From: Mike Stay <metaweta@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Sent: Wednesday, July 8, 2009 2:14:15 PM Subject: Re: [math-fun] math, existence, and God 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 _______________________________________________ Quantum theory provides an absolute test for indistinguishibility. Two particles are indistinguishable exactly when their wavefunction is required to be symmetric (bosons) or antisymmetric (fermions) under exchange. For composite particles this requires that their internal quantum states are identical. In the case of nuclei, this means that they must be the same isotope, and in the same excitation state. This has engineering consequences in the liquifaction of hydrogen. Hydrogen nuclei (protons) have spin 1/2, so the total nuclear wavefunction in the hydrogen molecule is antisymmetric. This total wavefunction is the product of three parts: vibration, rotation, and spin. The spins couple to form the symmetric spin 1 triplet state (orthohydrogen) or the antisymmetric spin 0 singlet state (parahydrogen). The lowest energy vibrational state is symmetric, and the lowest rotational state is symmetric. Thus the ground state of hydrogen at absolute zero must have the antisymmetric spin state, i.e. parahydrogen. The lowest state of orthohydrogen must be the first rotational excited state (this being antisymmetric and lower than the first excited vibrational state). At thermodynamic equilibrium at room temperature, hydrogen is 3 parts ortho to 1 part para. Now, the crucial property is that in pure hydrogen the conversion between ortho and para is very slow, on the order of days. So when hydrogen is liquified, it remains 3 parts ortho to 1 part para. As the liquid hydrogen slowly converts to the para form, the released rotational energy exceeds the heat of vaporization, so the liquid will boil away. Thus the liquifaction process must use a magnetic catalyst to speed the conversion to parahydrogen. See the Wikipedia article ( http://en.wikipedia.org/wiki/Spin_isomers_of_hydrogen ). -- Gene