Past exam of the mathematics course of the University of Cambridge / 2007 / iii / Paper 50 / 1

The Pauli matrix multiplication law is ${\sigma }_i{\sigma }_j={\delta }_{ij}{I}_2+i{ϵ}_{ijk}{\sigma }_k$, so $\{{\sigma }_i,{\sigma }_j\}=2{\delta }_{ij}{I}_2$. Block multiplication of the chiral gamma-matrix representation givesandThus the mixed anticommutators vanish and the spatial ones are $-2{\delta }_{ij}{I}_4$. With the Minkowski metric convention $\eta =\mathrm{diag}(1,-1,-1,-1)$,These matrices therefore represent the required Clifford algebra.

Apply $i{\gamma }^{\nu }{\partial }_{\nu }+m$ to the Dirac equation. Since partial derivatives commute, only the symmetric part of the gamma matrix product contributes:Here $\square ={\partial }_t^2-{\nabla }^2$ in the signature fixed above. The remaining operator is scalar in spinor indices, givingThis is the Klein-Gordon equation for each component, although the first-order Dirac equation additionally couples them.

For positive frequency take $p^0=E=\sqrt{|\mathbf{p}{|}^2+m^2}>0$, with $m\ge 0$. Lowering the index in $p_{\mu }{\sigma }^{\mu }$ is important: putThese are commuting Hermitian matrices with $AB=(E^2-|\mathbf{p}{|}^2)I_2=m^2{I}_2$. Their eigenvalues are $E\pm |\mathbf{p}|$, so they are positive definite for $m>0$ and positive semidefinite matrices for $m=0$. Simultaneous diagonalization defines their positive square roots of a matrix and gives $\sqrt{A}\sqrt{B}=m{I}_2$.

The plane-wave ansatz reduces the Dirac equation to $(/p-m)u=0$, whereFor $u_L=\sqrt{A}\xi$ and $u_R=\sqrt{B}\xi$, commutativity gives $A{u}_R=m{u}_L$ and $B{u}_L=m{u}_R$. ThereforeThis is the Hermitian square-root construction of Dirac plane waves.

For negative frequency retain the future-directed momentum label $p$ and use ${\psi }_{-}(x)=v(\mathbf{p})e^{+ip\cdot x}$. The equation is now $(/p+m)v=0$. Its general solution iswith arbitrary two-spinor $\zeta$. The minus between its two blocks gives $A{v}_R=-m{v}_L$ and $B{v}_L=-m{v}_R$. For $m>0$ these parametrizations span the two-dimensional positive- and negative-frequency solution spaces. At nonzero null momentum the square roots have complementary rank-one supports and still give the full two-dimensional kernels. If ${\xi }^{†}\xi ={\zeta }^{†}\zeta =1$, both spinors have norm squared $2E$; the positive one also has $\bar{u}u=2m$.

Both blocks of the given rotation matrix carry the same spin one-half representation, so a stationary spin-up state along the third axis uses the $+1$ eigenspinor ${\xi }_3=(1,0{)}^T$ of ${\sigma }_3$. At rest $A=B=m{I}_2$, givingFor a massless particle moving along the positive third axis, $p=(E,0,0,E)$ with $E>0$, andKeeping the same two-spinor label givesIt has positive helicity $+1/2$ and positive chirality in the Weyl representation.

For spin-up along the first axis, take the normalized $+1$ eigenspinor of ${\sigma }_1$, namely ${\xi }_1=(1,1{)}^T/\sqrt{2}$. At rest this givesFor momentum $(E,0,0,E)$, using the square roots displayed in part (i) with the same label givesThis is the massless limit of a transverse canonical-spin label: it is the equal superposition of the negative-helicity left-chiral mode and the positive-helicity right-chiral mode. It solves the massless Dirac equation and has norm squared $2E$.

A qualification is essential if “spin along the first axis” is interpreted literally at zero mass. The fixed transverse rotation generator is ${\Sigma }_1/2$, with ${\Sigma }_1=\mathrm{diag}({\sigma }_1,{\sigma }_1)$, and the massless Hamiltonian at this momentum is $H=E{\gamma }^0{\gamma }^3$. They anticommute. If $Hu=Eu$ and ${\Sigma }_{1}u=u$ both held, then $H{\Sigma }_{1}u=-{\Sigma }_{1}Hu=-Eu$, whereas $H{\Sigma }_{1}u=Hu=Eu$, forcing $u=0$. Thus there is no nonzero positive-energy massless solution with that literal transverse-spin eigenvalue. The displayed nonzero spinor is the transported rest-spin label requested by the massive-to-massless construction, rather than a transverse-spin eigenstate.