I have some questions about the BCS groundstate. This is given by
$$\prod_k (u_k + v_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger) |0\rangle$$
I've heard many times that $|v_k|^2$ is the probability that the pair $(k\uparrow, -k\downarrow$) is occupied and $|u_k|^2$ similarly for unoccupied. However, that does not seem obvious to me. If there was only one momentum state, this interpretation is clear; however, it seems that the product complicates this. Here is a calculation that explains what I mean. The probability that I have a cooper pair with one of the electrons having momentum $l$ is the mod squared of
$$\langle 0|c_{l\uparrow} c_{-l\downarrow} \prod_k (u_k + v_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger) |0\rangle$$
$$=\langle 0|[\prod_{k<l} u_k + v_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger] c_{l\uparrow} c_{-l\downarrow} (u_l + v_l c_{l\uparrow}^\dagger c_{-l\downarrow}^\dagger) [\prod_{k>l} u_k + v_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger] |0\rangle$$
$$=\langle 0|[\prod_{k<l} u_k + v_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger] (u_l c_{l\uparrow} c_{-l\downarrow} + v_l(1-n_{l\uparrow} - n_{-l\downarrow} + n_{-l\downarrow}n_{l\uparrow})) [\prod_{k>l} u_k + v_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger] |0\rangle$$
$$=(\prod_{k\neq l} u_k) v_l$$
But we just want the answer to be $v_l$. I realize that the state with a pair occupied is not only the state where there is only one pair, but I am not sure how to cleanly implement that.
I seems like when most people interpret $u_k$ and$v_k$, they make it sounds quite obvious, so there is probably a simple explanation. But if it would help if I was able to show this explicitly.