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Is This Formula Actually Right?

  1. Jun 27, 2006 #1
    I've attached the formula as a word document because I think if I try to type it I'll mess it up...
    Ok I tried to attach it but apparently it's too big so here goes:

    Spherical Harmonics:

    The complex conjugate of Y(l,m) = -1 to the power of m MULTIPLIED BY Y(l,m).

    Is this right? I thought that this formula would give Y(l,-m)?

    I can't find a decent explanation of where all this stuff comes from, I can find an explanation of where Y(l,m) comes from with the Legendre Polynomials etc. However I can't find an explanation for the complex conjugates and for Y(l,-m).

    Thank you,

  2. jcsd
  3. Jun 28, 2006 #2
    main properties of Spherical Harmonics

    The peculiar properties of Spherical Harmonics are sintetized as follows:

    [tex]Y_l ^{-m}=(-1)^m Y_l ^{m\ast}[/tex](orthonormality)

    [tex]\int_\Omega Y_{l_1} ^{m_1 \ast}Y_{l_2} ^{m_2} d\Omega=\delta_{l_1 l_2} \delta_{m_1 m_2}[/tex]

    [tex]\sum_{m=-l} ^lY_l ^{m\ast}Y_l ^m=1[/tex].

    You could also verify these statements from the currently used expression of Spherical Harmonics as:

    [tex]Y_l ^{m}(\theta,\varphi)=\epsilon\frac{\sqrt{2l+1}}{\sqrt{4\pi}}\frac{\sqrt{(l-\mid m\mid)!}}{\sqrt{(l+\mid m\mid)!}}P_l^{m}(cos\theta) \exp{(im\varphi)}[/tex]

    [tex]\epsilon=\left\{\begin{array}{cc}1,&\mbox{ if }
    m\leq 0\\(-1)^m, & \mbox{ if } m>0\end{array}\right[/tex]
    is a numerical coefficient;

    [tex]P_l^{m}(cos\theta)[/tex] represents the associated Legendre polynomials, given by:

    [tex]P_l^{m}(cos\theta)=\sin^{\mid m\mid}\theta \frac{d^{\mid m\mid}}{(d\cos)^{\mid m\mid}}P_l(cos\theta)= \frac{1}{2^l l!}\sin^{\mid m\mid}\theta \frac{d^{l+\mid m\mid}}{(d\cos)^{l+\mid m\mid}}(\cos^2\theta -1)^l[/tex]

    From these equations you will be able to get you needings. A detailed description would require additional time.

    Last edited: Jun 28, 2006
  4. Jun 28, 2006 #3


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    What is your reference for your equation (is it taken from a book?). There are several conventions for the overall phase of the Y_lm so this answer may be author dependent. But using the convention that I know (Cilestrino has posted all the main formula), you would indeed have

    Y*(l,m) = (-1)^m Y(l,-m)

    (the only complex dependence is through e^(i m phi). The only other dependence on m that is relevant is the epsilon factor. Everything else is real and depends only on the absolute value of m.

    So you need only to focus on [itex] \epsilon(m) e^{i m \phi} [/itex] where epsilon is defined in Cilestrino's post. Epsilon is real and [itex] \epsilon(-m) =(-1)^m \epsilon(m) [/itex]. So, finally,

    [tex] (\epsilon(m) e^{i m \phi})^* =\epsilon(m) e^{-i m \phi} = (-1)^m \epsilon(-m) e^{-i m \phi} = (-1)^m Y(l,-m) [/tex]
    Last edited: Jun 28, 2006
  5. Jun 28, 2006 #4

    Meir Achuz

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    This is a good summary, but needs several caveates:
    1. The 1st eq. is not "orthonormality". That describes the 2nd eq.
    2. The first 3 eqs. are standard, but the others vary from book to book.
    I don't think those given here for Plm with negative m are used be most physicists. They are OK if used consistently, though.,
    You must be careful in using Ylm to use the same convention throughout.
    Almost any Math physics text will give enough Ylm detail.
    Beware the web, though.
  6. Jun 28, 2006 #5
    Thanks to Meir Achuz for the correction. I confused the lines after the tex tags!
  7. Jun 28, 2006 #6
    Thanks everyone for the help. Much appreciated.
    Just one thing - I'm being a bit stupid - how does e(-m) = -1(m) * e(m) ?
  8. Jun 28, 2006 #7
    Nrged: The formula I got from a page of printed notes from one of my lecturers, it could easily have been wrong I think as they seem to make mistakes often.
    When you wrote earlier that e(-m) = -1(m) * e(m), were you saying e to the power of minus m = etc etc?
    Are they all to the power of?
    Thanks again,
  9. Jun 28, 2006 #8


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    You mean when I wrote [itex] \epsilon(-m) = (-1)^m \epsilon(m) [/itex]?

    when I write [itex] \epsilon(m) [/itex], what I mean is that epsilon can be seen as a function of m. So the (m) is just there to remind us that epsilon depends on m. So the above equation means that epsilon of minus m is equal to -1 raised to the mth power times epsilon of m.

    The reason for this is in the definition of epsilon (see Cilestrino's post). For negative m, epsilon is +1. For positive m, epsilon is. (-1)^m. So it is ether +1 (is m is even) or -1 (if n is odd).

    So if n is even, epsilon(-m) = epsilon(m) (no change since they are both +1)

    If n is odd, then epsilon(-m) and epsilon(m) will differ by a minus sign.

    All this is summarized by [itex] \epsilon(-m) = (-1)^m \epsilon(m) [/itex].

    Hope this makes sense.
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