# Eigenvalues of the exchange operator determined by the particle type (boson or fermion) in a two particle system

While dealing with a two particle system in QM (the particles are identical), the net wave function of the system would be simply the product of the wavefunctions of the individual particles in the given potential(assuming no interactions between the particles) if they were distinguishable. When the particles aren't distinguishable (as in this case), the eigenfunctions of the Hamiltonian must also be eigenfunctions of the exchange operator P (Griffiths section 5.1) The eigenvalues of P are 1 and -1 and based on whether the particles in question are bosons or fermions, we choose the corresponding eigenvalue of P(1 for bosons and -1 for fermions) and thus construct the correct 2 particle wavefunction.

I don't understand why it is the type of the particle(boson or fermion) that determines the eigenvalue. In other words, why must all fermions necessarily have an eigenvalue of -1 in all possible scenarios ( and bosons 1)? We group particles into bosons and fermions based on the eigenvalue they take, but why must a given kind of particle be forced to take a fixed eigenvalue in the first place?

For fermions, states must be antisymmetric so that $$\vert \Psi_-\rangle = \frac{1}{\sqrt{2}}\left(\vert \psi\rangle_1\vert\phi\rangle_2 - \vert \psi\rangle_2\vert\phi\rangle_1\right)\, .$$ In particular, if $\vert\phi\rangle=\vert\psi\rangle$, $\vert\Psi\rangle=0$, thus enforcing the Pauli condition that two fermions cannot occupy the same state.
Contrariwise, for bosons $$\vert\Psi_+\rangle=\frac{1}{\sqrt{2}}\left(\vert \psi\rangle_1\vert\phi\rangle_2 +\vert \psi\rangle_2\vert\phi\rangle_1\right)\,$$ permutation symmetry of identical bosonic states leads to a bunching effect demonstrated experimentally in Hanbury-Brown-Twiss-type of experiments.