Clarification on J and F (total angular momentum quantum numbers)

[Higgy]

Homework Statement

I am to consider the Zeeman Effect. I need to calculate the energy level shifts for a given magnetic field corresponding to different quantum numbers. I'm having a hard time knowing when a quantum number Q should be interpreted as just Q or as (Q, Q-1, ..., 0, ..., -Q).

Homework Equations

For low magnetic field values, the energy shift is given by:
\Delta E = \mu_{B} g_{F} B_{z} m_{F},
where \mu_{B} is the Bohr magneton, g_{F} is the Lande g-factor, B_{z} is the magnetic field, and m_{F} is the projection of the total angular momentum quantum number F.

The Lande g-factor g_{F} is<br /> g_{F} = g_{J} \frac{ F(F+1) + J(J+1) - I(I+1) }{ 2F(F+1) }
and g_{J}, is given by
g_{J} ≈ \frac{3}{2} + \frac{ 3/4 - L(L+1) }{ 2J(J+1) }
The quantum numbers above are:
L, "orbital angular momentum quantum number", depends on state
J, "total angular momentum quantum number", J = L + S
(S, "spin angular momentum quantum number", S = 1/2)
I, "nuclear spin quantum number", depends on isotope
F, "(super?) total angular momentum quantum number", F = J + I

3. The Attempt at a Solution , pt. 1
For concreteness, take potassium ( ^{39} K ). In this case, I = 3/2. The ground state is 4S_{1/2}, that is, n = 4, L = 0, and J = 1/2.

(a) Determining F:

Here is the first place that I run into trouble. The equation for F simply states that F = J + I, from which follows that F = 1/2 + 3/2 = 2. But I am looking at a diagram that shows two F states: F = 2 and F = 1. (That is, the hyperfine splitting.)

So why are there two states and not one? My best guess is that I made a mistake earlier in interpreting the state label 4S_{1/2}. The ``1/2&quot; is not J, but actually |J|, and J actually can take two values, J=+1/2,-1/2.

Of course, what is confusing about that is that J was given as L + S. But now I am supposing that I should interpret it as (what Wikipedia calls) the "main total angular momentum quantum number", whose values are determined by |L-S|≤J≤L+S.

Summary: Is it correct that the J in spectroscopic notation is really telling you |J|, and that in general when you use J to calculate F, the range of J values should be used?

(b) Determining g_{J}:

Here I run into trouble again. Should I interpret L and J to be 0 and 1/2, respectively? (Here L is easy, since it's 0, but if it were nonzero, such as if it were 1, I'm assuming that it would always be interpreted as 1 for this equation, and not its possible projections, right?)

My hunch is that the two possible values of J here should be considered, and hence there will be two different values of g_{J} in this particular case, one for J=+1/2 and another for J=-1/2.

(c) Determining g_{F}:

Again, should I expect to get several values of g_{F} corresponding to the different possible values of F (i.e. 2 and 1) and J (i.e. 1/2 and -1/2)?

4. The attempt at a solution, pt. 2
Here's my guess at how to calculate g_{F} m_{F} for the 4S_{1/2} state of ^{39} K. (Note that ultimately I want to calculate \Delta E = \mu_{B} g_{F} B_{z} m_{F}, but I don't care about B_{z} since it's the independent variable, and the Bohr magneton's just a number I can look up).

