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## A summary: Bivector form of quantum angular momentum operator

Posted by peeterjoot on September 6, 2009

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

Having covered a fairly wide range in the previous Geometric Algebra exploration of the angular momentum operator, it seems worthwhile to attempt to summarize what was learned.

The exploration started with a simple observation that the use of the spatial pseudoscalar for the imaginary of the angular momentum operator in its coordinate form

\begin{aligned}L_1 &= -i \hbar( x_2 \partial_3 - x_3 \partial_2 ) \\ L_2 &= -i \hbar( x_3 \partial_1 - x_1 \partial_3 ) \\ L_3 &= -i \hbar( x_1 \partial_2 - x_2 \partial_1 ) \end{aligned} \quad\quad\quad(69)

allowed for expressing the angular momentum operator in its entirety as a bivector valued operator

\begin{aligned}\mathbf{L} &= - \hbar \mathbf{x} \wedge \boldsymbol{\nabla} \end{aligned} \quad\quad\quad(72)

The bivector representation has an intrinsic complex behavior, eliminating the requirement for an explicitly imaginary $i$ as is the case in the coordinate representation.

It was then assumed that the Laplacian can be expressed directly in terms of $\mathbf{x} \wedge \boldsymbol{\nabla}$. This isn’t an unreasonable thought since we can factor the gradient with components projected onto and perpendicular to a constant reference vector $\hat{\mathbf{a}}$ as

\begin{aligned}\boldsymbol{\nabla} = \hat{\mathbf{a}} (\hat{\mathbf{a}} \cdot \boldsymbol{\nabla}) + \hat{\mathbf{a}} (\hat{\mathbf{a}} \wedge \boldsymbol{\nabla}) \end{aligned} \quad\quad\quad(73)

and this squares to a Laplacian expressed in terms of these constant reference directions

\begin{aligned}\boldsymbol{\nabla}^2 = (\hat{\mathbf{a}} \cdot \boldsymbol{\nabla})^2 - (\hat{\mathbf{a}} \cdot \boldsymbol{\nabla})^2 \end{aligned} \quad\quad\quad(74)

a quantity that has an angular momentum like operator with respect to a constant direction. It was then assumed that we could find an operator representation of the form

\begin{aligned}\boldsymbol{\nabla}^2 = \frac{1}{{\mathbf{x}^2}} \left( (\mathbf{x} \cdot \boldsymbol{\nabla})^2 - \left\langle{{(\mathbf{x} \cdot \boldsymbol{\nabla})^2}}\right\rangle + f(\mathbf{x}, \boldsymbol{\nabla}) \right) \end{aligned} \quad\quad\quad(75)

Where $f(\mathbf{x}, \boldsymbol{\nabla})$ was to be determined, and was found by subtraction. Thinking ahead to relativistic applications this result was obtained for the n-dimensional Laplacian and was found to be

\begin{aligned}\nabla^2 &= \frac{1}{{x^2}} \left( (n-2 + x \cdot \nabla) (x \cdot \nabla) - \left\langle{{(x \wedge \nabla)^2}}\right\rangle \right) \end{aligned} \quad\quad\quad(76)

For the 3D case specifically this is

\begin{aligned}\boldsymbol{\nabla}^2 &= \frac{1}{{\mathbf{x}^2}} \left( (1 + \mathbf{x} \cdot \boldsymbol{\nabla}) (\mathbf{x} \cdot \boldsymbol{\nabla}) - \left\langle{{(\mathbf{x} \wedge \boldsymbol{\nabla})^2}}\right\rangle \right) \end{aligned} \quad\quad\quad(77)

While the scalar selection above is good for some purposes, it interferes with observations about simultaneous eigenfunctions for the angular momentum operator and the scalar part of its square as seen in the Laplacian. With some difficulty and tedium, by subtracting the bivector and quadvector grades from the squared angular momentum operator $(x \wedge \nabla)^2$ it was eventually found that (76) can be written as

\begin{aligned}\nabla^2 &= \frac{1}{{x^2}} \left( (n-2 + x \cdot \nabla) (x \cdot \nabla) + (n-2 - x \wedge \nabla) (x \wedge \nabla) + (x \wedge \nabla) \wedge (x \wedge \nabla) \right) \end{aligned} \quad\quad\quad(78)

In the 3D case the quadvector vanishes and (77) with the scalar selection removed is reduced to

\begin{aligned}\boldsymbol{\nabla}^2 &= \frac{1}{{\mathbf{x}^2}} \left( (1 + \mathbf{x} \cdot \boldsymbol{\nabla}) (\mathbf{x} \cdot \boldsymbol{\nabla}) + (1 - \mathbf{x} \wedge \boldsymbol{\nabla}) (\mathbf{x} \wedge \boldsymbol{\nabla}) \right) \end{aligned} \quad\quad\quad(79)

In 3D we also have the option of using the duality relation between the cross and the wedge $\mathbf{a} \wedge \mathbf{b} = i (\mathbf{a} \times \mathbf{b})$ to express the Laplacian

\begin{aligned}\boldsymbol{\nabla}^2 &= \frac{1}{{\mathbf{x}^2}} \left( (1 + \mathbf{x} \cdot \boldsymbol{\nabla}) (\mathbf{x} \cdot \boldsymbol{\nabla}) + (1 - i (\mathbf{x} \times \boldsymbol{\nabla})) i(\mathbf{x} \times \boldsymbol{\nabla}) \right) \end{aligned} \quad\quad\quad(80)

Since it is customary to express angular momentum as $\mathbf{L} = -i \hbar (\mathbf{x} \times \boldsymbol{\nabla})$, we see here that the imaginary in this context should perhaps necessarily be viewed as the spatial pseudoscalar. It was that guess that led down this path, and we come full circle back to this considering how to factor the Laplacian in vector quantities. Curiously this factorization is in no way specific to Quantum Theory.

