The spin index in general form of BCS Hamiltonian 
I want to derive the general form of BCS Hamiltonian, and the original form is:$$H_{\mathrm{BCS}}=\sum_{k, \sigma} \xi_{k} c_{k, \sigma}^{\dagger} c_{k, \sigma}+\frac{1}{2} \sum_{k, k^{\prime}} \sum_{\sigma, \sigma^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime \prime \prime}} V_{k, k^{\prime} ; \sigma, \sigma^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime\prime \prime}} c_{k, \sigma}^{\dagger} c_{-k, \sigma^{\prime}}^{\dagger} c_{-k^{\prime}, \sigma^{\prime \prime}} c_{k^{\prime}, \sigma^{\prime \prime \prime}}$$
where $$V_{k,k^\prime;\sigma,\sigma^\prime,\sigma^{\prime\prime},\sigma^{\prime\prime\prime}}=\langle -k,\sigma;k,\sigma^{\prime}|V|-k^\prime,\sigma^{\prime\prime};k^\prime,\sigma^{\prime\prime\prime}\rangle$$
and introduce the order parameter:$$\Delta_{k ; \sigma, \sigma^{\prime}}=-\sum_{k^{\prime} ; \sigma^{\prime \prime}, \sigma^{\prime \prime\prime}} V_{k, k^{\prime} ; \sigma, \sigma^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime \prime \prime}}\left\langle c_{-k^{\prime}, \sigma^{\prime \prime}} c_{+k^{\prime}, \sigma^{\prime \prime \prime}}\right\rangle\\\begin{array}{l}{\Delta_{k, \sigma, \sigma^{\prime}}^{*}=-\sum_{k^{\prime} ; \sigma^{\prime \prime}, \sigma^{\prime \prime\prime}} V_{k, k^{\prime} ;  \sigma, \sigma^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime \prime\prime}}^{*}\left\langle c_{k^{\prime}, \sigma^{\prime \prime\prime}}^{\dagger} c_{-k^{\prime}, \sigma^{\prime \prime}}^{\dagger}\right\rangle} \\ {=-\sum_{k^{\prime} ; \sigma^{\prime \prime}, \sigma^{\prime \prime\prime}} V_{k^{\prime}, i ; \sigma^{\prime \prime}, \sigma^{\prime \prime \prime}, \sigma, \sigma^{\prime}}\langle c_{k^{\prime}, \sigma^{\prime \prime\prime}}^{\dagger} c_{-k^{\prime}, \sigma^{\prime \prime}}^{\dagger}\rangle}\end{array}$$
the second step origins from $$V_{k,k^\prime;\sigma,\sigma^\prime,\sigma^{\prime\prime},\sigma^{\prime\prime\prime}}^*=\langle-k^\prime,\sigma^{\prime\prime};k^\prime,\sigma^{\prime\prime\prime}|V|-k,\sigma;k,\sigma^{\prime}\rangle = V_{k^\prime,k;\sigma^{\prime\prime},\sigma^{\prime\prime\prime},\sigma,\sigma^\prime,}$$
then the pairing term can be written as:$$\frac{1}{2} \sum_{k, k^{\prime}} \sum_{\sigma, \boldsymbol{\sigma}^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime \prime \prime}} V_{k, k^{\prime} ; \sigma, \sigma^{\prime}, \sigma^{\prime \prime},\sigma^{\prime \prime\prime}}\left(\left\langle c_{-k^{\prime}, \sigma^{\prime \prime}} c_{k, \sigma^{\prime \prime\prime}}\right\rangle c_{k, \sigma}^{\dagger} c_{-k, \sigma^{\prime}}^{\dagger}+\left\langle c_{k, \sigma}^{\dagger} c_{-k, \sigma^{\prime}}^{\dagger}\right\rangle c_{-k^{\prime}, \sigma^{\prime \prime}} c_{k^{\prime}, \sigma^{\prime \prime \prime}}\right)$$
the first term is simple, but when I rewrite the second term, there is a question:
$$\frac{1}{2} \sum_{k, k^{\prime}} \sum_{\sigma, \sigma^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime \prime \prime}} V_{k^{\prime}, i ; \sigma^{\prime \prime}, \sigma^{\prime \prime \prime}, \sigma, \sigma^{\prime}}\left\langle c_{k^{\prime}, \sigma^{\prime \prime}}^{\dagger} c_{-k^{\prime}, \sigma^{\prime \prime \prime}}^{\dagger}\right\rangle c_{-k, \sigma} c_{k, \sigma^{\prime}}$$
compared with the expression of $\Delta_{k, \sigma, \sigma^{\prime}}^{*}$, the spin index is not consistent so that I cannot replace it as $\Delta_{k, \sigma, \sigma^{\prime}}^{*}$. As the result, I am confused that what's my problem? Or do I miss some constraints?