The two possible g_{J} values, running on the assumption that I need to calculate it for the whole range of J values:
g_{J=1/2} = \frac{3}{2} + \frac{\frac{3}{4} - 0(0+1)}{2(\frac{1}{2})(\frac{1}{2}+1)} = 2
g_{J=-1/2} = \frac{3}{2} + \frac{\frac{3}{4} - 0(0+1)}{2(-\frac{1}{2})(-\frac{1}{2}+1)} = 0
For F=2 (and J=1/2):
g_{F=2} = 2 \frac{2(2+1) + \frac{1}{2}(\frac{1}{2}+1) - \frac{3}{2}(\frac{3}{2}+1)}{2(2)(2+1)} = \frac{1}{2}
For F=2 (and J=-1/2):
g_{F=2} = 0 \frac{2(2+1) + -\frac{1}{2}(-\frac{1}{2}+1) - \frac{3}{2}(\frac{3}{2}+1)}{2(2)(2+1)} = 0
For F=1 (and J=1/2):
g_{F=1} = 2 \frac{1(1+1) + \frac{1}{2}(\frac{1}{2}+1) - \frac{3}{2}(\frac{3}{2}+1)}{2(1)(1+1)} = -\frac{1}{2}
For F=1 (and J=1/2):
g_{F=1} = 0 \frac{1(1+1) + -\frac{1}{2}(-\frac{1}{2}+1) - \frac{3}{2}(\frac{3}{2}+1)}{2(1)(1+1)} = 0
Finally, I multiply these (I think?) by all of their corresponding possible values of m_{F} (the projection of F, which ranges from F to -F in integer steps.)

So for F=2 (and J=1/2):
g_{F=2} m_{F=2} = (\frac{1}{2}) (2) = 1
g_{F=2} m_{F=2} = (\frac{1}{2}) (1) = \frac{1}{2}
g_{F=2} m_{F=2} = (\frac{1}{2}) (0) = 0
g_{F=2} m_{F=2} = (\frac{1}{2}) (-1) = -\frac{1}{2}
g_{F=2} m_{F=2} = (\frac{1}{2}) (-2) = -1

For F=2 (and J=-1/2):
g_{F=2} m_{F=2} = (0) (2) = 0
g_{F=2} m_{F=2} = (0) (1) = 0
g_{F=2} m_{F=2} = (0) (0) = 0
g_{F=2} m_{F=2} = (0) (-1) = 0
g_{F=2} m_{F=2} = (0) (-2) = 0

So for F=1 (and J=1/2):
g_{F=1} m_{F=1} = (-\frac{1}{2}) (1) = -\frac{1}{2}
g_{F=1} m_{F=1} = (-\frac{1}{2}) (0) = 0
g_{F=1} m_{F=1} = (-\frac{1}{2}) (-1) = \frac{1}{2}

For F=1 (and J=-1/2):
g_{F=1} m_{F=1} = (0) (1) = 0
g_{F=1} m_{F=1} = (0) (0) = 0
g_{F=1} m_{F=1} = (0) (-1) = 0

So, in this case, if you applied a magnetic field to the atom and measured its ground state energy levels, you would find five unique energy levels. If you had a way of counting the degeneracy of each level, you would find that the middle level had 10 degenerate states, while some of the others have 2 degenerate states (one for F=1 and one for F=2).

Does that look correct? The alternative would be to have only one value of g_{J} (not two), and only two values (or possibly one value) of g_{F} (not four). That would be if I took J in the above equations to mean |J_{max}| (and possibly F to be |F_{max}|).

 

[Higgy]

(Note: I would have edited the OP, but it doesn't appear that I can do that. Please excuse the bump.)

I found some significant mistakes in my original post. Eric Weisstein's World of Physics to the rescue! First, J cannot take negative values. That literally follows from something I stated (and I don't know how it got past me): |L-S|≤J≤L+S. The source of confusion was actually in how to calculate the possible values of F.

To quote,

A fine structure with fixed angular quantum number L and total angular momentum quantum number J is split into hyperfine structure components with F taking the possible values J+I,J+I-1,...,|J-I|.

So F is not simply J + I, it can takes on a range of values determined as stated in the quote.

For the case I gave, the 4S_{1/2} state of potassium, J is really just 1/2 (and emphatically never -1/2). Calculating the hyperfine states is straightforward:
J+I = \frac{1}{2} + \frac{3}{2} = 2
J+I-1 = \frac{1}{2} + \frac{3}{2} - 1 = 1
J + I - 2 = \frac{1}{2} + \frac{3}{2} - 2 = 0...
but |J-I| = 1 so we stop at 1, and have just two hyperfine states.

