Wow, that three electron case is surprisingly complicated.
I agree that there should be 27 states. @DrClaude suggested earlier that I should think about what I did in the first step. If you just add the number of states you get after the second step you get 7+5+3+5+3+1+3, which is 27. This is...
Sorry, I'm still somehow confused by this.
So do I just have 7 states for AM 3, 5 states for AM 2, 3 states for AM 1, and 1 state for AM 0?
Or do I add to get 7 for AM 3, 10 for AM 2, 9 for AM 1, and 1 for AM 0? (By adding the totals I got for the states at the end?)
My initial thought was to multiply the number of states you got from the AM 0, 1, 2 systems by the number of states you got by combining them, like:
There are 5 ##j = 2## states from combining ##j_1## and ##j_2##, so
## j = 3 ## = 7 states * 5 states = 35 states
## j = 2 ## = 5 states * 5 states...
I can solve the two particle system easily enough:
Using ##j_1 = 1## and ##j_2 = 1##, the possible total angular momentum values are ##j = 2, 1, 0##. With ## m = -j , -j+1, ..., j ##,
##j = 2: m = 2, 1, 0, -1, -2 ## (5 states)
##j = 1: m = 1, 0, -1## (3 states)
## j = 0: m = 0 ## (1 state)
I...
From what we did in class, I think we need to minimize the energy with respect to a.
Like ##E = \frac{\hbar ^2}{m} a^2 - 2 e^2 a + \frac{5}{8} e^2 a = \frac{\hbar ^2}{m} a^2 - \frac{11}{8} e^2 a ##, then minimize it
Finding the minimum value: ## - (\frac{11}{16})^2 \frac{m e^4}{\hbar^2} ##...