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## Bivector form of quantum angular momentum operator (derivatives of spherical polar unit vectors and bivectors)

Posted by peeterjoot on August 27, 2009

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

# Applying the vector derivative commutator (or not).

Let’s express the $\hat{\boldsymbol{\theta}}$ and $\hat{\boldsymbol{\phi}}$ unit vectors explicitly in terms of the standard basis. Starting with $\hat{\boldsymbol{\theta}}$ we have

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

Explicitly in vector form, eliminating the exponential, this is $\hat{\boldsymbol{\theta}} = \mathbf{e}_1 \cos\theta \cos\phi + \mathbf{e}_2 \cos\theta\sin\phi - \mathbf{e}_3\sin\theta$, but it is more convenient to keep the exponential as is.

For $\hat{\boldsymbol{\phi}}$ we have

\begin{aligned}\hat{\boldsymbol{\phi}} &= \tilde{R} \mathbf{e}_2 R \\ &= e^{-\mathbf{e}_{12}\phi/2} e^{-\mathbf{e}_{31}\theta/2} \mathbf{e}_2 e^{\mathbf{e}_{31}\theta/2} e^{\mathbf{e}_{12}\phi/2} \\ &= e^{-\mathbf{e}_{12}\phi/2} \mathbf{e}_2 e^{\mathbf{e}_{12}\phi/2} \\ &= \mathbf{e}_2 e^{\mathbf{e}_{12}\phi} \\ \end{aligned}

Again, explicitly this is $\hat{\boldsymbol{\phi}} = \mathbf{e}_2 \cos\phi - \mathbf{e}_1 \sin\phi$, but we’ll use the exponential form above. Last we want $\hat{\mathbf{r}}$

\begin{aligned}\hat{\mathbf{r}} &= \tilde{R} \mathbf{e}_3 R \\ &= e^{-\mathbf{e}_{12}\phi/2} e^{-\mathbf{e}_{31}\theta/2} \mathbf{e}_3 e^{\mathbf{e}_{31}\theta/2} e^{\mathbf{e}_{12}\phi/2} \\ &= e^{-\mathbf{e}_{12}\phi/2} \mathbf{e}_3 e^{\mathbf{e}_{31}\theta} e^{\mathbf{e}_{12}\phi/2} \\ &= 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}

Summarizing we have

\begin{aligned}\hat{\boldsymbol{\theta}} &= \mathbf{e}_1 \cos\theta e^{\mathbf{e}_{12}\phi} -\mathbf{e}_3 \sin\theta \\ \hat{\boldsymbol{\phi}} &= \mathbf{e}_2 e^{\mathbf{e}_{12}\phi} \\ \hat{\mathbf{r}} &= \mathbf{e}_3 \cos\theta + \mathbf{e}_1 \sin\theta e^{\mathbf{e}_{12}\phi} \end{aligned} \quad\quad\quad(22)

Or without exponentials

\begin{aligned}\hat{\boldsymbol{\theta}} &= \mathbf{e}_1 \cos\theta \cos\phi + \mathbf{e}_2 \cos\theta\sin\phi - \mathbf{e}_3\sin\theta \\ \hat{\boldsymbol{\phi}} &= \mathbf{e}_2 \cos\phi - \mathbf{e}_1 \sin\phi \\ \hat{\mathbf{r}} &= \mathbf{e}_3 \cos\theta + \mathbf{e}_1 \sin\theta \cos\phi + \mathbf{e}_2 \sin\theta \sin\phi \end{aligned} \quad\quad\quad(25)

Now, having worked out the cool commutator result, it appears that it will actually be harder to use it, then to just calculate the derivatives directly (at least for the $\hat{\boldsymbol{\phi}}$ derivatives). For those we have

\begin{aligned}\partial_\theta \hat{\boldsymbol{\phi}} &= \partial_\theta \mathbf{e}_2 e^{\mathbf{e}_{12}\phi} \\ &= 0 \end{aligned}

and

\begin{aligned}\partial_\phi \hat{\boldsymbol{\phi}} &= \partial_\phi \mathbf{e}_2 e^{\mathbf{e}_{12}\phi} \\ &= \mathbf{e}_2 \mathbf{e}_{12} e^{\mathbf{e}_{12}\phi} \\ &= -\mathbf{e}_{12} \hat{\boldsymbol{\phi}} \end{aligned}

This multiplication takes $\hat{\boldsymbol{\phi}}$ a vector in the $x,y$ plane and rotates it 90 degrees, leaving an inwards facing radial unit vector in the x,y plane.

