Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2018 / S18 / 3 / a

For a free particle the Dirac matrices are constant and the momenta commute. Squaring the Hamiltonian cancels all mixed terms by their anticommutators:Applying another time derivative to the Dirac equation therefore givesfor every component. This is the Klein-Gordon equation. Its componentwise satisfaction is a necessary consequence of the first-order Dirac equation, not a converse selecting all spinor solutions of the squared equation.

If $Av=\lambda v$ for $A={\alpha }_i$ or $\beta$, then $A^2=I$ implies ${\lambda }^2=1$. The polynomial $x^2-1$ has distinct roots, so such a matrix is diagonalizable andAnticommutation with another invertible Dirac matrix maps a $+1$ eigenspace to a $-1$ eigenspace, so both signs occur with equal multiplicities in a nontrivial representation. A Hermitian operator is appropriate for the physical Hamiltonian and additionally guarantees an orthonormal eigenbasis.

For any one of the four anticommuting matrices $A$, choose another $B$. Since $B^{-1}=B$ and $BAB=-A$, cyclic invariance of the trace givesThereforeEquivalently, the paired positive and negative unit eigenvalues cancel. This argument does not depend on a particular matrix representation.

The zero trace and unit eigenvalues imply equal numbers of $+1$ and $-1$ eigenvalues, so $N$ is even. A two-dimensional representation is impossible: all four matrices ${\alpha }_1,{\alpha }_2,{\alpha }_3,\beta$ would be traceless $2\times 2$ matrices. Anticommutation gives $\mathrm{Tr}(AB)=0$ for distinct matrices, whereas $\mathrm{Tr}(A^2)=2$. They would therefore be four linearly independent members of the three-dimensional complex space of traceless $2\times 2$ matrices, a contradiction.The usual irreducible $3+1$ dimensional Dirac representation has $N=4$. Larger representations can contain multiple copies; the requested lower bound is not itself a proof of irreducibility.