# Peeter Joot's (OLD) Blog.

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## Graphical representation of the associated Legendre Polynomials for l=1

Posted by peeterjoot on August 16, 2009

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# First observations.

In Bohm’s QT ([1], 14.17), the properties of $l=1$ associated Legendre polynomials are examined under rotation.

Those eigenfunctions are the normalized versions of following

\begin{aligned}\psi_1 &= \sin\theta e^{i\phi} \\ \psi_0 &= \cos\theta \\ \psi_{-1} &= \sin\theta e^{-i\phi} \end{aligned} \quad\quad\quad(1)

The normalization is provided by a surface area inner product

\begin{aligned}(u,v) &= \int_{\theta=0}^{\pi} \int_{\phi=0}^{2\pi} u v^{*} \sin\theta d\theta d\phi \end{aligned} \quad\quad\quad(4)

With the normalization discarded, there is a direct relationship between these normal eigenfunctions with a triple of vectors associated with a point on the unit sphere. Refering to figure (\ref{fig:L1Associated}), observe the three doubled arrow vectors, all associated with a point on the unit sphere $\mathbf{x} = (x,y,z) = (\sin\theta \cos\phi, \sin\theta \cos\phi, \cos\theta)$.

Vectorizing the l=1 associated Legendre polynomials.

The normal to the $x-y$ plane from $\mathbf{x}$, designated $\mathbf{n}$ has the vectorial value

\begin{aligned}\mathbf{n} = \cos\theta \mathbf{e}_3 \end{aligned} \quad\quad\quad(5)

From the origin to the point of of the $x-y$ plane intersection to the normal we have

\begin{aligned}\boldsymbol{\rho} = \sin\theta (\cos\phi \mathbf{e}_1 + \sin\phi \mathbf{e}_2) = \mathbf{e}_1 \sin\theta e^{\mathbf{e}_1 \mathbf{e}_2 \phi} \end{aligned} \quad\quad\quad(6)

and finally in the opposite direction also in the plane and mirroring $\boldsymbol{\rho}$ we have the last of this triplet of vectors

\begin{aligned}\boldsymbol{\rho}_{-} = \sin\theta (\cos\phi \mathbf{e}_1 - \sin\phi \mathbf{e}_2) = \mathbf{e}_1 \sin\theta e^{-\mathbf{e}_1 \mathbf{e}_2 \phi} \end{aligned} \quad\quad\quad(7)

So, if we choose to use $i=\mathbf{e}_1 \mathbf{e}_2$ (the bivector for the plane normal to the z-axis), then we can in fact vectorize these eigenfunctions. The vectors $\boldsymbol{\rho}$ (i.e. $\psi_1)$, and $\boldsymbol{\rho}_{-}$ (i.e. $\psi_{-1}$) are both normal to $\mathbf{n}$ (i.e. $\psi_0$), but while the vectors $\boldsymbol{\rho}$ and $\boldsymbol{\rho}_{-}$ are both in the plane one is produced with a counterclockwise rotation of $\mathbf{e}_1$ by $\phi$ in the plane and the other with an opposing rotation.

Summarizing, we can write the unnormalized vectors the relations

\begin{aligned}\begin{array}{l l l}\psi_1 &= \mathbf{e}_1 \boldsymbol{\rho} &= \sin\theta e^{\mathbf{e}_1\mathbf{e}_2 \phi} \\ \psi_0 &= \mathbf{e}_3 \mathbf{n} &= \cos\theta \\ \psi_{-1} &= \mathbf{e}_1 \boldsymbol{\rho}_{-} &= \sin\theta e^{-\mathbf{e}_1\mathbf{e}_2 \phi}\end{array} \end{aligned}

I have no familiarity yet with the $l=2$ or higher Legendre eigenfunctions. Do they also admit a geometric representation?

