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?
