1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Oct 19, 2013 #1
    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 [itex]Q[/itex] should be interpreted as just [itex]Q[/itex] or as [itex](Q, Q-1, ..., 0, ..., -Q)[/itex].

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

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

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

    (a) Determining [itex]F[/itex]:

    Here is the first place that I run into trouble. The equation for [itex]F[/itex] simply states that [itex]F = J + I[/itex], from which follows that [itex]F = 1/2 + 3/2 = 2[/itex]. But I am looking at a diagram that shows two [itex]F[/itex] states: [itex]F = 2[/itex] and [itex]F = 1[/itex]. (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 [itex] 4S_{1/2} [/itex]. The [itex]``1/2"[/itex] is not [itex]J[/itex], but actually [itex]|J|[/itex], and [itex]J[/itex] actually can take two values, [itex]J=+1/2,-1/2[/itex].

    Of course, what is confusing about that is that [itex]J[/itex] was given as [itex]L + S[/itex]. 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 [itex]|L-S|≤J≤L+S[/itex].

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

    (b) Determining [itex] g_{J} [/itex]:

    Here I run into trouble again. Should I interpret [itex] L [/itex] and [itex] J [/itex] to be [itex]0[/itex] and [itex] 1/2 [/itex], respectively? (Here [itex] L [/itex] 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 [itex]J[/itex] here should be considered, and hence there will be two different values of [itex] g_{J} [/itex] in this particular case, one for [itex]J=+1/2[/itex] and another for [itex]J=-1/2[/itex].

    (c) Determining [itex] g_{F} [/itex]:

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

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

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

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

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

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

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

    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 [itex]F=1[/itex] and one for [itex]F=2[/itex]).

    Does that look correct? The alternative would be to have only one value of [itex]g_{J}[/itex] (not two), and only two values (or possibly one value) of [itex]g_{F}[/itex] (not four). That would be if I took [itex]J[/itex] in the above equations to mean [itex]|J_{max}|[/itex] (and possibly [itex]F[/itex] to be [itex]|F_{max}|[/itex]).
    Last edited: Oct 19, 2013
  2. jcsd
  3. Oct 20, 2013 #2
    (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, [itex]J[/itex] cannot take negative values. That literally follows from something I stated (and I don't know how it got past me): [itex]|L-S|≤J≤L+S[/itex]. The source of confusion was actually in how to calculate the possible values of [itex]F[/itex].

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

    For the case I gave, the [itex]4S_{1/2}[/itex] state of potassium, [itex]J[/itex] is really just [itex]1/2[/itex] (and emphatically never [itex]-1/2[/itex]). Calculating the hyperfine states is straightforward:
    [tex]J+I = \frac{1}{2} + \frac{3}{2} = 2[/tex]
    [tex]J+I-1 = \frac{1}{2} + \frac{3}{2} - 1 = 1[/tex]
    [tex]J + I - 2 = \frac{1}{2} + \frac{3}{2} - 2 = 0...[/tex]
    but [itex]|J-I| = 1[/itex] 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 [itex]J[/itex] and [itex]F[/itex] 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.
  4. Oct 20, 2013 #3
    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 [itex]^{39} K[/itex] is the [itex]4s[/itex] state.

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

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

    (iii.) We calculate the [itex]``J"[/itex] Lande g-factor for our only value of [itex]J[/itex]:
    [tex]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[/tex]

    (iv.) We have two [itex]F[/itex] states, so we calculate [itex]g_{F}[/itex] separately for each.
    [tex]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}[/tex]

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

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

    I've attached a plot of these states.

    Attached Files:

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted