MHB Spherical Harmonics easy question

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The discussion centers on the normalization of spherical harmonics, specifically for the case where both $\ell$ and $m$ equal 1. The user calculates the spherical harmonic $Y_1^1$ and arrives at a result of $\frac{1}{2}\sqrt{\frac{3}{2\pi}}e^{i\varphi}\sin \theta$, while Mathematica provides a solution of $-\frac{1}{2} e^{i\varphi} \sqrt{\frac{3}{2\pi}} \sin\theta$. The discrepancy arises from differences in normalization conventions used in various texts. One participant suggests that their book incorporates a factor of $(-1)^m$ in the definition, which could explain the negative sign in Mathematica's output. Understanding these normalization differences is crucial for consistent results in spherical harmonics calculations.
Dustinsfl
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$$
Y_{\ell}^m = \sqrt{\frac{(2\ell + 1)(\ell - m)!}{4\pi(\ell + m)!}}P^m_{\ell}(\cos\theta)e^{im\varphi}
$$

For $\ell = m = 1$, we have
$$
\sqrt{\frac{(2 + 1)(0)!}{4\pi(2)!}}P^1_{1}(\cos\theta)e^{i\varphi} = \frac{1}{2}\sqrt{\frac{3}{2\pi}}e^{i\varphi}\sin \theta
$$

But Mathematica is telling me the solution is
$$
-\frac{1}{2} e^{i\varphi} \sqrt{\frac{3}{2\pi}} \sin\theta
$$

What is going wrong?
 
Last edited:
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dwsmith said:
$$
Y_{\ell}^m = \sqrt{\frac{(2\ell + 1)(\ell - m)!}{4\pi(\ell + m)!}}P^m_{\ell}(\cos\theta)e^{im\varphi}
$$

For $\ell = m = 1$, we have
$$
\sqrt{\frac{(2 + 1)(0)!}{4\pi(2)!}}P^1_{1}(\cos\theta)e^{i\varphi} = \frac{1}{2}\sqrt{\frac{3}{2\pi}}e^{i\varphi}\sin \theta
$$

But Mathematica is telling me the solution is
$$
-\frac{1}{2} e^{i\varphi} \sqrt{\frac{3}{2\pi}} \sin\theta
$$

What is going wrong?
I'm not sure about how your book normalizes spherical harmonics, but mine has
Y_l^m (\theta, \phi) = (-1)^m \sqrt{\frac{(2l+1)(l-m)!}{4 \pi (l+ m)!}} P_l^m(cos(\theta)) e^{im \phi}

-Dan
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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