📂偏微分方程式

球面調和関数：球面座標ラプラス方程式の極角、方位角に対する一般解

説明

ラプラス方程式を満たす関数を調和関数という。球面調和関数は球座標系でのラプラス方程式を満たす関数を言い、正確には径成分を除いた極角$\theta$と方位角$\phi$に対する一般解を意味する。

Proof

The Laplace equation in the spherical coordinate system is as follows.

$$\begin{equation} \nabla ^2 f = \frac{1}{r^2}\frac{\partial}{\partial r} \left( r^2\frac{\partial f}{\partial r} \right) + \frac{1}{r^2\sin\theta}\frac{\partial}{\partial\theta}\left( \sin\theta \frac{\partial f}{\partial \theta} \right) + \frac{1}{r^2\sin^2\theta}\frac{\partial^2 f}{\partial^2 \phi}=0 \label{eq1} \end{equation}$$

Assuming that $f$ is separable in variables,

$$f=R(r)\Theta (\theta)\Phi (\phi)$$

Substituting $f$ into the Laplace equation $(1)$ and multiplying both sides by $\dfrac{r^{2} }{R\Theta \Phi}$ yields the following.

$${\color{blue}\frac{1}{R}\frac{d}{d r} \left( r^2\frac{d R}{d r} \right)} + {\color{green}\frac{1}{\Theta\sin\theta}\frac{d}{d\theta}\left( \sin\theta \frac{d \Theta}{d \theta} \right) + \frac{1}{\Phi\sin^2\theta}\frac{d^2 \Phi}{d \phi^2}} =0$$

Now, the first term affects only $r$, and the rest affects only $\theta$, $\phi$. Therefore, each part is constant because if $r$ changes, affecting the first term, the rest remains unchanged, invalidating the equation. That’s why each term highlighted in colors is constant.

Let’s consider the term regarding $\theta$, $\phi$ as a constant $-l(l+1)$. Initially, it is assumed that $l$ is any complex number, but eventually, it must be an integer.

$$\frac{1}{\Theta\sin\theta}\frac{d}{d\theta}\left( \sin\theta \frac{d \Theta}{d \theta} \right) + \frac{1}{\Phi\sin^2\theta}\frac{d^2 \Phi}{d \phi^{2}}=-l(l+1)$$

Multiplying both sides by $\sin ^{2} \theta$ and rearranging gives the following.

$$\begin{equation} \frac{\sin \theta}{\Theta}\frac{d}{d\theta}\left( \sin\theta \frac{d \Theta}{d \theta} \right)+l(l+1)\sin^{2}\theta =- \frac{1}{\Phi}\frac{d^2 \Phi}{d \phi^{2}} \label{eq2} \end{equation}$$

Since the variables $\theta$, $\phi$ are separated on both sides, as before, each side is constant. Let’s set the right side as $m^{2}$. Then, we obtain the following.

$$\frac{d^2 \Phi}{d \phi^{2}}=-m^{2}\Phi$$

This is a simple second-order differential equation. In brief, a function that differentiates into itself is an exponential function. However, since the constant must become negative when squared, it must include $i$ in the exponent.

$$\begin{equation} \Phi (\phi)=e^{im\phi} \label{eq3} \end{equation}$$

The reason why $e^{-im\phi}$ is not considered is that solutions exist for both positive and negative values of the same magnitude for $m$. For example, when there’s a solution for $m=-1, 0,1$, both $e^{i\phi}$ and $e^{-i\phi}$ can be obtained, so it’s sufficient to express the solution with $(3)$ only. Since $\phi$ is the azimuthal angle, it must satisfy $\Phi (\phi)=\Phi (\phi+2\pi)$. Therefore, we obtain the following.

$$e^{im\phi}=e^{im(\phi+2\pi)}=e^{im\phi}e^{i2m\pi} \\ \implies e^{i2m\pi}=1$$

Using Euler’s formula, we obtain the following.

$$e^{i2m\pi}=\cos(2m\pi)+i\sin(2m\pi)=1$$

For this equation to hold, $m$ must be an integer.

$$m=0,\pm1,\pm2,\cdots$$

This result is related to the fact in quantum mechanics that the magnetic quantum number exists only as an integer value between $-l$ and $l$. Now, let’s solve the differential equation for $\theta$. From $(2)$, we obtain the following equation.

$$\sin \theta\frac{d}{d\theta}\left( \sin\theta \frac{d \Theta}{d \theta} \right)+[l(l+1)\sin^{2}\theta -m^2]\Theta=0$$

This is an associated Legendre differential equation, and its solution is as follows.

$$\begin{align} \Theta (\theta) &= P_{l}^{m}(\cos \theta) \nonumber \\ &= (1-\cos ^{2}\theta)^{\frac{|m|}{2}} \frac{ d^{|m|} }{ dx^{|m|} } P_{l}(x) \nonumber \\ &=(1-\cos ^{2}\theta)^{\frac{|m|}{2}} \frac{ d^{|m|} }{ dx^{|m|} }\left[ \dfrac{1}{2^l l!} \dfrac{d^l}{dx^l}(x^2-1)^l \right] \label{eq4} \end{align}$$

Here, $P_{l}(x)$ is a Legendre polynomial.

$$P_{l}(x)=\dfrac{1}{2^l l!} \dfrac{d^l}{dx^l}(x^2-1)^l$$

From the condition of the Legendre polynomial, $l$ is a non-negative integer.

$$l=0,1,2,\cdots$$

Where $\eqref{eq4}$, the polynomial of order $(x^{2}-1)^{l}$, is differentiated $l+|m|$ times, leading to $|m|>l$, rendering it meaningless. Therefore, $m$ must satisfy the condition of $-l\le m \le l$.

$$m=-l,-l+1,\cdots,-2,-1,0,1,2,\cdots,l-1,l$$

Finally, the spherical harmonic function $Y_{l}^{m}(\theta,\phi)$ is as follows.

$$Y_{l}^{m}(\theta,\phi)=e^{im\phi}P_{l}^{m}(\cos \theta)$$

Here, $l$ is $l=0,1,2\cdots$ and $m$ is an integer that satisfies $-l\le m \le l$.

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