MHB Approximating Solutions with Coefficients of $Y^m_\ell$

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The discussion focuses on approximating the solution for a function \( u(r,\theta,\phi) \) using coefficients from spherical harmonics. The approximation is derived by combining terms from various spherical harmonics, specifically for \( \ell = 0, 1, \) and \( 2 \). The final expression for the approximation is given as \( u(r,\theta,\phi) \simeq 25 + \frac{75\sqrt{2}}{2}r\sin\theta\cos\phi + \frac{125}{\pi}r^2\sin^2\theta\cos 2\phi \). Participants express confusion about integrating the spherical harmonics with the function, particularly regarding the absence of an exponential factor in the provided function. The final approximation successfully combines the contributions from the spherical harmonics to yield a coherent expression.
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Using the coefficients determined, combine the terms to arrive at the following approximation for the solution
$$
u(r,\theta,\phi)\simeq 25 + \frac{75\sqrt{2}}{2}r\sin\theta\cos\phi + \frac{125}{\pi}r^2\sin^2\theta\cos 2\phi.
$$

How do I combine them?

The coefficients are
For $\ell = 0$ and $m = 0$, we have
$$
Y^0_0(\theta,\varphi) = \frac{1}{2\sqrt{\pi}}.
$$
For $\ell = 1$ and $m = 1$, we have
\begin{alignat*}{3}
Y^{1}_1 & = & -\frac{1}{2}\sqrt{\frac{3}{2\pi}}(\cos^2\theta - 1)^{1/2}e^{i\varphi}\\
& = & -\frac{1}{2}\sqrt{\frac{3}{2\pi}}e^{i\varphi} \sin\theta
\end{alignat*}
From $\ell = 1$ and $m = 1$, we can obtain
$$
Y^{-1}_1 = \frac{1}{2}\sqrt{\frac{3}{2\pi}}e^{-i\varphi}\sin \theta.
$$
For $\ell = 1$ and $m = 0$, we have
$$
Y^0_1 = \frac{\sqrt{3}}{2\sqrt{\pi}}\cos\theta.
$$
For $\ell = 2$ and $m = 2$, we have
\begin{alignat*}{3}
Y^{2}_2 & = & \frac{3}{4}\sqrt{\frac{5}{6\pi}}(\cos^2\theta - 1)e^{2i\varphi}\\
& = & \frac{3}{4}\sqrt{\frac{5}{6\pi}}e^{2i\varphi}\sin^2\theta
\end{alignat*}
From $\ell = 2$ and $m = 2$, we can obtain
\begin{alignat*}{3}
Y^{-2}_2 & = & \frac{3}{4}\sqrt{\frac{5}{6\pi}}e^{-2i\varphi}\sin^2 \theta.
\end{alignat*}
For $\ell = 2$ and $m = 1$, we have
\begin{alignat*}{3}
Y^{1}_2 & = & -\frac{3}{2}\sqrt{\frac{5}{6\pi}}e^{i\varphi}(\cos^2\theta - 1)^{1/2}\cos\theta\\
& = & -\frac{3}{2}\sqrt{\frac{5}{6\pi}}e^{i\varphi}\sin \theta \cos\theta
\end{alignat*}
From $\ell = 2$ and $m = 1$, we can obtain
\begin{alignat*}{3}
Y^{-1}_2 & = & \frac{3}{2}\sqrt{\frac{5}{6\pi}}e^{-i\varphi}\sin \theta \cos\theta.
\end{alignat*}
For $\ell = 2$ and $m = 0$, we have
$$
Y^0_{\ell} = \frac{1}{4}\sqrt{\frac{5}{2\pi}}(3\cos^2\theta - 1).
$$
 
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dwsmith said:
Using the coefficients determined, combine the terms to arrive at the following approximation for the solution
$$
u(r,\theta,\phi)\simeq 25 + \frac{75\sqrt{2}}{2}r\sin\theta\cos\phi + \frac{125}{\pi}r^2\sin^2\theta\cos 2\phi.
$$

How do I combine them?

