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## Bivector grades of the squared angular momentum operator.

Posted by peeterjoot on September 6, 2009

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

# Motivation

The aim here is to extract the bivector grades of the squared angular momentum operator

\begin{aligned}{\left\langle{{ (x \wedge \nabla)^2 }}\right\rangle}_{2} \stackrel{?}{=} \cdots \end{aligned} \quad\quad\quad(1)

I’d tried this before and believe gotten it wrong. Take it super slow and dumb and careful.

# Non-operator expansion.

Suppose $P$ is a bivector, $P = (\gamma^k \wedge \gamma^m) P_{km}$, the grade two product with a different unit bivector is

\begin{aligned}{\left\langle{{ (\gamma_a \wedge \gamma_b) (\gamma^k \wedge \gamma^m) }}\right\rangle}_{2} P_{km} &= {\left\langle{{ (\gamma_a \gamma_b - \gamma_a \cdot \gamma_b) (\gamma^k \wedge \gamma^m) }}\right\rangle}_{2} P_{km} \\ &= {\left\langle{{ \gamma_a (\gamma_b \cdot (\gamma^k \wedge \gamma^m)) }}\right\rangle}_{2} P_{km} + {\left\langle{{ \gamma_a (\gamma_b \wedge (\gamma^k \wedge \gamma^m)) }}\right\rangle}_{2} P_{km} - (\gamma_a \cdot \gamma_b) (\gamma^k \wedge \gamma^m) P_{km} \\ &= (\gamma_a \wedge \gamma^m) P_{b m} -(\gamma_a \wedge \gamma^k) P_{k b} - (\gamma_a \cdot \gamma_b) (\gamma^k \wedge \gamma^m) P_{km} \\ &+ (\gamma_a \cdot \gamma_b) (\gamma^k \wedge \gamma^m) P_{km} - (\gamma_b \wedge \gamma^m) P_{a m} + (\gamma_b \wedge \gamma^k) P_{k a} \\ &= (\gamma_a \wedge \gamma^c) (P_{b c} -P_{c b})+ (\gamma_b \wedge \gamma^c) (P_{c a} -P_{a c} ) \end{aligned}

This same procedure will be used for the operator square, but we have the complexity of having the second angular momentum operator change the first bivector result.

# Operator expansion.

In the first few lines of the bivector product expansion above, a blind replacement $\gamma_a \rightarrow x$, and $\gamma_b \rightarrow \nabla$ gives us

\begin{aligned}{\left\langle{{ (x \wedge \nabla) (\gamma^k \wedge \gamma^m) }}\right\rangle}_{2} P_{km} &= {\left\langle{{ (x \nabla - x \cdot \nabla) (\gamma^k \wedge \gamma^m) }}\right\rangle}_{2} P_{km} \\ &= {\left\langle{{ x (\nabla \cdot (\gamma^k \wedge \gamma^m)) }}\right\rangle}_{2} P_{km} + {\left\langle{{ x (\nabla \wedge (\gamma^k \wedge \gamma^m)) }}\right\rangle}_{2} P_{km} - (x \cdot \nabla) (\gamma^k \wedge \gamma^m) P_{km} \end{aligned}

Using $P_{km} = x_k \partial_m$, eliminating the coordinate expansion we have an intermediate result that gets us partway to the desired result

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

An expansion of the first term should be easier than the second. Dropping back to coordinates we have

\begin{aligned}{\left\langle{{ x (\nabla \cdot (x \wedge \nabla)) }}\right\rangle}_{2} &={\left\langle{{ x (\nabla \cdot (\gamma^k \wedge \gamma^m)) }}\right\rangle}_{2} x_k \partial_m \\ &={\left\langle{{ x (\gamma_a \partial^a \cdot (\gamma^k \wedge \gamma^m)) }}\right\rangle}_{2} x_k \partial_m \\ &={\left\langle{{ x \gamma^m \partial^k }}\right\rangle}_{2} x_k \partial_m -{\left\langle{{ x \gamma^k \partial^m }}\right\rangle}_{2} x_k \partial_m \\ &=x \wedge (\partial^k x_k \gamma^m \partial_m )- x \wedge (\partial^m \gamma^k x_k \partial_m ) \end{aligned}

