Degeneracy for different energy states in Infinite cubic wellby Chelsea S Tags: cubic, degeneracy, energy, infinite, states 

#1
Dec212, 07:26 AM

P: 12

Alright, I'm back with yet another question...
So the prof was explaining that the energy in an infinite cubical well is E((h^{2}∏^{2})/2ma^{2}))(nx^{2}+ny^{2}+nz^{2}) Which is all well and good, and he gave us the example of: ψ1,2,1 = E = 6((h^{2}∏^{2})/2ma^{2})) And with little explanation mixed it up once switching one of the 1's with the 2. After some searching through my book I found out that this is the first energy level or E_{1}, and has 3fold degeneracy. Simply enough, E_{2} is ψ1,2,2 with 3 fold degeneracy also.. But after looking online, I find that E_{3} is not ψ2,2,2. How do you derive/find out what to use for the quantum numbers? As I found out, ψ2,2,2 is E_{4}. Is there a pattern? Or a formula? Any help is appreciated. 



#2
Dec212, 08:10 AM

Mentor
P: 10,791

Just sort by increasing energy:
111 (energy 3*const) 112 (energy 6*const) 122 (energy 9*const) 113 (energy 11*const) 222 (energy 12*const) 



#3
Dec212, 08:24 AM

Mentor
P: 15,571

E3 has quantum numbers 1,1,3, for a total energy of 1+1+9 = 11.
E4 has quantum numbers 2,2,2 for a total energy of 4+4+4 = 12. There is no pattern, but there is a formula  the very first one you wrote down. Plug in all the n's and then rank them by energies. 



#4
Dec212, 11:06 PM

P: 12

Degeneracy for different energy states in Infinite cubic well
Oh. Wow I was making that way too difficult. I.. don't even know why I didn't think of it that way. Geeze.
Well thanks for clarifying! hehe. I guess I should stop studying Q.M at 4 in the morning. 



#5
Dec312, 12:00 AM

Sci Advisor
P: 5,307

The interesting question is whether one can find two energies with
m_{x}² + m_{y}² + m_{z}² = n_{x}² + n_{y}² + n_{z}² with different ordered triples (m_{x}, m_{y}, m_{z}) and (n_{x}, n_{y}, n_{z}) 



#6
Dec312, 12:29 AM

P: 12

I think you should be able to...
that's exactly is what degeneracy is, right? Finding similar energy states for different wave functions. So I think it would still count as degeneracy even though your values for m and n (in your example) are different. Like: (3,3,3) = 27... and (5,1,1) would also be 27. I suppose it would work ... but hey just when you think you know something about QM, you don't. ;) 



#7
Dec312, 12:32 AM

Sci Advisor
P: 5,307

The intersting thing is that for different ordered triples the wave functions are not related by a symmetry transformation; for (1,1,2), (1,2,1) and (2,1,1) it's permutation symmetry, but for (3,3,3) and (5,1,1) there is no (no obvious ?) symmetry. One may find a hint when looking from geometrical aspects; the equation reads
n_{x}² + n_{y}² + n_{z}² = Z² which means that one tries to find integral lattice points with identical radius Z 



#8
Dec312, 12:46 AM

P: 12

So, beyond that it has not permutation symmetry (which, yes that is odd), does it make E14 special? Is there something particular about that energy state that makes it have those particular triples?




#9
Dec312, 12:48 AM

Sci Advisor
P: 5,307

Mathematically it means that there are two inequivalent ways to write Z² as a sum of three squares; physically I don't see anything special




#10
Dec312, 12:56 AM

P: 12

Oh, alight. On the inside I'm a little sad. I was half expecting the physical meaning to be something odd and hard to grasp; like figuring out how many times you'd have to throw a baseball at a wall before it tunneled through, just something odd.



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