# Creating multiparticle states from vacuum in QFT

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?

• It is just a convention. Different authors prefer different conventions. Commented Jun 15, 2021 at 17:15
• Note that since states are physically distinguishable only up to a phase, there can be no physical reason for adding prefactors like this. Commented Jun 15, 2021 at 18:11
• 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? Commented Jun 15, 2021 at 18:15
• hint: compute norm. Commented Jun 15, 2021 at 18:54