I guess I'll still assume that the two Lande g-factors need to be recalculated for the value of J and F corresponding to the state in question. I'll edit/reply when I have a chance to redo that calculation - it would, of course, be great to then have someone confirm my interpretation of the quantum numbers, as well as my math.

 

[Higgy]

Right, I think this is resolved. Here is my solution. I would appreciate it if someone looked it over and confirmed that it is correct.

Example: The ground state of ^{39} K is the 4s state.

(i.) To calculate J, observe that an s state has L = 0, and that the spin of the electron is S = 1/2:
|L-S| ≤ J ≤ L+S \\→ |0-\frac{1}{2}| ≤ J ≤ 0 + \frac{1}{2} \\→ \frac{1}{2} ≤ J ≤ \frac{1}{2} \\→ J = \frac{1}{2}So J only takes one value in this case, although it can take more values, just not negative values.


(ii.) To calculate the values of F, observe that the nucleus of the ^{39} K atom has spin of 3/2, so I = 3/2:
|J-I| ≤ F ≤ J+I \\→ |\frac{1}{2}-\frac{3}{2}| ≤ F ≤ \frac{1}{2} + \frac{3}{2} \\→ 1 ≤ F ≤ 2 \\→ F = 1, 2 So the ground state has two hyperfine states labelled by F=1 and F=2.


(iii.) We calculate the ``J&quot; Lande g-factor for our only value of J:
g_{J} ≈ \frac{3}{2} + \frac{ \frac{3}{4} - L(L+1) }{ 2J(J+1) } \\→ g_{J} ≈ \frac{3}{2} + \frac{ \frac{3}{4} - 0(0+1) }{ 2(\frac{1}{2})(\frac{1}{2}+1) } \\→ g_{J} = 2

(iv.) We have two F states, so we calculate g_{F} separately for each.
g_{F} = g_{J} \frac{ F(F+1) + J(J+1) - I(I+1) }{ 2F(F+1) } \\→ g_{F=2} = (2) \frac{ 2(2+1) + \frac{1}{2}(\frac{1}{2}+1) - \frac{3}{2}(\frac{3}{2}+1) }{ 2(2)(2+1) } = \frac{1}{2} \\→ g_{F=1} = (2) \frac{ 1(1+1) + \frac{1}{2}(\frac{1}{2}+1) - \frac{3}{2}(\frac{3}{2}+1) }{ 2(1)(1+1) } = -\frac{1}{2}

(v.) Finally, we find the slopes of the Zeeman states for each hyperfine state by multiplying by m_{F}.

For F=2:\Delta E_{F=2,m_{F}=2} = \mu _{B} (\frac{1}{2})(2) B _{z} = \mu _{B} B _{z}\Delta E_{F=2,m_{F}=1} = \mu _{B} (\frac{1}{2})(1) B _{z} = \frac{1}{2} \mu _{B} B _{z}\Delta E_{F=2,m_{F}=0} = \mu _{B} (\frac{1}{2})(0) B _{z} = 0\Delta E_{F=2,m_{F}=-1} = \mu _{B} (\frac{1}{2})(-1) B _{z} = -\frac{1}{2} \mu _{B} B _{z}\Delta E_{F=2,m_{F}=-2} = \mu _{B} (\frac{1}{2})(-2) B _{z} = -\mu _{B} B _{z}For F=1:\Delta E_{F=1,m_{F}=1} = \mu _{B} (-\frac{1}{2})(1) B _{z} = -\frac{1}{2} \mu _{B} B _{z}\Delta E_{F=1,m_{F}=0} = \mu _{B} (-\frac{1}{2})(0) B _{z} = 0\Delta E_{F=1,m_{F}=-1} = \mu _{B} (-\frac{1}{2})(-1) B _{z} = \frac{1}{2} \mu _{B} B _{z}

I've attached a plot of these states.