Now, having worked out the commutator method, lets at least verify that it works.

\begin{aligned}\partial_\theta \hat{\boldsymbol{\phi}} &= \left[{\hat{\boldsymbol{\phi}}},{\Omega_\theta}\right] \\ &= \hat{\boldsymbol{\phi}} \Omega_\theta - \Omega_\theta \hat{\boldsymbol{\phi}} \\ &= \frac{1}{{2}} (\hat{\boldsymbol{\phi}} \mathbf{e}_{31} e^{\mathbf{e}_{12}\phi} - \mathbf{e}_{31} e^{\mathbf{e}_{12}\phi} \hat{\boldsymbol{\phi}}) \\ &= \frac{1}{{2}} (\mathbf{e}_2 \mathbf{e}_3 \mathbf{e}_1 e^{-\mathbf{e}_{12}\phi} e^{\mathbf{e}_{12}\phi} - \mathbf{e}_{3} \mathbf{e}_{1} \mathbf{e}_{2} e^{-\mathbf{e}_{12}\phi} e^{\mathbf{e}_{12}\phi}) \\ &= \frac{1}{{2}} (-\mathbf{e}_3 \mathbf{e}_2 \mathbf{e}_1 - \mathbf{e}_{3} \mathbf{e}_{1} \mathbf{e}_{2} ) \\ &= 0 \end{aligned}

Much harder this way compared to taking the derivative directly, but we at least get the right answer. For the $\phi$ derivative using the commutator we have

\begin{aligned}\partial_\phi \hat{\boldsymbol{\phi}} &= \left[{\hat{\boldsymbol{\phi}}},{\Omega_\phi}\right] \\ &= \hat{\boldsymbol{\phi}} \Omega_\phi - \Omega_\phi \hat{\boldsymbol{\phi}} \\ &= \frac{1}{{2}} (\hat{\boldsymbol{\phi}} \mathbf{e}_{12} - \mathbf{e}_{12} \hat{\boldsymbol{\phi}}) \\ &= \frac{1}{{2}} (\mathbf{e}_2 e^{\mathbf{e}_{12}\phi} \mathbf{e}_{12} - \mathbf{e}_{12} \mathbf{e}_2 e^{\mathbf{e}_{12}\phi}) \\ &= \frac{1}{{2}} (-\mathbf{e}_{12} \mathbf{e}_2 e^{\mathbf{e}_{12}\phi} - \mathbf{e}_{12} \mathbf{e}_2 e^{\mathbf{e}_{12}\phi}) \\ &= -\mathbf{e}_{12} \hat{\boldsymbol{\phi}} \end{aligned}

Good, also consistent with direct calculation. How about our $\hat{\boldsymbol{\theta}}$ derivatives? Lets just calculate these directly without bothering at all with the commutator. This is

\begin{aligned}\partial_\phi \hat{\boldsymbol{\theta}} &= \mathbf{e}_1 \cos\theta \mathbf{e}_12 e^{\mathbf{e}_{12}\phi} \\ &= \mathbf{e}_2 \cos\theta e^{\mathbf{e}_{12}\phi} \\ &= \cos\theta \hat{\boldsymbol{\phi}} \end{aligned}

and

\begin{aligned}\partial_\theta \hat{\boldsymbol{\theta}} &= -\mathbf{e}_1 \sin\theta e^{\mathbf{e}_{12}\phi} -\mathbf{e}_3 \cos\theta \\ &= -\mathbf{e}_{12} \sin\theta \hat{\boldsymbol{\phi}} -\mathbf{e}_3 \cos\theta \\ \end{aligned}

Finally, last we have the derivatives of $\hat{\mathbf{r}}$. Those are

\begin{aligned}\partial_\phi \hat{\mathbf{r}} &= \mathbf{e}_2 \sin\theta e^{\mathbf{e}_{12}\phi} \\ &= \sin\theta \hat{\boldsymbol{\phi}} \end{aligned}

and

\begin{aligned}\partial_\theta \hat{\mathbf{r}} &= -\mathbf{e}_3 \sin\theta + \mathbf{e}_1 \cos\theta e^{\mathbf{e}_{12}\phi} \\ &= -\mathbf{e}_3 \sin\theta + \mathbf{e}_{12} \cos\theta \hat{\boldsymbol{\phi}} \\ \end{aligned}

