Questions around the BCS state

Question

Here is the BCS state:

$$ \left|\Psi_\mathrm{BCS}\right\rangle = \prod_k \left( u_k - v_ke^{i \phi} c_{k\uparrow}^{\dagger} c_{-k\downarrow}^{\dagger}\right) \left|0\right\rangle.$$

1. When I develop the BCS state to understand what it means, I will have a state like this : \begin{align} \left|\Psi_\mathrm{BCS}\right\rangle & = \prod_k u_k |0\rangle + \sum_{k_0}\prod_{k \neq k_0} u_k(-v_{k_0}e^{i \phi})|k_0\uparrow, -k_0\downarrow\rangle \\ & \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_1\uparrow, -k_1\downarrow, k_0\uparrow, -k_0\downarrow\rangle+ \cdots \end{align} So we have an infinite superposition of cooper pairs (one cooper pair + 2 pairs + 3 pairs + etc). Is this understanding correct?

2. 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.

Answer 1

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.