A few verifications of the Laplacian in (80) were made. First it was shown that the directional derivative terms containing $\mathbf{x} \cdot \boldsymbol{\nabla}$, are equivalent to the radial terms of the Laplacian in spherical polar coordinates. That is

\begin{aligned}\frac{1}{{\mathbf{x}^2}} (1 + \mathbf{x} \cdot \boldsymbol{\nabla}) (\mathbf{x} \cdot \boldsymbol{\nabla}) \psi &= \frac{1}{{r}} \frac{\partial^2}{\partial r^2} (r\psi) \end{aligned} \quad\quad\quad(81)

Employing the quaternion operator for the spherical polar rotation

\begin{aligned}R &= e^{\mathbf{e}_{31}\theta/2} e^{\mathbf{e}_{12}\phi/2} \\ \mathbf{x} &= r \tilde{R} \mathbf{e}_3 R \end{aligned} \quad\quad\quad(82)

it was also shown that there was explicitly no radial dependence in the angular momentum operator which takes the form

\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) \\ &= \hat{\mathbf{r}} \left( \hat{\boldsymbol{\theta}} \partial_\theta + \hat{\boldsymbol{\phi}} \frac{1}{{\sin\theta}} \partial_\phi \right) \end{aligned} \quad\quad\quad(84)

Because there is a $\theta$, and $\phi$ dependence in the unit vectors $\hat{\mathbf{r}}$, $\hat{\boldsymbol{\theta}}$, and $\hat{\boldsymbol{\phi}}$, squaring the angular momentum operator in this form means that the unit vectors are also operated on. Those vectors were given by the triplet

\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(86)

Using $I = \mathbf{e}_1 \mathbf{e}_2 \mathbf{e}_3$ for the spatial pseudoscalar, and $i = \mathbf{e}_1 \mathbf{e}_2$ (a possibly confusing switch of notation) for the bivector of the x-y plane we can write the spherical polar unit vectors in exponential form as

\begin{aligned}\begin{pmatrix}\hat{\boldsymbol{\phi}} \\ \hat{\mathbf{r}} \\ \hat{\boldsymbol{\theta}} \\ \end{pmatrix}&=\begin{pmatrix}\mathbf{e}_2 e^{i\phi} \\ \mathbf{e}_3 e^{I \hat{\boldsymbol{\phi}} \theta} \\ i \hat{\boldsymbol{\phi}} e^{I \hat{\boldsymbol{\phi}} \theta} \\ \end{pmatrix} \end{aligned} \quad\quad\quad(87)

These or related expansions were used to verify (with some difficulty) that the scalar squared bivector operator is identical to the expected scalar spherical polar coordinates parts of the Laplacian

\begin{aligned}-\left\langle{{ (\mathbf{x} \wedge \boldsymbol{\nabla})^2 }}\right\rangle =\frac{1}{{\sin\theta}} \frac{\partial {}}{\partial {\theta}} \sin\theta \frac{\partial {}}{\partial {\theta}} + \frac{1}{{\sin^2\theta}} \frac{\partial^2}{\partial \phi^2} \end{aligned} \quad\quad\quad(88)

Additionally, by left or right dividing a unit bivector from the angular momentum operator, we are able to find that the raising and lowering operators are left as one of the factors

\begin{aligned}\mathbf{x} \wedge \boldsymbol{\nabla} &= \mathbf{e}_{31} \left( e^{i\phi} (\partial_\theta + i \cot\theta \partial_\phi) + \mathbf{e}_{13} \frac{1}{{i}} \partial_\phi \right) \\ \mathbf{x} \wedge \boldsymbol{\nabla} &= \left( e^{-i\phi} (\partial_\theta - i \cot\theta \partial_\phi) - \mathbf{e}_{13} \frac{1}{{i}} \partial_\phi \right) \mathbf{e}_{31} \end{aligned} \quad\quad\quad(89)

Both of these use $i = \mathbf{e}_1 \mathbf{e}_2$, the bivector for the plane, and not the spatial pseudoscalar. We are then able to see that in the context of the raising and lowering operator for the radial equation the interpretation of the imaginary should be one of a plane.

Using the raising operator factorization, it was calculated that $(\sin\theta)^\lambda e^{i\lambda \phi}$ was an eigenfunction of the bivector operator $\mathbf{x} \wedge \boldsymbol{\nabla}$ with eigenvalue $-\lambda$. This results in the simultaneous eigenvalue of $\lambda(\lambda + 1)$ for this eigenfunction with the scalar squared angular momentum operator.

There are a few things here that have not been explored to their logical conclusion.

The bivector Fourier projections $I \mathbf{e}_k (\mathbf{x} \wedge \boldsymbol{\nabla} ) \cdot (-I \mathbf{e}_k)$ do not obey the commutation relations of the scalar angular momentum components, so an attempt to directly use these to construct raising and lowering operators does not produce anything useful. The raising and lowering operators in a form that could be used to find eigensolutions were found by factoring out $\mathbf{e}_{13}$ from the bivector operator. Making this particular factorization was a fluke and only because it was desirable to express the bivector operator entirely in spherical polar form. It is curious that this results in raising and lowering operators for the x,y plane, and understanding this further would be nice.

In the eigen solutions for the bivector operator, no quantization condition was imposed. I don’t understand the argument that Bohm used to do so in the traditional treatment, and revisiting this once that is done is in order.

I am also unsure exactly how Bohm knows that the inner product for the eigenfunctions should be a surface integral. This choice works, but what drives it. Can that be related to anything here?