Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Bivector form of quantum angular momentum operator (On the bivector and quadvector components of the squared angular momentum operator.)

Posted by peeterjoot on August 29, 2009

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

We found that we could by something similar to “factoring” the gradient into radial and non-radial terms express the Laplacian as

\begin{aligned}\nabla^2 &= \frac{1}{{x^2}} (x \cdot \nabla)^2 - \frac{1}{{x^2}} \left\langle{{(x \wedge \nabla)^2}}\right\rangle + (n - 2) \frac{1}{{x}} \cdot \nabla  \end{aligned} \quad\quad\quad(3)

The requirement for a scalar selection on all the (x \wedge \nabla)^2 terms is a bit ugly, but omitting it would be incorrect for two reasons. One reason is that this is a bivector operator and not a bivector (where the squaring operates on itself). The other is that we derived a result for arbitrary dimension, and the product of two bivectors in a general space has grade 2 and grade 4 terms in addition to the scalar terms. Without taking only the scalar parts, lets expand this product a bit more carefully, starting with

\begin{aligned}(x \wedge \nabla)^2 &=(\gamma_\mu \wedge \gamma_\nu) (\gamma^\alpha \wedge \gamma^\beta)x^\mu \partial^\nu x_\alpha \partial_\beta \end{aligned}

Just expanding the multivector factor for now, we have

\begin{aligned}2 (\gamma_\mu \wedge \gamma_\nu) (\gamma^\alpha \wedge \gamma^\beta)&=\gamma_\mu \gamma_\nu (\gamma^\alpha \wedge \gamma^\beta)- \gamma_\nu \gamma_\mu (\gamma^\alpha \wedge \gamma^\beta) \\ &=\gamma_\mu \left( {\delta_\nu}^\alpha \gamma^\beta - {\delta_\nu}^\beta \gamma^\alpha+ \gamma_\nu \wedge \gamma^\alpha \wedge \gamma^\beta \right)- \gamma_\nu \left({\delta_\mu}^\alpha \gamma^\beta - {\delta_\mu}^\beta \gamma^\alpha+ \gamma_\mu \wedge \gamma^\alpha \wedge \gamma^\beta \right)\\ &={\delta_\nu}^\alpha {\delta_\mu}^\beta - {\delta_\nu}^\beta {\delta_\mu}^\alpha-{\delta_\mu}^\alpha {\delta_\nu}^\beta + {\delta_\mu}^\beta {\delta_\nu}^\alpha \\ &+ \gamma_\mu \wedge \gamma_\nu \wedge \gamma^\alpha \wedge \gamma^\beta - \gamma_\nu \wedge \gamma_\mu \wedge \gamma^\alpha \wedge \gamma^\beta \\ &+ \gamma_\mu \cdot (\gamma_\nu \wedge \gamma^\alpha \wedge \gamma^\beta )-\gamma_\nu \cdot (\gamma_\mu \wedge \gamma^\alpha \wedge \gamma^\beta ) \\  \end{aligned}

Our split into grades for this operator is then, the scalar

\begin{aligned}\left\langle{{(x \wedge \nabla)^2 }}\right\rangle &= (x \wedge \nabla) \cdot (x \wedge \nabla) \\ &= \left( {\delta_\nu}^\alpha {\delta_\mu}^\beta - {\delta_\nu}^\beta {\delta_\mu}^\alpha \right)x^\mu \partial^\nu x_\alpha \partial_\beta \\  \end{aligned}

the pseudoscalar (or grade 4 term in higher than 4D spaces).

\begin{aligned}{\left\langle{{(x \wedge \nabla)^2 }}\right\rangle}_{4} &= (x \wedge \nabla) \wedge (x \wedge \nabla) \\ &= \left( \gamma_\mu \wedge \gamma_\nu \wedge \gamma^\alpha \wedge \gamma^\beta  \right)x^\mu \partial^\nu x_\alpha \partial_\beta \\  \end{aligned}

If we work in dimensions less than or equal to three, we’ll have no grade four term since this wedge product is zero (irrespective of the operator action), so in 3D we have only a bivector term in excess of the scalar part of this operator.

