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For a particle in a box (so that the momentum spectrum is discrete), we can write the identity operator as a sum over all momentum eigenstates of a projection to that eigenstate: $$I=\displaystyle\sum\limits_{p} |p\rangle\langle p|.$$

I was wondering what the corresponding form of the identity operator would be if we had multiple particles? In particular, a sum over both momenta appears to not lead to normalized states (when the particles are identical): $$A=\displaystyle\sum\limits_p\sum\limits_q |p,q\rangle\langle p,q|.$$ For example, given two possible momentum states for each particle 0 and 1, $$A=\sum\limits_{p=0}^1\sum\limits_{q=0}^1 |p,q\rangle\langle p,q|,$$ we find that $$A|0,1\rangle = |0,1\rangle +|1,0\rangle\langle 1,0|0,1\rangle=|0,1\rangle\pm|1,0\rangle$$ where the + sign is taken if the particles are both bosons and the minus sign if they are both fermions.

This looks like what we're supposed to get up to a normalization constant. I'm trying to figure out exactly how to modify the operator A so that the normalization constant works out. In particular, just dividing A by root 2 to fix this problem, $$A = \dfrac 1{\sqrt 2} \sum\limits_{p=0}^1\sum\limits_{q=0}^1 |p,q\rangle\langle p,q|,$$ we find that $$A|0,0\rangle = \dfrac 1{\sqrt 2}|0,0\rangle$$ which is not normalized.

Any help would be appreciated. Thanks!