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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.