Why is the creation operator of a particle in the conjugate field operator?

Question

I am learning QFT, and we discussed that to quantize a complex scalar field, we do this:
$$begin{align*} phi(x) &= int frac{d^3k}{(2pi)^3} frac{1}{sqrt{2omega_k}} big( a(vec{k}) e^{-ikx} + b^dagger(vec{k})e^{ikx}big)  phi(x)^ast &= int frac{d^3k}{(2pi)^3} frac{1}{sqrt{2omega_k}} big( b(vec{k}) e^{-ikx} + a^dagger(vec{k})e^{ikx}big) end{align*}. $$
To "motivate" this move in my own head, I told myself: "okay, since we have two fields, we need two different creation and annihilation operators. We can't use both $a$ and $a^dagger$ for the first field, or else the second field, being the conjugate, will only have $a$ and $a^dagger$ again. So maybe we use $a$ and $b$. But since one of them, say $b$, is a creation operator, we might as well call it $b^dagger$ instead (a dagger looks like plus-sign which means creation!)."
In any case, later we were told that $b^dagger$ creates an anti-particle while $a^dagger$ creates a normal particle. My question is why this is the case? While I admit it's nice that $a^dagger$ coincidentally still creates a normal particle just like the $a^dagger$ for a real scalar field, doesn't it seem like $b^dagger$ being a part of $phi$ and not $phi^ast$ should be the one to create normal particles?

jkb1603 · Accepted Answer

One motivation, which at least feels good for me is to consider a variable transformation to real fields $phi_1, phi_2$ through:
$$
phi = frac{1}{sqrt{2}} (phi_1 + i phi_2), qquad phi^* = frac{1}{sqrt{2}} (phi_1 - i phi_2).
$$
Then the Lagrangian becomes
$$
mathcal{L} = (partial_{mu} phi^*)(partial^{mu}phi) - m^2 phi^*phi = frac{1}{2} sum_{j=1}^2[(partial_{mu}phi_j)(partial^{mu} phi_j) - m^2 phi_j^2].
$$
Thus $mathcal{L}$ is just a sum two identical real scalar field Lagrangians (times a factor 1/2, which is irrelevant)! The usual quantised real scalar fields read:
$$
phi_j(x) = int frac{d^3p}{(2pi)^3} frac{1}{sqrt{2omega_p}} (a_{j,p} e^{ipx} + a_{j,p}^{dagger} e^{-ipx}).
$$
Now transforming back to the $phi, phi^{dagger}$ variables we get
$$
phi(x) = int frac{d^3p}{(2pi)^3} frac{1}{sqrt{2omega_p}} Big{(} frac{a_{1,p} + i a_{2,p}}{sqrt{2}} e^{ipx} + frac{a_{1,p}^{dagger} + ia_{2,p}^{dagger}}{sqrt{2}} e^{-ipx} Big{)},

phi^{dagger}(x) = int frac{d^3p}{(2pi)^3} frac{1}{sqrt{2omega_p}} Big{(} frac{a_{1,p} - i a_{2,p}}{sqrt{2}} e^{ipx} + frac{a_{1,p}^{dagger} - ia_{2,p}^{dagger}}{sqrt{2}} e^{-ipx} Big{)}.
$$
Now identifying $a_p equiv frac{a_{1,p} + i a_{2,p}}{sqrt{2}}$ and $b_p equiv frac{a_{1,p} - ia_{2,p}}{sqrt{2}}$ motivates the occurence of the operators and daggers.
For a better explanation Weinberg is certainly a great source.
To your second question. From the above it is not apparent, which particle should be considered the particle and which the anti-particle, in fact I think that this is just a convention (see e.g. Identification of particles and anti-particles). One can only show (quite readily) that the particles created by $a_p^{dagger}$ and $b_p^{dagger}$ have opposite charge (i.e. opposite eigenvalues of the conserved charge operator $Q$, corresponding to the symmetry $phi rightarrow e^{ialpha} phi$).

Why is the creation operator of a particle in the conjugate field operator?

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