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: $$\begin{align*} \phi(x) &= \int \frac{d^3k}{(2\pi)^3} \frac{1}{\sqrt{2\omega_k}} \big( a(\vec{k}) e^{-ikx} + b^\dagger(\vec{k})e^{ikx}\big) \\ \phi(x)^\ast &= \int \frac{d^3k}{(2\pi)^3} \frac{1}{\sqrt{2\omega_k}} \big( b(\vec{k}) e^{-ikx} + a^\dagger(\vec{k})e^{ikx}\big) \end{align*}. $$ 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?

Answer 1

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: $$ \phi_j(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} (a_{j,p} e^{ipx} + a_{j,p}^{\dagger} e^{-ipx}). $$ Now transforming back to the $\phi, \phi^{\dagger}$ variables we get $$ \phi(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} \Big{(} \frac{a_{1,p} + i a_{2,p}}{\sqrt{2}} e^{ipx} + \frac{a_{1,p}^{\dagger} + ia_{2,p}^{\dagger}}{\sqrt{2}} e^{-ipx} \Big{)}, \\ \phi^{\dagger}(x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} \Big{(} \frac{a_{1,p} - i a_{2,p}}{\sqrt{2}} e^{ipx} + \frac{a_{1,p}^{\dagger} - ia_{2,p}^{\dagger}}{\sqrt{2}} e^{-ipx} \Big{)}. $$ 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 $\phi \rightarrow e^{i\alpha} \phi$).