# I Spontaneous symmetry breaking of chiral symmetry

1. Aug 6, 2016

### CAF123

The quark sector of the QCD lagrangian can be written as (restricting to two flavours) $$\mathcal L = \sum_{i=u,d} \bar q_i ( i \gamma_{\mu} D^{\mu} + m) q_i .$$ Write $q = (u d)^T$ and $$M = \begin{pmatrix} m_u & 0 \\ 0 & m_d \end{pmatrix}$$

Given that the masses of the u and d quarks are much less than the hadronic confinement scale of QCD, we can consider $m_u \approx m_d =m$ so that M is proportional to identity matrix. In this way, we have a U(2) symmetry $q' = U q$ where U is a U(2) transformation, $U = \exp(i\alpha_i \sigma_i)$. (I think it's correct to say that this U(2) symmetry is a vector U(2) symmetry because using Noether's theorem we get vector currents as conserved quantities arising from this U(2) symmetry) Using properties of U(2), this is then equivalent to the direct product of vector U(1) and vector SU(2). In this form, the larger symmetry group U(2) from making the masses of u and d degenerate encompasses the global phase redefinition of the quark fields (which we had prior to assuming degeneracy of quark masses) and SU(2) is denoted an isospin symmetry.

Did I understand that part correctly? If we now make the masses of the quarks to be zero, then the symmetry of QCD is even larger and it turns out we can write the lagrangian as a sum of two terms, each involving only left and right handed fields. Therefore we have two independent U(2) symmetries, $U(2)_L$ and $U(2)_R$ forming the symmetry group $U(2)_L \otimes U(2)_R$. My first question is many books say this symmetry is $SU(2)_L \otimes SU(2)_R$. Why is this? We can write $U(2)_L \otimes U(2)_R = SU(2)_L \otimes SU(2)_R \otimes U(1)_V \otimes U(1)_A$ which also incorporate the previously discussed vector U(1) symmetry and the new axial U(1) symmetry now present given M=0.

My second question is if someone could explain this statement from Wikipedia (and also mentioned in many books): 'The quark condensate spontaneously breaks the $SU(2)_{L}\times SU(2)_{R}$ down to the diagonal vector subgroup $SU(2)_V,$ known as isospin.'

Thanks!

2. Aug 6, 2016

### vanhees71

Yes, you got the group-theoretical part correct. Note, however that the $\mathrm{U}(1)_{\mathrm{A}}$ is broken by an anomaly, i.e., although the classical field theory has this symmetry, the quantized version doesn't (see Adler-Bell-Jakiw anomaly and the decay $\pi^0 \rightarrow \gamma \gamma$).

Spontaneous symmetry breaking means that a theory has a symmetry and a corresponding conserved Noether charge but the ground state of the theory is not symmetric under the full symmetry group but only under a subgroup. Indeed, lattice QCD calculations show that that the vacuum expectation value $\langle \bar{q} q\rangle \neq 0$, but this is obviously only invariant under $\mathrm{SU}(2)_{\mathrm{V}}$ transformations but not under the axial-vector isovector transformations. Thus $\mathrm{SU}(2)_L \times \mathrm{SU}(2)_R$ is spontaneously broken to $\mathrm{SU}(2)_{\mathrm{V}}$.

This is related to the hadronic world (the low-energy phenomenology of QCD so to say) via the corresponding currents, and these again are related to the lowest-mass scalar and pseudo scalar bosons, the $\sigma$ meson and the three pions. Now you can build an effective theory, where the isovector symmetries are realized as an SO(4). You can write down the theory in terms of the real fields $\phi=(\sigma,\vec{\pi})^{\mathrm{T}}$. Since we want to have the symmetry spontaneously broken, i.e., the $\sigma$ field, which has the quantum numbers of the quark condensate, should have a non-zero vacuum expectation value. Thus the vector subgroup of SO(4) is realized as the SO(3) rotations of the pion fields. The Lagrangian is still symmetric. Writing down the renormalizable terms allowed by the SO(4) symmetry leads to
$$\mathcal{L}=\frac{1}{2} (\partial_{\mu} \phi) \cdot (\partial^{\mu} \phi) + \frac{\mu^2}{2} \phi^2 -\frac{\lambda}{8} (\phi^2)^2= \frac{1}{2} (\partial_{\mu} \phi) \cdot (\partial^{\mu} \phi)-V(\phi).$$
Note that the mass term has the "wrong sign", and $\phi=0$ is not the stable stationary solution of the field equations. The minimum of the "mexican-hat potential" $V(\phi)$ is determined up to an SO(4) transformation, i.e., the ground state of the theory is degenerate. To get a formulation that is treatable in the usual perturbative way of QFT, we choose the vacuum expectation value arbitrarily in $\sigma$ direction and write $\sigma=v+\tilde{\sigma}$. Then expanding the potential around $v$, where it has a minimum, leads to a theory for a massive $\sigma$ meson and three massless pions. Indeed the symmetry is broken from SO(4) to the SO(3) rotations among only the pion fields, and this rotations leave the vacuum filed $(v,0,0,0)$ invariant. This implies that without cost of energy you can do the SO(3) rotations, and thus you have three massless excitations, the Nambu-Goldstone bosons of the spontaneous breaking for the chiral symmetry.

