partial N}N=F-\mu N=-kT\log Z_{GC}$$

Since it can be proved that $G=\mu N$$G=\mu N$, we also have: $\Omega =-pV$$\Omega =-pV$. From where one can derive the equation of state for weakly quantum gases:

$$pV=kT\log Z_{GC}=nkT\left(1\pm {\frac {n}{n_{Q}2^{\frac {5}{2}}}}\right)$$

$$pV=kT\log Z_{GC}=nkT\left(1\pm \frac{n}{n_{Q}2^{\frac{5}{2}}}\right)$$

Where the plus sign is for fermions, while the minus sign is for bosons.

Electrons in a metal

Electrons in a metal can be idealized as fermions (electrons) in a box. Electrons will occupy energy levels according to the Fermi-Dirac distribution. In particular we have to note that:

$$f(\epsilon ){\underset {T\rightarrow 0}{\rightarrow }}\theta (\epsilon -\mu )={\begin{cases}0\;\;\epsilon >\mu \\1\;\;\epsilon <\mu \end{cases}}$$

$$f(\epsilon )\underset {T\rightarrow 0}{\rightarrow }\theta (\epsilon -\mu )=\begin{cases} 0\; \; \epsilon >\mu \\ 1\; \; \epsilon <\mu \end{cases}$$

We do define Fermi energy, the maximum of the energy of the occupied level: $\epsilon _{F}=\mu (T=0)$$\epsilon _{F}=\mu (T=0)$. If we define $D(\epsilon )d\epsilon$$D(\epsilon )d\epsilon$ as the number of states, per unit volume, with energy in the interval $(\epsilon ;\epsilon +d\epsilon )$$(\epsilon ;\epsilon +d\epsilon )$ we have:

$$\int _{0}^{\epsilon _{F}}D(\epsilon )d\epsilon =n$$

$$\int _{0}^{\epsilon _{F}}D(\epsilon )d\epsilon =n$$

The explicit expression of $D(\epsilon )$$D(\epsilon )$ can be derived, obtaining:

$$D(\epsilon )={\frac {1}{2\pi ^{2}}}\left({\frac {2m}{\hbar ^{2}}}\right)^{\frac {3}{2}}{\sqrt {\epsilon }}$$

$$D(\epsilon )=\frac{1}{2\pi ^{2}}\left(\frac{2m}{\hbar ^{2}}\right)^{\frac{3}{2}}\sqrt{\epsilon }$$

Meaning that the Fermi energy equals to:

$$\epsilon _{F}={\frac {\hbar ^{2}}{2m}}k_{F}^{2}={\frac {\hbar ^{2}}{2m}}\left(3\pi ^{2}n\right)^{\frac {2}{3}}$$

$$\epsilon _{F}=\frac{\hbar ^{2}}{2m}k_{F}^{2}=\frac{\hbar ^{2}}{2m}\left(3\pi ^{2}n\right)^{\frac{2}{3}}$$

Hence we have that $T_{f}={\frac {\epsilon _{f}}{k_{B}}}$$T_ f=\frac{\epsilon _ f}{k_ B}$. The chemical potential $\mu$$\mu$ depends on the temperature $T$$T$ with the relation:

$$\mu (T)=\epsilon _{F}\left(1-{\frac {\pi ^{2}}{12}}\left({\frac {T}{T_{F}}}\right)^{2}\right)$$

$$\mu (T)=\epsilon _{F}\left(1-\frac{\pi ^{2}}{12}\left(\frac{T}{T_{F}}\right)^{2}\right)$$

The specific heat, at constant volume $V$$V$, is defined as:

$$C_{V}={\frac {1}{V}}{\frac {\partial E}{\partial T}}={\frac {\pi ^{2}}{2}}nk{\frac {T}{T_{F}}}$$

$$C_{V}=\frac{1}{V}\frac{\partial E}{\partial T}=\frac{\pi ^{2}}{2}nk\frac{T}{T_{F}}$$

We now give three important relations that tell us what happens to the system when it happens a change in some quantity that describes it:

• $\Delta E^{2}=kT^{2}C_{V}$$\Delta E^{2}=kT^{2}C_{V}$ in a canonical ensemble;

• $\Delta V^{2}=-{\frac {\partial V}{\partial p}}kT$$\Delta V^{2}=-\frac{\partial V}{\partial p}kT$ in an isobaric ensemble;

• $\Delta N^{2}={\frac {\partial N}{\partial \mu }}kT$$\Delta N^{2}=\frac{\partial N}{\partial \mu }kT$ in a gran-canonical ensemble.

Quantum superfluids

Quantum superfluids can be idealized as bosons in a box. Bosons in a box occupy energy levels according to the Bose-Einstein distribution. Moreover, if we do assume that all the particles occupy the level at $\epsilon =0$$\epsilon =0$, i.e. the lowes energy level, we have:

$$0<f_{BE}(\epsilon =0)={\frac {1}{e^{-\beta \mu }-1}}$$

$$0<f_{BE}(\epsilon =0)=\frac{1}{e^{-\beta \mu }-1}$$

Where we do expect that $e^{-\beta \mu }>1$$e^{-\beta \mu }>1$, that is $\mu <0$$\mu <0$. Here we are assuming that a macroscopic number of particles occupy a certain energy level, more precisely the lowest possible! Actually, this is not a problem since bosons do not have to respect Pauli's exclusion principle. Meaning that the particles density is given by:

$$n=n_{0}+\int _{0}^{\infty }{\frac {D(\epsilon )}{e^{\beta (\epsilon -\mu )}-1}}d\epsilon$$

$$n=n_{0}+\int _{0}^{\infty }\frac{D(\epsilon )}{e^{\beta (\epsilon -\mu )}-1}d\epsilon$$

Where the contribute of $n_{0}$$n_{0}$ is foundamental for temperatures $T<T_{C}$$T<T_{C}$, the Bose-Einstein condensation temperature. We also have that:

$$n_{0}(T)=n\left(1-\left({\frac {T}{T_{C}}}\right)^{\frac {3}{2}}\right),\;T<T_{C}$$

$$n_{0}(T)=n\left(1-\left(\frac{T}{T_{C}}\right)^{\frac{3}{2}}\right),\; T<T_{C}$$

The condensation temperature can be derived and its value is:

$$kT_{C}=n^{\frac {2}{3}}{\frac {\hbar ^{2}}{2m}}4\pi \left({\frac {1}{2,612}}\right)^{\frac {2}{3}}$$

$$kT_{C}=n^{\frac{2}{3}}\frac{\hbar ^{2}}{2m}4\pi \left(\frac{1}{2,612}\right)^{\frac{2}{3}}$$

And the condition $T<T_{C}$$T<T_{C}$ is equivalent to:

$$n>n_{Q}(2,612),\;\;n_{Q}=\left({\frac {mkT}{2\pi \hbar ^{2}}}\right)^{\frac {3}{2}}$$

$$n>n_{Q}(2,612),\; \; n_{Q}=\left(\frac{mkT}{2\pi \hbar ^{2}}\right)^{\frac{3}{2}}$$

Bose-Einstein condensation is a necessary but not sufficient condition to the superfluidity. For such systems we can define the specific heat, at constant pressure $P$$P$: $C_{P}=T{\frac {\partial S}{\partial T}}$$C_{P}=T\frac{\partial S}{\partial T}$. We have that:

$$C_{P}=-T{\frac {\partial ^{2}G}{\partial T^{2}}}$$

$$C_{P}=-T\frac{\partial ^{2}G}{\partial T^{2}}$$