User:RPFB Maxwell Dirac/sandbox

TITLE: An Eight-Dimensional Matrix Representation of the Maxwell Field Equations and the Dirac Field Equations

The $8 × 8$ Spacetime Matrix Operator
The $8 × 8$ spacetime matrix operator $$\hat{M}$$ is defined by the following equation



\begin{align} \hat{M} \equiv \begin{pmatrix} -\partial_4 & 0 & 0 & 0 & 0 & -\partial_3 & +\partial_2 & -\partial_1\\ 0 & -\partial_4 & 0 & 0 & +\partial_3 & 0 & -\partial_1 & -\partial_2\\ 0 & 0 & -\partial_4 & 0 & -\partial_2 & +\partial_1 & 0 & -\partial_3\\ 0 & 0 & 0 & -\partial_4 & +\partial_1 & +\partial_2 & +\partial_3 & 0\\ 0 & +\partial_3 & -\partial_2 & +\partial_1 & +\partial_4 & 0 & 0 & 0\\ -\partial_3 & 0 & +\partial_1 & +\partial_2 & 0 & +\partial_4 & 0 & 0\\ +\partial_2 & -\partial_1 & 0 & +\partial_3 & 0 & 0 & +\partial_4 & 0\\ -\partial_1 & -\partial_2 & -\partial_3 & 0 & 0 & 0 & 0 & +\partial_4 \end{pmatrix} \end{align} $$

where the partial derivative symbols appearing in this matrix operator are defined by


 * $$\begin{align}

\partial_1 \equiv \frac{\partial}{\partial x} && \partial_2 \equiv \frac{\partial}{\partial y} && \partial_3 \equiv \frac{\partial}{\partial z} && \partial_4 \equiv \frac{\partial}{ic\partial t}. \end{align}$$

The symbol $$c$$ represents the speed of light in vacuum and the imaginary unit $$i$$ represents the positive square-root of minus one.

The spacetime matrix operator $$\hat{M}$$ may also be expressed as



\begin{align} \hat{M} = M_1\partial_1 + M_2\partial_2 + M_3\partial_3 + M_4\partial_4 \end{align} $$

where the following four $8 × 8$ matrices $$ M_\mu$$ for $$\mu$$ = 1, 2, 3, 4 are defined by



\begin{align} M_1 \equiv \begin{pmatrix} O & O & O & -\sigma_1\\ O & O & +\sigma_1 & O\\ O & +\sigma_1 & O & O\\ -\sigma_1 & O & O & O\end{pmatrix} \quad M_2 \equiv \begin{pmatrix} O & O & O & +\sigma_3\\ O & O & -\sigma_3 & O\\ O & -\sigma_3 & O & O\\ +\sigma_3 & O & O & O\end{pmatrix} \quad M_3 \equiv \begin{pmatrix} O & O & -i\sigma_2 & O\\ O & O & O & -i\sigma_2\\ +i\sigma_2 & O & O & O\\ O & +i\sigma_2 & O & O\end{pmatrix} \quad M_4 \equiv \begin{pmatrix} -I_2 & O & O & O\\ O & -I_2 & O & O\\ O & O & +I_2 & O\\ O & O & O & +I_2\end{pmatrix} \end{align} $$.

The $2 × 2$ matrices $$\sigma_1, \sigma_2, \sigma_3$$ are the Pauli matrices. The symbol $$O$$ represents the $2 × 2$ zero matrix and $$I_2$$ represents $2 × 2$ identity matrix.

Each matrix $$M_\mu$$ is a Hermitian matrix, that is


 * $$\begin{align} M_\mu = M^H_\mu \end{align}$$.

In addition, each matrix $$M_\mu$$ is equal to its own matrix inverse, namely


 * $$\begin{align} M_\mu = M^{-1}_\mu \end{align}$$.

These matrices also satisfy the anticommutator property



\begin{align} M_\mu M_\nu + M_\nu M_\mu = 2 \delta_{\mu \nu} I_8 \end{align} $$

where $$\delta_{\mu \nu}$$ represents the Kronecker delta and $$ I_8 $$ corresponds to the $8 × 8$ identity matrix.

It can be easily shown that the matrix product of the $8 × 8$ spacetime operator $$\hat{M}$$ with itself gives



\begin{align} \hat{M} \hat{M} = I_8\Box \end{align} $$

where the d'Alembert operator is defined by


 * $$\begin{align}

\Box \equiv {\partial_1^2} + {\partial_2^2} + {\partial_3^2} + {\partial_4^2} = \nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \end{align}$$

and $$\nabla^2$$ is the Laplace operator.

Maxwell's Equations
The symmetrical form of the four Maxwell's equations is given by

Ampere-Maxwell law

 * $$\begin{align}

\nabla \times \mathbf{B} - \frac{1}{c} \frac{\partial\mathbf E}{\partial t} &= + \frac{4\pi}{c}\mathbf{J_e} \end{align}$$

Gauss' law of magnetism

 * $$\begin{align}

\nabla \cdot \mathbf{B} &= +4{\pi}{\rho_m} \end{align}$$

Faraday's law of induction

 * $$\begin{align}

\nabla \times \mathbf{E} + \frac{1}{c} \frac{\partial\mathbf B}{\partial t} &= - \frac{4\pi}{c}\mathbf{J_m} \end{align}$$

Gauss' law of electricity

 * $$\begin{align}

\nabla \cdot \mathbf{E} &= +4{\pi}{\rho_e} \end{align}$$

Eight-Dimensional Maxwell Equation

 * $$\begin{align} \end{align}$$



\begin{align} \begin{pmatrix} -\partial_4 & 0 & 0 & 0 & 0 & -\partial_3 & +\partial_2 & -\partial_1\\ 0 & -\partial_4 & 0 & 0 & +\partial_3 & 0 & -\partial_1 & -\partial_2\\ 0 & 0 & -\partial_4 & 0 & -\partial_2 & +\partial_1 & 0 & -\partial_3\\ 0 & 0 & 0 & -\partial_4 & +\partial_1 & +\partial_2 & +\partial_3 & 0\\ 0 & +\partial_3 & -\partial_2 & +\partial_1 & +\partial_4 & 0 & 0 & 0\\ -\partial_3 & 0 & +\partial_1 & +\partial_2 & 0 & +\partial_4 & 0 & 0\\ +\partial_2 & -\partial_1 & 0 & +\partial_3 & 0 & 0 & +\partial_4 & 0\\ -\partial_1 & -\partial_2 & -\partial_3 & 0 & 0 & 0 & 0 & +\partial_4 \end{pmatrix}

\begin{pmatrix} iE_1\\ iE_2\\  iE_3\\  0\\  B_1\\  B_2\\  B_3\\ 0 \end{pmatrix} = \frac{4 \pi}{c}\begin{pmatrix} J_{e1}\\ J_{e2}\\ J_{e3}\\ c\rho_m\\ iJ_{m1}\\ iJ_{m2}\\ iJ_{m3}\\ -ic\rho_e \end{pmatrix} \end{align} $$

In compact matrix form


 * $$\begin{align}

\hat{M} |f \rangle = |j \rangle \end{align}$$

Dirac Equation
One symmetrical form of Dirac equation is the Dirac representation given by



\begin{pmatrix} +\partial_4 & 0 & -i\partial_3 & -\partial_2 - i\partial_1\\ 0 & +\partial_4 & +\partial_2 - i\partial_1 & +i\partial_3\\ +i\partial_3 & +\partial_2 + i\partial_1 & -\partial_4 & 0\\ -\partial_2 + i\partial_1 & -i\partial_3 & 0 & -\partial_4 \end{pmatrix}

\begin{pmatrix} \psi_1\\ \psi_2\\ \psi_3\\ \psi_4 \end{pmatrix} + \frac{m_o c}{\hbar}\begin{pmatrix} \psi_1\\ \psi_2\\ \psi_3\\ \psi_4 \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0\\ 0 \end{pmatrix} $$

or equivalently



\hat{\Gamma}|\psi \rangle + \kappa|\psi \rangle = |o \rangle $$



\kappa \equiv \frac{m_o c}{\hbar} $$.

The $4 × 4$ Dirac Matrix Operator
The Dirac matrix operator $$\hat{\Gamma}$$ may also be expressed as


 * $$\begin{align}

\hat{\Gamma} = \gamma_1\partial_1 + \gamma_2\partial_2 + \gamma_3\partial_3 + \gamma_4\partial_4 \end{align}$$

The four $4 × 4$ matrices $$ \gamma_\mu$$ for $$\mu$$ = 1, 2, 3, 4 are known as the Dirac Matrices. For the Dirac representation these four matrices are defined by


 * $$\begin{align}

\gamma_1 \equiv \begin{pmatrix} O & -i\sigma_1\\ +i\sigma_1 & O\end{pmatrix} \quad \gamma_2 \equiv \begin{pmatrix} O & -i\sigma_2\\ +i\sigma_2 & O\end{pmatrix} \quad \gamma_3 \equiv \begin{pmatrix} O & -i\sigma_3\\ +i\sigma_3 & O\end{pmatrix} \quad \gamma_4 \equiv \begin{pmatrix} +I_2 & O\\ O & -I_2\end{pmatrix} \end{align}$$.

