# Why must the Dirac equation multiplied by its complex conjugate give the KG equation?

This may be a simple question. I can show this is the case mathematically but cannot explain why it happens. It was only when asked why this happens when I realised I couldn't explain it intuitively/physically.

So I suppose there must be some way in which the Klein-Gordan equation can be recovered from the Dirac equation, just like a lot of things in Physics. But why does this $\textit{have}$ to be done by multiplying by the complex conjugate and not some other operation?

• Wasn't Dirac's equation inspired by an attempt to take the square root of the Klein Gordon equation? – John Rennie Jan 24 '14 at 11:26
• As I'm aware yes. Wasn't it an attempt not to have second order derivates and hence the Dirac equation is only first order in derivates. But why should taking the square root of the KG equation now describe relativistic particles with different spin? Luck? – Phibert Jan 24 '14 at 11:29

The reason is as follows.

The scalar Klein-Gordon equation is (and has to be) a second-order equation because the box $$\square = \partial_\mu \partial^\mu$$ is the simplest Lorentz-invariant differential operator that may act on scalar fields. However, Dirac wanted to find a first-order equation and it is indeed possible for spinors because $$\partial \!\!\!\! / = \gamma^\mu\partial_\mu$$ is also Lorentz-invariant (doesn't change the transformation properties of the object it acts upon), it is a first-order operator, and it is able to act on spinors (spinors are needed because the operator contains the Dirac matrices which must contract their indices with a spinor).

By special relativity, the wave equations must enforce the correct invariant mass. So the de Broglie waves $$\exp(ip_\mu x^\mu ) u$$ multiplied by a special form of the spinor (only in the Dirac case) must be a solution for $p_\mu p^\mu = m^2$. In other words, special relativity implies that a solution to the mass $m$ Dirac equation must be a solution to the mass $m$ Klein-Gordon equation, too.

However, the Dirac equation is stronger because it constrains the first derivatives; the Klein-Gordon equation allows you to choose the wave function and its first derivative(s) and it only constrains the second derivatives.

In the momentum basis, the Dirac equation says $$(p\!\!\!/ - m) \Psi = 0$$ which says that the field $\Psi$ must be an eigenstate of "p-slash" with the eigenvalue $+m$. The condition $$p\!\!\!/ -m = 0$$ implies that $p^2=m^2$ because $$p^2 \equiv p_\mu p^\mu = p\!\!\!/ \cdot p\!\!\!/$$ due to the anticommutator $$\{\gamma_\mu,\gamma_\nu\} = 2g_{\mu\nu}\cdot {\bf 1}$$ However, $p\!\!\!/ =+m$ (when acting on $\Psi$) is not equivalent to $p^2=m^2$ (when acting on $\Psi$). The former condition is stronger, as I said. We may very well have $p\!\!\!/ =-m$ and this will still imply $p^2=m^2$ (when acting on $\Psi$). This is nothing else than the claim that there are two solutions to the equation $x^2=m^2$, namely $x=+m$ and $x=-m$.

So the "Klein-Gordon" condition $(p^2-m^2)\Psi=0$ acting on $\Psi$ is simply equivalent to $$(p\!\!\!/ +m) (p\!\!\!/ -m)\Psi = 0$$ which isn't more controversial than the formula $a^2-b^2=(a-b)(a+b)$ and this product may be written as "p-slash plus m" times the Dirac equation.

In other, more physical words, the Klein-Gordon equation allows the positive-energy vector $p^\mu$ as well as $-p^\mu$, its negative-energy opposite. However, $(p\!\!\!/ -m)\Psi=0$, the Dirac equation, only allows one of the signs of the energy and this sign is correlated with the up/down polarization of the electron/positron.

The "other" or "reverted" Dirac equation, $(p\!\!\!/ +m)\Psi=0$, imposes the opposite sign of the energy as the function of the polarization. It's clear that the Klein-Gordon equation (acting on the spinor) which doesn't care about the sign of energy is equivalent to saying that the energy's sign is either what the Dirac equation demands (as a function of the polarization); or the opposite sign. The opposite sign is obtained by the solutions of the "reverted" Dirac equation.

So $\Psi$ obeys the Klein-Gordon equation iff it obeys the Dirac equation; or the reverted Dirac equation; or if it is a combination of these two types of solutions. This is mathematically equivalent to saying that the solution to the Klein-Gordon equation is annihilated by the product of the "Dirac operator" and the "reverted Dirac operator".

The reversal of the Dirac operator is also linked to some kind of complex conjugation, more precisely the charge conjugation (C), because the C-conjugated spinor simply obeys the reverted Dirac equation. This may be checked by defining C more carefully and counting the signs.

• Perfect explanation. Incidentally, why can we just work "in the momentum basis" and change from $\not{d_{\mu}}$ to $\not{p}$ when dealing with the Dirac equation. Even better, why do we want to do this? – Phibert Jan 24 '14 at 12:05
• We are working with the momentum basis because for the plane wave (with a well-defined direction and frequency), the differential equations become just simple algebraic equations for simple numbers $p_\mu$ and the components of the spinor $u$. The partial derivative $\partial_\mu$ is replaced by $ip_\mu$ because that's the factor that jumps out when the differential operator acts on the exponential $\exp(ip_\mu x^\mu)$. Similarly, you may add the slashes both to $\partial_\mu$ and $ip_\mu$. Sorry if $i$ should be $-i$, don't want to pick my convention and fix the errors now. – Luboš Motl Jan 24 '14 at 12:16
• In regular quantum mechanics, the momentum operator in the position representation is $\vec p = -i\hbar \nabla$. In the momentum representation, however, $\vec p$ is just the operator that multiplies the wave function $\tilde \psi(\vec p)$ by $\vec p$ itself, so it's simpler. The "position operator" is a differential one in the momentum basis but as long as you talk about things that completely avoid the notion of the "position operator for a particle", things simplify in the momentum representation. – Luboš Motl Jan 24 '14 at 12:19
• Particle physics generally talks about momentum eigenstates because the "position eigenstates" for particles with relativistic $v$ can't even be realized. For example, if you try to localize an electron to a space smaller than the Compton wave length (something in between the nuclear and atomic size), you squeeze so much energy to the electron that you start to spontaneously create electron-positron pairs etc. and things are more complex. On the other hand, the initial states as well as final states on accelerators are known to be particles with measurable momenta and energies (not positions). – Luboš Motl Jan 24 '14 at 12:21
• Last thing: I have here the wave function $\psi(x^{\mu})=u(p^{\mu})e^{-i p x}$ and substituting into the Dirac equation we get $(\gamma^{\mu}p_{\mu}-m)u=0$. I understand that $p_{\mu}$ is thrown out of $\partial_{\mu}e^{i p_{\mu} x^{\mu}}$ but this leaves the exponential as well. Have we been able to just forget about it (as in use the 0 on the RHS)? – Phibert Jan 24 '14 at 12:48