# Peeter Joot's (OLD) Blog.

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## Bivector form of quantum angular momentum operator (derivatives of the unit vectors)

Posted by peeterjoot on August 26, 2009

# Derivatives of the unit vectors

To properly evaluate the angular momentum square we’ll need to examine the $\partial_\theta$ and $\partial_\phi$ variation of the unit vectors $\hat{\mathbf{r}}$, $\hat{\boldsymbol{\theta}}$, and $\hat{\boldsymbol{\phi}}$. Some part of this question can be evaluated without reference to the specific vector or even which derivative is being evaluated. Writing $e$ for one of $\mathbf{e}_1$, $\mathbf{e}_2$, or $\mathbf{e}_k$, and $\sigma = \tilde{R} e R$ for the mapping of this vector under rotation, and $\partial$ for the desired $\theta$ or $\phi$ partial derivative, we have

\begin{aligned}\partial (\tilde{R} e R)&=(\partial \tilde{R}) e R + \tilde{R} e (\partial R) \end{aligned} \quad\quad\quad(18)

Since $\tilde{R} R = 1$, we have

\begin{aligned}0 &= \partial (\tilde{R} R) \\ &=(\partial \tilde{R}) R + \tilde{R} (\partial R) \end{aligned}

So substitution of $(\partial \tilde{R}) = -\tilde{R} (\partial R) \tilde{R}$, back into (18) supplies

\begin{aligned}\partial (\tilde{R} e R)&=-\tilde{R} (\partial R) \tilde{R} e R + \tilde{R} e (\partial R) \\ &=-\tilde{R} (\partial R) (\tilde{R} e R) + (\tilde{R} e R) \tilde{R} (\partial R) \\ &=-\tilde{R} (\partial R) \sigma + \sigma \tilde{R} (\partial R) \\ \end{aligned}

Writing the bivector term as

\begin{aligned}\Omega = \tilde{R} (\partial R) \end{aligned} \quad\quad\quad(19)

The change in the rotated vector is seen to be entirely described by the commutator of that vectors image under rotation with $\Omega$. That is

\begin{aligned}\partial \sigma = \left[{\sigma},{\Omega}\right] \end{aligned} \quad\quad\quad(20)

Our spherical polar rotor was given by

\begin{aligned}R = e^{\mathbf{e}_{31}\theta/2} e^{\mathbf{e}_{12}\phi/2} \end{aligned} \quad\quad\quad(21)

Lets calculate the $\Omega$ bivector for each of the $\theta$ and $\phi$ partials. For $\theta$ we have

\begin{aligned}\Omega_\theta &= \tilde{R} \partial_\theta R \\ &= \frac{1}{2} e^{-\mathbf{e}_{12}\phi/2} e^{-\mathbf{e}_{31}\theta/2} \mathbf{e}_{31} e^{\mathbf{e}_{31}\theta/2} e^{\mathbf{e}_{12}\phi/2} \\ &= \frac{1}{2} e^{-\mathbf{e}_{12}\phi/2} \mathbf{e}_{31} e^{\mathbf{e}_{12}\phi/2} \\ &= \frac{1}{2} \mathbf{e}_3 e^{-\mathbf{e}_{12}\phi/2} \mathbf{e}_{1} e^{\mathbf{e}_{12}\phi/2} \\ &= \frac{1}{2} \mathbf{e}_{31} e^{\mathbf{e}_{12}\phi} \\ \end{aligned}

Explicitly, this is the bivector $\Omega_\theta = (\mathbf{e}_{31} \cos\theta + \mathbf{e}_{32} \sin\theta)/2$, a wedge product of a vectors in $\hat{\mathbf{z}}$ direction with one in the perpendicular $x-y$ plane (curiously a vector in the $x-y$ plane rotated by polar angle $\theta$, not the equatorial angle $\phi$).

FIXME: picture. Draw this plane cutting through the sphere.

For the $\phi$ partial variation of any of our unit vectors our bivector rotation generator is

\begin{aligned}\Omega_\phi &= \tilde{R} \partial_\phi R \\ &= \frac{1}{2} e^{-\mathbf{e}_{12}\phi/2} e^{-\mathbf{e}_{31}\theta/2} e^{\mathbf{e}_{31}\theta/2} \mathbf{e}_{12} e^{\mathbf{e}_{12}\phi/2} \\ &= \frac{1}{2} \mathbf{e}_{12} \\ \end{aligned}

This one has no variation at all with angle whatsoever. If this is all correct so far perhaps that is not surprising given the fact that we expect an extra $\cot\theta$ in the angular momentum operator square, so a lack of $\phi$ dependence in the rotation generator likely means that any additional $\phi$ dependence will cancel out. Next step (for a different day) is to take these rotation generator bivectors, apply them via commutator products to the $\hat{\mathbf{r}}$, $\hat{\boldsymbol{\theta}}$, and $\hat{\boldsymbol{\phi}}$ vectors, and see what we get.