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?
 A: 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" $S_{ij}$.
Note added: Yes your state is normalized. It's just not the way I like to write it.
