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I Spherical harmonics

  1. Jul 25, 2017 #1
    How does one arrive at the equation

    $$\bigg( (1-z^2) \frac{d^2}{dz^2} - 2z \frac{d}{dz} + l(l+1) - \frac{m^2}{1-z^2} \bigg) P(z) = 0$$
    Solving this equation for ##P(z)## is one step in deriving the spherical harmonics "##Y^{m}{}_{l}(\theta, \phi)##".

    The problem is that the book I'm following doesn't show how to arrive at the above equation. It shows how to arrive at it only for the special case ##m=0##.

    I've tried googling "Associate Legendre's equation" and "Legendre's general equation derivation" but it seems there's no such derivation on web.
     
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  3. Jul 25, 2017 #2

    blue_leaf77

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    Are you trying to solve the hydrogen atom eigenfunction?
    As a matter of fact, that differential equation is the defining equation for the associated Legendre polynomial. Therefore, unless you encounter it in a different context, there is no way to derive it. It is simply defined such that the associated Legendre polynomial is the solution of that equation.
     
  4. Jul 25, 2017 #3

    vanhees71

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    The standard way to derive the equation is to apply the separation ansatz
    $$\psi(r,\vartheta,\varphi)=R(r) \Theta(\vartheta) \Phi(\varphi)$$
    for the eigenfunctions of the Laplacian in spherical coordinates.
    $$\Delta \psi=\frac{1}{r} \partial_r^2 (r \psi) + \frac{1}{r^2 \sin \vartheta} \partial_{\vartheta}(\sin \vartheta \partial_{\vartheta} \psi) + \frac{1}{r^2 \sin^2 \vartheta} \partial_{\varphi}^2 \psi.$$
     
  5. Jul 25, 2017 #4
    I'm trying to derive the spherical harmonics functions, often denoted by ##Y^{m}{}_{l}(\theta, \phi)##. It's part of the hydrogen atom problem.
    Maybe I should have put my question in the following way

    The book I'm reading shows how to arrive at the equation $$\bigg( (1-z^2) \frac{d^2}{dz^2} - 2z \frac{d}{dz} + A - \frac{m^2}{1-z^2} \bigg) P(z) = 0$$
    where one needs to find what ##A## is. The book shows how to find ##A## for the case ##m=0##; it's ##A = l(l+1)##. Then it states that ##A = l(l+1)## even for ##m \neq 0##, but does not proves that, as it proves for the former case.

    @vanhees71 By doing this way, can one show that ##A = l(l+1)## is the correct term?
     
  6. Jul 25, 2017 #5

    PeterDonis

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    Which book?
     
  7. Jul 25, 2017 #6

    vanhees71

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    If I think about it, an even simpler way is to look for harmonic functions, i.e., the Laplace equation
    $$\Delta \psi=0.$$
    You make the above separation ansatz. Plugging it into the Laplace Equation leads to
    $$\frac{\Theta \Phi}{r} (r R)''+\frac{R \Phi}{r^2 \sin \vartheta}(\sin \vartheta \Theta)'+\frac{R \Theta}{r^2 \sin^2 \vartheta} \Phi''=0.$$
    Multiply the entire expression with ##r^2/(R \Theta \Phi)##, and you get
    $$\frac{r}{R} (r R)''=-\frac{1}{\Theta \sin \vartheta}(\sin \vartheta \Theta)'-\frac{1}{\Phi \sin^2 \vartheta} \Phi''.$$
    Sine the left-hand side depends on ##r## only but the right-hand side on ##\vartheta## and ##\varphi## the entire expression must be some constant ##A##:
    $$r (r R)''=A R.$$
    The equation can obviously solved by the ansatz ##R=c r^{\lambda}##:
    $$r (r R)''=c r (r^{\lambda+1})''=c r (\lambda+1) \lambda r^{\lambda-1}=c A r^{\lambda} \; \Rightarrow\; \lambda(\lambda+1)=A.$$
    Then you repeat the argument with the right-hand side
    $$-\frac{1}{\Theta \sin \vartheta}(\sin \vartheta \Theta)'-\frac{1}{\Phi \sin^2 \vartheta} \Phi''=\lambda(\lambda+1)$$
    or, a bit rewritten,
    $$-\frac{\sin \vartheta }{\Theta}(\sin \vartheta \Theta')' - \lambda(\lambda+1) \sin^2 \vartheta = \frac{1}{\Phi} \Phi''=-m^2,$$
    where again ##m=\text{const}## since the left-hand side depends on ##\vartheta## only and the right-hand side on ##\varphi## only. You then find
    $$\Phi(\varphi)=\exp(\mathrm{i} m \varphi),$$
    and since the solution should be a ##2 \pi##-periodic function in ##\varphi## you get ##m \in \mathbb{N}##.

    Finally you are left with [Typos corrected]
    $$\sin \vartheta (\sin \vartheta \Theta')'+ \left [\lambda(\lambda+1) \sin^2 \vartheta - m^2 \right ]\Theta=0.$$
    Now substitute
    $$z=\cos \vartheta, \quad \Theta(\vartheta)=P(z).$$
    From this you get immediately your formula.
     
    Last edited: Jul 26, 2017
  8. Jul 25, 2017 #7
    McIntyre - Quantum Mechanics - A Paradigms Approach

    @vanhees71 Perfect. Thank you.
     
  9. Jul 25, 2017 #8
    Is there a typo here?
    $$ \Longrightarrow -\frac{\sin \vartheta }{\Theta}(\sin \vartheta \Theta)' - \lambda(\lambda+1) \sin^2 \vartheta = \frac{\Phi''}{\Phi}=-m^2$$

    And, also, here?

    $$ \Longrightarrow \sin \vartheta (\sin \vartheta \Theta)'+ \left [\lambda(\lambda+1) \sin^2 \vartheta - m^2 \right ]\Theta=0.$$
     
  10. Jul 25, 2017 #9

    vanhees71

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    I've corrected the typos in #6. I hope, now all signs are correct ;-).
     
  11. Jul 25, 2017 #10
    You are still keeping the derivative symbol on ##\Theta##. Is this so?

    Also, you have actually forgotten the first one that I have pointed out.
     
  12. Jul 26, 2017 #11

    vanhees71

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    Argh. Yes, I corrected it, but the ##\Theta'## must stay. You have a 2nd-order differential operator (it's more or less ##\hat{\vec{L}}^2##, and it's much more understandable what's coming out using the semi-algebraic way in my QM manuscript).
     
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