- #1
ghotra
- 53
- 0
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:
<br /> \binom{6+(4-1)}{4-1} = \frac{[6+(4-1)]!}{(4-1)!6!} = 84<br />
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
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:
<br /> \binom{6+(4-1)}{4-1} = \frac{[6+(4-1)]!}{(4-1)!6!} = 84<br />
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
Last edited:
