How Many Quantum States Can Four Bosons Have in a Harmonic Oscillator?

[ghotra]

Suppose I have 4 bosons in a one-dimensional harmonic oscillator potential and that the total energy is $$E_\text{tot} = 8 \hbar \omega$$. Recall, $$E_n = (n+1/2)\hbar\omega$$.

Question: How many quantum states exist? (assume no spin degeneracy)

After accounting for the ground state, we have 6 quanta of energy to distribute. The harmonic oscillators are indistinguishable, so we do not care about the ordering. Here is my count:

(6,0,0,0)
(5,1,0,0)
(4,2,0,0)
(4,1,1,0)
(3,3,0,0)
(3,2,1,0)
(3,1,1,1)
(2,2,2,0)
(2,2,1,1)

where, for example, (5,1,0,0) means we give one boson 5 quanta and another boson 1 quanta...the other two bosons are in the ground state.

So, there are 9 possible quantum states.

My question: What is the general formula that determines the number of quantum states?

Typical answer:

This situation is akin to have 6 balls and 4-1 walls, and then asking how many ways can you permute those objects. To which, the answer is:

$$\binom{6+(4-1)}{4-1} = \frac{[6+(4-1)]!}{(4-1)!6!} = 84$$

and is _clearly_ wrong. This formula comes from the question of asking how many ways can you put N indistinguishable bosons into d distinguishable degenerate energy levels. Essentially, it cares about the order. That is,

(6,0,0,0) 4!/3! permutations
(5,1,0,0) 4!/2! permutations
(4,2,0,0) 4!/2! permutations
(4,1,1,0) 4!/2! permutations
(3,3,0,0) 4!/2!/2! permutations
(3,2,1,0) 4! permutations
(3,1,1,1) 4!/3! permutations
(2,2,2,0) 4!/3! permutations
(2,2,1,1) 4!/2!/2! permutations

(yes that adds to 84)

For this problem, we are asking: How many ways can you put 6 indistinguishable quanta into 4 indistinguishable bosons? So the questions are not the same and the above formula does not apply. Here is my question once again:

How many ways can we divide M indistinquishable quanta into N indistinguishable bosons?

Notice, if N >= M, then the answer is given by p(M), the partition function of M.
http://en.wikipedia.org/wiki/Integer_partition

 

[ghotra]

Anyone? I know there are smart enough people here.

 

[denian]

But in this case, N < M, so the answer is not as straightforward. However, there is a general formula for finding the number of possible quantum states in a system of N indistinguishable particles with a total energy of E:

Number of states = \binom{N+E-1}{E}

Therefore, in this specific case, with 4 bosons and a total energy of 8 \hbar \omega, the number of possible quantum states is:

Number of states = \binom{4+8-1}{8} = \binom{11}{8} = \frac{11!}{8!3!} = 165

This formula takes into account the indistinguishability of the particles and the fact that the order does not matter. It is important to note that this formula only works for non-degenerate energy levels. If there is spin degeneracy, the formula would have to be modified accordingly.