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Angular momentum quantum numbers

  1. Apr 21, 2007 #1
    For angular momentum quantum numbers j, l, and s must it be true that [tex] m_s, m_l < m_j [/tex]?
    It would seem that it is true because I assume that [tex] m_s +m_l = m_j [/tex], but I have not actually seen that written down anywhere and am curious.
    Thanks.
     
  2. jcsd
  3. Apr 22, 2007 #2
    No. It is not generally true. Remember that m tells you the component of angular momentum along some certain direction. This means that it can be either positive or negative. Consider, then, the case where [tex]m_s = -\frac{1}{2}[/tex]. It is hopefully clear that [tex]m_l = m_j + \frac{1}{2}[/tex].
     
  4. Apr 22, 2007 #3
    ok, I think I meant the magnitude of the m's. i.e.
    [tex] |m_s|, |m_l| \leq |m_j| [/tex].
    For example, take the situation where [tex]m_j = -1/2[/tex], j=3/2, s=1/2. The orbital and spin angular momenta magnetic numbers can add to this for two cases:
    1. [tex] m_s = -1/2, m_l = 0 [/tex].
    2. [tex] m_s = 1/2, m_l = -1 [/tex].
    I would argue from above that case 2 is not a viable option because [tex]|m_l| \nleq |m_j| [/tex].
    What do you think?
     
    Last edited: Apr 22, 2007
  5. Apr 22, 2007 #4
    If you're using magnitudes, you can't have negative numbers. However, even if you want to compare the magnitudes of [tex]m_s[/tex], [tex]m_l[/tex], and [tex]m_j[/tex], you'll find that there are states where [tex]|m_j|[/tex] is smaller than either [tex]|m_s|[/tex] or [tex]|m_l|[/tex]. All this requires is that the spin and orbital angular momenta have their z components in opposite directions. This is something we should expect from normal vector analysis (i.e. it has nothing to do with quantum mechanics specifically). If I add two vectors which have projections in opposite directions along the z-axis, I should expect that the magnitude of the z component of the resultant vector must be smaller than at least that of one of the two vectors I added; and, it may be smaller than both.
     
  6. Apr 22, 2007 #5
    darn, I wrote that wrong. Hopefully this makes my question more clear:

    if j=3/2 then possible [tex] m_j [/tex] values are -3/2 to 3/2 by ones. if j=3/2 and we're dealing with an electron then l = 1 and s = 1/2. [tex] m_l [/tex] values are 1,0, and -1.
    If [tex] m_j = - 3/2 (+3/2) [/tex] then of course [tex] m_l = -1 (+1) [/tex] and [tex] m_s = -1/2 (+1/2) [/tex] respectively. Those are the only possibilities to form [tex] m_j [/tex] for the maximum values of [tex] m_j [/tex].
    Now when [tex] m_j = \pm 1/2[/tex] it seems ambiguous as to what [tex] m_{l,s} [/tex] are. For instance:
    1. [tex] m_s = -1/2, m_l = 0 [/tex]
    2. [tex] m_s = 1/2, m_l = -1 [/tex]
    both give [tex] m_j = -1/2[/tex].

    Is that ok or is one of them not correct?
     
  7. Apr 22, 2007 #6

    jtbell

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    Staff: Mentor

    The state with [itex]m_j = -1/2[/itex] is a linear combination of the two states with [itex](m_s = -1/2, m_l = 0)[/itex] and [itex](m_s = +1/2, m_l = -1)[/itex]. The coefficients of the linear combination are called Clebsch-Gordan coefficients:

    http://farside.ph.utexas.edu/teaching/qm/lectures/node47.html

    To relate the notation on that page to your example, substitute [itex]l[/itex] for its [itex]j_1[/itex], [itex]s[/itex] for its [itex]j_2[/itex], [itex]m_l[/itex] for its [itex]m_1[/itex], [itex]m_s[/itex] for its [itex]m_2[/itex], and finally [itex]m_j[/itex] for its [itex]m[/itex].
     
  8. Apr 23, 2007 #7
    Adding to what jtbell said, you should also be aware that there are [tex]j=3/2[/tex] states arising from [tex]l=2,\ s=1/2[/tex]. Here, there are 10 possible [tex]|m_l, m_s\!\!>[/tex] states; but, only 8 of them give allowed values of [tex]m_j[/tex].
     
    Last edited: Apr 23, 2007
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