Summarizing, all the derivatives we need to evaluate the square of the angular momentum operator are

\begin{aligned}\partial_\theta \hat{\boldsymbol{\phi}} &= 0 \\ \partial_\phi \hat{\boldsymbol{\phi}} &= -\mathbf{e}_{12} \hat{\boldsymbol{\phi}} \\ \partial_\theta \hat{\boldsymbol{\theta}} &= -\mathbf{e}_{12} \sin\theta \hat{\boldsymbol{\phi}} -\mathbf{e}_3 \cos\theta \\ \partial_\phi \hat{\boldsymbol{\theta}} &= \cos\theta \hat{\boldsymbol{\phi}} \\ \partial_\theta \hat{\mathbf{r}} &= -\mathbf{e}_3 \sin\theta + \mathbf{e}_{12} \cos\theta \hat{\boldsymbol{\phi}} \\ \partial_\phi \hat{\mathbf{r}} &= \sin\theta \hat{\boldsymbol{\phi}} \end{aligned} \quad\quad\quad(28)

Bugger. We actually want the derivatives of the bivectors $\hat{\mathbf{r}}\hat{\boldsymbol{\theta}}$ and $\hat{\mathbf{r}}\hat{\boldsymbol{\phi}}$ so we aren’t ready to evaluate the squared angular momentum. There’s three choices, one is to use these results and apply the chain rule, or start over and directly take the derivatives of these bivectors, or use the commutator result (which didn’t actually assume vectors and we can apply it to bivectors too if we really wanted to).

An attempt to use the chain rule get messy, but it looks like the bivectors reduce nicely, making it pointless to even think about the commutator method. Introducing some notational conveniences, first write $i = \mathbf{e}_{12}$. We’ll have to be a bit careful with this since it commutes with $\mathbf{e}_3$, but anticommutes with $\mathbf{e}_1$ or $\mathbf{e}_2$ (and therefore $\hat{\boldsymbol{\phi}}$). As usual we also write $I = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3$ for the Euclidean pseudoscalar (which commutes with all vectors and bivectors).

\begin{aligned}\hat{\mathbf{r}} \hat{\boldsymbol{\theta}}&= (\mathbf{e}_3 \cos\theta + i \sin\theta \hat{\boldsymbol{\phi}})(\cos\theta i \hat{\boldsymbol{\phi}} - \mathbf{e}_3 \sin\theta) \\ &= \mathbf{e}_3 \cos^2\theta i \hat{\boldsymbol{\phi}} -i \sin^2\theta \hat{\boldsymbol{\phi}} \mathbf{e}_3 +(i \hat{\boldsymbol{\phi}} i \hat{\boldsymbol{\phi}} -\mathbf{e}_3 \mathbf{e}_3 ) \cos\theta \sin\theta \\ &= i \mathbf{e}_3 (\cos^2\theta + \sin^2\theta) \hat{\boldsymbol{\phi}} +(-\hat{\boldsymbol{\phi}} i^2 \hat{\boldsymbol{\phi}} - 1) \cos\theta \sin\theta \\ \end{aligned}

This gives us just

\begin{aligned}\hat{\mathbf{r}} \hat{\boldsymbol{\theta}} = I \hat{\boldsymbol{\phi}} \end{aligned} \quad\quad\quad(34)

and calculation of the bivector partials will follow exclusively from the $\hat{\boldsymbol{\phi}}$ partials tabulated above.

Our other bivector doesn’t reduce quite as cleanly. We have

\begin{aligned}\hat{\mathbf{r}} \hat{\boldsymbol{\phi}} &=(\mathbf{e}_3 \cos\theta + i \sin\theta \hat{\boldsymbol{\phi}}) \hat{\boldsymbol{\phi}} \\ \end{aligned}

So for this one we have

\begin{aligned}\hat{\mathbf{r}} \hat{\boldsymbol{\phi}} = \mathbf{e}_3 \hat{\boldsymbol{\phi}} \cos\theta + i \sin\theta \end{aligned} \quad\quad\quad(35)

Tabulating all the bivector derivatives (details omitted) we have

\begin{aligned}\partial_\theta (\hat{\mathbf{r}} \hat{\boldsymbol{\theta}}) &= 0 \\ \partial_\phi (\hat{\mathbf{r}} \hat{\boldsymbol{\theta}}) &= \mathbf{e}_3 \hat{\boldsymbol{\phi}} \\ \partial_\theta (\hat{\mathbf{r}} \hat{\boldsymbol{\phi}}) &= -\mathbf{e}_3 \hat{\boldsymbol{\phi}} \sin\theta + i \cos\theta = i e^{I\hat{\boldsymbol{\phi}}\theta} \\ \partial_\phi (\hat{\mathbf{r}} \hat{\boldsymbol{\phi}}) &= -I \hat{\boldsymbol{\phi}} \cos\theta \end{aligned} \quad\quad\quad(36)

Okay, we are now armed to do the squaring of the angular momentum, but once again it’s late to start now. To be continued.