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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?

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

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$\langle{\hat c}_{k\sigma}^\dagger{\hat c}_{k\sigma}\rangle$ 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 \begin{equation}\label{numberavg}\langle{\hat n}_k\rangle=1-\frac{\xi_k}{\sqrt{\xi_k^2+|\Delta_k|^2}}\end{equation}

Suggestions and changes to answer are welcomed.

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