Questions around the BCS state

Question

Here is the BCS state:

$$ left|Psi_mathrm{BCS}rightrangle = prod_k left( u_k - v_ke^{i phi} c_{kuparrow}^{dagger} c_{-kdownarrow}^{dagger}right) left|0rightrangle.$$

When I develop the BCS state to understand what it means, I will have a state like this :
begin{align}
left|Psi_mathrm{BCS}rightrangle 
& = 
prod_k u_k |0rangle 
+ sum_{k_0}prod_{k neq k_0} u_k(-v_{k_0}e^{i phi})|k_0uparrow, -k_0downarrowrangle
 & quad+
sum_{k_0<k_1}prod_{k neq k_0,k_1}u_k(-1)^2e^{2i phi}v_{k_0}v_{k_1}|k_1uparrow, -k_1downarrow, k_0uparrow, -k_0downarrowrangle+ cdots
end{align}
So we have an infinite superposition of cooper pairs (one cooper pair + 2 pairs + 3 pairs + etc). Is this understanding correct?
In my course they say that $|v_k|^2$ is the probability to have a Fermi quasiparticle with wavevector $k$. But I don't understand this.

physshyp · Answer

think like this, you create a cooper pair at $k$ with a prob of $v_{k} ^{2}$ or you don't create a cooper pair at $k$ with a prob  $u_{k} ^{2}$. If you look carefully to your first equation you can see this.

In other words, there is a superposition of vacuum and cooper pair at each $k$. And the respective probabilities that you have a cooper pair or not at $k$ is $|v_{k} ^{2}|$.

Since whether you create a particle or not is probabilistic, a total number of particles is uncertain.

Questions around the BCS state

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