Before spontaneous symmetry breaking, what is the difference between the fermion generations?

Question

Before spontaneous symmetry breaking (SSB), all fermions have the same mass (0). Across the 3 different fermion generations, all the (left) quarks doublets have the same weak isospin, and all the up quarks have the same hypercharge (same for down quarks). This discussion also applies to the leptons.

After SBB you can distinguish between an up quark (generation I) and a charm quark (generation II) because of their mass. But here, we don't even have that difference. All quantum numbers are the same.

Can you actually distinguish the 3 fermions generation before SSB? If not, is there a symmetry at play here?

Answer 1

There are a couple common points of confusion which I would like to address.

Not all interactions are dictated by symmetries. Yes, symmetry is of course always important to physical understanding. And yes, the Standard Model is a gauge theory, and so the interactions of the fields with the gauge bosons are dictated by their representations under the gauge groups. For simplicity, just think about a 4d theory of some real scalar fields $\phi_i$, $i=1..n$. Their dynamics are governed by some Lagrangian of a general form $$ \mathcal{L} = m_i^2 \phi_i^2 + \rho_{ijk} \phi_i \phi_j \phi_k + \lambda_{ijkl} \phi_i \phi_j \phi_k \phi_l, $$ where I have used the freedom from the broken $O(n)$ to diagonalize the mass term (that is, to go to the mass basis). The coefficients $\rho_{ijk}$ and $\lambda_{ijkl}$ and symmetric tensors which parametrize some general interactions between these fields. The strength of these interactions are not governed by any 'quantum numbers', but are rather free parameters in this theory.

And indeed, this is just the situation with the Yukawa couplings in the Standard Model. Recall the Yukawa sector of the SM Lagrangian, $$ \mathcal{L} = y^u_{ij} \tilde{H} Q_i \bar u_j + y^d_{ij} H Q_i \bar d_j + y^e_{ij} H L_i \bar e_j, $$ where I prefer the convention of working with left-handed Weyl fermion fields $\lbrace Q,\bar u, \bar d, L, \bar e \rbrace$. The sizes of the Yukawa couplings $\lbrace y^u, y^d, y^e \rbrace$ are likewise not governed by any symmetry demands, but are instead free parameters of this theory which are to be fit by experiment. Indeed it is true that after the Higgs condenses and gets a vev $v$, these Yukawa interactions provide mass terms for the charged fermions $m^u_{ij} = y^u_{ij} v/\sqrt{2}$ etc. And this is how we fit the eigenvalues of the Yukawa matrices.

But to some extent that is getting ahead of ourselves as far as this question is concerned. In the unbroken phase, these couplings still have an independent life of their own! Indeed, even if we modified the theory such that electroweak symmetry were never broken, these are still real physical parameters. They control the interactions of two fermions, say $Q_i$ and $\bar u_j$ with the Higgs field $H$. For example, with these numbers as inputs into our Lagrangian we could calculate how likely it is if I collide a $Q_i$ and a $\bar u_j$, that I produce a Higgs boson. Or you can think about the early universe when there were lots of Higgses in the hot, dense plasma, and this coupling will tell you how often one of those Higgses will collide with a $Q_i$ to produce a $\bar u_j$.

Let me make a final comment about another more-elementary-particle-physics way one may like to think about this. After electroweak symmetry breaking, when the Higgs gets a vev and you go to the mass basis, the Yukawas still control the coupling of the fermions to the Higgs boson. The Higgs boson mediates a Yukawa force just like the pions do---it's just that the Higgs is much more massive, so the range of this Yukawa force is very small. So these Yukawa couplings control the size of the Yukawa force between the different fermions which is mediated by the Higgs field.

TL;DR: Before or after electroweak symmetry breaking, the different generations of SM fermions interact differently with the Higgs field, and so can be distinguished.

Answer 2

After SBB you can distinguish between an up quark (generation I) and a charm quark (generation II) because of their mass. But here, we don't even have that difference. All quantum numbers are the same.

All quantum numbers are the same, but you can tell the difference between a u and a c, because you effectively already put it in by hand, with malice aforethought, for that very purpose! You have observed the low mass of the u, and the higher mass of the c, and you have, plausibly, but arbitrarily, really, decided to assign them to the I and II generations, respectively.

You structure the arbitrary Yukawa couplings, then, in $$ -Y^d_{ij}\bar Q^i H d_R^j -Y^u_{ij}\bar Q^i \tilde H u_R^j + \hbox{h.c.}, $$ such that, after SSB, suppressing the generation indices of the Yukawa matrices, the mass part of the above collapses to the celebrated mass terms $$ -{v\over \sqrt 2} ( \bar d _L Y_d d_R + \bar u _L Y_u u_R ) + \hbox{h.c.}, $$ dialed to yield (postdict) the observed answers, as follows.

The two matrices $Y_d$ and $Y_u$ matrices are biunitarily diagonalizable, as your text probably explains, $$ Y_d= U_d M_d K_d^\dagger, \qquad Y_u= U_u M_u K_u^\dagger, $$ so the smallest and middle eigenvalue of the diagonal $M_u$ are then proportional to the masses of the u and c, after you have absorbed their adjoints into the definitions of the quarks, thus defining the mass basis $$ d_R\to K_d d_R; \qquad u_R\to K_u u_R;\qquad d_L\to U_d u_L;\qquad u_L\to U_u u_L. $$

(You then construct the CKM matrix $U^\dagger_u U_d$, but that outranges your question.)

• The takeaway is that the differentiation between the u and the c is already there from the very start, at the level of the Yukawa matrices Y, an input to be suitably (implicitly) fitted through post-SSB observation. Nobody has ever plausibly derived those from first principles (yet). But the Yukawa matrices' fitted eigenvalues are there before SSB.