Computing Spherical Harmonics

Orbital angular momentum

Let me start from the definition of orbital angular momentum.

\[\mathbf{L} = \mathbf{x} \times \mathbf{p}, \quad L_{i} = \epsilon_{ijk}x_{j}p_{k}\]

where $\epsilon_{ijk}$ denotes the Levi-Civita tensor.

Proposition.

\[|\mathbf{L}|^{2} = |\mathbf{x}|^{2}|\mathbf{p}|^{2} - (\mathbf{x} \cdot \mathbf{p})^{2} + i\hbar \mathbf{x}\cdot\mathbf{p}\]

Proof.

\[\begin{align*} |\mathbf{L}|^{2} &= \epsilon_{ijk}x_{j}p_{k}\epsilon_{ilm}x_{l}p_{m} = [\delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}]x_{j}p_{k}x_{l}p_{m} \\ &= \delta_{jl}\delta_{km}x_{j}(x_{l}p_{k} - i\hbar\delta_{kl})p_{m} - \delta_{jm}\delta_{kl}x_{j}p_{k}(p_{m}x_{l} + i\hbar\delta_{lm}) \\ &= |\mathbf{x}|^{2}|\mathbf{p}|^{2} - i\hbar\mathbf{x}\cdot\mathbf{p} - \delta_{jm}\delta_{kl}x_{j}p_{m}(x_{l}p_{k}-i\hbar\delta_{kl}) - i\hbar\mathbf{x}\cdot\mathbf{p} \\ &= |\mathbf{x}|^{2}|\mathbf{p}|^{2} - i\hbar\mathbf{x}\cdot\mathbf{p} - (\mathbf{x} \cdot \mathbf{p})^{2} + 3i\hbar\mathbf{x}\cdot\mathbf{p} - i\hbar\mathbf{x}\mathbf{p} \\ &= |\mathbf{x}|^{2}|\mathbf{p}|^{2} - (\mathbf{x} \cdot \mathbf{p})^{2} + i\hbar \mathbf{x}\cdot\mathbf{p} \quad \square \end{align*}\]

By substituting $\mathbf{p} = -i\hbar\nabla$, we get

\[|\mathbf{L}|^{2} = \hbar^{2}[-r^{2}\nabla^{2} + (\mathbf{x}\cdot\nabla)^{2} + \mathbf{x}\cdot\nabla]\]

Counting operator

Consider a monomial $x^{a}y^{b}z^{c} \; (a, b, c > 0)$. The action of operator $(\mathbf{x}\cdot\nabla)$ is

\[(\mathbf{x}\cdot\nabla)(x^{a}y^{b}z^{c}) = (x, y, z)\cdot(ax^{a-1}y^{b}z^{c}, bx^{a}y^{b-1}z^{c}, cx^{a}y^{b}z^{c-1}) = (a+b+c)x^{a}y^{b}z^{c}\]

Thus, $(\mathbf{x}\cdot\nabla)$ counts the number of $x$, $y$ and $z$’s in the monomial(or the degree of that monomial).

Degree-$l$ harmonic polynomials

The general degree-$l$ polynomials has the form

\[f_{l}(\mathbf{x}) = \sum_{a+b+c=l}c_{abc}x^{a}y^{b}z^{c}, \quad \mathbf{x}\cdot\nabla f_{l}(\mathbf{x}) = lf_{l}(\mathbf{x})\]

Hence,

\[|\mathbf{L}|^{2}f_{l}(x) = \hbar^{2}(-r^{2}\nabla^{2})f_{l}(\mathbf{x}) + \hbar^{2}l(l+1)f_{l}(\mathbf{x})\]

We will only consider harmonic polynomials: $\nabla^{2}f_{l}(\mathbf{x}) = 0$. Then we get familiar equation.

\[|\mathbf{L}|^{2}f_{l}(\mathbf{x}) = \hbar^{2}l(l+1)f_{l}(\mathbf{x})\]

We already know that spherical harmonics satisfy the same partial differential equation.
Let’s consider the number of parameters in $f_{l}$ satisfying (1) degree-$l$ condition and (2) harmonic condition.

• (1) We only use degree-$l$ polynomials: The number of possibilities of $(a, b, c)$ combination which satisfies $a+b+c=l$ is

  \[{}_{3}\mathrm{H}_{l} = {}_{2+l}\mathrm{C}_{2} = \frac{(l+2)(l+1)}{2}\]

• (2) Applying laplacian to $f_{l}$ yields constraints which contain degree $l-2$ polynomials.

  \[\nabla^{2}f_{l}(\mathbf{x}) = \sum_{a+b+c=l}c_{abc}[a(a-1)x^{a-2}y^{b}z^{c} + b(b-1)x^{a}y^{b-2}z^{c} + c(c-1)x^{a}y^{b}z^{c-2}] = 0\]

  The number of equations is

  \[{}_{3}\mathrm{H}_{l-2} = {}_{l}\mathrm{C}_{2} = l(l-1)/2\]

• Hence, the number of independent parameters is

  \[\frac{(l+2)(l+1)}{2} - \frac{l(l-1)}{2} = 2l+1\]

Therefore, we have constructed $2l+1$ independent polynomials satisfying the spherical harmonics equation. We will ignore the $r^{-l}$ factor and obtain $\displaystyle{f_{l}(\theta,\phi) \sim \frac{f_{l}(\mathbf{x})}{r^{l}}}$.

