Is there an exchange of electrons between two Cooper pairs in the BCS state?

The BCS state is $$\left|\psi\right>=\prod_k \left(u_k +v_k c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger\right)\left|0\right>$$ which pairs all the electrons into Cooper pairs. The interaction between these pairs is usually expressed as a process scattering one pair into another: $$V_1=\sum_{k,k'} V_{k-k'} c_{k\uparrow}^\dagger c_{-k\downarrow}^\dagger c_{-k'\downarrow} c_{k'\uparrow}$$

However, if we look at the two body interaction $$V_2=\int dr dr' \hat \rho(r) V(r-r') \hat\rho(r')=\sum_{\substack{k,k',q\\\sigma,\sigma'}}V_q c_{k+q,\sigma}^\dagger c_{k'-q,\sigma'}^\dagger c_{k',\sigma'} c_{k,\sigma}$$ then is is more general than $V_1$, as it includes more terms.

I've been told that under the BCS state, if we take $\left<\psi\middle|V_2\middle|\psi\right>$ then only the terms which are in $V_1$ as well will contribute. This makes sense, since any term which does not preserve the structure of Cooper pairs (i.e. breaks any apart) would vanish.

However, to me it seems that $V_2$ can also include terms that "mix" electrons between the different pairs, i.e. picking $q=k'-k$ to get $$V_{k'-k} c_{k',\sigma}^\dagger c_{k,\sigma}^\dagger c_{k',\sigma} c_{k,\sigma}$$ where the spins are identical. Note that $k,k'$ need not be related. Indeed even a term which would not mix them at all, would be for $q=0$ leave all particles as they are, namely, $$V_{0} c_{k,\sigma}^\dagger c_{k',\sigma'}^\dagger c_{k',\sigma'} c_{k,\sigma}$$ which even permits the spins to differ.

There I imagine in effect describe the "mean field" which the electrons feel (and correspond to roughly the average potential times the density squared). Shouldn't they be included in the BCS treatment as well? The two terms above would cancel each other due to the fermionic commutation relations, had $V_0=V_{k'-k}$, but I don't see why this should be true in general.