How does the Pauli Exclusion principle explain the incompressibility of metals? The Pauli Exclusion principles states that no two identical fermions can have the same quantum state. In my lecture notes, it is mentioned that this principle helps us explain the fact that metals are very hard to compress. Why does this principle explain the incompressibity of metals?
 A: To rather a good approximation one may regard metal as Fermi-Liquid, the gas of non-interacting fermions. These obey the Fermi-statistic and the corresponding distribution is Fermi-Dirac:
$$
n(\varepsilon) = \frac{1}{e^{(\varepsilon - \mu) / T} + 1}
$$
For normal metals Fermi energy $\varepsilon_F = \mu$ is of order $\simeq 10^{4}$ K. Therefore, the energy distribution, plotted in units of $ \mu$ $T / \mu$"> for room temperatures of order $\simeq 10^{2}$ K is very close to step function. 
Fermions occupy all states below the Fermi momentum $p_F$, with energy $< \varepsilon_F$. So the total energy of a system is:
$$
E = 2 \frac{V}{(2 \pi)^3} \int d^3 k \ n(\varepsilon) \frac{\hbar k^2}{2m} = 
2 \frac{V}{(2 \pi)^3} \frac{2 \pi \hbar^2}{m} \int_{0}^{k_F} d k \ k^4 = 
\frac{V \hbar^2 k_F^5}{10 \pi^2 m} = 
\frac{V \hbar^2}{10 \pi^2 m} \left(3 \pi^2 \frac{N}{V}\right)^{5/3}
$$
Where 2 is a factor due to the presence of spin degeneracy. And in the last equality I have substituted the expression for Fermi momentum in 3D. The pressure is given by:
$$
P = -\frac{\partial E}{\partial V} = \frac{(3 \pi^2)^{2/3}}{5} 
\frac{\hbar^2}{m} \left( \frac{N}{V}\right)^{5/3} = \frac{2}{5} n \varepsilon_F
\qquad n = N / V
$$
The obtained value is rather large, as the concentration is number of order $n \simeq 10^{22}$ cm$^{-3}$, which gives the values of Young's modulus of order $10-100$ GPa. 