The coefficients are
For $\ell = 0$ and $m = 0$, we have
$$
Y^0_0(\theta,\varphi) = \frac{1}{2\sqrt{\pi}}.
$$
For $\ell = 1$ and $m = 1$, we have
\begin{alignat*}{3}
Y^{1}_1 & = & -\frac{1}{2}\sqrt{\frac{3}{2\pi}}(\cos^2\theta - 1)^{1/2}e^{i\varphi}\\
& = & -\frac{1}{2}\sqrt{\frac{3}{2\pi}}e^{i\varphi} \sin\theta
\end{alignat*}
From $\ell = 1$ and $m = 1$, we can obtain
$$
Y^{-1}_1 = \frac{1}{2}\sqrt{\frac{3}{2\pi}}e^{-i\varphi}\sin \theta.
$$
For $\ell = 1$ and $m = 0$, we have
$$
Y^0_1 = \frac{\sqrt{3}}{2\sqrt{\pi}}\cos\theta.
$$
For $\ell = 2$ and $m = 2$, we have
\begin{alignat*}{3}
Y^{2}_2 & = & \frac{3}{4}\sqrt{\frac{5}{6\pi}}(\cos^2\theta - 1)e^{2i\varphi}\\
& = & \frac{3}{4}\sqrt{\frac{5}{6\pi}}e^{2i\varphi}\sin^2\theta
\end{alignat*}
From $\ell = 2$ and $m = 2$, we can obtain
\begin{alignat*}{3}
Y^{-2}_2 & = & \frac{3}{4}\sqrt{\frac{5}{6\pi}}e^{-2i\varphi}\sin^2 \theta.
\end{alignat*}
For $\ell = 2$ and $m = 1$, we have
\begin{alignat*}{3}
Y^{1}_2 & = & -\frac{3}{2}\sqrt{\frac{5}{6\pi}}e^{i\varphi}(\cos^2\theta - 1)^{1/2}\cos\theta\\
& = & -\frac{3}{2}\sqrt{\frac{5}{6\pi}}e^{i\varphi}\sin \theta \cos\theta
\end{alignat*}
From $\ell = 2$ and $m = 1$, we can obtain
\begin{alignat*}{3}
Y^{-1}_2 & = & \frac{3}{2}\sqrt{\frac{5}{6\pi}}e^{-i\varphi}\sin \theta \cos\theta.
\end{alignat*}
For $\ell = 2$ and $m = 0$, we have
$$
Y^0_{\ell} = \frac{1}{4}\sqrt{\frac{5}{2\pi}}(3\cos^2\theta - 1).
$$
I'm a bit confused about this one. The problem is very similar to using the dot product to separate terms. The usual
c_{nlm} = \int \psi^*_{nlm} u(r, \theta, \phi)~ d \tau

For example, the constant term is going deal with \psi _{000} since there are no radial or angular terms. But even \psi _{000} is going to contain a e^{-Zr/a_0} factor (assuming a Hydrogen-like wavefunction for the radial part) and there is no exponential factor in your function.

-Dan
 
topsquark said:
I'm a bit confused about this one. The problem is very similar to using the dot product to separate terms. The usual
c_{nlm} = \int \psi^*_{nlm} u(r, \theta, \phi)~ d \tau

For example, the constant term is going deal with \psi _{000} since there are no radial or angular terms. But even \psi _{000} is going to contain a e^{-Zr/a_0} factor (assuming a Hydrogen-like wavefunction for the radial part) and there is no exponential factor in your function.

-Dan

I am confuse too. I am not sure what I am supposed to do but that is the question verbatim.
 
dwsmith said:
I am confuse too. I am not sure what I am supposed to do but that is the question verbatim.
I figure it out.
 
dwsmith said:
I figure it out.
I'm curious about your solution. Could you post the general idea? (I had a thought this morning about expanding the "missing" exponential. I'm wondering if that was the right approach.)

-Dan
 
topsquark said:
I'm curious about your solution. Could you post the general idea? (I had a thought this morning about expanding the "missing" exponential. I'm wondering if that was the right approach.)

-Dan

I also had to use the spherical harmonics of the solution which wasn't apparent to me by the questions wording.

We ca expand our series out term by term
\begin{alignat*}{3}
u(r,\theta,\varphi) & = & \sum_{\ell = 0}^2\sum_{m = -\ell}^{\ell} A_{\ell,m}Y^m_{\ell}(\theta,\varphi).
\end{alignat*}
For $\ell = 0$, we have that $u(r,\theta,\varphi) = 25$.
For $\ell = 1$, we have that $u(r,\theta,\varphi) = 25 + r\frac{75}{\sqrt{2}}\sin\theta\frac{e^{i\varphi} + e^{-i\varphi}}{2}$.
Lastly, for $\ell = 2$, we have
$$
u(r,\theta,\varphi) = 25 + \frac{75r}{\sqrt{2}}\sin\theta\cos\varphi + \frac{125r^2}{\pi}\sin^2\theta\cos 2\varphi = 25 + \frac{75\sqrt{2}}{2}r\sin\theta\cos\phi + \frac{125}{\pi}r^2\sin^2\theta\cos 2\phi.
$$
That is, we can approximate $u(r,\theta,\varphi)$ when $\ell = 0,1,2$ to be
$$
u(r,\theta,\phi)\simeq 25 + \frac{75\sqrt{2}}{2}r\sin\theta\cos\phi + \frac{125}{\pi}r^2\sin^2\theta\cos 2\phi.
$$
 
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