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

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?

• This is not how anti-particles are treated in quantum field theories in general. Even electrons and positrons are treated as separate. This becomes more apparent in electro-weak theory where the components have manifest asymetries. In general any particle or anti-particle (represented usually as the complex conjugate field) is treated as separate and independent. In the degenerate case of charge-less particles (with any meaning for charge), it means a particle of exactly-same type. – Nikos M. Jun 1 '14 at 4:41
• Do you agree that we have only one term like $\bar{Q}(i\gamma_\mu D^\mu-m_Q) Q$ in the QCD lagrangian? This term will give the effective lagrangian that I have written. My question is where this new term with $\chi$ comes from? Do we need to put it by hand? Why? – Melquíades Jun 1 '14 at 12:05
• That is what i am saying, that anti-fields are treated as independant. So you add the "small component" $\chi$ by hand also. This is exactly like the original Dirac equation. – Nikos M. Jun 1 '14 at 12:59
• @NikosM.: While the conjugate fields, say $\bar{\psi}$, are suggestive of antiparticles they are not quite antiparticles, but instead each of $\psi$ and $\bar{\psi}$ contain both an antiparticle and particle piece in them? – JeffDror Jun 2 '14 at 19:03
• yes true, my comment was off. was thinking the non-relativistic treatment of the Dirac equation for the electron (where both the "large" and "small" components are entered manifestly). you are both right – Nikos M. Jun 4 '14 at 0:37

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.