4 ms·
As a former math-head, let me chime in. The parent poster is probably not looking for an explanation of what exchange operators are, which are rather abstract a
by cynicalkane 3y ago
As a former math-head, let me chime in. The parent poster is probably not looking for an explanation of what exchange operators are, which are rather abstract and uninteresting in a physical vacuum, but rather how they are related to physical reality.
I’ve learned a lot about both and still am not sure why exchange operators and physical meaning are related. Why does the fermion care if you do an exchange operator to it? I get that fields of fermions demand that behavior but it still feels abstract.
- apognwsi 3y agothe statistics of a collection of particles, ie fermions or bosons, depends on their symmetry under commutation. that is, the wavefunction of the collection of particles, which does have physical meaning isofar as its square is the real density of the particles, must also obey this symmetry, ie it must negate (-1*) under exchange of two identical fermions. so it's not that the 'fermion' cares, but that the consequence of (anti)symmetry under exchange affects the density of a collection of such particles, which leads to a measurable difference in their statistics (how likely they are to be close to each other) compared to non-symmetric (normal) counterparts.