Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Bivector form of quantum angular momentum operator (continued again)

Posted by peeterjoot on August 26, 2009

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

Explicit expansion of the angular momentum operator.

There’s a couple of things to explore before going forward. One is an explicit verification that \mathbf{x} \wedge \boldsymbol{\nabla} has no radial dependence (something not obvious). Another is that we should be able to compare the \mathbf{x}^{-2} (\mathbf{x} \wedge \boldsymbol{\nabla})^2 (as done for the \mathbf{x} \cdot \boldsymbol{\nabla} terms) the explicit r,\theta,\phi expression for the Laplacian to verify consistency and correctness.

For the spherical polar rotation we use the rotor

\begin{aligned}R = e^{\mathbf{e}_{31}\theta/2} e^{\mathbf{e}_{12}\phi/2} \end{aligned} \quad\quad\quad(11)

Our position vector in spherical polar coordinates is then

\begin{aligned}\mathbf{x} &= r \tilde{R} \mathbf{e}_3 R  \end{aligned} \quad\quad\quad(12)

\begin{aligned}\boldsymbol{\nabla} &= \hat{\mathbf{r}} \partial_r + \hat{\boldsymbol{\theta}} \frac{1}{r} \partial_\theta + \hat{\boldsymbol{\phi}} \frac{1}{{r \sin\theta}} \partial_\phi \end{aligned} \quad\quad\quad(13)

and our unit vectors translate from the standard basis as

\begin{aligned}\begin{pmatrix}\hat{\mathbf{r}} \\ \hat{\boldsymbol{\theta}} \\ \hat{\boldsymbol{\phi}} \\ \end{pmatrix}=\tilde{R}\begin{pmatrix}\mathbf{e}_3 \\ \mathbf{e}_1 \\ \mathbf{e}_2 \\ \end{pmatrix}R \end{aligned} \quad\quad\quad(14)

This last mapping can be used to express the gradient unit vectors in terms of the standard basis, as we did for the position vector \mathbf{x}. That is

\begin{aligned}\boldsymbol{\nabla} &= \tilde{R} \left( \mathbf{e}_3 R \partial_r + \mathbf{e}_1 R \frac{1}{r} \partial_\theta + \mathbf{e}_2 R \frac{1}{{r \sin\theta}} \partial_\phi \right) \end{aligned} \quad\quad\quad(15)

Okay, we’ve now got all the pieces collected, ready to evaluate \mathbf{x} \wedge \boldsymbol{\nabla}

\begin{aligned}\mathbf{x} \wedge \boldsymbol{\nabla} &=r {\left\langle{{\tilde{R} \mathbf{e}_3 R \tilde{R} \left( \mathbf{e}_3 R \partial_r + \mathbf{e}_1 R \frac{1}{r} \partial_\theta + \mathbf{e}_2 R \frac{1}{{r \sin\theta}} \partial_\phi \right) }}\right\rangle}_{2} \\ &=r {\left\langle{{\tilde{R} \left( R \partial_r + \mathbf{e}_3 \mathbf{e}_1 R \frac{1}{r} \partial_\theta + \mathbf{e}_3 \mathbf{e}_2 R \frac{1}{{r \sin\theta}} \partial_\phi \right) }}\right\rangle}_{2} \\  \end{aligned}

Observe that the {\mathbf{e}_3}^2 contribution is only a scalar, so bivector selection of that is zero. In the remainder we have cancellation of r/r factors, leaving just

\begin{aligned}\mathbf{x} \wedge \boldsymbol{\nabla} &=\tilde{R} \left( \mathbf{e}_3 \mathbf{e}_1 R \partial_\theta + \mathbf{e}_3 \mathbf{e}_2 R \frac{1}{{\sin\theta}} \partial_\phi \right)  \end{aligned} \quad\quad\quad(16)

Using (14) this is

\begin{aligned}\mathbf{x} \wedge \boldsymbol{\nabla} &=\hat{\mathbf{r}} \left( \hat{\boldsymbol{\theta}} \partial_\theta + \hat{\boldsymbol{\phi}} \frac{1}{{\sin\theta}} \partial_\phi \right)  \end{aligned} \quad\quad\quad(17)

As hoped, there is no explicit radial dependence here, taking care of the first of the desired verifications.

Next we want to square this operator. It should be noted that in the original derivation where we “factored” the Laplacian operator with respect to the reference vector \mathbf{x} we really used (\mathbf{x} \wedge \boldsymbol{\nabla})^2 \equiv \left\langle{{(\mathbf{x} \wedge \boldsymbol{\nabla})^2}}\right\rangle. That’s worth noting since a regular bivector would square to a negative constant, whereas the operator factors of the vectors in this expression do not intrinsically commute.

An additional complication for evaluating the square of \mathbf{x} \wedge \boldsymbol{\nabla} using the result of (17) is that \hat{\boldsymbol{\theta}} and \hat{\mathbf{r}} are functions of \theta and \phi, so we’d have to operate on those too. Without that operator subtlety we get the wrong answer

\begin{aligned}-(\mathbf{x} \wedge \boldsymbol{\nabla})^2&=\left\langle{{ \tilde{R} \left( \mathbf{e}_1 R \partial_\theta + \frac{\mathbf{e}_2 R}{\sin\theta}\partial_\phi \right)\tilde{R} \left( \mathbf{e}_1 R \partial_\theta + \frac{\mathbf{e}_2 R}{\sin\theta}\partial_\phi \right) }}\right\rangle \\ &\ne\partial_{\theta\theta} + \frac{1}{{\sin^2\theta \partial_{\phi\phi}}} \end{aligned}

Equality above would only be if the unit vectors were fixed. By comparison we also see that this is missing a \cot\theta \partial_\theta term. That must come from the variation of the unit vectors with position in the second application of \mathbf{x} \wedge \boldsymbol{\nabla}.


Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: