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Bivector form of quantum angular momentum operator (3D Quantum Hamiltonian)

Posted by peeterjoot on August 29, 2009

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

Going back to the quantum Hamiltonian we do still have the angular momentum operator as one of the distinct factors of the Laplacian. As operators we have something akin to the projection of the gradient onto the radial direction, as well as terms that project the gradient onto the tangential plane to the sphere at the radial point

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

Using the result of (7) and the radial formulation for the rest, we can write this

\begin{aligned}0 &= \left( \boldsymbol{\nabla}^2 - \frac{2m}{\hbar^2} (V - E) \right) \psi \\ &= \frac{1}{{r}}\frac{\partial}{\partial r} r \frac{\partial \psi}{\partial r}- \frac{1}{{r^2}} \left( \mathbf{x} \wedge \boldsymbol{\nabla} - \frac{1}{{2}} \right) (\mathbf{x} \wedge \boldsymbol{\nabla}) \psi - \frac{2m}{\hbar^2} (V - E) \psi \\ \end{aligned}

If $V = V(r)$, then a radial split by separation of variables is possible. Writing $\psi = R(r) Y$, we get

\begin{aligned}\frac{r}{R} \frac{\partial}{\partial r} r \frac{\partial R}{\partial r} - \frac{2m r^2}{\hbar^2} (V(r) - E) = \frac{1}{{Y}} \left( \mathbf{x} \wedge \boldsymbol{\nabla} - \frac{1}{{2}} \right) (\mathbf{x} \wedge \boldsymbol{\nabla}) Y = \text{constant} \end{aligned} \quad\quad\quad(49)

For the constant, lets use $c$, and split this into a pair of equations

\begin{aligned}r \frac{\partial}{\partial r} r \frac{\partial R}{\partial r} - \frac{2m r^2 R}{\hbar^2} (V(r) - E) = c R \end{aligned} \quad\quad\quad(50)

\begin{aligned}\left( \mathbf{x} \wedge \boldsymbol{\nabla} - \frac{1}{{2}} \right) (\mathbf{x} \wedge \boldsymbol{\nabla}) Y &= c Y \end{aligned} \quad\quad\quad(51)

In this last we can examine simultaneous eigenvalues of $\mathbf{x} \wedge \boldsymbol{\nabla}$, and $\left\langle{{(\mathbf{x} \wedge \boldsymbol{\nabla})^2}}\right\rangle$. Suppose that
$Y_\lambda$ is an eigenfunction of $\mathbf{x} \wedge \boldsymbol{\nabla}$ with eigenvalue $\lambda$. We then have

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

We see immediately that $Y_\lambda$ is then also an eigenfunction of $\left\langle{{(\mathbf{x} \wedge \boldsymbol{\nabla})^2}}\right\rangle$, with eigenvalue
$\lambda \left(\lambda - \frac{1}{{2}} \right)$.

Now, I was hoping to use an argument like this to confirm or deny the correctness of the $1/2$, by comparing to the scalar coordinate results in Bohm. Looking again though, I see this is something entirely different. There are three eigenvalue results. One for individual simultaneous eigenfunctions of $L_x, L_y$, or $L_z$ with $L^2$ (with matching eigenvalues), and the other was for the raising and lowering operators, $L_x \pm iL_y$ also eigenfunctions of $L^2$, but having $m(m \pm 1)$ eigenvalues.

So, if there is an error in the messy algebraic split of $(\mathbf{x} \wedge \boldsymbol{\nabla})^2$ into its scalar and bivector components, I’ll assume that this error is only a scalar factor. If there is no such numeric error, and even if there is, it appears the next order of business is figuring out how to solve the multivector eigenvalue problem

\begin{aligned}(\mathbf{x} \wedge \boldsymbol{\nabla}) Y_\lambda = \lambda Y_\lambda \end{aligned} \quad\quad\quad(52)