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Why does the l quantum number depend on n?

  1. Jul 23, 2011 #1
    Hi,

    I am a second year physicist and I have completed 2 courses on quantum mechanics. The first was a decent overview of wave mechanics and the second was only 12 lectures long and applied this knowledge to the hydrogen and helium atoms and introduced aspects of atomic and molecular physics.

    In the first course we were shown where the m and l quantum numbers came from when we began doing Quantum Mechanics in more than one dimension, however, in the second course we were told that the l quantum number is restricted to values between zero and n-1.

    I have attached our lecture notes of the first course. In the first attachment (lecture 21) we introduce m on page 4. In the second attachment (lecture 24) we introduce l and show how m is restricted by it (see page 3) as a result of trying to find the functional form of the eigenvalues of angular momentum by using ladder operators however nowhere in the course were we shown the restriction on l by n.

    Could someone please show me how this restriction arises?

    Thanks a lot for any help.
     

    Attached Files:

  2. jcsd
  3. Jul 23, 2011 #2
    Here's an intuitive answer: n corresponds to energy of orbits. l is the quantum number that corresponds to the magnitude of angular momentum. Think planet orbit around the sun: both its angular motion and radial motion contributes to its energy. So l < n is simply the requirement that the angular portion of the energy cannot exceeds the total kinetic energy.
     
    Last edited: Jul 23, 2011
  4. Jul 23, 2011 #3
    Hey, thanks for the reply.

    True, I understand that it should be less but why specifically restricted to n-1? If you see the derivations in the notes you'll see what I mean when I say that we were shown "where the restriction on m comes from".
     
  5. Jul 24, 2011 #4

    Vanadium 50

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    You're thinking about it backwards. l doesn't depend on n, but n depends on l. You have a radial quantum number k, and an orbital quantum number l. They can be anything positive. We define a new number n = k + l + 1.

    And that's it.
     
  6. Jul 24, 2011 #5
    Right, now I'm totally confused. I have never come across the quantum number "k".

    The way I have been taught is as follows:

    The first quantum number we saw in the course was n which came from solving the TISE for energy eigenvalues and we found that energy was quantised by an integer, n.

    The next quantum number,m, appeared when we did quantum mechanics in 2D and it came from the only component of angular momentum that can exist in 2 dimensions and was quantised as a result of the angular boundary condition on phi which is that rotating through 2pi radians puts you back at phi=0 this condition meant that m was restricted to be an integer (positive or negative) .

    Finally l was introduced as a result of finding angular momentum eigenvalues in 3 dimensions through the use of creation and annihilation operators. In finding the eigenvalues it was discovered that there is a relationship between l and m namely that m can take any integer value between -l and l.

    So perhaps my question was worded wrong. I don't mean to say that n depends on l or that l depends on n, the point of my question is: where does the relationship between the two numbers come from regardless of which quantity is the dependant one?
     
  7. Jul 24, 2011 #6
    I think you need to go through the derivation of solving Schrodinger eqn under coloumb potential, "n = k + l + 1" pops up natrually from the boundary condition.
     
  8. Jul 24, 2011 #7

    dextercioby

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    The full symmetry group for the H-atom is SO(4,2), a subgroup of whom is SO(4), which is the symmetry group for the bound states. An analysis performed by Pauli in 1926 led him to the solution of the discrete spectrum, alongside with the degeneracy. A modern short discussion on the <dynamical degeneracy> is to be found in Galindo & Pascual, page 250. An expanded discussion is provided by Biederharn in the first volume of his compendium on group theory.
     
    Last edited: Jul 24, 2011
  9. Jul 24, 2011 #8
    For any given [itex]l[/itex], the radial part of Schrodinger's equation for hydrogen atom is (after some manipulation):
    [tex]\left[ -\frac{1}{r}\frac{\partial^2}{\partial r^2} r + \frac{l(l+1)}{r^2}
    - \frac{1}{r a_0} - \frac{E}{E_0 a_0^2} \right] u(r) = 0 \,,[/tex]
    where [itex]a_0[/itex] is the Bohr radius and [itex]E_0[/itex] is the Rydberg constant.

    Consider two asymptotic behavior of [itex]u(r)[/itex]: when [itex]r\to0[/itex] and when [itex]r \to \infty[/itex]. In the form case, the wave function has to be regular at the origin, so try [itex]u(r) = r^a[/itex] for some non-negative a, and you get [itex]u(r) \to r^{l}[/itex]. In the latter case, since we are interested in bound state solution, we want something that decays exponentially. Just try a wave function like [itex]u(r) \simeq r^{b} e^{-q r}[/itex] for some unknown [itex]q,\;b[/itex], and you get [itex]u(r) \to r^{n-1} e^{-\frac{r}{n a_0}}[/itex],

    Near [itex]r=0[/itex] lower power dominates. Near [itex]r=\infty[/itex] higher power dominates. So it must be the case that [itex]n-1 \ge l[/itex].
     
    Last edited: Jul 24, 2011
  10. Jul 24, 2011 #9

    Vanadium 50

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    Whoa...let's not jump into SO(4) and the Runge-Lenz vector. The OP is having trouble with quantum numbers, so let's not jump into grad-level stuff.

    Like mathfeel says, an electron in a hydrogen atom can gain energy by being farther out (the k quantum number) or by carrying more angular momentum (the l quantum number). To get the total energy, it's easiest to define a new quantum number n, so that n = k + l + 1.
     
  11. Jul 24, 2011 #10

    dextercioby

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    Yes, but I thought about the roots of the degeneracy, not necessarily constraints from solving an ODE. Anyway, it's a forum, so the OP is free to pick his answer, if more than one is provided.
     
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