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,

1. 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?
2. 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).

• First we need to define what $\xi^P$ means as $\xi$ is a number ($\pm1$) and $P$ is a permutation. I'm quite sure that it's to be understood as $\xi^{N(P)},$ where $N(P)$ is the number of inversions in $P$. It's similar to the parity of a permutation. Aug 29 '21 at 17:25
• When you write $\sigma_0=i$ I suppose that $i$ is the identity element of $S_2,$ not the imaginary unit. And $\sigma_1\in S_2$ really means that $\sigma_1$ is the nonidentity element in $S_2,$ doesn't it? Aug 29 '21 at 17:27
• Yes, both comments correct. Aug 29 '21 at 17:40

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 $$\frac1{N!}\sum_{Q}\xi^{P'Q}\xi^{Q} = \operatorname{sgn}(P') = \xi^{P'}.$$

• I think that you should explain the notation $\sum_{P,Q | PQ^{-1} = P'}$. Aug 29 '21 at 17:55
• @md2perpe Sure, I mean $\sum_{(P,Q) \in S}$ where $S$ is the set $\{(P,Q)|PQ^{-1} = P'\}$. Aug 29 '21 at 17:57
• Differently worded: the sum is over $P$ and $Q$ such that $PQ^{-1}=P'.$ Aug 29 '21 at 17:59
• @md2perpe Yes, exactly. Aug 29 '21 at 18:06