Spin 1/2 Particles

Given a positive-definite operator, such as (m2c4 + P2c2), there is a mathematical theorem that guarantees that there is one, and only one, square root that is also positive definite, denoted by +(m2c4 + P2c2)1/2. Other square roots become possible if we give up positive definiteness. This may appear to spoil the theory by allowing negative energies; but, if the operator is Hermitean, states corresponding to negative energies will be orthogonal to positive-energy states and a sensible physical theory is obtained if we restrict ourselves to the latter. We can, moreover, ensure manifest covariance by looking for an equation not only linear in ∂_t but also linear in the space derivatives; that equation, we expect, will describe relativistic spin 1/2 particles, such as the electron1. We then use a multicomponent wave function2, $$\mathop \Psi \limits_ \sim $$, and look for an equation linear in the P _μ , the Dirac equation, 3.1.1$$ih{\partial _t}\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{\Psi } \left( {r,t} \right) = {\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{H} _0}\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{\Psi } = - ihc\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a} \nabla \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{\Psi } \left( {r,t} \right) + m{c^2}\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{\beta } \underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{\Psi } \left( {r,t} \right),$$ where the free Dirac Hamiltonian$$\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{H} _0 $$ satisfies 3.1.2$$\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{H} _0^2 = {m^2}{c^4} + {c^2}{P^2};$$3.1.3$${{\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{H} }_0} = ihc\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{a} \nabla + m{c^2}\underset{\raise0.3em\hbox{$\smash{\scriptscriptstyle\thicksim}$}}{\beta }$$