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.

Advertisements

Tags: , , , , , , ,

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s


%d bloggers like this: