10 Jun 2026
Dirac Equation
Linear relativistic dynamics, Dirac matrices, covariant form, and the origin of the four-component spinor.
The Klein–Gordon equation has the correct relativistic energy relation but is second order in time. Dirac sought an equation of Schrödinger form,
\[i\hbar\frac{\partial\psi}{\partial t}=H_D\psi,\]whose Hamiltonian is linear in momentum. Write
\[\boxed{H_D=c\,\boldsymbol\alpha\cdot\mathbf p+\beta mc^2},\]where $\alpha_i$ and $\beta$ are constant matrices. Consistency with
\[H_D^2=(p^2c^2+m^2c^4)I\]determines their algebra. Direct expansion gives
\[\begin{aligned} H_D^2 &= \left(c\alpha_i p_i+\beta mc^2\right) \left(c\alpha_j p_j+\beta mc^2\right)\\ &= c^2\alpha_i\alpha_jp_ip_j +mc^3(\alpha_i\beta+\beta\alpha_i)p_i +\beta^2m^2c^4. \end{aligned}\]The repeated spatial indices are summed. Since $p_ip_j=p_jp_i$ for a free particle, the first term contains only the symmetric part of the matrix product:
\[\alpha_i\alpha_jp_ip_j = \frac{1}{2} (\alpha_i\alpha_j+\alpha_j\alpha_i)p_ip_j =\frac{1}{2}\{\alpha_i,\alpha_j\}p_ip_j.\]To leave $p_i p_i I=p^2I$, remove the momentum–mass cross term, and recover the rest-energy term, the matrices must satisfy
\[\{\alpha_i,\alpha_j\}=2\delta_{ij}I, \qquad \{\alpha_i,\beta\}=0, \qquad \beta^2=I.\]Substituting these relations back into the expansion gives, step by step,
\[H_D^2 =c^2\delta_{ij}p_ip_jI+0+m^2c^4I =(p^2c^2+m^2c^4)I.\]Ordinary numbers cannot satisfy the required anticommutations, and $2\times2$ matrices are not sufficient for three spatial directions together with $\beta$. The smallest realization is four dimensional, so $\psi$ is a four-component spinor.
Dirac–Pauli representation
Using the Pauli matrices $\sigma_i$, a convenient representation is
\[\alpha_i= \begin{pmatrix} 0&\sigma_i\\ \sigma_i&0 \end{pmatrix}, \qquad \beta= \begin{pmatrix} I_2&0\\ 0&-I_2 \end{pmatrix}.\]The four components do not represent four independent particles. For a free field they accommodate two spin states on each of the particle and antiparticle branches.
The Hamiltonian form is therefore
\[\boxed{ i\hbar\frac{\partial\psi}{\partial t} = \left(c\,\boldsymbol\alpha\cdot\mathbf p+\beta mc^2\right)\psi }.\]Because the matrices $\alpha_i$ and $\beta$ are Hermitian, $H_D$ is Hermitian for the usual momentum boundary conditions.
Covariant form
Insert $\mathbf p=-i\hbar\nabla$ into the Hamiltonian equation and move every term to the left:
\[\left[ i\hbar\frac{\partial}{\partial t} +i\hbar c\,\alpha_i\partial_i -\beta mc^2 \right]\psi=0.\]Multiply by $\beta/c$. Since $\beta^2=I$,
\[\left[ i\hbar\beta\frac{1}{c}\frac{\partial}{\partial t} +i\hbar\beta\alpha_i\partial_i -mc \right]\psi=0.\]Define
\[\gamma^0=\beta, \qquad \gamma^i=\beta\alpha_i.\]Their products follow directly from the $\alpha_i,\beta$ algebra:
\[\{\gamma^0,\gamma^0\}=2\beta^2=2I_4,\] \[\begin{aligned} \{\gamma^0,\gamma^i\} &=\beta^2\alpha_i+\beta\alpha_i\beta\\ &=\alpha_i-\alpha_i=0, \end{aligned}\]and
\[\begin{aligned} \{\gamma^i,\gamma^j\} &=\beta\alpha_i\beta\alpha_j +\beta\alpha_j\beta\alpha_i\\ &=-(\alpha_i\alpha_j+\alpha_j\alpha_i)\\ &=-2\delta_{ij}I_4. \end{aligned}\]These three results combine into the Clifford algebra
\[\boxed{\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}I_4},\]with $g^{\mu\nu}=\operatorname{diag}(1,-1,-1,-1)$. Writing $x^0=ct$, so that $\partial_0=c^{-1}\partial_t$, the preceding equation becomes
\[\boxed{ \left(i\hbar\gamma^\mu\partial_\mu-mc\right)\psi=0 }.\]This compact expression is Lorentz covariant: the spinor and gamma matrices transform together so that the equation retains its form between inertial frames.
Relation to the Klein–Gordon equation
Act on the Dirac equation with the complementary operator
\[i\hbar\gamma^\nu\partial_\nu+mc.\]Let $D_-=i\hbar\gamma^\mu\partial_\mu-mc$ and $D_+=i\hbar\gamma^\nu\partial_\nu+mc$. Their product acting on $\psi$ is
\[\begin{aligned} D_+D_-\psi &= \left[ (i\hbar)^2\gamma^\nu\gamma^\mu \partial_\nu\partial_\mu +mc(i\hbar\gamma^\mu\partial_\mu) -mc(i\hbar\gamma^\nu\partial_\nu) -m^2c^2 \right]\psi\\ &= \left[ -\hbar^2\gamma^\nu\gamma^\mu \partial_\nu\partial_\mu -m^2c^2 \right]\psi. \end{aligned}\]The two mixed terms are the same operator with opposite signs and cancel. Since ordinary derivatives commute, their product is symmetric in $\mu,\nu$. The antisymmetric part of the gamma-matrix product therefore contributes nothing:
\[\begin{aligned} \gamma^\nu\gamma^\mu\partial_\nu\partial_\mu &= \frac{1}{2} \left(\gamma^\nu\gamma^\mu+\gamma^\mu\gamma^\nu\right) \partial_\nu\partial_\mu\\ &= \frac{1}{2}\{\gamma^\nu,\gamma^\mu\} \partial_\nu\partial_\mu\\ &=g^{\mu\nu}\partial_\mu\partial_\nu I_4 =\Box I_4. \end{aligned}\]Thus $D_+D_-\psi=0$ becomes
\[\left(-\hbar^2\Box-m^2c^2\right)\psi=0.\]After division by $-\hbar^2$, every component of a free Dirac spinor satisfies
\[\left(\Box+\frac{m^2c^2}{\hbar^2}\right)\psi=0.\]Thus the Dirac equation does not replace the relativistic dispersion relation; it factorizes it into a first-order equation. The matrix structure introduced by that factorization is also what permits a positive density and produces electron spin.
Maxima verification of the Dirac algebra and dispersion relation
Discussion