### Flavour physics and you, part I: Origins

14Jan08

This is the first of a few posts on flavour physics aimed at the level of a physics student.

Quarks and leptons come in [at least, but probably only] three “flavors.” The up and down quarks make up just about all visible matter in the universe. The electron and neutrino mediate just about all of chemistry. But nature seems to have provided two heavier copies of these particles for us, and nobody is sure why there should be three copies (or generations/families). The study of this apparent redundancy and the interplay between the different copies is referred to as flavour physics.

Fig. 1. The Standard Model: three generations… why? Image from Symmetry Magazine.

Naively, these heavy generations shouldn’t affect the physics of the light generation that dominates our visible universe. But then I wouldn’t be interested in this field. 🙂 At the heart of flavour physics is the CabibboKobayashiMaskawa (CKM) matrix. [1]

Where the CKM matrix comes from

Let’s focus on the quark sector. Before spontaneous symmetry breaking, the Standard Model is a theory of massless fermions. The masses come from the Yukawa couplings to the Higgs boson in the Lagrangian:

$\mathcal{L} \supset y^u_{ij} H \bar{q_i} u_j + y^d_{ij} H^* \bar{q_i} d_j$

Here the $y$s are the Yukawa couplings, $H$ is the Higgs doublet, $q$ is the left-handed quark doublet, $u$ is the right-handed up-type quark, and $d$ is the right-handed down-type quark. If you’re confused about why the left-handed and right-handed particles have different letters, take a break and read a bit about chiral models (it’s important).

The indices $i$ and $j$ label generations: for example $u_2$ refers to the charm quark. To keep things simple, I’ve left out the hermitian conjugate terms. See note [2] below if you’re suspicious about the SU(2) indices.

The quarks obtain their masses through electroweak symmetry breaking. This just meas that the Higgs doublet obtains a vacuum expectation value which we may choose to be:

$\langle H \rangle = (v \hspace{.1in},\hspace{.1in} 0)$

Inserting this into the Yukawa couplings, we see that the ground state of the theory now contains mass terms for the quarks:

$\mathcal{L} \supset (M_u)_{ij} \bar{u_{Li}} u_{Rj} + (M_d)_{ij} \bar{d_{Li}} d_{Rj}$

I’ve explicitly labelled the chirality of the quarks so we can keep track of those that came from the left-handed doublet $Q$ and the right handed singlets $u,d$. (If you’re worried about how SU(2) indices worked out, this is because of the technical detail’ mentioned in note [2].)

Where the mass matrices are:

$M_u = vy^u$
$M_d = vy^d$

The first glaring thing is that these matrices are not, in general, flavor-diagonal. In order to find the physical states of a theory (up to perturbations), one needs to diagonalize the mass terms. The mass matrices can be diagonalized by unitary transformations $U$ on the fields:

$\bar{q_i} M_{ij}q_j = \bar{q_a} U^\dag_{ai} M^\mathrm{diag}_ij U_{jb}q_b$

For $q = u_L, u_R, d_L, d_R$ as appropriate. We have at our disposal the freedom to rotate our fields in flavour-space, so it seems like we can do this trivially by defining the ‘physical’ fields:

$\bar{u_{Li}}' = \bar{u_{Li}}[U^\dag(u_L)]_{ai}$
$\bar{d_{Li}}' = \bar{d_{Li}}[U^\dag(d_L)]_{ai}$
$u_{Rj}' = [U(u_R)]_{jb}u_{Rb}$
$d_{Rj}' = [U(d_R)]_{jb}d_{Rb}$

(Recall from linear algebra that we can diagonalize complex matrices with unitary transformations in this way.)

But alas! We’ve made a mistake. We do not actually have this much freedom. Recall that the left-handed up- and down-type quarks actually belong to the same SU(2) doublet! This means that the SU(2) symmetry of the Standard Model forces $u_L$ and $d_L$ to transform with the same unitary flavor matrix $U(Q_L)$.

Said in another way, the flavor symmetry of the quark sector is $SU(3)_Q\times SU(3)_u \times SU(3)_d$. We only have the freedom to make three rotations, not the four required to diagonalize the two (up and down) mass matrices!

By convention, we go ahead and diagonalize the up-type quarks. But this means the down-type mass matrix is not flavour diagonal, and the physical down-type quarks mix flavours. The mixing matrix is called the CKM matrix (after the folks mentioned above) and is, explicitly:

$V_{\mathrm{CKM}} = U(d_L)U^{-1}(Q_L)$

Where $U(d_L)$ is the matrix that would have diagonalized the down-type mass matrix.

Next time, we’ll start discussing how the CKM matrix affects your everyday life (kinda). See note [3] for furthe reading suggestions.

Appendix A

The unitarity of the CKM matrix allows you to write down some nice relations that can be represented graphically as the unitarity triangle.’ Bee and Stefan over at Backreaction wrote a very informative post about the unitarity triangle as part of the Backreaction Advent calendar.

I’ve been planning a few posts on the CKM matrix for some time now, but they got to it first. Perhaps noteably, this wasn’t the last time I’d be scooped on Backreaction’s Advent calendar. 🙂

Notes and References

[1] For those taking bets on Nobel prizes, by the way, the CKM trio is long overdue.

[2] To be completely accurate, $H^*$ should be $i\sigma_2 H^*$, since this is in the $\bar{\mathbf{3}}$ representation of SU(2). But this is a `technical detail’ for those not familiar with representation theory.

[3] Further References:

#### 4 Responses to “Flavour physics and you, part I: Origins”

1. 1 andy.s

I get the impression that if the CKM matrix was the unit matrix, then the generation of a quark would be a conserved quantity. The strange could go the charmed by the weak interaction, but never to the up quark.

If that were true, then would there be three kinds of matter? With ccs and css making a sort of heavy proton and neutron for example?

2. Hi Andy!

Indeed, the flavour-changing currents are proportional to the CKM matrix. So if the CKM matrix were equal to the unit matrix, then there would be no transitions between the generations. I.e. the different generations of quarks would indeed decouple*.

As you said, then the lowest-energy ccs, css, etc. baryons would then be stable particles. And we’d have three copies of the hadrons with different masses.

An interesting note: the up-down generation is the only one where the charge +2/3 quark is lighter than the charge -1/3 quark. It is because of this that the charge +1 proton is the only stable (it’s the lightest) baryon in the Standard Model. And hence we end up with atoms. (No hydrogen.) If the masses for the up-down generation were reversed, then the neutron would be the lightest and only stable baryon. And hence there’d be no atoms, and we wouldn’t exist.

So, in the hypothetical case you presented, the other copies of matter wouldn’t quite be the same as the matter that we see around us. You wouldn’t have charm-strange atoms because the ccs ‘proton’ would decay into a css ‘neutron.’

* – a caveat: if the CKM matrix were equal to the unit matrix, there would still be some inter-generational mixing in the leptonic sector via the analogous MNS matrix (responsible for neutrino oscillations).