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

1. Oct 19, 2013

### Higgy

1. The problem statement, all variables and given/known data
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 $(Q, Q-1, ..., 0, ..., -Q)$.

2. Relevant 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$$g_{F} = g_{J} \frac{ F(F+1) + J(J+1) - I(I+1) }{ 2F(F+1) }$$
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 $1/2"$ 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 $|F_{max}|$).

Last edited: Oct 19, 2013
2. Oct 20, 2013

### 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,
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.

3. Oct 20, 2013

### 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 $J"$ 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.

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