# Expressing Legendre eigenfunctions using rotations.

We can express a point on a sphere with a pair of rotation operators. First rotating $\mathbf{e}_3$ towards $\mathbf{e}_1$ in the $z,x$ plane by $\theta$, then in the $x,y$ plane by $\phi$ we have the point $\mathbf{x}$ in figure (\ref{fig:L1Associated})

Writing the result of the first rotation as $\mathbf{e}_3'$ we have

\begin{aligned}\mathbf{e}_3' = \mathbf{e}_3 e^{\mathbf{e}_{31}\theta} = e^{-\mathbf{e}_{31}\theta/2} \mathbf{e}_3 e^{\mathbf{e}_{31}\theta/2} \end{aligned} \quad\quad\quad(8)

One more rotation takes $\mathbf{e}_3'$ to $\mathbf{x}$. That is

\begin{aligned}\mathbf{x} = e^{-\mathbf{e}_{12}\phi/2} \mathbf{e}_3' e^{\mathbf{e}_{12}\phi/2} \end{aligned} \quad\quad\quad(9)

All together, writing $R_\theta = e^{\mathbf{e}_{31}\theta/2}$, and $R_\phi = e^{\mathbf{e}_{12}\phi/2}$, we have

\begin{aligned}\mathbf{x} = \tilde{R_\phi} \tilde{R_\theta} \mathbf{e}_3 R_\theta R_\phi \end{aligned} \quad\quad\quad(10)

It’s worth a quick verification that this produces the desired result.

\begin{aligned}\tilde{R_\phi} \tilde{R_\theta} \mathbf{e}_3 R_\theta R_\phi &= \tilde{R_\phi} \mathbf{e}_3 e^{\mathbf{e}_{31}\theta} R_\phi \\ &= e^{-\mathbf{e}_{12}\phi/2} (\mathbf{e}_3 \cos\theta + \mathbf{e}_1 \sin\theta) e^{\mathbf{e}_{12}\phi/2} \\ &= \mathbf{e}_3 \cos\theta + \mathbf{e}_1 \sin\theta e^{\mathbf{e}_{12}\phi} \\ \end{aligned}

This is the expected result

\begin{aligned}\mathbf{x} = \mathbf{e}_3 \cos\theta + \sin\theta (\mathbf{e}_1 \sin\theta + \mathbf{e}_2 \cos\theta) \end{aligned} \quad\quad\quad(11)

The projections onto the $\mathbf{e}_3$ and the $x,y$ plane are then, respectively,

\begin{aligned}\mathbf{x}_z &= \mathbf{e}_3 (\mathbf{e}_3 \cdot \mathbf{x}) = \mathbf{e}_3 \cos\theta \\ \mathbf{x}_{x,y} &= \mathbf{e}_3 (\mathbf{e}_3 \wedge \mathbf{x}) = \sin\theta (\mathbf{e}_1 \sin\theta + \mathbf{e}_2 \cos\theta) \end{aligned} \quad\quad\quad(12)

So if $\mathbf{x}_{\pm}$ is the point on the unit sphere associated with the rotation angles $\theta,\pm\phi$, then we have for the $l=1$ associated Legendre polynomials

\begin{aligned}\psi_0 &= \mathbf{e}_3 \cdot \mathbf{x} \\ \psi_{\pm 1} &= \mathbf{e}_1 \mathbf{e}_3 (\mathbf{e}_3 \wedge \mathbf{x}_{\pm}) \end{aligned} \quad\quad\quad(14)

Note that the $\pm$ was omitted from $\mathbf{x}$ for $\psi_0$ since either produces the same $\mathbf{e}_3$ component. This gives us a nice geometric interpretation of these eigenfunctions. We see that $\psi_0$ is the biggest when $\mathbf{x}$ is close to straight up, and when this occurs $\psi_{\pm 1}$ are correspondingly reduced, but when $\mathbf{x}$ is close to the $x,y$ plane where $\psi_{\pm 1}$ will be greatest the $z$-axis component is reduced.

# References

[1] D. Bohm. Quantum Theory. Courier Dover Publications, 1989.