# Micro-canonical Ensemble of Ideal Bose Gas

1. Oct 6, 2013

### Ang Han Wei

Hi,
can I know why the number of partitions separating different states have to be taken into account for the derivation of number of states in an ideal Bose Gas but not in the Fermi Gas?

What is the physical significance of this "partition"? In what ways can they vary?

2. Oct 6, 2013

### dauto

That's just a math trick to easily calculate the total number of states. The partitions have no physical significance whatsoever.

3. Oct 6, 2013

### Ang Han Wei

I see that there are some occasions when such partitions are called into use and they have to be taken into the total number of possible states in a factorial, while at other times, they do not appear.

How would one know when such partitions are to be taken into account?

4. Oct 6, 2013

### dauto

The partitions should be used when they help solve the problem at hand. They have no physical significance and it is always possible to solve a problem without mentioning them. But sometimes they make it much easier to solve a problem. For instance, suppose you are calculating the number of ways that you can put two Fermions in three boxes, since each box can contain either zero or one fermion, than you can count them as follows
1: (1|1|0)
2: (1|0|1)
3: (0|1|1)
for a total of three states.
You can see that the first fermion can go in any one of the three boxes, and second fermion can go on any of the remaining 2 boxes. 3*2=6= 3!/1!. At the ed you must divide by 2! because the fermions are identical so permutations among them wont create new states. You have then a total of 3!/(2!*1!)= 3 states. That can be generalized to

# of states = N!/[(N-n)!*n!] for n fermions on N boxes.

Now suppose you have two bosons that need to be placed in three boxes. since you can have more than one boson per boxe (no exclusion principle applies to bosons), there will be more states. Lets count them
1: (2|0|0)
2: (1|1|0)
3: (1|0|1)
4: (0|2|0)
5: (0|1|1)
6: (0|0|2)
for a total of six states. How can we generalize that? Doesn't seem obvious at first.
Here comes the math trick. Lets represent the particles by a '*'. the six states can now be represented as
1: (**||)
2: (*|*|)
3: (*||*)
4: (|**|)
5: (|*|*)
6: (||**)
Physically the '*' represents a particle while the '|' represents nothing whatsoever, it is just part of the notation. But that doesn't matter. Mathematically you have four symbols, Two '*'s and two '|'s. This symbols can be placed in 4!=24 different orders, but since the '*'s are identical we must divide by 2!, and the '|'s are also identical so we also divide by 2!. You have then a total of 4!/(2!*2!)= 6 states. That can be generalized to

# of states = (N+n-1)!/[(N-1)!*n!] for n bosons in N boxes.

Voila!

Last edited: Oct 6, 2013