**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 *H*⊗*H* of the joint wavefunction, the two-particle wave functions for one particle in state *a* and one in state *b* is

for bosons, and

for fermions (the *C*s are normalization constants).

What happens, then, if we put the two particles into the same state, here state *a*? Our boson wavefunction becomes

(assuming the single particle state is properly normalized), so that two (or more) bosons may be placed in the same quantum state.

However, our fermion wavefunction becomes

,

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.

### Like this:

Like Loading...

*Related*

Tags: Boson, Exchange Symmetry, Fermion, Friday Physics, Pauli Exclusion Principle, physics, Quantum Mechanics, Spin-Statistics Theorem

This entry was posted on July 3, 2009 at 12:01 am and is filed under Math/Science. You can follow any responses to this entry through the RSS 2.0 feed.
You can leave a response, or trackback from your own site.

## Leave a Reply