Spherical harmonics

Spherical harmonics $Y_l^{\vert m_l\vert}(\theta,\phi)$ are mathematical functions that arise in problems with spherical symmetry, particularly in the study of atomic orbitals.

The normalised form of spherical harmonics can be expressed as:

$Y_l^{\vert m_l\vert}(\theta,\phi)=N_l^{\vert m_\vert}P_l^{\vert m_l\vert}(cos\theta)e^{im_l\phi}\;\;\;\;\;\;\;\;400$

where:

$\theta$ is the polar angle.
$\phi$ is the azimuthal angle.
$l$ is the orbital angular momentum quantum number (also known as azimuthal quantum number).
$m_l$ is the magnetic quantum number.
$P_l^{\vert m_l\vert}(cos\theta)=\frac{1}{2^ll!}(1-cos^2\theta)^{\frac{\vert m_l\vert}{2}}\frac{d^{l+\vert m_l\vert}}{dcos\theta^{l+\vert m_l\vert}}(cos^2\theta-1)^l$ are the associated Legendre polynomials.
$N_l^{\vert m_l\vert}=\frac{1}{\sqrt{2\pi}}\biggr\[\biggr$\frac{2l+1}{2}\biggr$\frac{(l-\vert m_l\vert)!}{(l+\vert m_l\vert)!}\biggr\]^{\frac{1}{2}}$ is the normalisation constant of the spherical harmonics.

To derive eq400, consider the Schrödinger equation of an electron that is confined to a spherical surface:

$\biggr\[\frac{1}{sin^2\theta}\frac{\partial^2}{\partial\phi^2}+\frac{1}{sin\theta}\frac{\partial}{\partial\theta}\biggr$sin\theta\frac{\partial}{\partial\theta}\biggr$\biggr\]\psi=-\frac{2IE}{\hbar^2}\psi\;\;\;\;\;\;\;\;401$

where $I=mr^2$ .

Assuming $\psi(\theta,\phi)=\Theta(\theta)\Phi(\phi)$ , we can multiply and divide eq401 by $sin^2\theta$ and $\Theta\Phi$ , respectively, to give

$\frac{1}{\Phi}\frac{d^2\Phi}{d\phi^2}+\frac{sin\theta}{\Theta}\frac{d}{d\theta}\biggr$sin\theta\frac{d\Theta}{d\theta}\biggr$+sin^2\theta\frac{2IE}{\hbar^2}=0\;\;\;\;\;\;\;\;402$

The first term on the left of eq402 is independent of $\theta$ . If $\theta$ is varied, only the second and third terms are affected. But the sum of these terms is a constant given by the right-hand side of the equation when $\theta$ is varied. Therefore, we can write

$\frac{sin\theta}{\Theta}\frac{d}{d\theta}\biggr$sin\theta\frac{d\Theta}{d\theta}\biggr$+sin^2\theta\frac{2IE}{\hbar^2}=\vert m_l\vert^2\;\;\;\;\;\;\;\;403$

By a similar argument, the first term is a constant when $\phi$ changes, and

$\frac{1}{\Phi}\frac{d^2\Phi}{d\phi^2}=-m_l^2\;\;\;\;\;\;\;\;404$

The constants $\vert m_l\vert^2$ and $- m_l^2$ are chosen to be consistent with subsequent derivations. The general solution of eq404 is $\Phi=e^{a\phi}$ , where $a$ is a constant. To find the specific solution, we substitute $\Phi=e^{a\phi}$ in eq404 to give $e^{a\phi}(a^2+m_l^2)=0$ . An eigenfunction must be a non-zero function to have non-trivial solutions. It must also be single-valued to satisfy the Born interpretation. Since $e^{a\phi}\neq 0$ , we have $a=im_l$ and $\Phi=e^{im_l\phi}$ , which after normalisation, becomes $\Phi=\frac{1}{\sqrt{2\pi}}e^{im_l\phi}$ . If $\Phi$ is single-valued, then $\Phi(\phi+2\pi)=\Phi(\phi)$ .