Okay, a bit closer. Backpedaling with the reinsertion of the complete vector quantities we have

\begin{aligned}{\left\langle{{ x (\nabla \cdot (x \wedge \nabla)) }}\right\rangle}_{2} &= x \wedge (\partial^k x_k \nabla ) - x \wedge (\partial^m x \partial_m ) \end{aligned} \quad\quad\quad(3)

Expanding out these two will be conceptually easier if the functional operation is made explicit. For the first

\begin{aligned}x \wedge (\partial^k x_k \nabla ) \phi&=x \wedge x_k \partial^k (\nabla \phi)+x \wedge ((\partial^k x_k) \nabla) \phi \\ &=x \wedge ((x \cdot \nabla) (\nabla \phi))+ n (x \wedge \nabla) \phi \end{aligned}

In operator form this is

\begin{aligned}x \wedge (\partial^k x_k \nabla ) &= n (x \wedge \nabla) + x \wedge ((x \cdot \nabla) \nabla ) \end{aligned} \quad\quad\quad(4)

Now consider the second half of (3). For that we expand

\begin{aligned}x \wedge (\partial^m x \partial_m ) \phi&=x \wedge (x \partial_m \partial^m \phi)+ x \wedge ((\partial^m x) \partial_m \phi) \end{aligned}

Since $x \wedge x = 0$, and $\partial^m x = \partial^m x_k \gamma^k = \gamma^m$, we have

\begin{aligned}x \wedge (\partial^m x \partial_m ) \phi&=x \wedge (\gamma^m \partial_m ) \phi \\ &=(x \wedge \nabla) \phi \end{aligned}

Putting things back together we have for (3)

\begin{aligned}{\left\langle{{ x (\nabla \cdot (x \wedge \nabla)) }}\right\rangle}_{2} &= (n-1) (x \wedge \nabla) + x \wedge ((x \cdot \nabla) \nabla ) \end{aligned} \quad\quad\quad(5)

This now completes a fair amount of the bivector selection, and a substitution back into (2) yields

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

The remaining task is to explicitly expand the last vector-trivector dot product. To do that we use the basic alternation expansion identity

\begin{aligned}a \cdot (b \wedge c \wedge d)&= (a \cdot b) (c \wedge d)-(a \cdot c) (b \wedge d)+(a \cdot d) (b \wedge c) \end{aligned} \quad\quad\quad(7)

To see how to apply this to the operator case lets write that explicitly but temporarily in coordinates

\begin{aligned}x \cdot ((\nabla \wedge (x \wedge \nabla)) \phi&=(x^\mu \gamma_\mu) \cdot ((\gamma^\nu \partial_\nu ) \wedge (x_\alpha \gamma^\alpha \wedge (\gamma^\beta \partial_\beta))) \phi \\ &=x \cdot \nabla (x \wedge \nabla) \phi-x \cdot \gamma^\alpha \nabla \wedge x_\alpha \nabla \phi+x^\mu \nabla \wedge x \gamma_\mu \cdot \gamma^\beta \partial_\beta \phi \\ &=x \cdot \nabla (x \wedge \nabla) \phi-x^\alpha \nabla \wedge x_\alpha \nabla \phi+x^\mu \nabla \wedge x \partial_\mu \phi \end{aligned}

Considering this term by term starting with the second one we have

\begin{aligned}x^\alpha \nabla \wedge x_\alpha \nabla \phi&=x_\alpha (\gamma^\mu \partial_\mu) \wedge x^\alpha \nabla \phi \\ &=x_\alpha \gamma^\mu \wedge (\partial_\mu x^\alpha) \nabla \phi +x_\alpha \gamma^\mu \wedge x^\alpha \partial_\mu \nabla \phi \\ &=x_\mu \gamma^\mu \wedge \nabla \phi +x_\alpha x^\alpha \gamma^\mu \wedge \partial_\mu \nabla \phi \\ &=x \wedge \nabla \phi +x^2 \nabla \wedge \nabla \phi \end{aligned}