For a very readable introduction to this, leading to "chiral perturbation theory" as the effective hadronic low-energy theory of QCD, see

http://inspirehep.net/record/444848

3. Aug 7, 2016

### CAF123

Hi vanhees71, thanks for the detailed response.

I am just wondering in what sense it is a spontaneously broken symmetry though - the theory in which we have a $U(2)_L \otimes U(2)_R = SU(2)_L \otimes SU(2)_R \otimes U(1)_V \otimes U(1)_A$ symmetry is one in which the masses of the quarks are set to zero. So, there is no quark operator of the form $\bar q q$ in the lagrangian to begin with. The addition of masses then couples together the left and right handed components of the fields so the symmetry is no longer the direct product of two U(2)'s - would it therefore not make sense to say that the mass term softly breaks the chiral symmetry? (ie. we have a form of an explicit symmetry breaking rather than spontaneous). My reasoning went wrong somewhere so thanks if you could clarify! Or, is it the case that the ground state of the theory corresponds to the situation in which all the masses of the quarks are identically zero?

Thanks!

4. Aug 7, 2016

### vanhees71

Sure, in Nature chiral symmetry is only approximate and broken by the quark masses, but this explicit breaking is much less than the spontaneous breaking. While the quark masses are of order of some MeV the mass differences of chiral partners like the vector meson $\rho$ with a mass of around 770 MeV and the axialvector meson $a_1$ which is around 1200 MeV. Note that also the mass difference between u and d quark are of the order of magnitude of some MeV, and this means that isospin breaking is of the same order of magnitude of chiral symmery. That we don't see so obviously the chiral symmetry is rather to the spontaneous breaking than the explicit breaking.

5. Aug 7, 2016

### CAF123

I see. I suppose what I am not understanding is why there is any spontaneous symmetry breaking at all - in chiral QCD we have the symmetry group $SU(2)_L \otimes SU(2)_R$. In this theory, the chiral lagrangian has no operator $\bar q q$ (because M=0) so I'm not seeing the relevance of the non vanishing VEV of this operator to imply spontaneous symmetry breaking of $SU(2)_L \otimes SU(2)_R \rightarrow SU(2)_V$.

Thanks!

6. Aug 8, 2016

### vanhees71

The spontaneous symmetry breaking is due to the dynamical formation of the quark condensate, which is a non-perturbative property of QCD. It's due to the attractive nature of the strong interaction in this quark-antiquark channel. Now since $\langle \overline{\psi} \psi \rangle \neq 0$ the QCD vacuum is not invariant under the full chiral group. The "axial rotations" $\exp(-\mathrm{i} \vec{\alpha} \cdot \vec{\tau} \gamma_5)$ do not leave this VEV invariant, while the "vector rotations" $\exp(-\mathrm{i} \vec{\beta} \cdot \vec{\tau})$ do (here $\vec{\tau}=\vec{\sigma}/2$ are the generators of isospin rotations, i.e., matrices in flavor space).

7. Aug 8, 2016

### CAF123

I am not sure if the dynamical spontaneous symmetry breaking that is occuring here is understood in terms of a description outwith the lattice QCD remit but can the SSB be thought of in the following way? -

SSB of a symmetry means that the lagrangian of the theory exhibits the symmetry but the ground state of the theory does not. In this case our theory is chiral QCD described by $\mathcal L = \sum_{i=u,d} i\bar q_i \gamma_{\mu} D^{\mu} q_i$ . Are we then adding the term $\bar q q m$ onto the chiral lagrangian which can be seen as the analogous term of the Higgs potential like in SSB of free scalar field theories? The VEV of all terms are then taken and $\langle \bar q q \rangle$ is seen to be non vanishing? The ground state VEV is thus non zero and writing the $\bar q q$ in terms of left and right handed components we see that this term couples them thus the symmetry is no longer the direct product of two $SU(2)$'s.

Or perhaps in other language, $\langle \bar \psi \psi \rangle m$ is the order parameter for the phase transition such that for m identically zero it is zero but for any finite m it is non zero.

Is any of those descriptions correct?

Thanks!

8. Aug 8, 2016

### vanhees71

No, the term $\sum_{i} m_i \overline{q}_i q_i$ is explicitly breaking the chiral symmetry. Spontaneous breaking occurs for an exact symmetry and is indeed the case that the ground state is not invariant under the full symmetry group but the Lagrangian/Hamiltonian is. Indeed $\overline{q} q$ is an order parameter and $\propto$ the $\sigma$ field, which has a non-vanishing VEV. Have a look at the above cited review by Volker Koch which is really very pedagogical and nicely readable. Here's the direct link to the freely available preprint:

http://arxiv.org/abs/nucl-th/9706075

9. Aug 8, 2016

### CAF123

Thanks, that does indeed look quite tractable - do you know of a reference where I might find more details on why the non vanishing of $\langle \bar q q \rangle$ implies SSB of the chiral symmetry? I've only seen SSB in the context of a scalar field theory and gauge theories so this might be why I'm getting a little confused in tying things together with how formations of condensates can lead to SSB. I read a little in Schwartz's book too but I still feel I didn't fully understand.

Thanks!

10. Aug 9, 2016

### vanhees71

As I said, try Volker Koch's review on chiral symmetry. There it becomes quite clear, how you get from QCD to chiral effective models. The trick is to identify the currents in both models. After all a lot can be learnt from only using "current algebra" which was the way to address hadrons before QCD. Note that the idea of chiral models goes far back before QCD and even quarks were on the agenda.