Addition of Permutations Sorry, if the title doesn't provide any clarity, but I didn't really know how to call it. Anyways, I've been studying quantum field theory from Blundell's book and during the derivation of the formula for second quantizing an operator I'm getting a little bit confused about a step he makes. I'll walk you through:
We first write down an expression of an $N$-particle state
$$\left|\psi_1\cdots\psi_N\right>=\frac{1}{\sqrt{N!}}\sum_P\xi^P\prod_{i=1}^N\left|\psi_{P(i)}\right>$$
where the sum is taken over all $N!$ permutations of the single particle states $\left|\psi_i\right>$. We take $\xi=+1$ for bosons while $\xi=-1$ is for fermions. Now, the inner product of two such states (either both boson states or fermion states) is:
$$\left<\chi_1\cdots\chi_N|\psi_1\cdots\psi_N\right>=\frac1{N!}\sum_P\sum_Q\xi^{P+Q}\prod_{i=1}\left<\chi_{Q(i)}|\psi_{P(i)}\right>\label{before}\qquad (1)$$
Then comes the confusing part. The author says that we can rewrite the above expression using $P'=P+Q$ which spans all the permutations $N!$ times and hence:
$$\left<\chi_1\cdots\chi_N|\psi_1\cdots\psi_N\right>=\sum_{P'}\xi^{P'}\prod_{i=1}^N\left<\chi_i|\psi_{P'(i)}\right>\qquad (2)$$
Writting down examples for $N=2$ and $N=3$ I can see what he means and then the generalisation seems kind of intuitive. So for $N=2$ one would write down the 2-particle state as:
$$\left|\psi_1\psi_2\right>=\frac1{\sqrt{2!}}(\xi^{\sigma_0}\left|\psi_1\right>\left|\psi_2\right>+\xi^{\sigma_1}\left|\psi_2\right>\left|\psi_1\right>)$$
Where $\sigma_0=i,\sigma_1\in S_2$ (the elements of the symmetry group). Then the inner product with another state $\left|\chi_1\chi_2\right>$ would be:
$$ \frac12(\xi^{\sigma_0+\sigma_0}\left<\chi_1|\psi_2\right>\left<\chi_2|\psi_1\right> +
+\xi^{\sigma_0+\sigma_1}\left<\chi_1|\psi_1\right>\left<\chi_2|\psi_2\right>
+\xi^{\sigma_1+\sigma_0}\left<\chi_1|\psi_1\right>\left<\chi_2|\psi_2\right>
+\xi^{\sigma_1+\sigma_1}\left<\chi_1|\psi_2\right>\left<\chi_2|\psi_1\right>)$$
$$ =\frac122(\xi^0\left<\chi_1|\psi_2\right>\left<\chi_2|\psi_1\right>+\xi^{1}\left<\chi_1|\psi_1\right>\left<\chi_2|\psi_2\right>) $$
So, as it seems to me, for $N=2$:
$$\sigma_0+\sigma_0=\sigma_1+\sigma_1=\sigma_1$$
$$\sigma_0+\sigma_1=\sigma_1+\sigma_0=\sigma_0$$
However, the author from the example of $N=3$ seems to have defined in general for $\sigma_n,\sigma_m\in S_3$
$$\sigma_n+\sigma_m=\sigma_{(n+m)mod(3!)}$$
Well, now to make my confusion a bit more precise,

*

*Why is there a disagreement with my definition of addition and the authors one. Is it maybe because he might consider $\sigma_1$ for example a different permutation in the symmetry group?

*Does the generalisation come from noticing the common theme in specific examples or is there a more formal and elegant way of transitioning from $(1)$ to $(2)$?

The question is also posted on PSE (link).
 A: Here you aren't really "adding permutations".  As mentioned in the comments, $\xi^{P}$ is a shorthand notation that means $\operatorname{sgn}(P)$ when $\xi = -1$, and is a constant $1$ when $\xi = 1$.
Now it is a matter of doing some substitutions to re-arrange the sum.  We can do this by noting
$$\prod_{i}\left<\chi_{Q(i)}|\psi_{P(i)}\right> = \prod_{i}\left<\chi_{i}|\psi_{PQ^{-1}(i)}\right>$$
by the commutativity of scalar multiplication: we've just ordered the factors by the index of $\chi$.  So we have
\begin{align*}
\left<\chi_1\cdots\chi_N|\psi_1\cdots\psi_N\right> &=\frac1{N!}\sum_P\sum_Q\xi^{P+Q}\prod_{i}\left<\chi_{Q(i)}|\psi_{P(i)}\right>\\
&=\frac1{N!}\sum_P\sum_Q\xi^{P+Q}\prod_{i}\left<\chi_{i}|\psi_{P(Q^{-1}(i)}\right>.\end{align*}
Now we again re-arrange, by commutativity of addition, letting $P' = PQ^{-1}$. The factor $\prod_{i}\left<\chi_{i}|\psi_{P'(i)}\right>$ appears in every summand corresponding to  $P,Q$ such that $PQ^{-1} = P'$. So we get
\begin{align*}
\frac1{N!}\sum_P\sum_Q\xi^{P+Q}\prod_{i}\left<\chi_{i}|\psi_{P(Q^{-1}(i)}\right> &= \frac1{N!}\sum_{P'} \sum_{P,Q | PQ^{-1} = P'}\xi^{P}\xi^{Q} \prod_{i}\left<\chi_{i}|\psi_{P'(i)}\right> \\
&= \frac1{N!}\sum_{P'} \prod_{i}\left<\chi_{i}|\psi_{P'(i)}\right> \sum_{P,Q | PQ^{-1} = P'}\xi^{P}\xi^{Q}\\
&=\sum_{P'} \prod_{i}\left<\chi_{i}|\psi_{P'(i)}\right> \frac1{N!}\sum_{Q}\xi^{P'Q}\xi^{Q}.\\
\end{align*}
Now, if $\xi = 1$, then $\frac1{N!}\sum_{Q}\xi^{P'Q}\xi^{Q} = 1 = \xi^{P'}$ and we are done. If $\xi = -1$, then we have $$\xi^{P'Q}\xi^{Q} = \operatorname{sgn}(P'Q) \operatorname{sgn}(Q) = \operatorname{sgn}(P')\operatorname{sgn}(Q) \operatorname{sgn}(Q) = \operatorname{sgn}(P')$$ and so $$\frac1{N!}\sum_{Q}\xi^{P'Q}\xi^{Q} = \operatorname{sgn}(P') = \xi^{P'}.$$
