Fourier Coefficents in general solution to Klein-Gordon Dirac-equation?

The most general solution to the Klein-Gordon equation is written as

\begin{equation} \Phi(x)= \int \mathrm{d }k^3 \frac{1}{(2\pi)^3 2\omega_k} \left( a(k){\mathrm{e }}^{ -i(k x)} + a^\dagger(k) {\mathrm{e }}^{ i(kx)}\right) \end{equation}

where I guess the second part was added to make the solution real, i.e. $c+c^\dagger= 2Re(c)$, is this correct?

The general solution to the Dirac equation is written \begin{equation} \Psi = \sum_r \sqrt{\frac{m}{(2\pi)^3}} \int \frac{ d^3p}{\sqrt{w_p}} \left(c_r(p) u_r(p) {\mathrm{e }}^{-ipx}+ d_r^\dagger (p) v_r(p) {\mathrm{e }}^{+ipx} \right) \end{equation}

and to the Dirac-adjoint equation \begin{equation} \bar \Psi = \sum_r \sqrt{\frac{m}{(2\pi)^3}} \int \frac{ d^3p}{\sqrt{w_p}} \left(c_r^\dagger(p) \bar u_r(p) {\mathrm{e }}^{+ipx}+ d_r (p) \bar v_r(p) {\mathrm{e }}^{-ipx} \right) \end{equation} and I would be interested in how the naming of the Fourier coefficents is justified in the first place. I know that they are interepreted in terms of creating and annihilating particles and anti-particles in QFT, but why do we name the coefficents here $c$ and $d^\dagger$ and not for example $c$ and $c^\dagger$ for $\Psi$, just as for the scalar case?

You didn't write the correct general solution of the KG equation for a complex scalar field. Your $a^\dagger$ should be $b^\dagger$. See, for example, Eq. 3-37 in this PDF book by R. Klauber.

That book also motivates that the operators $a$ & $a^\dagger$ are destruction & creation operators for particles, while $b$ and $b^\dagger$ are the same for the corresponding antiparticles. These are essential for a theory describing our world.

Also, if your solution $\Phi$ solves KG, so does $\Phi^\dagger$. One typically uses both of these solutions in spin-zero relativistic QFT.

Hopefully this addresses your Dirac equation solution question.

• Thanks for your answer. What I wrote is, I think, the general solution for a real scalar field. A complex scalar field is equivalent to two real scalar fields and if one wants to consider just one scalar field, a real scalar field is used. Your answer helps me because now I understand, that the general solution would have $a$ and $b$ in it, but with the constraint for a real scalar field the $b$ can't be arbirary but must be $a^\dagger$. Nevertheless I'm still unsure, why its $a$ and $b^\dagger$ and not $a$ and $b$.I will have a look at the pdf – jak May 24 '14 at 8:37
• Hindsight. You could call it $a$ and $b$ in the solution, but in the end you'd interpret $a$ as a lowering operator and $b$ as a raising operator. So it makes sense to call the coefficient $b^\dagger$ for consistency. (Also, I'll update my answer to be more specific about complex vs real.) – BMS May 24 '14 at 8:39
• Klauber simply states: "... and a complex conjugate form for the coefficient of the last term above, i.e., $b^\dagger$ , is used because it will prove advantageous later". Is this because in general we have simply $a$ and $b$, but if we use this, we will see, using the commutator of QFT that $b$ creates antiparticles, whereas $b^\dagger$ destroys antiparticles, which would be a bit confusion. With choosing the second coefficent to be $b^\dagger$, we assure that $b^\dagger$ creates, just like $a^\dagger$?! – jak May 24 '14 at 8:42
• That's my understanding. (It's a good book for self-studying QFT.) – BMS May 24 '14 at 8:43
• Choosing $b^\dagger$ as coefficient of the negative energy 1-particles solution instead of $b$ is related to the Feynman-Stueckelberg interpretation of the negative energy (-p) solutions of KG- and Dirac equation. "The absorption of a particle of 4-momentum $-p$ is equivalent to the emission of a particle of 4-momentum $p$". This way one gets rid of the problem of the negative energy solutions. To express the emission the $b^\dagger$ has to be used. – Frederic Thomas Aug 12 '17 at 23:56