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Angular momentum polar form, factoring out the raising and lowering operators, and simultaneous eigenvalues.

Posted by peeterjoot on August 30, 2009

Continuation of ‘Bivector form of quantum angular momentum operator’ notes.

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After a bit more manipulation we find that the angular momentum operator polar form representation, again using $i = \mathbf{e}_1 \mathbf{e}_2$, is

\begin{aligned}\mathbf{x} \wedge \boldsymbol{\nabla} = I \hat{\boldsymbol{\phi}} ( \partial_\theta + i \cot\theta \partial_\phi + \mathbf{e}_{23} e^{i\phi} \partial_\phi ) \end{aligned} \quad\quad\quad(54)

Observe how similar the exponential free terms within the braces are to the raising operator as given in Bohm’s equation (14.40)

\begin{aligned}L_x + i L_y &= e^{i\phi} (\partial_\theta + i \cot\theta \partial_\phi ) \\ L_z &= \frac{1}{{i}} \partial_\phi \end{aligned} \quad\quad\quad(55)

In fact since $\mathbf{e}_{23}e^{i\phi} = e^{-i\phi} \mathbf{e}_{23}$, the match can be made even closer

\begin{aligned}\mathbf{x} \wedge \boldsymbol{\nabla} = I \hat{\boldsymbol{\phi}} e^{-i\phi} ( \underbrace{e^{i\phi} (\partial_\theta + i \cot\theta \partial_\phi)}_{= L_x + i L_y} + \mathbf{e}_{13} \underbrace{\frac{1}{{i}} \partial_\phi}_{=L_z} ) \end{aligned} \quad\quad\quad(57)

This is a surprising factorization, but noting that $\hat{\boldsymbol{\phi}} = \mathbf{e}_2 e^{i\phi}$ we have

\begin{aligned}\mathbf{x} \wedge \boldsymbol{\nabla} = \mathbf{e}_{31} ( e^{i\phi} (\partial_\theta + i \cot\theta \partial_\phi) + \mathbf{e}_{13} \frac{1}{{i}} \partial_\phi ) \end{aligned} \quad\quad\quad(58)

It appears that the factoring out from the left of a unit bivector (in this case $\mathbf{e}_{31}$) from the bivector angular momentum operator, leaves as one of the remainders the raising operator.

Similarily, noting that $\mathbf{e}_{13}$ anticommutes with $i = \mathbf{e}_{12}$, we have the right factorization

\begin{aligned}\mathbf{x} \wedge \boldsymbol{\nabla} = ( e^{-i\phi} (\partial_\theta - i \cot\theta \partial_\phi) - \mathbf{e}_{13} \frac{1}{{i}} \partial_\phi )\mathbf{e}_{31} \end{aligned} \quad\quad\quad(59)

Now in the remainder, we see the polar form representation of the lowering operator $L_x - i L_y = e^{-i\phi}(\partial_\theta - i\cot\theta \partial_\phi)$.

I wasn’t expecting the raising and lowering operators “to fall out” as they did by simply expressing the complete bivector operator in polar form. This is actually fortunitous since it shows why this peculiar combination is of interest.

If we find a zero solution to the raising or lowering operator, that is also a solution of the eigenproblem $(\partial_\phi - \lambda) \psi = 0$, then this is neccessarily also an eigensolution of $\mathbf{x} \wedge \boldsymbol{\nabla}$. A secondary implication is that this is then also an eigensolution of $\left\langle{{(\mathbf{x} \wedge \boldsymbol{\nabla})^2}}\right\rangle \psi = \lambda' \psi$. This was the starting point in Bohm’s quest for the spherical harmonics, but why he started there wasn’t clear to me.

Saying this without the words, let’s look for eigenfunctions for the non-raising portion of (58). That is

\begin{aligned}\mathbf{e}_{31} \mathbf{e}_{13} \frac{1}{{i}} \partial_\phi f = \lambda f \end{aligned} \quad\quad\quad(60)

Since $\mathbf{e}_{31} \mathbf{e}_{13} = 1$ we want solutions of

\begin{aligned}\partial_\phi f = i \lambda f \end{aligned} \quad\quad\quad(61)

Solutions are

\begin{aligned}f = \kappa(\theta) e^{i\lambda \phi} \end{aligned} \quad\quad\quad(62)

A demand that this is a zero eigenfunction for the raising operator, means we are looking for solutions of

\begin{aligned}\mathbf{e}_{31} e^{i\phi} (\partial_\theta + i \cot\theta \partial_\phi) \kappa(\theta) e^{i\lambda \phi} = 0 \end{aligned} \quad\quad\quad(63)

It is sufficient to find zero eigenfunctions of

\begin{aligned}(\partial_\theta + i \cot\theta \partial_\phi) \kappa(\theta) e^{i\lambda \phi} = 0 \end{aligned} \quad\quad\quad(64)

Evaluation of the $\phi$ partials and rearrangement leaves us with an equation in $\theta$ only

\begin{aligned}\frac{\partial \kappa }{\partial \theta} = \lambda \cot\theta \kappa \end{aligned} \quad\quad\quad(65)

