Computing a Spherical Harmonic

Calculus Level 1

Which of the following is the formula for the spherical harmonic Y 3 − 2 ( θ , ϕ ) ? Y^{-2}_3 (\theta, \phi)?

− 105 32 π cos ⁡ 2 θ sin ⁡ θ e − 2 i ϕ -\sqrt{\frac{105}{32 \pi}} \cos^2 \theta \sin \theta e^{-2i \phi} 105 32 π cos ⁡ 2 θ sin ⁡ θ e 2 i ϕ \sqrt{\frac{105}{32 \pi}} \cos^2 \theta \sin \theta e^{2i \phi} 105 32 π sin ⁡ 2 θ cos ⁡ θ e − 2 i ϕ \sqrt{\frac{105}{32 \pi}} \sin^2 \theta \cos \theta e^{-2i \phi} 105 32 π sin ⁡ 2 θ cos ⁡ θ e 2 i ϕ \sqrt{\frac{105}{32 \pi}} \sin^2 \theta \cos \theta e^{2i \phi}

Matt DeCross by Matt DeCross , 27, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

This is a part of orthonormalized spherical harmonics that employ the Condon-Shortley phase up to degree l = 10 l = 10 .

Y 2 − 3 ( θ , ϕ ) = 1 4 105 2 π e − 2 i ϕ sin ⁡ 2 ( θ ) cos ⁡ ( θ ) = 1 4 105 2 π ⋅ ( x ⋅ i y ) 2 z r 3 \begin{aligned} Y^{-3}_2 (\theta, \phi) &= \frac{1}{4} \sqrt{\frac{105}{2\pi}}e^{-2i\phi}\sin^2(\theta)\cos(\theta) \\&= \frac{1}{4}\sqrt{\frac{105}{2\pi}} \cdot\frac{(x \cdot iy)^{2}z}{r^3} \end{aligned} .

This makes you 105 32 π sin ⁡ 2 ( θ ) cos ⁡ ( θ ) e − 2 i ϕ □ \sqrt{\frac{105}{32\pi}}\sin^2(\theta)\cos(\theta)e^{-2i\phi} \space \space \square

ADIOS!!! \LARGE \text{ADIOS!!!}

7 Helpful 1 Interesting 0 Brilliant 10 Confused

https://mathworld.wolfram.com/SphericalHarmonic.html

Luiz Gustavo Gustavo - 9 months, 4 weeks ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...