I am confused about Chiral symmetry breaking and the terminology we use. First of all, I think the symmetry is started with taking quark masses zero and writing the Lagrangian as; $$ L=-\frac{1}{4}(F_{\mu \nu}^a)^2 + \bar{u}(i \not{D})u+\bar{d}(i \not{D})d $$ without any mass term. There is SU(2) isospin symmetry here, for sure. But if one write down quark fields as doublets, one conclude with $$ SU(2)_L\times SU(2)_R $$

Question1 : Why we start with SU(2) isospin symmetry and then enlarge it $$ SU(2)_L\times SU(2)_R? $$ (approach in many textbooks.) Why we need to declare the fields as doublets? The Lagrangian was already invariant under isospin transformations. And is this the so-called Chiral symmetry?(multiplets of left and right handed particles?)

Question2 If it is so, where is hypercharge conservation? How about $$ SU(2)_L \times U(1)_\Upsilon $$ symmetry? Is this still conserved while chiral SSB is broken?

Question3 If $$ SU(2)_L\times SU(2)_R \to SU(2)_F $$ is the isospin symmetry, the phrase "only left handed fermions interact via weak interactions" is only phenomenological or what? Do we see it in nature?

I may be confused electroweak symmetry breaking and chiral SSB. Anyway, correct me if I am wrong in any postulate.


2 Answers 2


Question 1

First of all, when you discuss chiral symmetry spontaneous breaking, you need to assume pure QCD theory.

QCD lagrangian with $u-,d-,s-$quarks (they have relatively small masses in compare with $b, t, c$-quarks) has the form $$ \tag 1 L_{QCD} = \bar{q}_{i}i\gamma_{\mu}D^{\mu}_{ij}q_{j} - \frac{1}{4}G_{\mu \nu}^{a}G^{\mu \nu}_{a} - \bar{q}_{ij}M^{ij}q_{j} $$ In high energy limis, when we may neglect the mass term in compare with kinetic term, $(1)$ is invariant under combined vector-axial $SU_{V}(3)\times SU_{A}(3)$ global transformation, precisely $$ q \to e^{iq_{V}t_{a}\epsilon_{a} + iq_{A}t_{a}\theta_{a}}q, $$ where $t_{a}$ is $3\times 3$ Gell-Mann matrices. Here I don't take into account broken by QCD anomaly vector-axial $U(1)$ part of restored symmetry.

By introducing left-right chirality projectors, we may state that $$ \tag 2 SU_{V}(3)\times SU_{A}(3)\sim SU_{L}(3)\times SU_{R}(3) $$ To summarize, we have important statement that $SU_{L}(3)\times SU_{R}(3)$ becomes exact symmetry at high energies. $SU_{L}(3)\times SU_{R}(3)$ group is called chiral symmetry group of QCD lagrangian.

This statement implies existense of extra physical states which have same values of spin, strangeness and the baryon number as experimentally observed ones, but with opposite parity. Since we don't see these states, but below energy scale $E\sim \Lambda_{QCD} \sim 0.1 \text{ GeV}$ we see approximately massless meson states, we have to require that aproximate $SU_{L}(3)\times SU_{R}(3)$ symmetry is spontaneously broken in QCD by itself down to $SU_{\text{diag}}(3)$ at scales $\sim \Lambda_{QCD}$. It is broken by quark bilinear form VEV, namely $\langle |\bar{u}u|\rangle \sim \Lambda_{QCD}^3$. In the result, we have pseudogoldstone bosons octet - $\pi^{0, \pm}, \eta^{0} , K^{0}, \bar{K}^{0, \pm}$.

Suppose now we neglect the $s-$quark contribution into such picture. Then we deal with SSB $SU_{L}(2)\times SU_{R}(2)$ down to $SU_{\text{diag}}(2)$. $SU_{\text{diag}}(2)$ in this case is the group of isospin transformation. It is not exact, however since masses of $u-,d-$quarks are not the same: for example, this drives the pure QCD reaction $$ d + d \to He^{4} + \pi^{0} $$ which doesn't conserve the isospin.

Questions 2 and 3

QCD chiral symmetry in principle has nothing common to electroweak symmetry. First of all, QCD chiral symmetry group is global symmetry of quark triplets transformation, left and right, while electroweak symmetry group is combined local $SU(2)$ transformation of left quark weak doublets and their local $U_{Y}(1)$ trasformation: $$ \begin{pmatrix} u \\ d\end{pmatrix} \to e^{i\frac{1-\gamma_{5}}{2}Q_{L}\sigma_{a}\epsilon_{a}(x) + iQ_{Y}\theta (x)}\begin{pmatrix} u \\ d\end{pmatrix} $$ Electroweak symmetry possesses interactions, while chiral QCD symmetry just state global conservation of chiral charges (i.e., there isn't associated interaction). But if we include electroweak symmetry to $(1)$, then even if we completely neglect the quark masses, the chiral symmetry becomes explicitly (on the level of action), not spontaneously (on the level of states which is described by action) broken because of chiral nature of electroweak group. For example, electromagnetic breaking of isospin symmetry, which breaks $SU(2)$ isospin symmetry down to symmetry transformation generated by $\sigma_{3} = \begin{pmatrix} 1 & 0 \\ 0 & -1\end{pmatrix}$, makes the contribution in the masses of $\pi^{\pm}$ mesons, so that they become different from the mass of $\pi^{0}$ meson, while QCD predicts precisely exact equality of these masses.

