NRQCD: Why are quarks and anti-quarks treated independently?

Question

I am studying these lectures on effective field theories and I am having some problems to understand how the Non-Relativistic QCD (NRQCD) Lagrangian is constructed.

This theory is often used to describe $c\bar{c}$ and $b\bar{b}$ mesons. The idea is to decompose the heavy quark spinor $Q$ in a large and small component:

$$Q=e^{-i m_Q t}(\psi+\chi)$$ where $$\psi=e^{i m_Q t}\frac{1+\gamma_0}{2}Q \qquad \rm{and}\qquad \chi=e^{i m_Q t}\frac{1-\gamma_0}{2}Q.$$

Making this substitution, I can show that the QCD Lagrangian is given by

$$\mathcal{L}_{\rm QCD} \ni \bar{Q}(i\gamma_\mu D^\mu-m_Q) Q=\psi^\dagger\left(i{D_0}+\frac{\mathbf{D}}{2 m_Q}\right)\psi + \mathcal{O}\left(\frac{1}{m_Q^2}\right),$$ where I have neglected terms of order $1/{m_Q^2}$ and I have used the equation of motion to eliminate the small component $\chi$.

Now, my question:

After these manipulations, the author of arXiv:0308266 adds a second term with exactly the same form, but with a spinor $\zeta$ in the place of $\psi$. In his words, "$\zeta$ is the large component of the anti-quark field" (cf. eq. 75). The same anti-quark field appears later in other interaction terms (cf. eq. 77).

In my opinion, the insertion of an anti-quark field has no sense at all, because quark and anti-quarks are just different excitation of the same field. So, I think we should only keep the term with $\psi$. Where do I get wrong?

Answer 1

In a relativistic theory quarks and anti-quarks are described by the same four-component spinor field. Once we pass to a non-relativistic EFT these two physical excitations are described by two separate two-component Schroedinger fields. One is obtained by expanding $Q=e^{-im_Qt}(\psi+\chi)$, the other by expanding $Q=e^{im_Qt}(\zeta+\phi)$. Charge conjugation invariance of the original relativistic theory leads to some obvious relations between the couplings of $\psi$ and $\zeta$ (they have opposite charge). The possibility of annihilation implies that there must be higher order terms like $\psi^\dagger\zeta\zeta^\dagger\psi$.

Physically it is pretty clear what we are doing here: Consider a heavy $Q\bar{Q}$ bound state. If the life-time is sufficiently long we can view both the $Q$ and the $\bar{Q}$ as non-relativistic particles. Obviously, the $\bar{Q}$ is not some kind far-off-shell $Q$, but a separate field, just like the $Q$ but with the opposite charge.