L-S Coupling Problem Solutions - and More QM

We left off our quantum mechanics with a problem to do with L-S coupling, so let's examine the solution for that first. This was for the case: L = 3 , S = 1/2, and J = 5/2.


We see that from the diagram(Fig. 1) , if L = 3 we get two possible values of l. So l1 = 1 and l2 = 2 and l1 + l2 = 1 + 2 = 3.

Meanwhile, S can be defined by only one value of s, or s = ½

The possible j-values are:

j = l + s = l1 + s = 1 + ½ = 3/2

j = l2 +s = 2 + ½ = 5/2

j = l - s = l1 - s =1 - ½ = ½

j = l2 - s = 2 - ½ = 3/2

So in total: j = ½, 3/2 and 5/2

Note that, conforming to j-selection rules, all the j's differ by an integral amount, though they are half-integral (e.g. 3/2, 5/2) themselves.

To obtain any J (total angular momentum) we need an L-S coupling vector that yields J = 5/2. Two possible L-S couplings are available: [L + S] and [L + S – 1] and it is the last that yields the appropriate result: [3 + ½ - 1] = 5/2

Which means we need the values: So l1 = 1 and l2 = 2 and s = ½ to make this work. (See Fig. 1 and how the right side discloses the J result)

Now, another problem:

Let's enumerate all the possible values of j and the quantum number m_j, for which l = 3 and s = ½, but not connected at all to the original problem. Then vector solutions (which the reader can do) following the same tack as the left side of Fig. 1, will show:

j = l + s = 3 + ½ = 7/2

and:

m_j = -7/2, -5/2, -3/2, -½, ½, 3/2, 5/2 and 7/2

meanwhile:

j = l - s = 3 -½ = 5/2

and so:

m_j = -5/2, -3/2, -½, ½, 3/2, 5/2

Note that these last m_j quantum numbers would be tjhe ones applicable to the original problem for which: L = 3 , S = ½, and J = 5/2.



Another problem: Enumerate the possible values of j and m_j for the states in which l = 1 and s = 1/2 and draw the associated vector diagrams.

We have:

j = l + s = 1 + ½ = 3/2

then: m_j = - 3/2, -½, ½, 3/2

and

j = l + s = 1 + (- ½ ) = 1/2

so:

m_j = -½, ½,


The diagram for the vectors is shown in Fig. 2.

Applications to more specific atomic configurations:

We can apply the preceding rules to a particular atomic configuration, say: 2p 3d then go through a procedure of taking differences, assigning values etc. based on selection rules. For the configuration given, one particular energy level occurs for:

s’ = 1 (s’ = ½ + ½) and:

l’ = 1 (l’ = [1-2])

for which j’ = 2, 1 and 0. A ‘triplet’ spectral line appears – which might be depicted (with an appropriate diagram) with energy states 3P2, 3P1, and 3P0. The capital letter (P) denotes one of several energy levels given by S, P, D and F, etc., corresponding to values of the quantum number l = (0, 1, 2, 3 etc.) in an ascending alphabetical order from F.

Hence, l’ = 1 implies P-level, and the subscripts denote J’ corresponding to 0, 1, 2 values. The superscript ‘3’ is the multiplicity, derived from the relation: 2s’

Problem for next time:

Find the possible values of s', l' and j' for an atomic configuration with 2 optically active electrons with quantum numbers: l1 = 2, s1 = ½, l2 = 3, s2 = ½.

Specifiy which j' go with which l' and s' combination.