Covariant Form of the Dirac Equation

We now cast the Dirac equation into a more apparent covariant form. Multiplying (5.3) by $\beta/c$ and defining

$$\gamma^0\equiv\beta \quad\textrm{and}\quad \gamma^i\equiv\beta\alpha_i,$$ (5.96)

where $i=1,2,3$, we have

$$i\hbar \left( \gamma^0\frac{\partial}{\partial x^0} + \gamma^1\fr... ...al}{\partial x^2} + \gamma^3\frac{\partial}{\partial x^3} \right) \psi - mc\psi = 0 ,$$

$$\left( i\hbar \gamma^\mu\frac{\partial}{\partial x^\mu} \right) \psi - mc\psi = 0 ,$$ (5.97)

where

$$\gamma^\mu\frac{\partial}{\partial x^\mu} = \frac{\gamma^0}{c} \frac{\partial}{\partial t} + \vec{\gamma}\cdot\vec{\nabla} .$$ (5.98)

In terms of the momentum operator we write

$$(\gamma^\mu \hat{p}_\mu - mc)\psi = 0 .$$ (5.99)

Introducing the Feynman dagger, or slash notation, for 4-vector $A_\mu$, we have

$$\not{\!\!A} \equiv \gamma^\mu A_\mu = g_{\mu\nu}\gamma^\mu A^\nu = \gamma^0A^0 - \vec{\gamma}\cdot\vec{A} .$$ (5.100)

Also notice that

$$\not{\!\!A}\not{\!\!A}= A^2 .$$ (5.101)

We write

$$\fbox{$\displaystyle (\not{\!\hat{p}} - mc)\psi = 0 $}\ .$$ (5.102)

We introduce the electromagnetic interaction by the usual minimal substitution

$$\left( \not{\!p} - \frac{e}{c}\not{\!\!A} - mc \right) \psi = 0.$$ (5.103)

Let us study the properties of the $\gamma^\mu$ matrices.

$$\alpha_i\alpha_k + \alpha_k\alpha_i = 2\delta_{ik}$$

$$\beta\alpha_i\alpha_k + \beta\alpha_k\alpha_i = 2\beta\delta_{ik}$$

$$-\alpha_i\beta\alpha_k - \alpha_k\beta\alpha_i = 2\beta\delta_{ik}$$

$$-\beta\alpha_i\beta\alpha_k - \beta\alpha_k\beta\alpha_i = 2\delta_{ik}$$

$$\gamma^i\gamma^k + \gamma^k\gamma^i = -2\delta_{ik} .$$ (5.104)

And

$$\alpha_i\beta + \beta\alpha_i = 0$$

$$\beta\alpha_i\beta + \beta\beta\alpha_i = 0$$

$$\gamma^i\gamma^0 + \gamma^0\gamma^i = 0 .$$ (5.105)

Since

$$\beta^2 = 1 \Rightarrow \gamma^0\gamma^0 + \gamma^0\gamma^0 = 2 .$$ (5.106)

Therefore

$$\fbox{$\displaystyle \gamma^\mu\gamma^\nu + \gamma^\nu\gamma^\mu = 2g^{\mu\nu} $}\ ,$$ (5.107)

where the matrices are $4\times 4$ and $\mu, \nu=0,1,2,3$. Although the Dirac matrices $\gamma^\mu$ are written with Greek indices, they are not four vectors. Rather, they have the same value in every frame.

The $\gamma^i$ matrices are anti-hermitian and $\gamma^0$ is hermitian:

$$(\gamma^i)^\dagger = (\beta\alpha_i)^\dagger = \alpha_i^\dagger\beta^\dagger = \alpha_i\beta = -\beta\alpha_i = -\gamma^i ,$$ (5.108)

$$(\gamma^0)^\dagger = \beta^\dagger = \beta = \gamma^0 .$$ (5.109)

This can be summarized by writing

$$(\gamma^\mu)^\dagger = \gamma^0 \gamma^\mu \gamma^0 .$$ (5.110)

Using our previous representation (equation 5.10),

$$\gamma^i \equiv \beta\alpha_i = \left( \begin{array}{cc} I & 0 \\ 0 & -I \end{ar... ...) = \left( \begin{array}{cc} 0 & \sigma_i \\ -\sigma_i & 0 \end{array}\right),$$ (5.111)

$$\gamma^0 \equiv \beta = \left( \begin{array}{cc} I & 0 \\ 0 & -I \end{array}\right) .$$ (5.112)