Fetter & Walecka's derivation of second quantised potential term in many-particle TDSE For the potential term in the Hamiltonian, I understand that we go through the same process as with the kinetic energy term. On the RHS of the TDSE, we get something like $\frac{1}{2}\sum_{i}\sum_{j\neq i}\sum_{k}\sum_{l\neq k}\langle ij|V|kl\rangle C(\dots E_{i-1}E_kE_{i+1}\dots E_{j-1}E_lE_{j+1}\dots;t)$.
Reordering the argument list of $C(\dots)$ on the  RHS incurs a phase factor of $(-1)^P$, where $P$ depends on the order of $i, j, k, l$.
Some examples: if their order of ascent is ...
$kilj$: $P = n_{k+1}+\dots+n_{i-1}+n_{l+1}+\dots+n_{j-1}$
$klij$: $P = n_{k+1}+\dots+n_{i-1}+n_{l+1}+\dots+n_{j-1}$
$likj$: $P=n_{i+1}+\dots+n_{k-1}+n_{l+1}+\dots+n_{j-1}+1$ [the extra $+1$ comes from the fact that $i$ and $j$ are positioned with opposite 'polarity' to $k$ and $l$, therefore the $E_k,E_l$ have to be swapped past each other to get into the correct positions]
As above, by my reckoning, $P(klij)=P(kilj)$, however if we calculate $(-1)^P$ by operating on $|\lbrace n_m\rbrace\rangle$ like $a_i^{\dagger} a_j^{\dagger} a_l a_k|\lbrace n_m\rbrace\rangle$, I get $(-1)^{S_k+S_{l}-1+S_{j}-2+S_{i}-2}|\dots n_k-1\dots n_l-1 \dots n_j+1\dots n_i+1\rangle \\ = (-1)^{S_k+S_{l}+S_{j}+S_{i}-1}|\dots n_k-1\dots n_l-1\dots n_j+1\dots n_i+1\dots\rangle$
if the order is $klij$ but 
$(-1)^{S_k+S_{l}-1+S_{j}-2+S_{i}-1}|\dots n_k-1\dots n_l-1\dots n_j+1\dots n_i+1\dots\rangle\\= (-1)^{S_k+S_{l}+S_{j}+S_{i}}|\dots n_k-1\dots n_l-1\dots n_j+1\dots n_i+1\dots\rangle$
if the order is $kilj$. The order of $i$ and $l$ seems to matter when I calculate the phase factor by successive application of the $a, a^{\dagger}$ but not when I calculate it by reordering the arguments of the $C(\dots)$ coefficients. Where am I going wrong?
 A: Let the ordering of the occupation numbers $n_m$ in the Fock states be such that $\left( klij \Leftrightarrow k<l<i<j \right)$ and let $S_m = n_1 + n_2 + \dots + n_{m-1}$ be phase factors calculated before applying any creation/annihilation operators. 
You do take into account, correctly, that applying an annihilation operator lowers the phase factor for the action of the next operator by $1$, but should also account for the fact that applying a creation operator raises the phase factor for the next operator by 1. 
Take the $klij$, $k<l<i<j$ case, step by step: 
$$
a_j^{\dagger} a_i^{\dagger} a_l a_k|\dots n_k\dots n_l \dots n_i\dots n_j\dots\rangle = (-1)^{S_k}\delta_{n_k,1}a_j^{\dagger} a_i^{\dagger} a_l |\dots n_k-1\dots n_l \dots n_i\dots n_j\dots\rangle =\\
= (-1)^{S_k+S_l-1}\delta_{n_k,1}\delta_{n_l,1}a_j^{\dagger} a_i^{\dagger}|\dots n_k-1\dots n_l-1 \dots n_i\dots n_j\dots\rangle = \\
= (-1)^{S_k+S_l-1+S_i-2}\delta_{n_k,1}\delta_{n_l,1}\delta_{n_i,0}a_j^{\dagger}|\dots n_k-1\dots n_l-1 \dots n_i+1\dots n_j\dots\rangle = \\
= (-1)^{S_k+S_l-1+S_j-2+S_i-2+1}\delta_{n_k,1}\delta_{n_l,1}\delta_{n_i,0}\delta_{n_j,0}|\dots n_k-1\dots n_l-1 \dots n_i+1\dots n_j+1\dots\rangle =\\
= (-1)^{S_k+S_l+S_j+S_i}\delta_{n_k,1}\delta_{n_l,1}\delta_{n_i,0}\delta_{n_j,0}|\dots n_k-1\dots n_l-1 \dots n_i+1\dots n_j+1\dots\rangle
$$
Now take the $kilj$, $k<i<l<j$ case:
$$
a_j^{\dagger} a^{\dagger}_l a_i a_k|\dots n_k\dots n_i \dots n_l\dots n_j\dots\rangle = (-1)^{S_k}\delta_{n_k,1}a^{\dagger}_j a^{\dagger} _l a_i|\dots n_k-1\dots n_i \dots n_l\dots n_j\dots\rangle =\\
= (-1)^{S_k+S_i-1}\delta_{n_k,1}\delta_{n_i,1}a^{\dagger}_j a^{\dagger}_l|\dots n_k-1\dots n_i-1 \dots n_l\dots n_j\dots\rangle = \\
= (-1)^{S_k+S_i-1+S_l-2}\delta_{n_k,1}\delta_{n_i,1}\delta_{n_l,0}a^{\dagger}_j|\dots n_k-1\dots n_i-1 \dots n_l+1\dots n_j\dots\rangle = \\
= (-1)^{S_k+S_i-1+S_l-2+S_j-2+1}\delta_{n_k,1}\delta_{n_i,1}\delta_{n_l,0}\delta_{n_j,0}|\dots n_k-1\dots n_i-1 \dots n_l+1\dots n_j+1\dots\rangle =\\
= (-1)^{S_k+S_l+S_j+S_i}\delta_{n_k,1}\delta_{n_i,1}\delta_{n_l,0}\delta_{n_j,0}|\dots n_k-1\dots n_i-1 \dots n_l+1\dots n_j+1\dots\rangle
$$
It checks out. 
A: Worked out where I've been going wrong. Referring to the content of my question, in the $klij$ case, i.e. $k<l<i<j$, when reordering the $C(\dots)$ argument list on the RHS, $E_l$ has to be swapped past the space where the vacant $E_i$ would be, hence in this case
$\begin{split}
P(klij) &= n_{k+1} + \dots + n_{i-1} + \dots + n_{l+1} + \dots + n_{j-1} - 1 \\
&= S_i - S_k - n_k + S_j - S_l - n_l - 1
\end{split}$
and so the phase factor, $(-1)^{P(klij)} = (-1)^{S_i + S_k - n_k + S_j + S_l - n_l - 1}$, which, with the $\delta_{n_k0}\delta_{n_l0}$ on the RHS, gives  $(-1)^{S_i+S_k+S_j+S_l-1}$. This is the same phase factor you get when you operate on the state like $a^{\dagger}_ia^{\dagger}_j a_la_k|\lbrace n_m\rbrace\rangle$ as I mentioned in the original question.
