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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.
     

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