Each matrix $$\gamma_\mu$$ is a Hermitian matrix, that is


 * $$\begin{align} \gamma_\mu = \gamma^H_\mu \end{align}$$.

In addition, each matrix $$\gamma_\mu$$ is equal to its own matrix inverse, namely


 * $$\begin{align} \gamma_\mu = \gamma^{-1}_\mu \end{align}$$.

These matrices also satisfy the anticommutator property


 * $$\begin{align}

\gamma_\mu \gamma_\nu + \gamma_\nu \gamma_\mu = 2 \delta_{\mu \nu} I_4 \end{align}$$

where $$\delta_{\mu \nu}$$ represents the Kronecker delta and $$ I_4 $$ corresponds to the $4 × 4$ identity matrix.

It can be easily shown that the matrix product of the $4 × 4$ Dirac matrix operator $$\hat{\Gamma}$$ with itself gives

\hat{\Gamma} \hat{\Gamma} = I_4\Box $$

where the d'Alembert operator is defined by


 * $$\begin{align}

\Box \equiv {\partial_1^2} + {\partial_2^2} + {\partial_3^2} + {\partial_4^2} = \nabla^2 - \frac{1}{c^2} \frac{\partial^2}{\partial t^2} \end{align}$$

and $$\nabla^2$$ is the Laplace operator.

Eight-Dimensional Dirac Equation

 * $$\begin{align} \end{align}$$



\begin{pmatrix} -\partial_4 & 0 & 0 & 0 & 0 & -\partial_3 & +\partial_2 & -\partial_1\\ 0 & -\partial_4 & 0 & 0 & +\partial_3 & 0 & -\partial_1 & -\partial_2\\ 0 & 0 & -\partial_4 & 0 & -\partial_2 & +\partial_1 & 0 & -\partial_3\\ 0 & 0 & 0 & -\partial_4 & +\partial_1 & +\partial_2 & +\partial_3 & 0\\ 0 & +\partial_3 & -\partial_2 & +\partial_1 & +\partial_4 & 0 & 0 & 0\\ -\partial_3 & 0 & +\partial_1 & +\partial_2 & 0 & +\partial_4 & 0 & 0\\ +\partial_2 & -\partial_1 & 0 & +\partial_3 & 0 & 0 & +\partial_4 & 0\\ -\partial_1 & -\partial_2 & -\partial_3 & 0 & 0 & 0 & 0 & +\partial_4 \end{pmatrix}

\begin{pmatrix} \phi_1\\ \phi_2\\  \phi_3\\  \phi_4\\  \phi_5\\  \phi_6\\  \phi_7\\ \phi_8 \end{pmatrix} + \frac{m_o c}{\hbar}\begin{pmatrix} \phi_1\\ \phi_2\\  \phi_3\\  \phi_4\\  \phi_5\\  \phi_6\\  \phi_7\\ \phi_8 \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0 \end{pmatrix} $$

In compact matrix form


 * $$\begin{align}

\hat{M} |\phi \rangle + \kappa |\phi \rangle = |o \rangle \end{align}$$

The Connection Between the Dirac Equation and the Eight-Dimensional Dirac Equation
The $4 × 8$ linear transformation matrix $$Z$$ defined by the equation



Z \equiv \begin{pmatrix} 0 & 0 & 0 & 0 & -i & -1 & +1 & -i\\ 0 & 0 & 0 & 0 & +1 & +i & +i & -1\\ +i & +1 & -1 & +i & 0 & 0 & 0 & 0\\ -1 & -i & -i & +1 & 0 & 0 & 0 & 0 \end{pmatrix} $$

establishes a connection between the $8 × 8$ spacetime matrix operator $$\hat{M}$$ and the $4 × 4$ Dirac matrix operator $$\hat{\Gamma}$$. In particular, it can be shown in a straight forward manner that


 * $$\begin{align}

Z \hat{M} = \hat{\Gamma} Z \end{align}$$.

The reason this is important is that $$Z$$ can be used to transform solutions satisfying the eight-dimensional Dirac equation to those satisfying the traditional Dirac equation. To demonstrate this, first start with the eight-dimensional Dirac equation, namely


 * $$\begin{align}

\hat{M} |\phi \rangle + \kappa |\phi \rangle = |o \rangle \end{align}$$.

