Symmetric BCS state The ground state wavefunction for the BCS can be written $$|\Psi_{G}\rangle\equiv\prod_{\textbf{k}}[u_{k}+v_{k}c_{\textbf{k}1}^{+}c_{\textbf{-k}-1}^{+}]|\phi_{0}\rangle,$$ where $|\phi\rangle$ denotes the vacuum, $|u_{k}|^{2}+|v_{k}|^{2}=1$ and $c,c^{+}$ are the usual creation and annihilation operators in second quantization. Also, $1\equiv$ up spin, and $-1\equiv$ down spin. This wavefunction is not manifestly invariant under rotation of spin. However, we know we must be able to write it that way, in singlet form, for instance. Is there a convenient way to arrive to such an expression? 
 A: Choose a single value of $\mathbf{k}$, and consider the product of the $\mathbf{k}$ and $-\mathbf{k}$ factors. (All of the factors in the product can be paired this way.) The linear-in-$v$ term involves
$$
 c^+_{\mathbf{k},1}c^+_{-\mathbf{k},-1}
 +
 c^+_{-\mathbf{k},1}c^+_{\mathbf{k},-1}.
$$
Use the fact that the creation operators anticommute to rewrite this as
$$
 c^+_{\mathbf{k},1}c^+_{-\mathbf{k},-1}
 -
 c^+_{\mathbf{k},-1}c^+_{-\mathbf{k},1},
$$
which is manifestly invariant under rotation of spin because it's antisymmetric under the exchange of spins $1$ and $-1$. The $v^2$ term involves
$$
 c^+_{\mathbf{k},1}c^+_{-\mathbf{k},-1}
 c^+_{-\mathbf{k},1}c^+_{\mathbf{k},-1}.
$$
Use the fact that the creation operators anticommute to see that this is proportional to
$$
 (c^+_{\mathbf{k},1}
 c^+_{\mathbf{k},-1})
 (c^+_{-\mathbf{k},1}
 c^+_{-\mathbf{k},-1}).
$$
Each factor in parentheses is antisymmetric under the exchange of spins $1$ and $-1$, so this is again manifestly invariant under rotation of spin. The $u^2$ term doesn't involve spin at all, so we're done.
