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Q on solution of Spherical Harmonic

  1. Dec 26, 2005 #1
    I read the solution for spherical harmonic using associated Legendre polynomials, and am wondering....

    For example, a solution is written in here, but I wonder why the constant in forumula (3) can be determined as [tex]-m^2[/tex], as a negative value of square of an integer.

    Similar thing applies to the [tex]r[/tex] variable (in the above page it doesn't appear, though),

    [tex]{r(\frac {\partial^2} {{\partial r}^2})(rR(r))} / {R(r)} = l(l+1)[/tex]

    Here, I can understand this should be a constant, but cannot understand why it's a form of [tex]l(l+1)[/tex] where [tex]l[/tex] is an integer....

    Will anyone tell me??

    Thanks in advance!
     
    Last edited: Dec 26, 2005
  2. jcsd
  3. Dec 26, 2005 #2
    We know "the constant" in eqn (3) is constant, so we have the freedom to call it *whatever* we want, as long as it is indeed constant. Calling it -m^2 is just a trick because it simplifies the algebra later on (when you get the equation (4)).

    Another way to explain it is that the differential equation:

    (d^2 X)/(dY^2) = (-c^2)*X

    Is a pretty standard one (ie, it crops up a lot in various bits of physics and maths), and we know the solutions to it (given by eqn (4) on your link). So when you come across the equation:

    (d^2 X)/(dY^2) = S*X , S=const

    You recall the other differential equation and see it's better to write S as -m^2 because then we immediatly have the solutions because it's just a standard differential equation.

    As for the other question, I suspect it's for the same reason, but I'm not sure.
     
  4. Dec 26, 2005 #3
    Good question. Well, actually, you can take any constant that you want. The fact that -m² is taken, is a direct consequence of , err, some mathematical "playing"...Here it goes :

    [tex]L_z \Phi_m( \phi) = m \hbar \Phi_m( \phi)[/tex]
    [tex]-i \frac {\partial}{\partial \phi} \Phi_m( \phi) = m \Phi_m( \phi)[/tex]
    [tex]-i \frac {\partial ^2}{{\partial \phi}^2} \Phi_m( \phi) = m \frac {\partial}{\partial \phi}\Phi_m( \phi) = \frac {m^2}{-i} \Phi_m ( \phi)[/tex]

    Thus

    [tex]\frac{1}{\Phi_m (\phi)} \frac{\partial ^2}{{\partial \phi}^2} \Phi_m ( \phi) = \frac{m^2}{i^2} = -m^2[/tex]

    QED

    marlon

    edit : if i can make a suggestion : you should not be studying this from a website, nomatter how reliable it is. What books are you using for your QM course ?
     
    Last edited: Dec 26, 2005
  5. Dec 26, 2005 #4
    Correct. There are several ways to explain this :

    1) structure of the Legendre Polynomials

    2) symmetry of the physics at hand (ie the orbitals) : grouptheory

    3) combining following two aspects:
    a) the eigen equations for [tex]L^2[/tex] and [tex]L_z[/tex]
    b) [tex] <L^2> \geq <L_z^2>[/tex] because [tex]L_x[/tex] and [tex]L_y[/tex] are Hermitian

    regards
    marlon

    edit : again, do not use an internet site as your primary source of study
     
    Last edited: Dec 26, 2005
  6. Dec 26, 2005 #5
    Thanks all! I think I've understood: My understanding is, all the solutions of Laplace equation in spherial coordiates are represented with these m and l using associated polynomials, so these "discrete constants" are proved to be enough to solve this equation.

    As you all say, if these discrete values are just for convenience (just to simplify), it's enough to me.

    I'm now mainly reading Greiner's "Quantum Mechanics: an introduction" because it has a lot of calculations and exercises.
     
    Last edited: Dec 26, 2005
  7. Dec 26, 2005 #6
    You mean Schrödinger equation, right ?
    Well, be careful with this vocabularium. It is not just about "convenience". These values are used for a specific reason : because they are building blocks of the correct description of nature. One can, and one DOES, prove them using one of the three systems that i outlined in my previous post. Which of the two is your book using ? Probably the second one if it is introductory.

    regards
    marlon
     
    Last edited: Dec 26, 2005
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