The curl of a gradient is zero, since summing over an product of antisymmetric and symmetric indexes $\gamma^\mu \wedge \gamma^\nu \partial_{\mu\nu}$ is zero. Only one term remains to evaluate in the vector-trivector dot product now

\begin{aligned}x \cdot (\nabla \wedge x \wedge \nabla) &=(-1 + x \cdot \nabla )(x \wedge \nabla) +x^\mu \nabla \wedge x \partial_\mu \end{aligned} \quad\quad\quad(8)

Again, a completely dumb and brute force expansion of this is

\begin{aligned}x^\mu \nabla \wedge x \partial_\mu \phi&=x^\mu (\gamma^\nu \partial_\nu) \wedge (x^\alpha \gamma_\alpha) \partial_\mu \phi \\ &=x^\mu \gamma^\nu \wedge (\partial_\nu (x^\alpha \gamma_\alpha)) \partial_\mu \phi +x^\mu \gamma^\nu \wedge (x^\alpha \gamma_\alpha) \partial_\nu \partial_\mu \phi \\ &=x^\mu (\gamma^\alpha \wedge \gamma_\alpha) \partial_\mu \phi +x^\mu \gamma^\nu \wedge x \partial_\nu \partial_\mu \phi \end{aligned}

With $\gamma^\mu = \pm \gamma_\mu$, the wedge in the first term is zero, leaving

\begin{aligned}x^\mu \nabla \wedge x \partial_\mu \phi&=-x^\mu x \wedge \gamma^\nu \partial_\nu \partial_\mu \phi \\ &=-x^\mu x \wedge \gamma^\nu \partial_\mu \partial_\nu \phi \\ &=-x \wedge x^\mu \partial_\mu \gamma^\nu \partial_\nu \phi \end{aligned}

In vector form we have finally

\begin{aligned}x^\mu \nabla \wedge x \partial_\mu \phi &= -x \wedge (x \cdot \nabla) \nabla \phi \end{aligned} \quad\quad\quad(9)

The final expansion of the vector-trivector dot product is now

\begin{aligned}x \cdot (\nabla \wedge x \wedge \nabla) &=(-1 + x \cdot \nabla )(x \wedge \nabla) -x \wedge (x \cdot \nabla) \nabla \phi \end{aligned} \quad\quad\quad(10)

This was the last piece we needed for the bivector grade selection. Incorporating this into (6), both the $x \cdot \nabla x \wedge \nabla$, and the $x \wedge (x \cdot \nabla) \nabla$ terms cancel leaving the surprising simple result

\begin{aligned}{\left\langle{{ (x \wedge \nabla)^2 }}\right\rangle}_{2}&=(n-2) (x \wedge \nabla) \end{aligned} \quad\quad\quad(11)

The power of this result is that it allows us to write the scalar angular momentum operator from the Laplacian as

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

The complete Laplacian is

\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 - (n-2) (x \wedge \nabla) - (x \wedge \nabla) \wedge (x \wedge \nabla) \right) \end{aligned} \quad\quad\quad(12)

In particular in less than four dimensions the quad-vector term is necessarily zero. The 3D Laplacian becomes

\begin{aligned}\boldsymbol{\nabla}^2 &= \frac{1}{{\mathbf{x}^2}} (1 + \mathbf{x} \cdot \boldsymbol{\nabla})(\mathbf{x} \cdot \boldsymbol{\nabla})+ \frac{1}{{\mathbf{x}^2}} (1 - \mathbf{x} \wedge \boldsymbol{\nabla}) (\mathbf{x} \wedge \boldsymbol{\nabla}) \end{aligned} \quad\quad\quad(13)

So any eigenfunction of the bivector angular momentum operator $\mathbf{x} \wedge \boldsymbol{\nabla}$ is necessarily a simultaneous eigenfunction of the scalar operator.