Is the Dirac equation equivalent to the Klein-Gordon equation for its left handed component?

Question

The Dirac equation $$(i\gamma^a\partial_a - m)\psi=0\tag{0}$$ is given by a first order operator acting on a Dirac spinor, which is the direct sum of a left handed spinor and a right handed spinor. The fact that it is first order makes sense if we want to treat it as the Schrodinger equation for some particle, but if we want to treat it as the equation of motion of a field then it would be nice if it were second order like the Klein-Gordon equation.

We can write the gamma matrices as: $$\gamma^a=\begin{pmatrix} &\sigma^a\\ \bar{\sigma}^a& \end{pmatrix}$$ and so the Dirac equation can be split into two: $$\begin{align} i\sigma^a\partial_a\psi_R - m\psi_L&=0\tag{1}\\ i\bar{\sigma}^a\partial_a\psi_L - m\psi_R&=0\tag{2} \end{align}$$ Then we can substitute $(2)$ into $(1)$ to get $$i\sigma^a\partial_a(i\bar{\sigma}^b\partial_b\psi_L) - m^2\psi_L=0$$ which simplifies to $$(\partial^2+m^2)\psi_L=0\tag{3}$$ the Klein-Gordon equation for the left handed component! Conversely, if we started with $(3)$ then we could take $(2)$ as our definition of $\psi_R$ and derive $(1)$.

It seems to me that $(3)$ is much simpler than $(0)$. For example counting the degrees of freedom is much more straightforward. So why do all textbooks use the Klein-Gordon equation for scalar fields but then switch over to the Dirac equation when they come to talk about spinors? Do the two versions of the equations make different predictions somehow? (Perhaps after quantizing?) If so how do they differ?

EDIT: To put what I'm saying in other words: Is the field theory of a Dirac spinor obeying the Dirac equation equivalent to the field theory of a left handed spinor obeying the KG equation?

Answer 1

The Dirac equation is more restrictive than the Klein-Gordon equation. For every solution to the Dirac equation, its components will be a solution of the Klein-Gordon equation, but the converse isn't true: if you form a spinor whose components are solutions of the Klein-Gordon equation, it might not solve the Dirac equation.

If we start with the Klein-Gordon equation for the whole spinor $\psi$ $$(\partial^2 +m^2)\psi =0 $$ the solution is $$\psi = u(\vec{p}) e^{-ipx} \quad \text{or} \quad \psi = v(\vec{p}) e^{+ipx}$$ where $u$ ans $v$ are arbitrary spinors. But, what happens if we plug these solutions in the original Dirac equation? $$(\partial^2 + m^2)u(\vec{p})e^{-ipx} = -e^{-ipx}(\gamma^\mu p_\mu - m)u(\vec{p}) = 0$$ $$(\partial^2 + m^2)v(\vec{p})e^{+ipx} = -e^{+ipx}(\gamma^\mu p_\mu + m)v(\vec{p}) = 0$$ $u$ and $v$ are not arbitrary anymore! Instead, they must obey the stronger restriction $$(\gamma^\mu p_\mu - m)u(\vec{p}) = 0 \qquad\qquad (\gamma^\mu p_\mu + m)v(\vec{p}) = 0$$

If you consider only the Klein-Gordon equation, you're introducing extra "solutions" that don't really solve the Dirac equation.

Why does this happen? You can regard the Klein-Gordon equation as the "squared" version of the Dirac equation. And when you square an equation, you always get this nasty false solutions: if you have e.g the equation $(x-3)=5$ the solution is $x=8$, but if you square it $(x-3)^2 = 5^2$ then you have two solutions, $x=8$ and $x=-2$. The first equation implies the second, but the converse isn't true.