# Pauli Exclusion Principle in Landau Fermi's liquid theory

I do not understand how Pauli exclusion principle helps us to understand the excitations in Landau Fermi's liquid theory. In Landau Fermi liquid theory, Pauli exclusion principle and adiabatic continuity are the central ingredients. Adiabatic continuity allows us that the ground state of non-interacting system connects to the ground state of interacting system so long as the interaction switches on slow enough. During this process, quantum numbers are conserved quantities(i.e. spins/momentum/charge). Consider there is a quasi-particle and quasi-hole with momenta $$k_1$$ and $$k_2$$ respectively, where $$k_1 \geq k_F$$ and $$k_2 \leq k_F$$. When the interaction turns on, the quasiparticle and quasihole will interact and scatter to particles with momenta $$k_3$$ and $$k_4$$. My confusion is that why Pauli exclusion principle only allows both $$k_3, k_4 \geq k_F$$ (which are both quasiparticles) but not 1 quasiparticle with 1 quasihole ?

## 1 Answer

Maybe I understand the reason why both the momenta of new quasiparticles $$k_3, k_4 \geq k_F$$ after scattering. During the scattering process, the momenta of the original quasiparitcle and quasihole exchange and we know that $$k_1$$ must loses its momentum and $$k_2$$ gains momentum. Under the scattering process, it is impossible that $$k_1 = k_3$$ and $$k_2 = k_4$$ since one must loses its momentum. Besides, it is also impossible for $$k_1 \rightarrow k_3 \leq k_F$$ and $$k_2 \rightarrow k_4 \geq k_F$$ since there is no available state with $$k_3 \leq k_F$$ inside the filled Fermi sea. Or by Pauli exclusion principle, it is impossible that you put two fermions in same energy state with if you have a quasiparitcle with momentum $$k_3 \leq k_F$$. Therefore, this scattering channel is prohibited. The remaining scattering channel is $$k_1 \rightarrow k_3 \geq k_F$$, $$k_2 \rightarrow k_4 \geq k_F$$, where $$k_1 \geq k_F \geq k_2$$.