Complex transformations

Define $\omega \equiv x + iy$ and $\omega^{\ast} \equiv x - iy$. Then

\[\partial_{\omega} = \frac{1}{2}[\partial_{x} - i\partial_{y}], \quad \partial_{\omega^{\ast}} = \frac{1}{2}[\partial_{x} + i\partial_{y}], \quad \nabla^{2} = 4\partial_{\omega}\partial_{\omega^{\ast}} + \partial_{z}^{2}\]

Now let’s re-write $z$-component of orbital angular momentum operator with $(\omega, \omega^{\ast})$.

\[L_{z} = -i\hbar[x\partial_{y}-y\partial_{x}] = \hbar[\omega\partial_{\omega} - \omega^{\ast}\partial_{\omega^{\ast}}]\]

Following the counting operator analogy, $L_{z}$ operator can be understood as the operator that counts (number of $\omega$’s) - (number of $\omega^{\ast}$’s).
In conclusion, the $(2l+1)$ polynomials $f_{l}(\mathbf{x}) = f_{l}(\omega, \omega^{\ast}, z)$ can be further classified by the eigenvalues of $L_{z}$!

\[L_{z}f_{l}^{m}(\mathbf{x}) = \hbar[(\#\;\text{of}\;\omega\text{'s}) - (\#\;\text{of}\;\omega^{\ast}\text{'s})]f_{l}^{m}(\mathbf{x}) \equiv m\hbar f_{l}^{m}(\mathbf{x})\]

Computing spherical harmonics

Case 1: $l = 0$

The zeroth-order harmonic polynomial is a scalar: $f_{0}(\mathbf{x}) = 1$. Multiplying $r^{-l}$ gives $Y_{0}^{0}(\theta,\phi) = 1$(ignoring the normalization). This is the spherically symmetric $s$-wave.

Case 2: $l = 1$

The first-order harmonic polynomials’ bases are $x, y$ and $z$. Note that all of them are harmonic. Instead of them, we use complex transformation variables: $\omega, \omega^{\ast}, z$. Order them by the eigenvalue of $L_{z}$.

\[\begin{align*} \omega: &\quad m = 1, \; x + iy = r\sin\theta e^{i\phi} \\ z: &\quad m = 0, \; z = r\cos\theta \\ \omega^{\ast} &\quad m = -1, \; x - iy = r\sin\theta e^{-i\phi} \end{align*}\]

Here, we used relation between Cartesian and spherical polar coordinates. Multiplying $r^{-l}$ yields the $l=1$ spherical harmonics.

\[Y_{1}^{1} \sim \sin\theta e^{i\phi}, \quad Y_{1}^{0} \sim \cos\theta, \quad Y_{1}^{-1} \sim \sin\theta e^{-i\phi}\]

Case 3: $l = 2$

The second-order harmonic polynomials’s bases are

\[x^{2}, \quad y^{2}, \quad z^{2}, \quad xy, \quad yz, \quad zx\]

However, the first three polynomials are not harmonic. Therefore, we need to take appropriate linear combination to make harmonic polynomials. Therefore, we get only five $l=2$ spherical harmonics, rather than six. Now, let’s consider all degree 2 polynomials of $\omega, \omega^{\ast}$ and $z$.

\[\omega^{2} (m = 2), \; \omega z (m = 1), \; z^{2} \;\&\; \omega\omega^{\ast} (m = 0) \; , \omega^{\ast}z (m = -1), \; (\omega^{\ast})^{2} (m = -2)\]

To make them harmonic ($\nabla^{2}f_{2} = [4\partial_{\omega}\partial_{\omega^{\ast}} + \partial_{z}^{2}]f_{2} = 0$), we need to take appropriate linear combinations to make the laplacian vanish. In $l=2$ case, only $m = 0$ case is problematic.

Taking the linear combination yields

\[\begin{align*} \frac{\omega^{2}}{r^{2}}&:\; Y_{2}^{2} \sim \sin^{2}\theta e^{2i\phi} \\ \frac{\omega z}{r^{2}}&:\; Y_{2}^{1} \sim \sin\theta\cos\theta e^{i\phi} \\ \frac{\omega\omega^{\ast} - 2z^{2}}{r^{2}}&:\; Y_{2}^{0} \sim \sin^{2}\theta - 2\cos^{2}\theta = 1 - 3\cos^{2}\theta \\ \frac{\omega^{\ast}z}{r^{2}}&:\; Y_{2}^{-1} \sim \sin\theta\cos\theta e^{-i\phi} \\ \frac{(\omega^{\ast})^{2}}{r^{2}}&:\; Y_{2}^{-2} \sim \sin^{2}\theta e^{-2i\phi} \end{align*}\]

Note that this methodology is only suitable for low $l$ values, and you need to normalize the spherical harmonics by yourself.