Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Graphical representation of the associated Legendre Polynomials for l=1

Posted by peeterjoot on August 16, 2009

[Click here for a PDF of this post with nicer formatting]

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.

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.


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


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