# Many body BCS theory related question

I am trying to find the mean number of particles in the BCS ground state $$|\psi>$$, but I am stuck on a step.

$$|\psi> = \Pi_{k}(u_{k} + v_{k}c^{{\dagger}}_{{k}{\uparrow}}c^{\dagger}_{{-k}{\downarrow}})$$

where $$u_{k}$$, $$v_{k}$$ are complex numbers. So to begin, we start with a specific spin state $$<\psi|N_{{p}{\uparrow}}|\psi> =<\psi|c^{{\dagger}}_{{p}{\uparrow}}c_{{p}{\uparrow}}|\psi>$$

This is what I get

$$\Pi_{k} |v_{k}|^{2} (c_{{k}{\uparrow}}c_{{-k}{\downarrow}}c^{{\dagger}}_{{p}{\uparrow}}c_{{p}{\uparrow}}c^{{\dagger}}_{{k}{\uparrow}}c^{\dagger}_{{-k}{\downarrow}})$$

I don't know how to simplify this or if I am even on right track. Any hints?

• put the orders in normal order using the anti-commutation rule for fermion operators. Search Wick's theorem. – wcc Apr 30 at 13:32

Here is a calculation based on a general Bogoluibov transformation $$a_i= u_{i\beta}b_\beta +v^*_{i\beta}b^\dagger_\beta\\ a^\dagger_i= v_{i\alpha} b_\alpha +u^*_{i\alpha}b^\dagger_\alpha.$$ With $$\vert 0 \rangle_b$$ being the vacuum state for the $$b_\alpha$$'s we have
$$N=\sum_i { _b\langle{0}\vert{ a^\dagger_i a_i}\vert{0}\rangle _b} \\ ={\sum_{i,\alpha,\beta} { _b\langle{0}\vert{( v_{i\alpha} b_\alpha +u^*_{i\alpha}b^\dagger_\alpha)(u_{i\beta}b_\beta +v^*_{i\beta}b^\dagger_\beta)}\vert{0}\rangle_b}}\\ = \sum_{i,\alpha,\beta} (v_{i\alpha} v^*_{i\beta}) {_b\langle{0}\vert{ b_\alpha b^\dagger_\beta }\vert{0}\rangle_b}\\ =\sum_{\alpha,\beta} v_{i\alpha} v^*_{i\beta} \delta_{\alpha\beta}\\ = \sum_\alpha |v_\alpha|^2.$$ We can also write $$N= \frac 12 \sum_{\alpha=1}^N\left((|v_\alpha|^2-|u_\beta|^2) + (|u_\alpha|^2+|v_\alpha|^2)\right)\\ =\frac 12 \sum_{\alpha=1}^N\left((|v_\alpha|^2-|u_\beta|^2) + 1\right),$$ which looks like the contribution of the Dirac sea negative-energy Bogoliubov-de Gennes equation eigenstates $$(v^*_\alpha, u^*_\alpha)$$, but corrected by an infinite normal-ordering counterterm (the sum of all the +1''s).
As an aside, note I'm not sure that your BCS ground state is normalized. I like to write $$\vert{0}\rangle _b ={\mathcal N} \exp\left\{\frac 12 a^\dagger_i a^\dagger_jS_{ij}\right\} \vert{0}\rangle_a$$ where $$S_{ij}= v^*_{i\alpha}(u^*)^{-1}_{\alpha j}$$ is skew symmetric and $${\mathcal N}= {\rm det}^{-1/4}(I+S^\dagger S)$$ is derived by using the fermionic version of MacMahon's master theorem. This way makes clear that the BCS ground state is a coherent state superposition of Cooper pairs with pair "wavefunction" $$S_{ij}$$.