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).

  • 1
    $\begingroup$ 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. $\endgroup$
    – md2perpe
    Commented Aug 29, 2021 at 17:25
  • $\begingroup$ 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? $\endgroup$
    – md2perpe
    Commented Aug 29, 2021 at 17:27
  • $\begingroup$ Yes, both comments correct. $\endgroup$ Commented Aug 29, 2021 at 17:40

1 Answer 1


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

  • $\begingroup$ I think that you should explain the notation $\sum_{P,Q | PQ^{-1} = P'}$. $\endgroup$
    – md2perpe
    Commented Aug 29, 2021 at 17:55
  • $\begingroup$ @md2perpe Sure, I mean $\sum_{(P,Q) \in S}$ where $S$ is the set $\{(P,Q)|PQ^{-1} = P'\}$. $\endgroup$ Commented Aug 29, 2021 at 17:57
  • 1
    $\begingroup$ Differently worded: the sum is over $P$ and $Q$ such that $PQ^{-1}=P'.$ $\endgroup$
    – md2perpe
    Commented Aug 29, 2021 at 17:59
  • $\begingroup$ @md2perpe Yes, exactly. $\endgroup$ Commented Aug 29, 2021 at 18:06

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