Quantum harmoinic oscillator degeneracy

[nabeel17]

Ok, I know this question will sound really stupid but I'm just not following the derivation for the formula of degeneracy's given by

1/2(n+1)(n+2)

This is what I get n1+n2+n3=n
so for a given n1, n2+n3=n-n1
Then this is the line I don't understand (and I'm sure its something simple I'm overlooking)
There are n-n1+1 possible pairs of (n2,n3). I don't understand why this is. After this
its a Ʃ n-n1+1 from n1=0 to n1=n which is a bit confusing too. I can solver the sum, but I don't know why the limits are like that. Also it says n2 can take on the values 0 to n-1 which makes sense, but what if n=0? Then n2=-1? But that's is just an exception I'm guessing?

Thanks for the help :)

 

[Mute]

Perhaps it will help if you think about the problem a bit differently.

You have 3 oscillators and n units of energy. How many different ways can you distribute the energy units among the 3 oscillators (taking into account that all of the energy units are indistinguishable and the 3 oscillators are indistinguishable)?

 

[nabeel17]

Hmm, If both are indistinguishable I believe that would be 3^N/N! ?

 

[Mute]

Hmm, If both are indistinguishable I believe that would be 3^N/N! ?

Nope. The formula you wrote would describe, say the number of possible states a system has if you have N indistinguishable particles each of which can be in one of three states.

What you have here is three oscillators to which you assign dump energy units, and you have N energy units to distribute among the three boxes. How many ways can you do it?

To figure this out, there's a better way than viewing the oscillators as boxes into which you are placing energy units. Suppose you wrote down a list of the energy units

$$\underbrace{EEE\dots EE}_{N~\mbox{units}}$$

How many ways can you partition this list into three partions, with each parition corresponding to an oscillator?

 

[nabeel17]

Hmm, I had to search this up but I would have to use the Stirling formula?

 

[Mute]

Nope, Stirling's formula doesn't come into this, if I recall correctly.

You have a list of energy units $EEE\dots EE$ that you want to partition into 3 sets, with each corresponding to the energy held by one of the oscillators. Say, for example, you had 7 energy units, and you divided them like so:

$$EE|EEEE|E,$$
where the partitions "|" split the list into lists of EE, which we say belongs to oscillator 1, EEEE which belongs to oscillator 2, and E which belongs to oscillator 3. This is just one possible distribution of the energy among the three oscillators. How many ways can we partition this list into 3 sets? (Hint: think about the placement of the partitions).

 

[nabeel17]

Right so we want the number of ways to partition a set of n objects into k subsets? In this case N energy units corresponding to 3 oscillators. So it can be given by Stirling numbers of a second kind equation no? Or am I still not seeing something..

 

[Mute]

Right so we want the number of ways to partition a set of n objects into k subsets? In this case N energy units corresponding to 3 oscillators. So it can be given by Stirling numbers of a second kind equation no? Or am I still not seeing something..

Ah, I see what you're asking about now. When you said "Stirling's formula" in your previous post, I thought you meant the approximation to the factorial/Gamma function. In any case, no, I don't think the answer will exactly be a "Stirling number". The wikipedia article states that a stirling number (of the second kind) is the number of ways to partition a set of n objects into k non-empty sets. However, we are allowing empty partitions here: one of the oscillators can have all N energy units.

The way to calculate the result is to think about it like this: We can split the list $EE\dots E$ up into two partitions by inserting a "|" somewhere in the list. How many places are there that we can place that partition? Then, we want to insert another "|" so that we have three partitions in total, with the possibility of an empty partition. How many places are there that we can insert this second partition "|"?

 

[nabeel17]

Ohhh! I think I get what you're saying now. At least I think i do. I can partion the the energy units by putting a | somewhere in the EE..E but there would only be (n+1) places I can do that. Then the next partition would have (n+2) ways right? That is if I'm allowed subsets with zero E units. So now I see my factors of (n+1)and (n+2) but why are they being multiplied and where does the 1/2 come in?

 

[Mute]

Ohhh! I think I get what you're saying now. At least I think i do. I can partion the the energy units by putting a | somewhere in the EE..E but there would only be (n+1) places I can do that. Then the next partition would have (n+2) ways right? That is if I'm allowed subsets with zero E units. So now I see my factors of (n+1)and (n+2) but why are they being multiplied and where does the 1/2 come in?

Yes! That's it! The factors are being multiplied because you have two independent choices to make for where to put the partitions. Say you put the first parition between energy units 2 and 3 in the list. You still have n+2 options for where you can place the second partition. But what if you put the first partition between energy units 27 and 28? For that placement you still have n+2 choices of where to put the partition. We can say this for each one of the n+1 placements of the first partition, so we have (n+1)*(n+2) choices - at first glance. Now to deal with the factor of 1/2. Would you know where it came from if it had been written as "1/2!"?

 

[nabeel17]

Ok, I think I understand now up until the 1/2! If I relate it to finding the number of states, it divides by N! to account for duplicate counts. In this case, is it accounting for any duplicates...? Ohhhh right its not accounting for duplicate energy units, its accounting for duplicate positions I can put the | right?

 

[Mute]

Ok, I think I understand now up until the 1/2! If I relate it to finding the number of states, it divides by N! to account for duplicate counts. In this case, is it accounting for any duplicates...? Ohhhh right its not accounting for duplicate energy units, its accounting for duplicate positions I can put the | right?

Yes! The 1/2 accounts for the degeneracy in where you can put the partitions. In this abstraction of the problem, we're not moving the energy units around - they just sit there - so we need only worry about the position of the partitions and the degeneracy associated with them.

 

[nabeel17]

Ahh, thank you so much! I really appreciate the help.

 

[Mute]

No problem.