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

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.

In two spatial dimensions there are only two kinetic matrices and one mass matrix. The three Pauli matrices square to identity and anticommute pairwise, sois a two-component Dirac Hamiltonian with $H_{2+1}^2=c^2(p_x^2+p_y^2)+m^2{c}^4$. It therefore reproduces the stated first-order equation and its Klein-Gordon square. Requiring a fourth anticommuting matrix in three spatial dimensions is precisely what prevented this two-component construction above.

Printed angular-momentum qualification. Define the dimensionless spin $s_i=-i{\epsilon }_{ijk}{\alpha }_j{\alpha }_k/4$ and the physical orbital angular momentum ${\ell }_i={\epsilon }_{ijk}{x}_j{p}_k$, with $[x_i,p_j]=i\hslash {\delta }_{ij}$. ThenThe source prints $L_i=i{\ell }_i$ instead. Taken literally, it is anti-Hermitian, has incompatible units with the dimensionless spin when they are added, and givesIt cannot yield the requested usual angular-momentum algebra or conservation law. For a concrete counterexample set $\hslash =1$ and take a wave with momentum along $z$. Sincethe printed $J_i=i{\ell }_i+s_i$ obeys $[H_D,J_x]=(1+i)c{\alpha }_y{p}_z\ne 0$.

The physically consistent operator is ${\mathbf{J}}_{\mathrm{phys}}=\mathbf{\ell }+\hslash \mathbf{s}$. It satisfiesThe cancellation follows directly from the two mixed commutators above. In a spherical basis the free spinor can be labelled by energy, $j=1/2,3/2,\dots$, and $m_j=-j,\dots ,j$, together with a parity or equivalent Dirac angular label distinguishing the two orbital couplings $\ell =j\pm 1/2$. Alternatively one can choose a momentum and helicity basis. One cannot simultaneously assign all components of momentum and all angular labels as independent commuting observables.

Use $\mathbf{p}=-i\hslash \nabla$ and regard $e$ as the signed charge in the printed minimal coupling. Multiplying the first Dirac equation by $c{\gamma }^0$ gives $H=c\mathbf{\alpha }\cdot \mathbf{\pi }+\beta m{c}^2+e\Phi$, with $\mathbf{\pi }=\mathbf{p}-e\mathbf{A}/c$. Removing the rest-energy phase yieldsFor the alternative equation use its specified field conventions literally. They giveConsequently its Hamiltonian is $c\mathbf{\alpha }\cdot \mathbf{p}+\beta m{c}^2-\kappa \beta \mathbf{\Sigma }\cdot \mathbf{B}+(i\kappa /c)\beta \mathbf{\alpha }\cdot \mathbf{E}$. The two shifted equations areThese electric-field factors depend on the source's convention $E_i=c{F}_{0i}$; substituting a different Gaussian or SI definition without changing the equation would give incorrect powers of $c$.

For kinetic and field energies small compared with $m{c}^2$, the minimally coupled lower spinor is $\tilde{\chi }\simeq (\mathbf{\sigma }\cdot \mathbf{\pi })\tilde{\phi }/(2mc)$. Inserting it into the upper equation and using $[{\pi }_i,{\pi }_j]=i(e\hslash /c){\epsilon }_{ijk}{B}_k$ gives the Pauli equationThe spin moment is $e\hslash /(2mc)$ multiplying $\mathbf{\sigma }$, or gyromagnetic factor two relative to physical spin $\hslash \mathbf{\sigma }/2$.

For the alternative equation, elimination givesTo leading kinetic order and leading magnetic interaction this yieldsKeeping the source electric coupling during the same elimination writes the kinetic part more explicitly asThis product includes derivatives acting on the field and must not be treated as an ordinary scalar product of commuting variables. The parameter $\kappa$ is the intrinsic magnetic-moment coefficient: the moment operator is $\kappa \mathbf{\sigma }$. As printed, the alternative equation has no minimal charge coupling, so it can describe a neutral particle carrying a magnetic moment. If a Pauli term were added to the charged equation instead, its coefficient would represent an additional, anomalous contribution to the charge-generated Dirac moment.