Spherical Harmonics easy question

In summary, the conversation is discussing the normalization of spherical harmonics, specifically for the case where $\ell = m = 1$. The first equation provided shows the general formula for $Y_{\ell}^m$, while the second equation shows the specific case for $\ell = m = 1$. However, there is a discrepancy between the solution provided by Mathematica and the expected solution. The speaker is unsure about the normalization used in the book and provides their own version of the formula for $Y_{\ell}^m$.
  • #1
Dustinsfl
2,281
5
$$
Y_{\ell}^m = \sqrt{\frac{(2\ell + 1)(\ell - m)!}{4\pi(\ell + m)!}}P^m_{\ell}(\cos\theta)e^{im\varphi}
$$

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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  • #2
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
[tex]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}[/tex]

-Dan
 

FAQ: Spherical Harmonics easy question

What are spherical harmonics?

Spherical harmonics are a set of mathematical functions that are used to describe the spatial variations of a spherical object. They are commonly used in physics and engineering to represent the shape and orientation of objects in three-dimensional space.

What is the purpose of using spherical harmonics?

The purpose of using spherical harmonics is to simplify the representation of complex three-dimensional objects. They allow us to break down a complex shape into simpler components, making it easier to analyze and manipulate.

How are spherical harmonics calculated?

Spherical harmonics are calculated using a mathematical formula that involves trigonometric functions and Legendre polynomials. The exact formula depends on the specific type of spherical harmonics being used.

What are some practical applications of spherical harmonics?

Spherical harmonics have a wide range of practical applications, including computer graphics, geophysics, and quantum mechanics. They are also commonly used in the analysis and synthesis of audio and image signals.

Are spherical harmonics easy to understand?

This is a subjective question and the answer may vary from person to person. However, with a basic understanding of mathematics and physics, spherical harmonics can be understood and applied in various fields relatively easily.

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