This has solutions $\kappa = A(\phi) (\sin\theta)^\lambda$, where because of the partial derivatives in (65) we are free to make the integration constant a function of $\phi$. Since this is the functional dependence that is a zero of the raising operator, including this at the $\theta$ dependence of (62) means that we have a simultaneous zero of the raising operator, and an eigenfunction of eigenvalue $\lambda$ for the remainder of the angular momentum operator.

\begin{aligned}f(\theta,\phi) = (\sin\theta)^\lambda e^{i\lambda \phi} \end{aligned} \quad\quad\quad(66)

This is very similar seeming to the process of adding homogeneous solutions to specific ones, since we augment the specific eigenvalued solutions for one part of the operator by ones that produce zeros for the rest.

As a check lets apply the angular momentum operator to this as a test and see if the results match our expectations.

\begin{aligned}(\mathbf{x} \wedge \boldsymbol{\nabla} ) (\sin\theta)^\lambda e^{i\lambda \phi}&=\hat{\mathbf{r}} \left( \hat{\boldsymbol{\theta}} \partial_\theta + \hat{\boldsymbol{\phi}} \frac{1}{{\sin\theta}} \partial_\phi \right) (\sin\theta)^\lambda e^{i\lambda \phi} \\ &=\hat{\mathbf{r}} \left( \hat{\boldsymbol{\theta}} \lambda (\sin\theta)^{\lambda-1} \cos\theta + \hat{\boldsymbol{\phi}} \frac{1}{{\sin\theta}} (\sin\theta)^\lambda (i\lambda)\right) e^{i\lambda \phi} \\ &=\lambda \hat{\mathbf{r}} \left( \hat{\boldsymbol{\theta}} \cos\theta + \hat{\boldsymbol{\phi}} i \right) e^{i\lambda \phi} (\sin\theta)^{\lambda-1} \\ \end{aligned}

From (38) we have

\begin{aligned}\hat{\mathbf{r}} \hat{\boldsymbol{\phi}} i &= \mathbf{e}_3 \hat{\boldsymbol{\phi}} i \cos\theta - \sin\theta \\ &= \mathbf{e}_{32} i e^{i\phi} \cos\theta - \sin\theta \\ &= \mathbf{e}_{13} e^{i\phi} \cos\theta - \sin\theta \\ \end{aligned}

and from (37) we have

\begin{aligned}\hat{\mathbf{r}} \hat{\boldsymbol{\theta}}&= I \hat{\boldsymbol{\phi}} \\ &= \mathbf{e}_{31} e^{i\phi} \end{aligned}

Putting these together shows that $(\sin\theta)^\lambda e^{i\lambda \phi}$ is an eigenfunction of $\mathbf{x} \wedge \boldsymbol{\nabla}$,

\begin{aligned}(\mathbf{x} \wedge \boldsymbol{\nabla} ) (\sin\theta)^\lambda e^{i\lambda \phi} = -\lambda (\sin\theta)^\lambda e^{i\lambda \phi} \end{aligned} \quad\quad\quad(67)

This negation suprised me at first, but I don’t see any errors here in the arithmetic. Observe that if this is correct, then it provides a demonstration that the previous suspected calculation leading to (7) is in fact wrong as guessed. That suspected incorrect result, a product of very messy calculation, was

\begin{aligned}\left\langle{{(\mathbf{x} \wedge \boldsymbol{\nabla})^2}}\right\rangle \stackrel{?}{=} \left( \mathbf{x} \wedge \boldsymbol{\nabla} - \frac{1}{{2}} \right) (\mathbf{x} \wedge \boldsymbol{\nabla}) \end{aligned} \quad\quad\quad(68)

the one half factor seemed unasthetic, with the following somehow preferable

\begin{aligned}\left\langle{{(\mathbf{x} \wedge \boldsymbol{\nabla})^2}}\right\rangle \stackrel{?}{=} \left( \mathbf{x} \wedge \boldsymbol{\nabla} - 1 \right) (\mathbf{x} \wedge \boldsymbol{\nabla}) \end{aligned} \quad\quad\quad(69)

If (67) is the correct version then calculating the operator effect of $\left\langle{{(\mathbf{x} \wedge \boldsymbol{\nabla})^2}}\right\rangle$ for the eigenvalue we have

\begin{aligned}\left\langle{{(\mathbf{x} \wedge \boldsymbol{\nabla})^2}}\right\rangle (\sin\theta)^\lambda e^{i\lambda \phi}&=\left( \mathbf{x} \wedge \boldsymbol{\nabla} - 1 \right) (\mathbf{x} \wedge \boldsymbol{\nabla}) (\sin\theta)^\lambda e^{i\lambda \phi} \\ &=((-\lambda)^2 - (-\lambda)) (\sin\theta)^\lambda e^{i\lambda \phi} \\ \end{aligned}

So the eigenvalue is $\lambda(\lambda + 1)$. This we do know to be the case in fact, so a second look at the messy algebra leading to (68) is justified (or an attempt at a coordinate free expansion).