Particle in 3D Box: Degeneracy of E14 & Why Interesting?

In summary, the conversation is about solving an infinite cubical well and determining the degeneracies of the energy levels. The degeneracy of E14 is 6 and there is nothing special about it. The quantum numbers for E14 are (3,3,3) and (5,5,1). The special case comes from the fact that the energy level can be formed by two different combinations of quantum numbers, resulting in a degeneracy of 4.
  • #1
eep
227
0
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.
 
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  • #2
eep said:
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.

I seem to recall something special about that level. What are the possible values of the quantum numbers giving that energy?

If two of the three quantum numbers of a certain state are eqaul to one another but different from the third, one expects a 3-fold degeneracy (like 112, 121, 211). If the three quantum numbers are different, one expects a 6-fold 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
It's the 14th energy level. I have the quantum numbers as (4,3,1).
 
  • #4
eep said:
It's the 14th energy level. I have the quantum numbers as (4,3,1).

I tried to verify that this was the 14th energy level but this seemed to be the 12th one to me...But I might have missed a couple. There is nothing special about that one, no.

Pat
 
  • #5
nrqed said:
I tried to verify that this was the 14th energy level but this seemed to be the 12th one to me...But I might have missed a couple. There is nothing special about that one, no.

Pat

oops, I find that it's the 13th...

I probably missed one


111

211 plus permutations

221 plus perms..

222

311

321

322

331

332

411

421

422

431
 
  • #6
eep said:
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.

You know what, if it was the *next* level, the 14th... then it would be either 333 or 511, so it would have a 4 -fold degeneracy, which is special
 
  • #7
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
hello,

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

thanks
 
  • #9
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
 

1. What is a particle in a 3D box?

A particle in a 3D box is a simplified model used in quantum mechanics to represent a particle confined to a finite region in three-dimensional space. It is often used to study the behavior and properties of atoms, molecules, and other small particles.

2. What is degeneracy in the context of a particle in a 3D box?

Degeneracy refers to the phenomenon of multiple quantum states having the same energy level. In the case of a particle in a 3D box, degeneracy occurs when multiple energy levels have the same energy value, resulting in degenerate states.

3. Why is the degeneracy of E14 interesting in a particle in a 3D box?

The degeneracy of E14 in a particle in a 3D box is interesting because it violates the general rule that energy levels in a system are non-degenerate, meaning each energy level only corresponds to one unique state. The degeneracy of E14 suggests the existence of additional symmetries in the system.

4. How is the degeneracy of E14 calculated in a particle in a 3D box?

The degeneracy of E14 can be calculated using the formula (2n+1)^2, where n is the principal quantum number. In the case of E14, n=3, so the degeneracy is (2(3)+1)^2 = 49.

5. What are the practical applications of studying degeneracy in a particle in a 3D box?

Studying the degeneracy of energy levels in a particle in a 3D box can help us understand the underlying symmetries and properties of the system. This information can then be used to make predictions and calculations in various fields, such as quantum chemistry, solid-state physics, and materials science.

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