## Physics Friday 79

Part 4 of 4

In the previous part, we introduced the spin-statistics theorem, which tells us that identical bosons, particles with integer spin, have joint wavefunctions symmetric with respect to exchange of the particles; while fermions, with half-integer spin, have wavefunctions antisymmetric under exchange. In our first part, we noted that in terms of the tensor product Hilbert space HH of the joint wavefunction, the two-particle wave functions for one particle in state a and one in state b is
$\Psi_b=C_1\left(|a\rangle|b\rangle+|b\rangle|a\rangle\right)$
for bosons, and
$\Psi_f=C_2\left(|a\rangle|b\rangle-|b\rangle|a\rangle\right)$
for fermions (the Cs are normalization constants).
What happens, then, if we put the two particles into the same state, here state a? Our boson wavefunction becomes
$\Psi_b=C_1\left(|a\rangle|a\rangle+|a\rangle|a\rangle\right)=|a\rangle|a\rangle$
(assuming the single particle state $|a\rangle$ is properly normalized), so that two (or more) bosons may be placed in the same quantum state.
However, our fermion wavefunction becomes
$\Psi_f=C_2\left(|a\rangle|a\rangle-|a\rangle|a\rangle\right)=0$,
which means it is a state with zero probability; it is impossible for any two indistinguishable fermions to exist in the same quantum state simultaneously. This is the Pauli exclusion principle.

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