The bivector term deserves some reduction. As expanded so far it is

\begin{aligned}{\left\langle{{(x \wedge \nabla)^2 }}\right\rangle}_{2} &= \frac{1}{{2}} \left(  \gamma_\mu \cdot (\gamma_\nu \wedge \gamma^\alpha \wedge \gamma^\beta )-\gamma_\nu \cdot (\gamma_\mu \wedge \gamma^\alpha \wedge \gamma^\beta ) \right)x^\mu \partial^\nu x_\alpha \partial_\beta \end{aligned}

Lets expand this bivector term more completely.

\begin{aligned}2 {\left\langle{{(x \wedge \nabla)^2 }}\right\rangle}_{2} &=\gamma_\mu \cdot \gamma_\nu (\gamma^\alpha \wedge \gamma^\beta ) x^\mu \partial^\nu x_\alpha \partial_\beta-\gamma_\mu \cdot \gamma^\alpha (\gamma_\nu \wedge \gamma^\beta ) x^\mu \partial^\nu x_\alpha \partial_\beta+\gamma_\mu \cdot \gamma^\beta (\gamma_\nu \wedge \gamma^\alpha ) x^\mu \partial^\nu x_\alpha \partial_\beta \\ &- \gamma_\nu \cdot \gamma_\mu ( \gamma^\alpha \wedge \gamma^\beta ) x^\mu \partial^\nu x_\alpha \partial_\beta+ \gamma_\nu \cdot \gamma_\alpha ( \gamma_\mu \wedge \gamma^\beta ) x^\mu \partial^\nu x_\alpha \partial_\beta- \gamma_\nu \cdot \gamma_\beta ( \gamma_\mu \wedge \gamma^\alpha ) x^\mu \partial^\nu x_\alpha \partial_\beta \\ &=(\gamma^\alpha \wedge \gamma^\beta ) x^\mu \partial_\mu x_\alpha \partial_\beta-(\gamma_\nu \wedge \gamma^\beta ) x^\mu \partial^\nu x_\mu \partial_\beta+(\gamma_\nu \wedge \gamma^\alpha ) x^\mu \partial^\nu x_\alpha \partial_\mu \\ &- ( \gamma^\alpha \wedge \gamma^\beta ) x^\mu \partial_\mu x_\alpha \partial_\beta+ ( \gamma_\mu \wedge \gamma^\beta ) x^\mu \partial^\nu x_\nu \partial_\beta- ( \gamma_\mu \wedge \gamma^\alpha ) x^\mu \partial^\nu x_\alpha \partial_\nu \\  \end{aligned}

The first and fourth terms cancel, and a change of dummy indexes on the remainder, eliminating \beta, gives us

\begin{aligned}2 {\left\langle{{(x \wedge \nabla)^2 }}\right\rangle}_{2} &=-(\gamma_\nu \wedge \gamma^\alpha ) x^\mu \partial^\nu x_\mu \partial_\alpha+(\gamma_\nu \wedge \gamma^\alpha ) x^\mu \partial^\nu x_\alpha \partial_\mu + ( \gamma_\mu \wedge \gamma^\alpha ) x^\mu \partial^\nu x_\nu \partial_\alpha- ( \gamma_\mu \wedge \gamma^\alpha ) x^\mu \partial^\nu x_\alpha \partial_\nu \\ &=(\gamma_\nu \wedge \gamma^\alpha ) x^\mu \partial^\nu (x_\alpha \partial_\mu - x_\mu \partial_\alpha)+( \gamma_\mu \wedge \gamma^\alpha ) x^\mu \partial^\nu (x_\nu \partial_\alpha - x_\alpha \partial_\nu) \\ &=(\gamma_\nu \wedge \gamma^\alpha ) x^\mu (x_\alpha \partial^\nu \partial_\mu - x_\mu \partial^\nu \partial_\alpha)+( \gamma_\mu \wedge \gamma^\alpha ) x^\mu (x_\nu \partial^\nu \partial_\alpha - x_\alpha \partial^\nu \partial_\nu) \\ &+(\gamma_\nu \wedge \gamma^\nu ) x^\mu \partial_\mu -(\gamma_\mu \wedge \gamma^\alpha ) x^\mu \partial_\alpha+n ( \gamma_\mu \wedge \gamma^\alpha ) x^\mu \partial_\alpha -( \gamma_\mu \wedge \gamma^\nu ) x^\mu \partial_\nu \\ &=(\gamma_\nu \wedge \gamma^\alpha ) x^\mu x_\alpha \partial^\nu \partial_\mu +( \gamma_\mu \wedge \gamma^\alpha ) x^\mu x_\nu \partial^\nu \partial_\alpha -(\gamma_\mu \wedge \gamma^\alpha ) ( x^2 \partial^\mu \partial_\alpha + x^\mu x_\alpha \nabla^2 ) \\ &+(n-1)(\gamma_\mu \wedge \gamma^\alpha ) x^\mu \partial_\alpha-( \gamma_\mu \wedge \gamma^\nu ) x^\mu \partial_\nu \\ &=(\gamma_\nu \wedge \gamma^\mu ) ( x^\alpha x_\mu \partial^\nu \partial_\alpha + x^\nu x_\alpha \partial^\alpha \partial_\mu )- ( x^2 (\nabla \wedge \nabla) + ( x \wedge x ) \nabla^2 ) +(n-2) x \wedge \nabla\\  \end{aligned}