 A: Your analysis is mostly correct, but you need to carry it one step further.
Let's consider the term of confusion:
$$\sum_{k, k^{\prime}} \sum_{\sigma, \sigma^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime \prime \prime}} V_{k, k^{\prime} ; \sigma, \sigma^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime\prime \prime}} c_{k, \sigma}^{\dagger} c_{-k, \sigma^{\prime}}^{\dagger} c_{-k^{\prime}, \sigma^{\prime \prime}} c_{k^{\prime}, \sigma^{\prime \prime \prime}}\tag{1}$$
The $c$s are fermionic operators and as such they satisfy anticommutation relations. This implies some symmetries of the matrix elements of the potential. In particular, anticommute the two $c^\dagger$, then send $k \to -k$ and swap $\sigma \leftrightarrow \sigma'$. We get:
$$-\sum_{k, k^{\prime}} \sum_{\sigma, \sigma^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime \prime \prime}} V_{-k, k^{\prime} ; \sigma', \sigma, \sigma^{\prime \prime}, \sigma^{\prime\prime \prime}} c_{k, \sigma}^{\dagger} c_{-k, \sigma'}^{\dagger} c_{-k^{\prime}, \sigma^{\prime \prime}} c_{k^{\prime}, \sigma^{\prime \prime \prime}}$$
Therefore we must have:
$$V_{k, k^{\prime} ; \sigma, \sigma^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime\prime \prime}}= -V_{-k, k^{\prime} ; \sigma', \sigma, \sigma^{\prime \prime}, \sigma^{\prime\prime \prime}}$$
We can do the same with the two $c$ and we get the similar relationship
$$V_{k, k^{\prime} ; \sigma, \sigma^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime\prime \prime}}= -V_{k, -k^{\prime} ; \sigma, \sigma', \sigma^{\prime \prime\prime}, \sigma^{\prime\prime}}$$
Applying these two identities one after the other we get,
$$V_{k, k^{\prime} ; \sigma, \sigma^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime\prime \prime}}= V_{-k, -k^{\prime} ; \sigma', \sigma, \sigma^{\prime \prime\prime}, \sigma^{\prime\prime}}\tag{2}$$
This will be useful later.
Now apply the mean field approximation to $(1)$. We get a constant term, which does not concern us, and the interesting part:
$$\sum_{k, k^{\prime}} \sum_{\sigma, \sigma^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime \prime \prime}} V_{k, k^{\prime} ; \sigma, \sigma^{\prime}, \sigma^{\prime \prime},\sigma^{\prime \prime\prime}}\left(\left\langle c_{-k^{\prime}, \sigma^{\prime \prime}} c_{k^\prime, \sigma^{\prime \prime\prime}}\right\rangle c_{k, \sigma}^{\dagger} c_{-k, \sigma^{\prime}}^{\dagger}+\left\langle c_{k, \sigma}^{\dagger} c_{-k, \sigma^{\prime}}^{\dagger}\right\rangle c_{-k^{\prime}, \sigma^{\prime \prime}} c_{k^{\prime}, \sigma^{\prime \prime \prime}}\right)$$
This can be rewritten as
$$\sum_{k, \sigma, \sigma^{\prime}} c_{k, \sigma}^{\dagger} c_{-k, \sigma^{\prime}}^{\dagger} \sum_{k^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime \prime \prime}} V_{k, k^{\prime} ; \sigma, \sigma^{\prime}, \sigma^{\prime \prime},\sigma^{\prime \prime\prime}}\left\langle c_{-k^{\prime}, \sigma^{\prime \prime}} c_{k^\prime, \sigma^{\prime \prime\prime}}\right\rangle +\\
+\sum_{k^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime \prime \prime}}c_{-k^{\prime}, \sigma^{\prime \prime}} c_{k^{\prime}, \sigma^{\prime \prime \prime}}\sum_{k, \sigma, \sigma^{\prime}} V_{k, k^{\prime} ; \sigma, \sigma^{\prime}, \sigma^{\prime \prime},\sigma^{\prime \prime\prime}}\left\langle c_{k, \sigma}^{\dagger} c_{-k, \sigma^{\prime}}^{\dagger}\right\rangle \tag{3}$$
Now let 
