It's known that Fermi-Dirac and Bose-Einstein statistics describe non-interacting indistinguishable particles at thermodynamic equilibrium. However, if we solve some problems wtih using such statistics, we get curious results, which are interpreted as an effective interaction presence. Let's consider few examples:

1. Fermi gas free expansion

$N$ fermions with spin $\frac{1}{2}$ are concluded in a small box with volume $V_1$ and have energy $E_1=\frac{3}{5}N\varepsilon_F$, where $\varepsilon_F=(3\pi^2)^{\frac{2}{3}}\frac{\hbar^2}{2m}(\frac{N}{V_1})^{\frac{2}{3}}$ is the Fermi energy. Their temperature $T_1=0$ ($T_1\ll \varepsilon_F$, the gas is degenerate). The small box is separated from a big one with volume $V_2\gg V_1$ by a small partition. After removing the partition fermions occupy the big box. The problem is to find their final temperature $T_2$ knowing that $E_2=E_1$ (an isolated system).

If we suppose that the gas is still degenerate and described by Fermi-Dirac statistics we will obtain $T_2\propto \tilde{\varepsilon}_F\left(\frac{V_2}{V_1}\right)^{\frac{1}{3}}\gg \tilde{\varepsilon}_F$, where $\tilde{\varepsilon}_F=\varepsilon_F\left(\frac{V_1}{V_2}\right)^{\frac{2}{3}}$ is the new Fermi energy. It means the assumption is wrong.

Then we can use classical expressions for gas values in the final state, which are derived from Boltzmann statistics: $E_2=\frac{3}{2}NT_2$. From energy conservation we obtain $T_2=\frac{2}{5}\varepsilon_F\propto \left(\frac{N}{V_1}\right)^{\frac{2}{3}}$, which is obviously much greater than $T_1$ whilst we consider macroscopical concentration.

So, Fermi gas is heated during free expansion high enough to be non-degenerate. At the same time Boltzmann ideal gas is neither heated or cooled during free expansion. To explain this some people claim that there is a kind of repulsion between fermions, which is related to the Pauli exclusion principle.

2. Bose gas condensation in a harmonic trap

$N$ bosons sit in a 3D harmonic trap with frequencies $\omega_1, \omega_2, \omega_3 \ll \frac{T}{\hbar}$. It can be shown that density of states $g(\varepsilon)=\frac{\varepsilon^2}{2\hbar^3\tilde{\omega}^3}$, where $\tilde{\omega}=(\omega_1\omega_2\omega_3)^{\frac{1}{3}}$ is a characteristic frequency. After some calculations it's possible to obtain the following relations: $$\Omega(T,V,\mu)\equiv F(T,V,N)-\mu N=-\frac{1}{3}E(S,V,N)$$ $$\frac{\partial E}{\partial \mu}=-3\frac{\partial \Omega}{\partial \mu}=3N$$ $$E=E_0+3\mu N$$ $$\mu=\begin{cases} 0 & ,\quad T<T_0\\ \frac{6\hbar^3\tilde{\omega}^3N}{\pi^2T^2}\left(1-\left(\frac{T}{T_0} \right )^3 \right ) & ,\quad T_0\leq T<T_0(1+\tau),\ \tau\ll1 \end{cases}$$ $$C_V=\begin{cases} \frac{2\pi^4T^3}{15\hbar^3\tilde{\omega}^3} & ,\quad T<T_0\\ \frac{2\pi^4T^3}{15\hbar^3\tilde{\omega}^3}-\frac{18\hbar^3\tilde{\omega}^3N^2}{\pi^2T_0^3}\left(1+2\left(\frac{T_0}{T} \right )^3 \right ) & ,\quad T_0\leq T<T_0(1+\tau),\ \tau\ll1 \end{cases}$$ where $T_0=\left(\frac{N}{\zeta(3)}\right)^{\frac{1}{3}}\hbar\tilde{\omega}$ is the critical temperature.

As we can see there is a negative jump of $C_V$ at $T=T_0$, which means that for some reason we spend more energy to heat up the condensate. A similar behaviour of $C_V$ is observed in superconductors, but it's clear why it happens there: we need some energy to break Cooper pairs. But Bose ideal gas does't have any bonds between particles and remains Bose ideal gas after "evaporation".

Nevertheless, some people claim that there is an effective attraction between bosons in the condensate.

3. Quantum gas at high temperature

In "Statistical Physics, Part 1: Volume 5" L. D. Landau demonstrates how to obtain the following expression of the equation of state for ideal quantum gases at high temperature: $$PV=NT\left ( 1\pm \frac{\pi^{\frac{3}{2}}N\hbar^3}{2V(mT)^{\frac{3}{2}}} \right )$$ where the sign of the second term depends on statistics. In the limit $T\rightarrow \infty$ it turns into the equation of state for a classical ideal gas.

Then Landau gives a comment:

Thus we see that the deviations of an ideal gas from classical properties, occuring when the temperature is lowered at constant density (the gas then being said to become degenerate), cause in Fermi statistics an increase in pressure as compared with its value in an ordinary gas; we may say that in this case the quantum exchange effects lead to the occurrence of an additional effective repulsion between the particles.

In Bose statistics, on the other hand, the value of the gas pressure changes in the opposite direction, becoming less than the classical value; we may say that here there is an effectife attraction between the particles.

Unfortunately, he doesn't explain why there is such opportunity to consider initially non-interacting particles as interacting.

4. Quantum gas at zero temperature

It's the last and brief example.

For fermions at zero temperature their pressure isn't zero: $$P=\frac{(3\pi^2)^{\frac{2}{3}}\hbar^2}{5m}\left(\frac{N}{V}\right)^{\frac{5}{3}}$$ (according to Landau)

For bosons it is, because all particles sit on the ground level with energy $\varepsilon=0$.

The difference is often explained by the same effective interaction.

Now, let me remind my question. Unfortunately, I'm not able to understand how it happens when we consider non-interacting particles and obtain many consequences, which apparently need an interction in the system. Those qualitative explanations with help of the Pauli principle and bonds in the condensate don't satisfy me. They seem contrary to the initital problem statement.

Is it possible to show mathematically on what step of statistics deriving we take into account interactions after all?


3 Answers 3


I would recommend you take a look at this paper on the subject, which is referred to in the "Introduction to Quantum Mechanics" book by D.J. Griffiths on the subject. There they show how all the phenomena you go through can be understood without reference to any effective interaction.

The short answer is that the "effective interaction" is another way to express the requirement that the relative fermionic wavefunction must be zero for all configurations which leave two fermions in the same location. This means that squeezing two fermions together requires a larger investment of momentum than in the bosonic case. This might look like an interaction, but the paper gives a number of reasons for why such an interpretation is, if not wrong then somewhat dangerous.


A compound and strongly interacting system at low temperatures may often be represented as an ensebmle of nearly non interacting quasi-particles - normal modes of the system. It is the strongly interacting system that dictates the quasi-particle properties. We should not take the quasi-particles for "literally particles". In QM there are always some "boundary conditions" created by a strongly interacting system, let us not forget it.


According to my understanding, those interactions are due to 'exchange force', which happens due to symmetry properties of wave functions. Fermions should always have anti symmetric wave function and bosons should always have symmetric wave function. It will lead to a kind of repulsive/attractive interactions which can be shown mathematically. See derivations in Griffiths Quantum Mechanics, Identical Particles>exchange force. I tried to explain about identical particles here: https://youtu.be/aq5tMYlzFjo


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