Dirac Equation

This is a compact lookup card. Lorentz covariance, Klein–Gordon factorization, the adjoint current, initial data, and the free Hamiltonian domain are derived on the Covariant Dirac Equation owner.

Formula

With $\eta=\operatorname{diag}(1,-1,-1,-1)$ and gamma matrices satisfying

$$\{\gamma^\mu,\gamma^\nu\} = 2\eta^{\mu\nu}I,$$

the free Dirac equation is

$$\left( i\hbar c\gamma^\mu\partial_\mu - mc^2 \right)\psi=0.$$

In natural units $\hbar=c=1$, this is

$$\left( i\gamma^\mu\partial_\mu-m \right)\psi=0.$$

The Hamiltonian form is

$$i\hbar\frac{\partial\psi}{\partial t} = \left[ c\,\boldsymbol\alpha\cdot(-i\hbar\nabla) + \beta mc^2 \right]\psi,$$

where

$$\alpha^i=\gamma^0\gamma^i, \qquad \beta=\gamma^0.$$

For a free plane-wave spinor in natural units, define

$$p\!\!\!/ \equiv \gamma^\mu p_\mu.$$

Then the momentum-space equation is

$$(p\!\!\!/-m)u(p)=0.$$

Assumptions

• The gamma-matrix convention and metric signature are fixed.
• $\psi$ is a four-component spinor in four-dimensional spacetime.
• The displayed equation is free; electromagnetic coupling requires a stated minimal-coupling convention.
• A one-particle Dirac equation is useful but not the full interacting relativistic quantum theory.

Validity

The Dirac equation is first order in time and space and squares to the Klein–Gordon dispersion relation because of the Clifford algebra. It naturally includes spin-$1/2$ structure and both frequency sectors. Their consistent antiparticle interpretation belongs to the quantized Dirac field; the one-particle equation alone is not the full interacting relativistic theory.

Common Mistakes

• Using gamma matrices from one metric signature with formulas from another.
• Forgetting that $p_\mu$ and $p^\mu$ differ by signs in the spatial components for the $(+---)$ convention.
• Treating the four spinor components as four independent nonrelativistic wavefunctions.
• Using slash notation without defining the metric and contraction convention.

Quick Check

Why does the Dirac equation imply the relativistic dispersion relation?

Solution

In natural units, multiply $(p\!\!\!/-m)u=0$ on the left by $(p\!\!\!/+m)$. Since

$$(p\!\!\!/)^2 = \gamma^\mu\gamma^\nu p_\mu p_\nu = p^\mu p_\mu I,$$

the result is

$$(p^\mu p_\mu-m^2)u=0.$$

For a nonzero spinor, $p^\mu p_\mu=m^2$, which is $E^2=\mathbf p^2+m^2$ in natural units.

Canonical owner and released companions

• Covariant Dirac Equation
• Gamma-Matrix Identities
• Pauli Equation
• Minimal Coupling

References

• P. A. M. Dirac, “The quantum theory of the electron”, Proceedings of the Royal Society A 117, 610-624, 1928.
• J. D. Bjorken and S. D. Drell, Relativistic Quantum Mechanics, McGraw-Hill, 1964.
• J. J. Sakurai, Advanced Quantum Mechanics, Addison-Wesley, 1967.
• S. Weinberg, The Quantum Theory of Fields, Volume I, Cambridge University Press, 1995.