Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2012 / S18 / 1

Work with $\eta =\mathrm{diag}(1,-1,-1,-1)$ and $/\partial ={\gamma }^{\mu }{\partial }_{\mu }$. Multiplying the free Dirac equation by the operator with the opposite mass sign givesThe ordinary derivatives commute, so only the symmetric part of ${\gamma }^{\mu }{\gamma }^{\nu }$ contributes. Requiring this operator to factor the Klein-Gordon equation, $({\partial }^{\mu }{\partial }_{\mu }+m^2)\psi =0$, for arbitrary free momenta givesThus the Dirac matrices represent the corresponding Clifford algebra. The inference uses factorization for the general free equation; the fact that one specially chosen wavefunction happens to solve both equations would not alone determine every matrix anticommutator.

Let $U(x)=e^{ie\alpha (x)}$, ${\psi }^{\prime }=U\psi$, and $A_{\mu }^{\prime }=A_{\mu }+{\partial }_{\mu }\alpha$. Direct differentiation in the gauge covariant derivative givesThe gradient of the local phase is exactly cancelled by the potential transformation. Since $U$ is scalar in spinor space and commutes with the Dirac matrices,Consequently a solution is mapped into a solution: the minimally coupled Dirac equation is covariant under the local U(1) gauge symmetry. The operator and wavefunction both transform; ordinary differentiation of the transformed wavefunction alone would leave an extra phase-gradient term.

Use $F_{\mu \nu }={\partial }_{\mu }{A}_{\nu }-{\partial }_{\nu }{A}_{\mu }$. The stated derivative convention gives $[D_{\mu },D_{\nu }]=-ie{F}_{\mu \nu }$. Also $[{\gamma }^{\mu },{\gamma }^{\nu }]=-2i{\sigma }^{\mu \nu }$. Splitting products into symmetric and antisymmetric parts therefore yieldsMultiplying the first-order equation by $i{\gamma }^{\mu }{D}_{\mu }+m$ now givesThe antisymmetric derivative commutator is the electromagnetic field strength. It is the extra term absent when squaring an uncoupled Dirac operator with commuting derivatives, and it generates the spin–magnetic-field interaction in the nonrelativistic limit.

With $A^{\mu }=(0,\mathbf{A})$, the lowered components are $A_{\mu }=(0,-\mathbf{A})$. Define $\mathbf{\pi }=-i\nabla +e\mathbf{A}$ and $\mathbf{B}=\nabla \times \mathbf{A}$. This convention describes physical charge $q=-e$. Since $F_{ij}=-{ϵ}_{ijk}{B}_k$,Writing $\psi =e^{-i(m+E)t}\varphi$ makes the squared equationDrop $E^2$ relative to $2mE$ and write $\varphi =(\phi ,\chi {)}^T$. Both two-component blocks obey the Pauli equationThe gauge-covariant kinetic term contains ${\mathbf{p}}^2/(2m)$, $e(\mathbf{A}\cdot \mathbf{p}+\mathbf{p}\cdot \mathbf{A})/(2m)$, and $e^2{A}^2/(2m)$: free kinetic, orbital magnetic, and diamagnetic contributions. The spin term is $-\mathbf{\mu }\cdot \mathbf{B}$ with $\mathbf{\mu }=-e\mathbf{\sigma }/(2m)$, giving spin gyromagnetic factor $g=2$ at this order. In uniform $\mathbf{B}$ and symmetric gauge, the orbital term is $e\mathbf{B}\cdot \mathbf{L}/(2m)$.

Squaring introduces extra formal solutions unless the original first-order equation is retained. In the usual positive-energy Dirac basis it relates the small lower component to the upper one by $\chi =(\mathbf{\sigma }\cdot \mathbf{\pi })\phi /(2m+E)\simeq (\mathbf{\sigma }\cdot \mathbf{\pi })\phi /(2m)$. Thus these are not two independent nonrelativistic particles. Restoring units places $\hslash$ in the spin term, $e\hslash \mathbf{\sigma }\cdot \mathbf{B}/(2m)$.