Question

Energy levels and their degenercies for the cubic potential.

I do not understand the middle column on page 5 on the top of the chart in the link provided. How do I know when to switch from 1 to 2 and to 3 ect ect?

http://zeus.phys.uconn.edu/~mcintyre/workfiles/Transfer-Stuff/Cap%206.pdf

Answer 1

its the possible set of quantum numbers

Answer 2

I don't know. There is something I'm missing. But assuming it is the possible set of quantum numbers then how do I know how to determine them?

Answer 3

how to determine the possible set of quantum numbers?

Answer 4

the amount of possible sets depends on the n values, see, if you have n = n = n = 1, then there is only one possible way to display the set (1, 1, 1). no other order can be made that is different.
where as if you have 1, 1, and 2, there are 3-ways you can arrange the sequence of numbers to create the set.

Answer 5

so then why do you go from 122 to 113 instead of 112 to 222?

Answer 6

also how do you determine the degeneracy?

Answer 7

eq 6.36

Answer 8

thats all I got man, good luck!

Answer 9

The quantum numbers come from the fact that you're quantizing - assuming discrete levels - your angular momentum, wavenumber, electron orbit... They start at 1, called the ground level, and then only go up by integer values. See eq. 6.23

Eq. 36 doesn't determine the degeneracies of the cubic potential - that's for the "Isotropic Harmonic Oscillator." In the cubic potential (very specifically the cubic potential) the degeneracies come about through adding up the squares of quantum numbers together in the energy equation eq. 6.27:

\[E_{n_xn_yn_z} = \frac {\hbar^2\pi^2}{2mL^2} (n_x^2 +n_y^2+n_z^2) \]

this part of the term:
\[ (n_x^2 +n_y^2+n_z^2)\]
plugging in (1,1,2), (1,2,1), (211), gives you the same sum. That equivalence is the degeneracy, and since there are 3 ways to arrange those quantum numbers, The first excited state, corresponding to energy 6E1, has a degeneracy of 3. It is three fold degenerate. Reiterating, degenerate energies are identical energies - you mathematically can't tell E(1,1,2), E(1,2,1), and E(2,1,1) apart.

**"so then why do you go from 122 to 113 instead of 112 to 222?"

That table shows the energy levels ranked from lowest energy to highest energy. Looking at
\[ (n_x^2 +n_y^2+n_z^2)\]

1st excited state: (1,1,2): (1+1+4) = (1+4+1) = (4+1+1) = 6 ; Degeneracy of 3
2nd excited state: (1,2,2): (1+4+4) = (4+1+4) = (4+4+1) = 9 ; Degeneracy of 3
3rd excited state: (1,1,3): (1+1+9) = (1+9+1) = (9+1+1) = 11 ; Degeneracy of 3
4th excited state: (2,2,2): ----------------- (4+4+4) = 12 ; Degeneracy of 1

Each new excited state is achieved by adding 1 quantum number at a time, starting with the lowest possible energy..
1+1+1 = 3 ---------> Degeneracy of 1 (1,1,1)
1+1+2 = 4 ---------> Degeneracy of 3 (2,1,1) , (1,2,1) , (1,1,2)
1+2+2 = 5 ---------> Degeneracy of 3
1+1+3 = 5 ----------> Degeneracy of 3
2+2+2 = 6 -----------> Degeneracy of 1

Lastly, you know that the 3rd excited state (1,1,3)(and degeneracies) follows the 2nd excited state (1,1,2)(...) because even though in both cases their quantum numbers add up to 5, E(1,1,3) has a higher energy

E(1,2,2) = 9 E1
E(1,1,3) = 11 E1

Hope this helps :)