# Boson Counting

1. Oct 13, 2005

### ghotra

A typical stat mech. question is the following: If I have 5 bosons and energy E to divide among the bosons, what is the total number of possible configurations?

I can't remember this answer, so if someone reading this can post it that would be appreciated.

Now, I want to ask a slightly more difficult question.

Suppose I have n bosons and I do not allow any boson to be excited past the kth energy level. How many possible configurations are there?

Example: 2 bosons and I restricted to ground, first and second excited states.
If we assume they bosons are indistinguishable, then there are 3^2 possible configurations:

00
01
02
10
11
12
20
21
22

Of course, the bosons are indistinguishable...so the possible configurations is reduced to six:

00
01
02
11
12
22

If there are 3 bosons restricted to the ground, first and second energy levels, then it turns out that there are 10 possible configurations.

What is the general formula that counts the number of possible configurations for n bosons restricted to the kth excitation?

I posted this same question in the probability forum....but I didn't refer to any physics....so one seemed to know (thus far anyway). I know that this is also a standard question...so the answer is definitely out there is well-known.

Last edited: Oct 13, 2005
2. Oct 13, 2005

### ghotra

Well, someone just posted the solution in the probability forum. I am lost as to why it is the same exact answer as the "typical problem".

If I have M units of energy to divide among N particles, then the number of ways to do this is:

(M+N-1)!/(N-1)!/M!

In my question, I do not have a set amount of energy to divide up among the M particles. Rather, I have maximum amount of energy that each particle can have. So the collectin of combinations will involve sets of particles that have different amounts of total energy.

For example

00 has two particles in the ground state
01 has a single particle in the first excited state
11 has two particles in the first excited state

So why does this question have the same exact answer. Namely, if I restrict each particle to the kth energy level and have n particles, then the total number of combinations is:

(k+n-1)!/(k-1)!/n!

Intuitively, it seems that if I don't require each combination to have the same total energy, then there should be more combinations possible. Obviously not, but why.

3. Oct 14, 2005

### ghotra

Here is my question stated much more clearly:

If I have k quanta of energy to divide into n distinguishable harmonic oscillators, what is the total number of possible configurations?

If I have n indistinguishable harmonic oscillators and I do not allow any of them to be excited above the (k-1)th energy level, then what is the total number of possible configurations.

The answer to BOTH questions is:

(n+k-1)! / n! / (k-1)!

These questions seem quite different, but yet the have the same answer. Can someone provide an explanation as to why? In the first question, each configuration has the same total energy. In the second question, the various configurations do not have to have the same total energy.

It seems like the additional states I pick up in question 2 by loosing the restriction on the total energy is exactly cancelled by making the oscillators indistinguishable.

4. Oct 14, 2005

### Gokul43201

Staff Emeritus
I'm not so sure it is the same answer for both cases. Let's take another example and work both cases.

You have 2 different fermions and 3 total quanta. Using the labeling notation (E1,E2) the possible states are (3,0), (0,3), (2,1), (1,2) - that's 4 states.

Now take 2 bosons and allow a maximum of 2 levels (called levels 1 and 2). Labeling by occupation number (ie : (n1,n2)), the possible states are (2,0), (1,1) and (0,2) - that's only 3 states.

5. Oct 14, 2005

### ghotra

Obviously I have butchered this problem.

I'm gonna have to sort through this and respond again later.

Last edited: Oct 14, 2005
6. Oct 14, 2005

### ghotra

I swear I have tested this. This should work.

Question:

If I have k quanta of energy to divide into n distinguishable harmonic oscillators, what is the total number of possible configurations?

(k+n-1)! / k! / (n-1)!

Question:

If I have k indistinguishable harmonic oscillators and I do not allow any oscillator to be excited above the (n-1)th energy level, what is the total number of possible configurations?

(k+n-1)! / k! / (n-1)!

So I have finally straightened it out. My problem was that I was interchanging the variables. Now, this is what I would like an explanation on:

These situations seem wildly different but they are governed by the same combinatoric relation (with swapped variables). Why is this so?

It is a standard argument to cast the first problem into a permutation problem: How many ways can you permute k identical balls and (n-1) identical dividing lines? This recasting makes sense since the harmonic oscillators are distinguishable (and since the dividing lines make n distinct boxes).

However, I am having trouble making sense of this recasting when I think about the second problem that involves indistinguishable harmonic oscillators.

Any insights would be appreciated.

7. Oct 14, 2005

### ghotra

I think I've found why these are related.

In the first question, it is the quanta that are indistinguishable while in the second question, it is the oscillators that are indistinguishable.

In other words, it makes sense that the roles of k and n swapped between the questions. Now it also makes some sense that the underlying combinatorics is the same.