# Number of quasi-particle in $|\psi_{BCS}\rangle$

What would the average number of quasi-particle in the superconducting state in BCS theory $$\langle{\hat n}_k\rangle=\langle{\hat \gamma}_{k\sigma}^\dagger{\hat \gamma}_{k\sigma}\rangle$$ or $$\langle{\hat n}_k\rangle=\langle{\hat c}_{k\sigma}^\dagger{\hat c}_{k\sigma}\rangle$$ Or they are the same thing. Because we have changed fermionic basis after Bogoliubov transformation?

• By “number of superconducting states” do you mean “number of quasiparticles in |BCS>”? Apr 25, 2020 at 5:15
• @ragnar Oh yes. I will correct my question. Apr 25, 2020 at 6:43

$$\langle{\hat c}_{k\sigma}^\dagger{\hat c}_{k\sigma}\rangle$$ will give you average number of quasi-partice. $$\langle{\hat \gamma}_{k\sigma}^\dagger{\hat \gamma}_{k\sigma}\rangle$$ will give you average number of Bogoliv particle. The calculation is straight forward.
\begin{align*} \langle{\hat c}_{k\sigma}^\dagger{\hat c}_{k\sigma}\rangle&=|u_k|^2\langle{\hat \gamma}_{k\sigma}^\dagger{\hat \gamma}_{k\sigma}\rangle+|v_k|^2\langle{\hat \gamma}_{-k\bar\sigma}{\hat \gamma}_{-k\bar\sigma}^\dagger\rangle=2|u_k|^2f_k+2|v_k|^2(1-f_k) \end{align*} where $$f_k=\langle{\hat \gamma}_{k\sigma}^\dagger{\hat \gamma}_{k\sigma}\rangle=\frac{1}{e^{\beta E_{k}}+1}$$. As $$\beta\to\infty$$, $$\langle{\hat c}_{k\sigma}^\dagger{\hat c}_{k\sigma}\rangle=2|v_k|^2$$ and $$$$\label{numberavg}\langle{\hat n}_k\rangle=1-\frac{\xi_k}{\sqrt{\xi_k^2+|\Delta_k|^2}}$$$$