On Wed, Jul 8, 2009 at 1:56 PM, Allan Wechsler<acwacw@gmail.com> wrote:
Clark, I'm not sure I understand what I'm reading, but I think there are "particles" that obey Bose statistics, and are indistinguishable; those are called bosons; then there are the distinguishable fermions which obey Fermi statistics.
No, all bosons of a particular kind are indistinguishable, as are all fermions of a particular kind. You can distinguish a boson from a fermion by looking at its spin (fermions have half-integer spin, bosons have integer spin). You can distinguish an electron from a proton by looking at either its charge or mass, even though both are fermions.
some isolated atoms are bosons and some are fermions. There was some element that they couldn't Bose-condense because the atoms were fermions, but diatomic molecules of the same element were bosons and condensed just fine. Who knew?
You can pair up fermions into bosons. Superconductivity can be explained by taking the fundamental particle to be a system of two electrons interacting via phonons (quantized sound waves). Helium-3 and Helium-4 behave very differently when you cool them. He-4 becomes a superfluid fairly quickly, because it's a boson (even number of nuclei) and superfluid is a big Bose-Einstein condensate. He-3 only becomes a superfluid at much lower temperatures when the nuclei pair up into Cooper pairs (six nuclei interacting via phonons). When you swap two indistinguishable fermions, you negate the wave function, so the amplitude for having two indistinguishable fermions in the same state is zero. If electrons were bosons, chemistry would be very boring; as it is, the quantum state of a bound electron consists of spin and orbital angular momentum. So for any given orbital, you can have at most two electrons inhabiting it, because there are only two spin states for an electron. -- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike http://reperiendi.wordpress.com