Question

How does $\Phi(\phi+2\pi)=\Phi(\phi)$ show that $\Phi$ is single-valued?

Answer

A function $\Phi$ is considered single-valued if, for every point in its domain, there is a unique value of the function. In mathematical terms, for a given argument $\phi$ , the function $\Phi(\phi)$ should yield a single, well-defined output. Since $\phi$ is the azimuthal angle, which ranges from zero to $2\pi$ , the only way to satisfy the single-valued condition is for $\Phi(\phi+2\pi)=\Phi(\phi)$ .

Substituting $\phi=\phi+2\pi$ in $\Phi$ yields ${\Phi(\phi+2\pi)=\frac{1}{\sqrt{2\pi}}e^{im_l\phi}e^{im_l2\pi}=(-1)^{2m_l}\Phi(\phi)}$ . As $\Phi(\phi+2\pi)=\Phi(\phi)$ , we have $(-1)^{2m_l}=1$ and therefore, $m_l=0,\pm 1,\pm 2,\cdots$ . In other words, the normalised solution to eq404 is

$\Phi=\frac{1}{\sqrt{2\pi}}e^{im_l\phi}\;\;\;\;\;where\;m_l=0,\pm 1,\pm2,\cdots$

To solve eq403, we begin by carrying out the differentiation to give

$sin\theta cos\theta\frac{d\Theta}{d\theta}+sin^2\theta\frac{d^2\Theta}{d\theta^2}+\Theta sin^2\theta\frac{2IE}{\hbar^2}=\Theta\vert m_l\vert^2\;\;\;\;\;\;\;\;405$

Like the derivation for the quantum harmonic oscillator, we use a change of variable method, with $x=cos\theta$ , to solve eq405.

Question

Show that $\frac{d^2\Theta}{d\theta^2}=(1-x^2)\frac{d^2P_l^{\vert m_l\vert}}{dx^2}-x\frac{dP_l^{\vert m_l\vert}}{dx}$ .

Answer

Since $x=cos\theta$ , we have $\Theta(\theta)=P_l^{\vert m_l\vert}(x)$ . Using $\frac{dx}{d\theta}=-sin\theta$ and the chain rule,

$\frac{d\Theta}{d\theta}=\frac{dP_l^{\vert m_l\vert}}{dx}\frac{dx}{d\theta}=-\frac{dP_l^{\vert m_l\vert}}{dx}sin\theta\;\;\;\;\;\;\;\;406$

$\frac{d^2\Theta}{d\theta^2}=-\biggr$sin\theta\frac{d}{d\theta}\frac{dP_l^{\vert m_l\vert}}{dx}+\frac{dP_l^{\vert m_l\vert}}{dx}\frac{d}{d\theta}sin\theta\biggr$$

Applying the chain rule again on the first term of RHS of the above equation, $\frac{d^2\Theta}{d\theta^2}=-\biggr\[sin\theta(-sin\theta)\frac{d^2P_l^{\vert m_l\vert}}{dx^2}+x\frac{dP_l^{\vert m_l\vert}}{dx}\biggr\]$ or

$\frac{d^2\Theta}{d\theta^2}=(1-x^2)\frac{d^2P_l^{\vert m_l\vert}}{dx^2}-x\frac{dP_l^{\vert m_l\vert}}{dx}\;\;\;\;\;\;\;\;407$

Substituting $x=cos\theta$ , eq406 and eq407 in eq405 yields

$(1-x^2)\frac{d^2P_l^{\vert m_l\vert}}{dx^2}-2x\biggr$\frac{dP_l^{\vert m_l\vert}}{dx}\biggr$+\biggr\[\frac{2IE}{\hbar^2}-\frac{\vert m_l\vert^2}{1-x^2}\biggr\]P_l^{\vert m_l\vert}=0$

Comparing the term $\frac{2IE}{\hbar^2}$ with eq50 and eq131a,

$(1-x^2)\frac{d^2P_l^{\vert m_l\vert}}{dx^2}-2x\biggr$\frac{dP_l^{\vert m_l\vert}}{dx}\biggr$+\biggr\[l(l+1)-\frac{\vert m_l\vert^2}{1-x^2}\biggr\]P_l^{\vert m_l\vert}=0\;\;\;\;\;\;\;\;408$

