How many ways two particles can be arranged in three phase cells according to BE statistics?

Next: Formulation of the statistical Up: Quantum statistics Previous: Symmetry requirements in quantum Consider a very simple gas made up of two identical particles. Suppose that each particle can be in one of three possible quantum states,

. Let us enumerate the possible states of the whole gas according to Maxwell-Boltzmann, Bose-Einstein, and Fermi-Dirac statistics, respectively.

For the case of Maxwell-Boltzmann (MB) statistics, the two particles are considered to be distinguishable. Let us denote them

and

. Furthermore, any number of particles can occupy the same quantum state. The possible different states of the gas are shown in Tab. 6.

Table 6: Two particles distributed amongst three states according to Maxwell-Boltzmann statistics.

1	2	3

There are clearly 9 distinct states.

For the case of Bose-Einstein (BE) statistics, the two particles are considered to be indistinguishable. Let us denote them both as

. Furthermore, any number of particles can occupy the same quantum state. The possible different states of the gas are shown in Tab. 7.

Table 7: Two particles distributed amongst three states according to Bose-Einstein statistics.

1	2	3

There are clearly 6 distinct states.

Finally, for the case of Fermi-Dirac (FD) statistics, the two particles are considered to be indistinguishable. Let us again denote them both as

. Furthermore, no more than one particle can occupy a given quantum state. The possible different states of the gas are shown in Tab. 8.

Table 8: Two particles distributed amongst three states according to Fermi-Dirac statistics.

1	2	3

There are clearly only 3 distinct states.

It follows, from the above example, that Fermi-Dirac statistics are more restrictive (i.e., there are less possible states of the system) than Bose-Einstein statistics, which are, in turn, more restrictive than Maxwell-Boltzmann statistics. Let

(575)

For the case under investigation,

			(576)
			(577)
			(578)

We conclude that in Bose-Einstein statistics there is a greater relative tendency for particles to cluster in the same state than in classical statistics. On the other hand, in Fermi-Dirac statistics there is less tendency for particles to cluster in the same state than in classical statistics.

Next: Formulation of the statistical Up: Quantum statistics Previous: Symmetry requirements in quantum

Richard Fitzpatrick 2006-02-02

First of all, the distributions are actually the macroscopical result from certain microsocopical assumptions. As your system is microscopic, I assume you refer to these assumptions, instead of the actual distributions. (The word distributions makes (in physics) only sense for a large number of particles.)

(If particles A is in state 1 and particle B in state 2, I'll write it in the following as A1B2 etc.)

These assumtions are: a) Maxwell-Boltzman: Particles are distinguishable. (As your question sounds a lot like it's from a problem sheet, I guess you understand the meaning of this word in the context of statistical physics.)

For a a 2 particle system with two states each, the possible states are: A1B1, A1B2, A2B1, A2B2

b) Bose-Einstein: Particles are indistinguishable and an exchange of particles gives us a + in the wave function. This implies that the wave function needs to be symmetric.

For a a 2 particle system with two states each, the possible states are: A1B1, A1B2 + A2B1, A2B2 (I didn't put normailzation factors.)

c) Fermi-Dirac: Particles are indistinguishable and an exchange of particles gives us a - in the wave function. This implies that the wave function needs to be anti-symmetric.

For a a 2 particle system with two states each, the possible states are: A1B2 - A2B1 (I didn't put normailzation factors.)

This should help you to solve you excercise. I didn't fully answer the question on purpose, as I assume you would like to actually underrstand it yourself. I hope this answer brings you on the way. If you need more details, just write a comment with a precise question and I'll edit the answer.