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Magnetic quantum number

  1. Jun 13, 2013 #1
    Hi,

    I am going through the hydrogen atom at the moment, and I saw that the magnetic quantum number must run between ±L. I cannot find the reason in a simple enough form for me to understand, so I was hoping someone here could quickly (or slowly!) explain why ml can't be greater than L.

    Thanks
     
  2. jcsd
  3. Jun 13, 2013 #2

    Nugatory

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    (Warning - some hand-waving follows)

    Suppose that I told you that an airplane was flying at 1000 km/hr, but didn't tell you in which direction. Even without knowing the direction, you would know that it couldn't be moving north at any more than 1000 km/hr (if the direction of travel was due north) nor at any less than -1000 km/hr (if the direction of travel was due south). And of course it could be anywhere between those two extremes.

    The L quantum number is the magnitude of the angular momentum. The magnetic quantum number is the component of the angular momentum around a particular axis, so can't be more than the total.
     
  4. Jun 13, 2013 #3
    Ah, that's very simple. I thought it was more complicated than that. However I am still a little confused, only because I was told that the magnitude of the angular momentum is given by sqrt(l(l+1)) (in units of hbar), which should be a little bit more than l.

    Is it just because ml has to be an integer, and so it has to be less than sqrt(l(l+1)) and so also l?
     
  5. Jun 13, 2013 #4

    dextercioby

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    Well, there's no integer between l and sqrt(l(l+1)), so if m is an integer, it can't be bigger than l, so that its maximum value is l.
     
  6. Jun 13, 2013 #5
    I suppose it has to do with Heisenberg Uncertainty Principle. For angular momentum, it says
    [tex]\sigma_{L_x} \sigma_{L_y} \ge \frac{\hbar}{2} \left< | L_z | \right> [/tex]
    So, you can never quite align the angular momentum all into the z direction because there's some uncertainty in the x and y angular momenta.
     
    Last edited: Jun 14, 2013
  7. Jun 13, 2013 #6

    Nugatory

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    Hey, I did warn you that there was some hand-waving in my answer, didn't I?

    Dextercioby and Khashishi have given you good answers (although Khashishi could have appealed to the non-commutativity of the angular momentum components instead of the HUP).

    If you want to visualize what's going on... Imagine that the z-axis is pointing straight up. The angular momentum L cannot be pointing straight up as well (##L_z = L##) because then we'd be in an eigenstate of all three ##L_i## (with eigenvalues 0, 0, and L), and that's impossible because they don't commute.

    Instead, the angular momentum vector L lies in the surface of a cone with its apex at the origin and its axis around the z-axis. Each value of ##L_z## corresponds to a cone of different height ##L_z## and width; in any cone the value of L and ##L_z## is fixed and ##L_z < L##.
     
    Last edited: Jun 13, 2013
  8. Jun 13, 2013 #7
    What do you know about differential equations? If you're good with the math, then the answer is fairly clear from the derivation of the quantum numbers from the Schrodinger equation.

    Eventually, in solving for the wave function, we get a Legendre polynomial, which is in terms of l and m. The series only converges if m is within ±l.
     
  9. Jun 14, 2013 #8
    Well that all seems fairly straightforward. Is there an answer reminiscent of

    for why l also has to be less than the principal quantum number? If it just pops out of the math that's fair enough, but that explanation was very enlightening (somehow I missed the physical connection between ml and l!)
     
  10. Jun 14, 2013 #9

    Nugatory

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    Yes, although it comes with the same disclaimer about hand-waving. :smile:

    The principal quantum number is roughly analogous to the radius of the classical electron orbit. The angular momentum of an orbiting particle is greatest in a circular orbit, and smaller values of the angular momentum correspond to more elliptical orbits. Thus, there can be no angular momentum values greater than that corresponding to a circular orbit with the energy given by the principal quantum number.

    This is a completely heuristic hand-wavy argument... Useful only because it puts some intuition behind the results that just pop out of the math.
     
  11. Jun 14, 2013 #10
    That's very helpful, thanks.

    I think the last conceptual hurdle for me to get over is just that these numbers are only integers linking to real variables - thinking of them in terms of "angular momentum has to be less than energy" (and similar nonsense) is what has gotten me confused.
     
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