Eq408 is known as the associated Legendre differential equation, whose solutions are the associated Legendre polynomials $P_l^{\vert m_l\vert}$ . The product of $P_l^{\vert m_l\vert}$ and $e^{im_l\phi}$ gives the spherical harmonics. If $m_l=0$ , eq408 becomes the Legendre differential equation:

$(1-x^2)\frac{d^2P_l}{dx^2}-2x\biggr(\frac{dP_l}{dx}\biggr)+l(l+1)P_l=0$

where $P_l^{\vert 0\vert}=P_l$ .

The derivations of the explicit forms of the Legendre polynomials and the associated Legendre polynomials, along with their respective normalisation constants, are described in previous articles. In this article, we offer an alternative approach to demonstrating how Legendre polynomials and the associated Legendre polynomials are related to each other.

Differentiating the Legendre differential equation $\vert m_l\vert$ times using Leibniz’s theorem gives

$\sum_{k=0}^{\vert m_l\vert}\begin{\pmatrix}\vert m_l\vert\\k\end{pmatrix}\frac{d^k(1-x^2)}{dx^k}\frac{d^{\vert m_l\vert-k+2}P_l}{dx^{\vert m_l\vert-k+2}}-2\sum_{k=0}^{\vert m_l\vert}\begin{\pmatrix}\vert m_l\vert\\k\end{pmatrix}\frac{d^kx}{dx^k}\frac{d^{\vert m_l\vert-k+1}P_l}{dx^{\vert m_l\vert-k+1}}+l(l+1)-\frac{d^{\vert m_l\vert}P_l}{dx^{\vert m_l\vert}}=0\;\;\;\;\;\;\;\;409$

Assuming that $\vert m_l\vert\geq 2$ , only the first three terms and the first two terms of the first sum and the second sum, respectively, survive. So, eq409 becomes

$(1-x^2)z''-2x(\vert m_l\vert+1)z'+(l-\vert m_l\vert)(l+\vert m_l\vert+1)z=0\;\;\;\;\;\;\;\;410$

where $z=\frac{d^{\vert m_l\vert}}{dx^{\vert m_l\vert}}P_l$ .

If we let $z=\frac{d^{\vert m_l\vert}}{dx^{\vert m_l\vert}}P_l=(1-x^2)^{-\frac{\vert m_l\vert}{2}}P_l^{\vert m_l\vert}$ , and substitute $z$ , $z'$ and $z''$ in eq410, we get eq408, where $P_l^{\vert m_l\vert}=(1-x^2)^{\frac{\vert m_l\vert}{2}}\frac{d^{\vert m_l\vert}}{dx^{\vert m_l\vert}}P_l$ . In other words, we can derive $P_l^{\vert m_l\vert}$ , and hence $\Theta(\theta)$ , if we know $P_l$ .

Question

If we have assumed that $\vert m_l\vert\geq 2$ when we differentiate the Legendre differential equation $\vert m_l\vert$ times to obtain the associated Legendre differential equation, does it mean that $\vert m_l\vert$ in the associated Legendre differential equation are restricted to values greater than or equal to two?

Answer

No, the procedure merely demonstrates that the solutions to the associated Legendre differential equation are related to those of the Legendre differential equation by $P_l^{\vert m_l\vert}=(1-x^2)^{\frac{\vert m_l\vert}{2}}\frac{d^{\vert m_l\vert}}{dx^{\vert m_l\vert}}P_l$ . The allowed values of $\vert m_l\vert$ must be consistent with the solutions to the associated Legendre differential equation. To avoid the trivial solution of $P_l^{\vert m_l\vert}=0$ , we require that $\vert m_l\vert\leq l$ . Thus, $\vert m_l\vert$ can take values of $0,1,2,\cdots,l$ , and while we may choose to differentiate with $\vert m_l\vert\geq 2$ , the full set of associated Legendre functions exists for $\vert m_l\vert\leq l$ .

Spherical harmonics

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