### Introduction to the Dirac Equation

The Dirac equation describes the quantum mechanical state of a relativistic, massive, spin-$1\over2$ particle, such as the electron. The Dirac equation in 4 dimensions is usually given as $$(i\hbar\gamma^\mu\partial_\mu - mc) \Psi = 0 \label{diraceq}$$ What do these symbols mean? First of all, $m$ is the mass of the particle described by $\Psi$, $c$ is the speed of light (which we normally set to $1$) and $\hbar$ is Planck's constant (divided by $2\pi$). The notation $\partial_\mu$ is short for $\partial\over\partial x^\mu$, and the repeated index $\mu$ implies a summation over the spacetime index $\mu=0,1,2,3$, where $x^0=ct$, $x^1=x$, $x^2=y$, and $x^3=z$. But what is $\gamma^\mu$? We will briefly postpone this question.

The form ($\ref{diraceq}$) assumes that the “squared length” of the spacetime vector whose components are $x^\mu$ is given by 1) $$g_{\mu\nu} x^\mu x^\nu = (x^0)^2 - (x^1)^2 - (x^2)^2 - (x^3)^2$$ so that the spacetime metric is given by the matrix $$(g_{\mu\nu}) = \begin{pmatrix} 1& 0& 0& 0\cr 0& -1& 0& 0\cr 0& 0& -1& 0\cr 0& 0& 0& -1\cr \end{pmatrix}$$

What happens if we multiply the Dirac equation ($\ref{diraceq}$) by the differential operator obtained by replacing $m$ with $-m$? We get $$0 = (i\hbar\gamma^\mu\partial_\mu + mc) (i\hbar\gamma^\nu\partial_\nu - mc) \Psi = - (\hbar^2 \gamma^\mu\partial_\mu \gamma^\nu\partial_\nu + m^2 c^2) \Psi \label{factor}$$ where we have been careful to use different dummy indices in the two implicit sums. The right-hand-side looks almost like the Klein-Gordon equation, $$(\hbar^2 \square + m^2 c^2) \phi = 0$$ a wave equation which describes the state of a relativistic, massive, spin-$0$ particle. Here, $\square$ is the d'Alembertian operator, the spacetime version of the Laplacian, and is given by $$\square := g^{\mu\nu} \partial_\mu\partial_\nu = \partial_t^2 - \triangle = \partial_t^2 - \nabla^2$$ where $(g^{\mu\nu})$ is the inverse metric, that is, the inverse of the matrix $(g_{\mu\nu})$.

The Klein-Gordon equation was known to Dirac, but there were difficulties (later resolved) interpreting the squares of its solutions as probability densities, arising from the equation being second order. Dirac was led to his first-order equation by reversing the argument just given, thus “factoring” the Klein-Gordon equation. To make this work, he had to show that ($\ref{factor}$) really is the Klein-Gordon equation.

Assume that each $\gamma^\mu$ is constant. Then we must solve the equation $$\gamma^\mu \gamma^\nu \partial_\mu \partial_\nu = g^{\mu\nu} \partial_\mu \partial_\nu \label{geq1}$$ Writing out all the terms of ($\ref{geq1}$) and comparing coefficients of $\partial_\mu\partial_\nu$, remembering that partial derivative operators commute ($\partial_\mu\partial_\nu=\partial_\nu\partial_\mu$), we obtain $$\{\gamma^\mu,\gamma^\nu\} := \gamma^\mu \gamma^\nu + \gamma^\nu \gamma^\mu = g^{\mu\nu} + g^{\nu\mu} = 2 g^{\mu\nu} \label{geq2}$$ where we have introduced the curly bracket notation for anticommutators. It is easily seen that ($\ref{geq2}$) has no solutions if each $\gamma^\mu$ is a number. Dirac's brilliant idea was to use matrices instead, which we will discuss in the next chapter.

Suppose that $\Psi$ is a plane-wave of the form $$\Psi = e^{-ip_\nu x^\nu/\hbar} \Psi_0 \label{wave}$$ where $p_\nu$ is the 4-momentum of the particle and where $\Psi_0$ does not depend on $x^\mu$. Inserting ($\ref{wave}$) into ($\ref{diraceq}$) leads to the momentum space Dirac equation $$( \gamma^\mu p_\mu - mc ) \Psi_0 = 0 \label{diracmom}$$ which is purely algebraic. 2) (We will usually write $\Psi$ rather than $\Psi_0$.)

Note the absence of any explicit factors of $i$ in the momentum-space Dirac equation ($\ref{diracmom}$)! We view momentum space as more fundamental than position space, and deliberately chose the metric signature to achieve this. It is now straightforward to generalize ($\ref{diracmom}$) to higher dimensions using the other division algebras, which would not have been the case had there been factors of $i$ to worry about.

1) This is called a choice of signature. The other choice is to use $-g_{\mu\nu}$ for the metric components, which eliminates the $i$ in ($\ref{diraceq}$) (and multiplies all the $\gamma$ matrices by $i$).
2) One can also view ($\ref{diracmom}$) as the Fourier transform of ($\ref{diraceq}$). Equivalently, one obtains ($\ref{diracmom}$) from ($\ref{diraceq}$) by the formal substitution $i\hbar\partial_\mu\mapsto p_\mu$.