2
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ If you have access to it, the discussion of antiparticles in Weinberg Volume I, in section 5.1, is quite excellent in motivating this form for the fields when one wishes to write down a Lagrangian field theory for a system with a conserved charge. In particular, the discussion surrounding Eq. (5.1.33). $\endgroup$ – Seth Whitsitt Oct 7 '20 at 18:57
1
$\begingroup$

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 $\phi \rightarrow e^{i\alpha} \phi$).

$\endgroup$
1
  • $\begingroup$ Thank you! This is extremely helpful! $\endgroup$ – UrsaCalli79 Oct 9 '20 at 0:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.