Now multiply both sides of this equation by the linear transformation matrix $$Z$$. This gives


 * $$\begin{align}

Z\hat{M} |\phi \rangle + \kappa Z|\phi \rangle = Z|o \rangle \end{align}$$

which can be rewritten in the form


 * $$\begin{align}

\hat{\Gamma} Z|\phi \rangle + \kappa Z|\phi \rangle = |o \rangle \end{align}$$.

Now define



$$
 * \psi \rangle \equiv Z |\phi \rangle

and we obtain


 * $$\begin{align}

\hat{M} |\phi \rangle + \kappa |\phi \rangle = |o \rangle \end{align}$$

which is just the traditional Dirac equation.

Thus, solutions of the eight-dimensional version of the Dirac equation can be transformed into solutions satisfying the traditional Dirac equation through the $4 × 8$ linear transformation matrix $$Z$$.

Macroscopic Maxwell's Equations
Macroscopic Maxwell's equations in the absence of free charges and currents for class A dielectrics


 * $$\begin{align}

\nabla \cdot \mathbf{E} = 0 \quad \quad \nabla \cdot \mathbf{H} = 0 \quad \quad \nabla \times \mathbf{E} + \frac{\mu}{c} \frac{\partial\mathbf H}{\partial t} = 0 \quad \quad \nabla \times \mathbf{H} - \frac{\epsilon}{c} \frac{\partial\mathbf E}{\partial t} = 0 \end{align}$$

Wave equations


 * $$\begin{align}

\nabla^2 \mathbf{E} - \frac{1}{v^2} \frac{\partial^2 \mathbf{E}}{\partial t^2} = 0 \quad \quad \nabla^2 \mathbf{H} - \frac{1}{v^2} \frac{\partial^2 \mathbf{H}}{\partial t^2} = 0 \end{align}$$

Monochromatic electromagnetic plane wave solutions


 * $$\begin{align}

\mathbf{E}(\mathbf{r},t) = \mathbf{E}_o e^{i(\mathbf{k}\cdot\mathbf{r} - \omega t)} \quad \quad \mathbf{H}(\mathbf{r},t) = \mathbf{H}_o e^{i(\mathbf{k}\cdot\mathbf{r} - \omega t)} \end{align}$$

Substitution yields


 * $$\begin{align}

\mathbf{k} \cdot \mathbf{E_o} = 0 \quad \quad \mathbf{k} \cdot \mathbf{H_o} = 0 \quad \quad \mathbf{k} \times \mathbf{E_o} - \mu     \frac{\omega}{c} \mathbf{H_o} = 0 \quad \quad \mathbf{k} \times \mathbf{H_o} + \epsilon \frac{\omega}{c} \mathbf{E_o} = 0 \end{align}$$

Define two constants


 * $$\begin{align}

\Delta^- \equiv \mu     \frac{\omega}{c} \quad \quad \Delta^+ \equiv \epsilon \frac{\omega}{c} \end{align}$$

Implies


 * $$\begin{align}

\mathbf{k} \cdot \mathbf{E_o} = 0 \quad \quad \mathbf{k} \cdot \mathbf{H_o} = 0 \quad \quad \mathbf{k} \times \mathbf{E_o} - \Delta^- \mathbf{H_o} = 0 \quad \quad \mathbf{k} \times \mathbf{H_o} + \Delta^+ \mathbf{E_o} = 0 \end{align}$$

Ancillary results


 * $$\begin{align}

\mathbf{k} \perp \mathbf{E_o} \quad \quad \mathbf{k} \perp \mathbf{H_o} \quad \quad \mathbf{E_o} \perp \mathbf{H_o} \end{align}$$


 * $$\begin{align}

v = \frac{c}{n} \quad \quad n = \sqrt{\mu \epsilon} \quad \quad \frac{E_o}{H_o} = \sqrt{\frac{\mu}{\epsilon}} \end{align}$$


 * $$\begin{align}

\lambda f = v \quad \quad \omega = 2 \pi f \quad \quad k = \frac{2 \pi}{\lambda} \end{align}$$


 * $$\begin{align}

E = \frac{hc}{\lambda} \quad \quad p = \frac{h}{\lambda} \quad \quad E = p c \end{align}$$

Multiple Quantum Well Structures

 * $$\begin{align} \end{align}$$



\partial_4 \rightarrow \partial_4 + \frac{e}{\hbar c} V $$

Monochromatic Plane Waves


\mathbf{F}(\mathbf{r},t) = \mathbf{F}_o e^{i(\mathbf{k}\cdot\mathbf{r} - \omega t)} \quad \quad \nabla \cdot \mathbf{F} = i \mathbf{k} \cdot \mathbf{F} \quad \quad \nabla \times \mathbf{F} = i \mathbf{k} \times \mathbf{F} \quad \quad \frac{1}{c} \frac{\partial}{\partial t} \mathbf{F} = - i \frac{\omega}{c} \mathbf{F} $$



