Addition of Angular Momentum in a Helium Atom

Hello, I have attached a homework problem I need help with as a Picture file. With my exam only a day away, I'm unfortunately stuck trying to get to the solutions to these problems before I can fully attempt them myself, so that I can study them for the exam and get as much preparation possible. There were seven total, but I have finished three of them myself. This is the third of four I need help with in order to have time to study. Thank you for your assistance.

See attachment for better symbol representation:
Consider a Helium atom with two electrons. Suppose you know that one of the electrons is in the l1=3 state, while the other is in the l2 = 2 state. What are the possible values of l1z and l2z? So how many different quantum states describing the orbital angular momentum configuration of the two electrons are possible? Suppose L = L1 + L2 represents the total orbital angular momentum. What are the possible values of l, the quantum number associated with the total orbital angular momentum of the two electrons. For each possible value of l, list the possible values of lz, the total z-component of the orbital angular momentum. Show that counting the states in this l, lz basis agrees with that of the product basis.

© SolutionLibrary Inc. 9836dcf9d7

Solution Preview

...(using (3) = -i (Lx Ly + Ly Lx)

[Ly^2, Lz] = Ly [Ly,Lz] + [Ly,Lz] Ly = (using (2) = i (Ly Lx + Lx Ly)

So, we see that [Lx^2 + Ly^2,Lz] = 0 and obviously [Lz^2,Lz] = 0, so we have [L^2,Lz] = 0.

Note that from

[Lx^2, Lz] = = -i (Lx Ly + Ly Lx)

you could have invoked rotational invariance. If you apply a rotation in the x-y plane so that x ----> y and y---> -x, then the equation should transform covariantly to:

[Ly^2, Lz] = = -i (Ly L_{-x} +L_{-x}Ly)

Now, L_{-x} is the angular momentum operator for the -x direction and measuring the angular momentum in that diorection is obvipously the same as measuring the angular momentum in the x-direction, except that the results change sign. Therefore this is just -Lx, leading to the same result:

[Ly^2, Lz] = = i (Ly Lx + Lx Ly)

Next we consider the so-called ladder operators:

L+ = Lx + i Ly

L- = Lx - i Ly

Since L^2 commutes with Lx, Ly and Lz, L^2 will also commute with L+ and L-. The commutation relations with Lz are easily derived:

[Lz, L+] = [Lz, Lx] + i [Lz, Ly] = Lx + i Ly = L+ (4)

[Lz, L-] = [Lz, Lx] - i [Lz, Ly] =- Lx + i Ly =- L- (5)

Suppose that we have some simultaneous eigenstate L^2 and Lz. We may label the state using its eigenvalue for these two operators, it's convenient to denote the state as |l,m> when the eigenvalue for L^2 is l(l+1) while the eigenvalue for Lz is m.

Now consider the state L+|l,m>

Using (4) we see that:

LzL+|l,m> = (L+Lz + L+) |l,m> = L+(Lz + 1)|l,m> = (m+1)L+|l,m>

And similarly, using (5)

LzL-|l,m> = (L-Lz - L-) |l,m> = L-(Lz - 1)|l,m> = (m-1)L-|l,m>

So, L+ produces a new state that raises the eigenvalue of Lz by one while L- lowers it by one, provided, of course that the state is normalizable. The eignevalue of L^2 doesn't change as the ladder operators commute with L^2. Let's consider the norm of the state L+|l,m>

||L+|l,m>||^2 = equal to the ...