$$\Delta_{k, \sigma, \sigma^{\prime}}\equiv-\sum_{k^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime \prime\prime}} V_{k, k^{\prime} ; \sigma, \sigma^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime \prime \prime}}\left\langle c_{-k^{\prime}, \sigma^{\prime \prime}} c_{+k^{\prime}, \sigma^{\prime \prime \prime}}\right\rangle$$
Then the first line of $(3)$ becomes
$-\sum_{k, \sigma, \sigma^{\prime}} c_{k, \sigma}^{\dagger} c_{-k, \sigma^{\prime}}^{\dagger} \Delta_{k, \sigma, \sigma^{\prime}}$
The overall result must be Hermitian, so in the end the second line of $(3)$ will have to become the hermitian conjugate of what we just got, which is
$$-\sum_{k, \sigma, \sigma^{\prime}} c_{-k, \sigma^{\prime}} c_{k, \sigma} \Delta_{k, \sigma, \sigma^{\prime}}^*$$
In order to compare this with the second line of $(3)$, swap $k\leftrightarrow k'$, $\sigma \leftrightarrow \sigma^{\prime\prime\prime}$, $\sigma^{\prime}\leftrightarrow \sigma^{\prime\prime}$  in the latter to get,
$$\sum_{k, \sigma, \sigma^{\prime}}c_{-k, \sigma^{\prime}} c_{k, \sigma}\sum_{k', \sigma^{\prime\prime}, \sigma^{\prime\prime\prime}} V_{k^\prime, k; \sigma^{\prime \prime\prime}, \sigma^{\prime\prime}, \sigma^{\prime},\sigma}\left\langle c_{k', \sigma^{\prime\prime\prime}}^{\dagger} c_{-k', \sigma^{\prime\prime}}^{\dagger}\right\rangle$$
So to get the correct final expression it must be the case that 
$$\Delta_{k, \sigma, \sigma^{\prime}}^*=-\sum_{k', \sigma^{\prime\prime}, \sigma^{\prime\prime\prime}} V_{k^\prime, k; \sigma^{\prime \prime\prime}, \sigma^{\prime\prime}, \sigma^{\prime},\sigma}\left\langle c_{k', \sigma^{\prime\prime\prime}}^{\dagger} c_{-k', \sigma^{\prime\prime}}^{\dagger}\right\rangle$$
However from the definition of $\Delta$, 
$$\Delta_{k, \sigma, \sigma^{\prime}}^*\equiv-\sum_{k^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime \prime\prime}} V^*_{k, k^{\prime} ; \sigma, \sigma^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime \prime \prime}}\left\langle c_{k^{\prime}, \sigma^{\prime \prime \prime}}^\dagger c_{-k^{\prime}, \sigma^{\prime \prime}}^\dagger\right\rangle$$
So in order to get the correct result in the end, the matrix elements of the potential must satisfy
$$V^*_{k, k^{\prime} ; \sigma, \sigma^{\prime}, \sigma^{\prime \prime}, \sigma^{\prime \prime \prime}}=V_{k^\prime, k; \sigma^{\prime \prime\prime}, \sigma^{\prime\prime}, \sigma^{\prime},\sigma}\tag{4}$$
To prove this relation, start from the (correct) definition of the matrix elements of $V$:
$$V_{k,k^\prime;\sigma,\sigma^\prime,\sigma^{\prime\prime},\sigma^{\prime\prime\prime}}=\langle k,\sigma;-k,\sigma^{\prime}|\,\,V\,\,|-k^\prime,\sigma^{\prime\prime};k^\prime,\sigma^{\prime\prime\prime}\rangle$$
Note that the OP's question contains a small mistake in the definition of the matrix elements. We can take the complex conjugate remembering that $V$ is hermitian, 
$$V^*_{k,k^\prime;\sigma,\sigma^\prime,\sigma^{\prime\prime},\sigma^{\prime\prime\prime}}=\langle -k^\prime,\sigma^{\prime\prime};k^\prime,\sigma^{\prime\prime\prime}|\,\,V\,\,|k,\sigma;-k,\sigma^{\prime}\rangle=V_{-k',-k;\sigma^{\prime\prime},\sigma^{\prime\prime\prime},\sigma,\sigma^{\prime}}$$
Now apply $(2)$ and you get precisely $(4)$.
Therefore the non-constant term of $(1)$ in the mean-field approximation is
$$-\sum_{k, \sigma, \sigma^{\prime}} (c_{k, \sigma}^{\dagger} c_{-k, \sigma^{\prime}}^{\dagger} \Delta_{k, \sigma, \sigma^{\prime}}+c_{-k, \sigma^{\prime}} c_{k, \sigma} \Delta_{k, \sigma, \sigma^{\prime}}^*)$$