\nabla \times (\nabla \times \mathbf{A}) = \nabla (\nabla \cdot \mathbf{A}) - \nabla^2 \mathbf{A} $$

Matrix Representation
Free-Space



\begin{pmatrix} -\partial_4 & 0 & 0 & 0 & 0 & -\partial_3 & +\partial_2 & -\partial_1\\ 0 & -\partial_4 & 0 & 0 & +\partial_3 & 0 & -\partial_1 & -\partial_2\\ 0 & 0 & -\partial_4 & 0 & -\partial_2 & +\partial_1 & 0 & -\partial_3\\ 0 & 0 & 0 & -\partial_4 & +\partial_1 & +\partial_2 & +\partial_3 & 0\\ 0 & +\partial_3 & -\partial_2 & +\partial_1 & +\partial_4 & 0 & 0 & 0\\ -\partial_3 & 0 & +\partial_1 & +\partial_2 & 0 & +\partial_4 & 0 & 0\\ +\partial_2 & -\partial_1 & 0 & +\partial_3 & 0 & 0 & +\partial_4 & 0\\ -\partial_1 & -\partial_2 & -\partial_3 & 0 & 0 & 0 & 0 & +\partial_4 \end{pmatrix}

\begin{pmatrix} iU_1\\ iU_2\\  iU_3\\  0\\  L_1\\  L_2\\  L_3\\ 0 \end{pmatrix} + \kappa\begin{pmatrix} iU_1\\ iU_2\\  iU_3\\  0\\  L_1\\  L_2\\  L_3\\ 0 \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0\\ 0 \end{pmatrix} $$

Vector Representation
Free-Space


 * $$\begin{align}

\nabla \cdot \mathbf{U} = 0 \quad \quad \nabla \cdot \mathbf{L} = 0 \quad \quad \nabla \times \mathbf{U} + \frac{1}{c} \frac{\partial\mathbf L}{\partial t} = -i \kappa \mathbf{L} \quad \quad \nabla \times \mathbf{L} - \frac{1}{c} \frac{\partial\mathbf U}{\partial t} = -i \kappa \mathbf{U} \end{align}$$

Free-Space to Constant Scalar Potential V



\quad \quad \frac{\partial}{\partial t} \rightarrow \frac{\partial}{\partial t} + \frac{i e}{\hbar} V $$

Constant Scalar Potential V
 * $$\begin{align}

\nabla \cdot \mathbf{U} = 0 \quad \quad \nabla \cdot \mathbf{L} = 0 \quad \quad \nabla \times \mathbf{U} + \frac{1}{c} \frac{\partial\mathbf L}{\partial t} = -i \frac{m_o c^2 + eV}{\hbar c} \mathbf{L} \quad \quad \nabla \times \mathbf{L} - \frac{1}{c} \frac{\partial\mathbf U}{\partial t} = -i \frac{m_o c^2 - eV}{\hbar c} \mathbf{U} \end{align}$$

Monochromatic Matter Plane Wave Solutions


 * $$\begin{align}

\mathbf{k} \cdot \mathbf{U_o} = 0 \quad \quad \mathbf{k} \cdot \mathbf{L_o} = 0 \quad \quad \mathbf{k} \times \mathbf{U_o} - \frac{\hbar \omega - m_o c^2 - eV}{\hbar c} \mathbf{L_o} = 0 \quad \quad \mathbf{k} \times \mathbf{L_o} + \frac{\hbar \omega + m_o c^2 - eV}{\hbar c} \mathbf{U_o} = 0 \end{align}$$

Two Constants


 * $$\begin{align}

\Delta^- \equiv \frac{\hbar \omega - m_o c^2 - eV}{\hbar c} \quad \quad \Delta^+ \equiv \frac{\hbar \omega + m_o c^2 - eV}{\hbar c} \end{align}$$

Implies


 * $$\begin{align}

\mathbf{k} \cdot \mathbf{U_o} = 0 \quad \quad \mathbf{k} \cdot \mathbf{L_o} = 0 \quad \quad \mathbf{k} \times \mathbf{U_o} - \Delta^- \mathbf{L_o} = 0 \quad \quad \mathbf{k} \times \mathbf{L_o} + \Delta^+ \mathbf{U_o} = 0 \end{align}$$

Further Implications


 * $$\mathbf{k} \perp \mathbf{U_o}$$
 * $$\mathbf{k} \perp \mathbf{L_o}$$