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In creating multiparticle states from the vacuum we apply the creation operators, \begin{align} &|p_1,s_1\rangle=a^{\dagger}_{p_1,s_1}|0\rangle,\\ &|p_1,s_1;p_2,s_2\rangle=a^{\dagger}_{p_1,s_1}a^{\dagger}_{p_2,s_2}|0\rangle,\\ & ...\quad ...\quad ...\quad ...\quad ...\quad ... \\ &|p_1,s_1;p_2,s_2;...\,\,...;p_n,s_n\rangle=a^{\dagger}_{p_1,s_1}a^{\dagger}_{p_2,s_2}...\,...\, ...a^{\dagger}_{p_n,s_n}|0\rangle \qquad \qquad \text{etc.} \end{align}

for particles having momentum $p$ and spin $s$. Some authors also include combinatorial prefactors like $\frac{1}{\sqrt{2!}}$, ... $\frac{1}{\sqrt{n!}}$ on the r.h.s. My question is that is it just a convention, to be compensated later by some other similar normalization prefactor, e.g. in the definition of inner product, amplitude, etc? Or does it have a concrete physical reason, in order to incorporate the indistinguishability of fermions/bosons, since the ordering of the creation-operators is ambiguous up to a sign?

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  • $\begingroup$ It is just a convention. Different authors prefer different conventions. $\endgroup$ Commented Jun 15, 2021 at 17:15
  • $\begingroup$ Note that since states are physically distinguishable only up to a phase, there can be no physical reason for adding prefactors like this. $\endgroup$
    – Charlie
    Commented Jun 15, 2021 at 18:11
  • $\begingroup$ Thank you for your comment! Can you please highlight where exactly the combinatorial pre-factor for indistinguishability is accounted for in QFT if the states are defined without them? $\endgroup$ Commented Jun 15, 2021 at 18:15
  • $\begingroup$ hint: compute norm. $\endgroup$ Commented Jun 15, 2021 at 18:54

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