The self wedges for both \nabla and x are zero, and we are left with

\begin{aligned}2 {\left\langle{{(x \wedge \nabla)^2 }}\right\rangle}_{2} &=(\gamma_\nu \wedge \gamma^\mu ) ( x_\mu (x \cdot \nabla) \partial^\nu + x^\nu (x \cdot \nabla) \partial_\mu )+(n-2) x \wedge \nabla\\  \end{aligned}

Swapping \mu with \nu and raising and lowering we have exact cancellation of the first two terms

\begin{aligned}2 {\left\langle{{(x \wedge \nabla)^2 }}\right\rangle}_{2} &=(\gamma_\nu \wedge \gamma^\mu ) x_\mu (x \cdot \nabla) \partial^\nu (\gamma^\mu \wedge \gamma_\nu ) x_\mu (x \cdot \nabla) \partial^\nu +(n-2) x \wedge \nabla\\  \end{aligned}

So are left, almost mystically simplified, with just

\begin{aligned}{\left\langle{{(x \wedge \nabla)^2 }}\right\rangle}_{2} &=\frac{n-2}{2} x \wedge \nabla \end{aligned} \quad\quad\quad(4)

We can now write for the squared operator

\begin{aligned}(x \wedge \nabla)^2 &=\frac{n-2}{2} (x \wedge \nabla)+(x \wedge \nabla) \wedge (x \wedge \nabla) +(x \wedge \nabla) \cdot (x \wedge \nabla)  \end{aligned} \quad\quad\quad(5)

and then eliminate the scalar selection from the (3)

\begin{aligned}\nabla^2 &= \frac{1}{{x^2}} (x \cdot \nabla)^2 + (n - 2) \frac{1}{{x}} \cdot \nabla - \frac{1}{{x^2}} \left((x \wedge \nabla)^2 - \frac{n-2}{2} (x \wedge \nabla) - (x \wedge \nabla) \wedge (x \wedge \nabla) \right) \end{aligned} \quad\quad\quad(6)

In 3D this is

\begin{aligned}\boldsymbol{\nabla}^2 &= \frac{1}{{\mathbf{x}^2}} (\mathbf{x} \cdot \boldsymbol{\nabla})^2 + \frac{1}{{\mathbf{x}}} \cdot \boldsymbol{\nabla} - \frac{1}{{\mathbf{x}^2}} \left( \mathbf{x} \wedge \boldsymbol{\nabla} - \frac{1}{{2}} \right) (\mathbf{x} \wedge \boldsymbol{\nabla})  \end{aligned} \quad\quad\quad(7)

Wow, that was an ugly mess of algebra, and presented lots of potential possibilities for error. There is likely a more clever coordinate free way to do the same expansion, and some verification of correctness of this result seems justified. I suspect that the factor of 1/2 will allow for verification by considering of the Quantum eigenvalue problem, and simultaneous eigenvalues of \mathbf{x} \wedge \boldsymbol{\nabla}, and \left\langle{{\mathbf{x} \wedge \boldsymbol{\nabla})^2}}\right\rangle.

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