# Quantum gases

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

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

Which is equivalent to:

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

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:

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

Since it can be proved[1] that ${\displaystyle G=\mu N}$, we also have: ${\displaystyle \Omega =-pV}$. From where one can derive the equation of state for weakly quantum gases:

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

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:

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

We do define Fermi energy, the maximum of the energy of the occupied level: ${\displaystyle \epsilon _{F}=\mu (T=0)}$. If we define ${\displaystyle D(\epsilon )d\epsilon }$ as the number of states, per unit volume, with energy in the interval ${\displaystyle (\epsilon ;\epsilon +d\epsilon )}$ we have[2]:

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

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

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

Meaning that the Fermi energy[3] equals to:

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

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

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

The specific heat, at constant volume ${\displaystyle V}$, is defined as:

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

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:

• ${\displaystyle \Delta E^{2}=kT^{2}C_{V}}$ in a canonical ensemble;
• ${\displaystyle \Delta V^{2}=-{\frac {\partial V}{\partial p}}kT}$ in an isobaric ensemble;
• ${\displaystyle \Delta N^{2}={\frac {\partial N}{\partial \mu }}kT}$ 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 ${\displaystyle \epsilon =0}$, i.e. the lowes energy level, we have:

${\displaystyle 0

Where we do expect that ${\displaystyle e^{-\beta \mu }>1}$, that is ${\displaystyle \mu <0}$. Here we are assuming that a macroscopic number of particles occupy a certain energy level, more precisely the lowest possible[4]! 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:

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

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

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

The condensation temperature can be derived[5] and its value is:

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

And the condition ${\displaystyle T is equivalent to:

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

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 ${\displaystyle P}$: ${\displaystyle C_{P}=T{\frac {\partial S}{\partial T}}}$. We have that:

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

1. From ${\displaystyle G=G(P,T,N)}$ and ${\displaystyle G=E-TS+pV}$, using ${\displaystyle dE=TdS-pdV+\mu dN}$
2. For a gas, given its density ${\displaystyle \rho }$, its mass number ${\displaystyle A}$ (expressed in u.a.m) and its atomic number ${\displaystyle Z}$ we can write ${\displaystyle n={\frac {N}{V}}={\frac {Z\rho }{Am_{p}}}}$.
3. We can also note that:
${\displaystyle {\frac {E}{V}}={\frac {N}{V}}\left\langle \epsilon \right\rangle =\int _{0}^{\epsilon _{F}}D(\epsilon )\epsilon d\epsilon ={\frac {3}{5}}n\epsilon _{F}}$
Meaning that ${\displaystyle \left\langle \epsilon \right\rangle ={\frac {3}{5}}\epsilon _{F}}$
4. This condition is called Bose-Einstein condition.
5. From
${\displaystyle n=\int _{0}^{\infty }d\epsilon D(\epsilon ){\frac {1}{e^{\frac {\epsilon }{T_{C}k}}-1}}}$