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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?

Phibert
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  • 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

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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.

Luboš Motl
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  • 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
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    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
  • Well, you're not really "forgetting it". The $p$ or $\partial$ is an operator, and you want to know the value of the operator - it multiplies the "state vector", in this case $\Psi$, by something. It has some eigenvalues. The point is that we replace $\partial_\mu$ by $ip_\mu$. The exponential isn't replaced, it isn't erased, and it isn't newly created, either. It was there at the beginning and it's there at the end. Do you understand the concept of eigenstates and eigenvalues - in linear algebra and quantum mechanics? – Luboš Motl Jan 24 '14 at 12:59
  • You could say that the eigenstate equation $({\rm Operator}-{\rm Eigenvalue})V = 0$ may be treated by "forgetting" $V$ i.e. by dividing it by $V$ except one can't really divide by vectors. The vector is defining the state we study and it always stands on the right end of the product. What we're interested in is what the operators are doing with the vector. The equation "operator is equal to the eigenvalue" is only true if both terms act on a particular vector, the eigenstate, it is never true in general (as an operator equation). – Luboš Motl Jan 24 '14 at 13:02
  • Yes, the problem is more about physical understanding of aspects of QFT. Our "state vector" $\psi$ is $u(p^{\mu})e^{-ipx}$. The general eigenvalue equation is $O \psi = E \psi$ (for operator O and eigenvalue E of course), so why aren't we getting $\psi$ back out the other end and only $u$? Essentially, why is it we have $\partial_{\mu}\psi = p_{\mu}u$? (Some factors of i floating around also) – Phibert Jan 24 '14 at 13:13
  • We are getting both the exponential and $u$ on both sides! For reasons that seem irrational to me, you just decided to write the LHS in terms of $\psi$ and RHS using $u$ but if you use the same variables, you will either have $\psi$ on both sides or $u(p)\exp(...)$ on both sides, it's the same thing, after all. – Luboš Motl Jan 24 '14 at 13:49
  • This is exactly what I thought. It isn't me writing these equations; I'm taking them from lecture notes and trying to understand why they are irrational like you say and not writing $\psi$ on both sides. – Phibert Jan 24 '14 at 14:29
  • OK but there's no sharp problem here, is there? If $\psi=u\cdot \exp(..)$, you may always replace one by the other and vice versa, right? – Luboš Motl Jan 24 '14 at 15:20
  • Yes, that is very far from the problem if you're just saying I can write $\psi$ or $u \cdot exp(..)$ in the equation. My problem was/is: for the Dirac equation $(i\gamma^{\mu}\partial_{\mu}-m)\psi=0$ and wave function for a free fermion $\psi = u(p^{\mu})e^{-i p \cdot x}$, how do we then get the equation $(\gamma^{\mu}p_{\mu}-m)u(p )=0$? – Phibert Jan 24 '14 at 17:33
  • Dear @user13223423, they seem like manifestly identical equations. In the Dirac equation, the action of $i\partial_\mu$ has the same effect on the exponential hiding in $\psi$ as $p_\mu$, so what you called the Dirac equation is clearly equivalent to $(\gamma^\mu p_\mu - m)\psi=0$. Substitute $\psi=u\cdot \exp(\dots)$ and divide the equation by the exponential, if you wish, to get the last equation you wrote down. Clear? – Luboš Motl Jan 24 '14 at 18:05
  • Yes, that is what I meant when I said above "forgot about it". As in we can just use the 0 on the RHS to eliminate it from the equation. – Phibert Jan 24 '14 at 18:32
  • You may surely divide an equation by a nonzero number (scalar), if that was the question. Do we really need to make this issue controversial? Note that the "ket vector" on which the operators act isn't just the exponential but it also contains the $u$, so we are not dividing by the whole vector. – Luboš Motl Jan 24 '14 at 18:55
  • I completely understand everything I asked about and did so after my 2nd comment. I think was a misunderstanding and my last comment intended to drop the issue. Your answer to the original question was very good. – Phibert Jan 24 '14 at 19:09
  • Excellent, sorry if I could have been more efficient. – Luboš Motl Jan 24 '14 at 19:17