Next, electroweak group symmetry is also spontaneously broken at scale $\approx 240\text{ GeV}$, namely $$ SU_{L}(2)\times U_{Y}(1) \to U_{EM}(1), $$ by VEV of Higgs field doublet $\langle|H |\rangle \sim v$. The scales $\Lambda_{QCD}, v$ are different, and in general has different nature.

Finally, the fact that only left charged currents (i.e., $\bar{u}\gamma_{\mu}\left(\frac{1- \gamma_{5}}{2}\right)d$) interact with electroweak sector comes from the experiment. Your statement that only left fermions interact electroweakly is incorrect, since all electrically charged right particles interact with EM field as well as left, and all right particles interact with $Z-$boson too.

  • $\begingroup$ A small comment: eqn (2) is bound to confuse, and has confused. Please see Gell-Mann's original resolution of the current algebra. L and R currents commute, but never Vs with As, so there is no such thing as an SU(N)_A group. The coset space in χSB is not a group. $\endgroup$ Commented Jan 16, 2017 at 15:20
  • $\begingroup$ SU(2) singlets with zero charge ("sterile" right-handed neutrinos / left-handed anti-neutrinos) do not interact with the Z-boson. $\endgroup$ Commented Oct 1, 2019 at 3:51
  • $\begingroup$ what is $q_A$ and $q_V$? Are $\epsilon_a$ and $\theta_a$ only real numbers? $\endgroup$ Commented Jul 25, 2021 at 21:20
  1. There is no need of starting with $SU(2)$ symmetry and then extending it to $SU(2)_L\otimes SU(2)_R$. In fact, the moment you write down the Lagrangian, the symmetry is by default $U(2)_L\otimes U(2)_R$ with $U(2)_L$ acting on the left handed quark doublet $(u , d)_L$ and $U(2)_R$ acting on the right handed quark doublet $(u , d)_R$, which can be extended to $SU(2)_L\otimes SU(2)_R \otimes U(1)_L \otimes U(1)_R$. This is due to the fact that the left handed and right handed transformations on the respective doublets independently leave the Lagrangian invariant since the mass term has been taken to zero. The quark doublets are taken because the fundamental representation of $SU(2)$ acts upon the doublets, whereas $U(1)$ symmetry can be enforced by a simple phase factor. So $SU(2)_L\otimes SU(2)_R$ is an inbuilt feature and not an enforced one.

Chiral symmetry is referred to as invariance under parity. Here the Lagrangian is parity-invariant, so it has chiral symmetry.

  1. The electroweak symmetry is taken to be already broken here, although it is not relevant that much for this context. The up quark and down quark masses are of the order $2-4$ MeV and can be safely ignored and hence the chiral symmetry of the Lagrangian can be assumed. For the energy scale of phenomena we are considering, we can even set the $s$ quark mass to zero, and get the symmetry group to be $U(3)_L\otimes U(3)_R$ but it is not taken. For that case, we would have the fundamental representation of $SU(3)$ matrices acting on triplets. In a nutshell, since this is a low-energy limit, therefore electroweak symmetry breaking has already taken place at a much higher scale of about $250 GeV$, but the masses of quarks are not that significant and so they don't matter that much. So this description is there keeping in mind that EWSB has already taken place.

  2. Here $SU(2)_L$ doesn't refer to the weak interaction but the symmetry arises due to the structure of the chosen Lagrangian. As said earlier, if you take the $s$ quark into consideration too, then what you will get is the invariance under $SU(3)_L$. Moreover, here we are considering the structure of the strong interactions and that is why the Lagrangian is parity- invariant, which is not the case for the weak interaction. Along with this, the addition of the left and the right hand Dirac currents of this Lagrangian give the baryon number and isospin currents. We then assume the axial vector currents to be spontaneously broken thereby giving rise to Goldstone bosons, i.e the pions. The pions should be massless, but we can argue that since the initial premise of quarks being massless was not entirely true therefore they have some residual masses. As you can see, this process is different from electroweak symmetry breaking.

So we see pions in nature, which is the out product of chiral symmetry breaking, and which is different from electroweak symmetry breaking's product of vector bosons getting masses from Goldstone modes. Notice that EWSB utilizes a scalar potential in order to break the symmetry. However, in the QCD vacuum, since the quarks and antiquarks have strong attractive interactions, the energy cost of creating a quark-antiquark pair is very small. As a result, the vacuum state with a quark pair condensate is characterized by a non-zero expectation value, which signals the symmetry breaking of the full symmetry group down to the group of vector symmetries only, therefore resulting in the symmetry breaking of isospin currents and the conservation of baryon number. See Peskin and Schroeder Chapter 19 for more details, as it provides a very crisp description.


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