From one point of view I have to admit from the start that this question might be a trivial calculation problem but the truth is, I'm quite stuck. I am studying chiral perturbation theory with the aim of constructing the $\chi PT$ Lagrangian. For this reason I examine first the appearance of the quark condensate.

In more details, we have the pseudoscalar density $P_i= \bar q \gamma_5 \tau_i q$ and we are working with the up and down quarks only, so the full symmetry of the QCD Lagrangian is $SU(2) \times SU(2) $.

My problem is on how should I calculate the VEV of P. That is, the fact that the Pauli matrices $\tau_i$ are $2 \times 2$ dimensional and the $\gamma_5$ is $4 \times 4$. I work on the Dirac basis for the $\gamma_5$ so that it is zero on the diagonal and $I_{2 \times 2} $ in the other entries. I don't know if I must somehow use the fact that the u and d are Dirac spinors or, as I find in page 81 of https://arxiv.org/abs/hep-ph/0210398 use a relation similar to relation 4.17 - which in fact I don't understand.


To put it on a more mathematical picture, what we wish to calculate is something of the form $$\bar q \gamma_5 \tau_3 q $$ where q is $$q= \left( \begin{array}{c} {u \\ d } \end{array} \right) = \left( \begin{array}{c} u_1 \\ u_2 \\ u_3 \\ u_4 \\ d_1 \\ d_2 \\ d_3 \\ d_4 \end{array} \right),$$ that is, both u and d are Dirac spinors. But I have in mind that the gamma matrix is $4 \times 4 $ and the Pauli matrix is $2 \times 2 $, which seems ok to me because it acts only on the isospin u and d components.

So, how should I realize the action on the $\gamma_5 $ matrix on the $q$ vector, which is in fact a two-component Dirac spinor?

Thank you.

  • $\begingroup$ Ummm... the Ps are QCD interpolating fields for the pseudo scalar mesons, the pions, while the S's for the scalar mesons... the sigma.... so (4.17-18) connect the Ps with the Ss through the action of an axial charge, which does not annihilate the vacuum.... It is the r.h.s. of (4.17-18) that has a nontrivial v.e.v., the chiral condensate, and you don't calculate it, unless you do lattice gauge theory. The table of p 81 tells you what it is, and you correlate it with properties of the pseudoscalars... $\endgroup$ Commented Jul 20, 2017 at 22:10

1 Answer 1


As I indicated, I don't wish to make faces at your review article, but the text by TP Chang & LF Li , Ch 5.4, 5.5, is less unfriendly.

In any case, you appear lost, and a trail map of the lay of the land is in order so I'll be schematic, as all the correct formulas are there, but their connectivity appears lost... All γ matrix indices in the direct product $\gamma \otimes \tau_a$ are saturated by spinors, so you are only dealing with flavored scalar and pseudoscalar combinations Sa, Pa , and flavorless such: S, P. (So, to answer your edit side-question, the $\gamma_5$ action is flavor diagonal: it connects us with us and ds with ds.)

The point is non-perturbative QCD produces a scalar condensate, that is, only the Ss pick up a v.e.v. and never the Ps, the last line of table 4.1, p. 81, so $\langle S\rangle = \langle \bar{q} q\rangle \neq 0$, what normal people call the σ.

So, to start with, you are wondering how to compute a 0, which is actually a given. QCD, as a vectorlike theory, cannot break parity. For starters, it cannot break isospin either (which is broken explicitly by the small quark masses), so relate the vector components of $\langle S_a\rangle=0$, to produce additional constraints on bilinears, as he does in (4.15), for isospin just $\langle \bar{u} u \rangle = \langle \bar{d} d \rangle $.

You are not computing the v.e.v. of pions, but, effectively, of bilinears of such objects (each a quark bilinear). Without summation over repeated isospin indices, (4.17,8), $\langle \delta_a P_a\rangle\neq 0$, yields the non-vanishing unintegrated version of (4.19). The axial charge cannot annihilate the vacuum, and neither can the pseudscalar density Pa.

Now, the pseudoscalar bilinears $P_a$ are the QCD interpolating fields of the 3 pions, π, the Goldstone fields of this spontaneous/dynamical chiral symmetry breaking, and they definitely do not have non-vanishing v.e.v.s--remind yourself of the chiral σ model in previous sections, even though they do not annihilate the vacuum, by above.

But they do know about the vacuum... Instead of killing it, excluded above, they slip in and out of it, via the most important equation of the article--and any article on the subject--(4.19), PCAC, the mother of all soft pion theorems in current algebra, $$ \langle 0| A_\mu^a (0) |\pi^b (p)\rangle= ip_\mu F \delta _{ab}. $$

The axial field in the middle is the chiral charge current, so, in Fourier transform, its 0th component integrated over space would give 0 acting on the left, If the vacuum were chirally invariant---but it isn't! (Actually, the chiral charge has infrared troubles in its definition, as per the Fabry-Picasso theorem, but let's not fuss here...) What it does is generate a pion in this nonlinear realization, which annihilates the pion it acts on, on the right.

So, taking the divergence of this current in Fourier space, we get $m_\pi^2$ on the right-hand side, and that should vanish in an ideal world (CAC) in which $\partial^\mu A^a_\mu=0$. But this is the dirty real world in which the chiral symmetry is approximate up to quark masses, so the pion mass may survive to be nonzero (P for "partially") as long as there are small "seed" quark masses to prevent it from being a goldston, as required by Goldstone's theorem: Pions are pseudo-goldstons.

Their v.e.v. is vanishing in a single vacuum, |0>, but the chiral charge jams into the vacuum and creates almost (pseudo) zero frequency states of them.


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