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Understand some quantum numbers in a problem

  1. Nov 9, 2011 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Consider the following states of the hydrogen atom corresponding to [itex]l=2[/itex] whose quantum numbers corresponding to [itex]L_z[/itex] and [itex]S_z[/itex] are given by [itex]m_l=2[/itex], [itex]m_s=-1/2[/itex] and [itex]m_l=1[/itex], [itex]m_s=1/2[/itex]. What are the possible values for the quantum number j for the states [itex]m_j=3/2[/itex]?


    2. Relevant equations
    This is where the problem lies. In my class notes, I noted that [itex]m_j=m_l+m_s[/itex] but for the second cases this makes no sense.
    I also have noted [itex]j=l+s, l+s-1,...,|l-s|[/itex].

    3. The attempt at a solution
    I tried to find some information on j in hyperphysics and wikipedia but I'm still stuck. I don't really understand what is the j. Is it just an index used in the "[itex]m_j[/itex]"? And [itex]m_j[/itex] is the quantum number for the total angular momentum of the atom? I don't really understand what it means.
     
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  3. Nov 10, 2011 #2

    Redbelly98

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    j is the index for the total angular momentum. mj gives the z-component of j, i.e. it's the quantum number corresponding to Jz.

    If you're still stuck, I would make a list of:
    • Possible values of j, given that l=2
    • For each j, what are the possible values of mj?
     
  4. Nov 10, 2011 #3

    fluidistic

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    Hey redbelly, I'm a bit confused.
    Thanks for your description. Isn't [itex]J_z[/itex] equal to [itex]S_z+L_z[/itex]?
    When [itex]l=2[/itex], [itex]m_l[/itex] runs from -2 to 2 and [itex]m_s[/itex] runs from [itex]-1/2[/itex] to [itex]1/2[/itex].
    I think that [itex]L_z=\frac{\hbar}{2}[/itex] no matter what [itex]l[/itex] is worth. And [itex]S_z[/itex] could be worth [itex]-\hbar /2[/itex] or [itex]\hbar /2[/itex].
    Hmm I'm sure I'm wrong, I need some sleep I think. I'm getting back to it right after breakfast. Feel free to correct me meanwhile :)
     
  5. Nov 11, 2011 #4

    Redbelly98

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    I am now remembering how confusing it was for me to get a handle on a lot of quantum mechanics concepts when I was first learning them.

    Yes.
    Yes.
    Well, no. Do you really mean Lz here? Lz is the value of the z-component of orbital angular momentum, and equal to [itex]m_l \hbar[/itex].
    Yes.
    .​
    Hey, I just realized that I did not answer an important question of yours from Post #1:
    No, not really.

    Since the orbital and spin angular momenta represent the same type of physical quantity -- namely, angular momentum -- they can be added together to get a total angular momentum. This total angular momentum is denoted by J, which is the vector sum of orbital and spin, L+S. The quantum number associated with J is j (not mj as you said).

    By straightforward vector addition, the magnitude of J must be from |L-S| (minimum) to |L+S| (maximum). It's quantum number j takes on values from |l-s| to |l+s|, in increments of 1.

    Just like we do for the other angular momentum quantities L and S, we can talk about the z-component of J, which we call Jz. The quantum number associated with Jz is mj, which takes on values from -j to +j in increments of 1. Note that mj works the same way that ml and ms do with respect to l and s.
     
  6. Nov 11, 2011 #5

    fluidistic

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    Ok thank you very much for your last post, I've learned much from it.
    What I've done:
    j is either 5/2 or 3/2 in both cases.
    This gives me [itex]m_j[/itex] could be either [itex]-5/2[/itex], [itex]-3/2[/itex], [itex]-1/2[/itex], [itex]1/2[/itex], [itex]3/2[/itex] and [itex]5/2[/itex] for [itex]j=5/2[/itex] and [itex]m_j=-3/2[/itex], [itex]-1/2[/itex], [itex]1/2[/itex], [itex]3/2[/itex] for [itex]j=3/2[/itex].
    Now I don't know how to determine the value of [itex]m_j[/itex] given the values of [itex]m_l[/itex] and [itex]m_s[/itex]
     
  7. Nov 12, 2011 #6

    fluidistic

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    Hmm looking back at the original question, I'd answer that both j=5/2 and j=3/2 can give [itex]m_j=3/2[/itex].
    But I didn't use [itex]m_l[/itex] nor [itex]m_s[/itex]. I don't feel like I'm doing the things right.
     
  8. Nov 13, 2011 #7

    Redbelly98

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    Yes, I agree.
    You're answer is good. ml and ms were used in the actual problem statement, to come up with mj (=ml and ms) = 3/2.

    Just like the z-components of vectors are added to calculate the z-component of a vector sum, ml and ms add up to mj.
     
  9. Nov 13, 2011 #8

    fluidistic

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    Ok Redbelly. You've been so helpful, thank you very much!
     
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