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

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?

• Hmm? All the textbooks mention this relationship between Dirac and KG. Except yours, it seems. – arivero Aug 25 '15 at 11:58
• @arivero They all mention that the Dirac equation implies the Klein-Gordon equation (but not necessarily that KG implies Dirac). Either way they then go on to use the Dirac equation for everything, even though they are still using the KG equation for scalar fields (of course they could also turn the KG equation into a first order equation, but no one does this either). – Oscar Cunningham Aug 25 '15 at 12:11
• Note: Your argument does not show equivalence, it only shows a one-way implication. – ACuriousMind Aug 25 '15 at 23:55

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 $$(\gamma^\mu p_\mu - m)u(\vec{p}) = 0 \qquad\qquad (\gamma^\mu p_\mu + m)v(\vec{p}) = 0$$
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.
• @arivero Indeed! These restrictions are needed because the components of the spinor will mix under a Lorentz transform, and the Dirac equation links them together. Without them, every component of $\psi$ would be independent in every reference frame, so the spinor would be "4 objects in the $(0,0)$ rep" rather than "1 object in the $(1/2, 0) \oplus (0, 1/2)$" rep. – Bosoneando Aug 25 '15 at 15:58
• @Bosoneando I'm not suggesting applying the KG equation to the whole Dirac spinor! I'm just applying it to a single left handed spinor. Then we may construct a Dirac spinor by defining the right handed component by $\psi_R=\frac{i}{m}\bar{\sigma}^a\partial_a\psi_L$. This Dirac spinor will then obey the Dirac equation. To put what I'm saying in other words: The field theory of a Dirac spinor obeying the Dirac equation is the same as the field theory of a left handed spinor obeying the KG equation. – Oscar Cunningham Aug 25 '15 at 18:34
• @OscarCunningham lets consider the $m=0$ case since it is free of the component-mixing so we can avoid confusion. Then you have this Weyl equation: $\sigma^{\mu} \partial_{\mu} \psi_R = 0$. How is it equivalent to the KG equation? For starters, it is first-order in space-time derivatives. – Prof. Legolasov Aug 26 '15 at 17:21