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

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 $\bar{u}u=2m$.