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Particle in a 3D Boxby eep
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#1
Mar2806, 08:06 PM

P: 228

Hi,
This comes from Griffiths Intro to Quantum Mechanics Prob. 4.2 We're asked to solve an infinite cubical well, which I have no problem with. The next part asks you to call the distinct energy levels E1, E2, E3... etc. in order of increasing energy and determine their degeneracies. It then asks what the degeneracy of E14 is and why is this case interesting. I think the degeneracy of E14 is 6, however I don't see why this case is interesting. 


#2
Mar2806, 08:26 PM

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If two of the three quantum numbers of a certain state are eqaul to one another but different from the third, one expects a 3fold degeneracy (like 112, 121, 211). If the three quantum numbers are different, one expects a 6fold degeneracy (123, 132, 213, 231, 321, 312). But there was something special about that state. (Does E14 means that n_x^2 + n_y^2 +n_z^2 = 14? or is it the 14th energy level?) Patrick 


#3
Mar2806, 08:39 PM

P: 228

It's the 14th energy level. I have the quantum numbers as (4,3,1).



#4
Mar2806, 08:51 PM

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Particle in a 3D Box
Pat 


#5
Mar2806, 08:58 PM

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I probably missed one 111 211 plus permutations 221 plus perms.. 222 311 321 322 331 332 411 421 422 431 


#6
Mar2806, 09:00 PM

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#7
Mar2806, 09:01 PM

P: 228

Oops, I wasn't putting the levels in order. So I have the quantum numbers as (3,3,3) then but I still don't see what's special about this state besides that it has a degeneracy of 1...
EDIT: Ah, I just saw your post. Okay, now I see why that's interesting. 


#8
Feb1410, 01:46 PM

P: 5

hello,
i didn't understand why 333 or 551 would be 4 fold degenerate. can you please explain it to me. thanks 


#9
Mar2512, 12:08 PM

P: 1

Know this thread is old, but I want to "end" it.
333 AND 511 do indeed make up energy lvl 14, as both give a value of 27. Thus, it has a degenerancy of 1 + 3 = 4 


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