I am currently studding theory for Chiral Perturbation Theory( ChPT) and I have stumbled at two point, having a difficulty at understanding. I will present the problem at first and then phrase the question which will be again summarized at the end.

In these two previews of the subject (http://arxiv.org/abs/1001.3229 and http://arxiv.org/abs/1105.2919) the introduction to constructing the effective Lagrangians is almost the same. So, after discussing the unbroken chiral symmetry of QCD in the limit of two (up and down) massless quarks the writers turn their attention to the procedure of constructing the effective Lagrangian. Both papers show that we need to expand the effective Lagrangian in terms of pionic degrees of freedom, the scalar fields that describe the particles created as Goldstone bosons from the spontaneous symmetry breaking.

My problem now comes: to discuss the effective Lagrangian both writers introduce a matrix $U$ that collects the pionic fields and use this matrix (2x2 if I understand correctly) to write down the Lagrangian, where the matrix $U$ is something of the form:

$$U= 1+ {{i} \over {f_π}} τ \cdot π – {{1} \over {2f_π ^2}} π^2 – {{ia} \over {f_π ^3}} ( τ \cdot π)^3 + …. $$

So the Lagrangian in Leading Order(LO) is $$L_{ππ} ^{(2)} = {f_π ^2 \over 4} tr[\partial _μ U \partial ^ μ U^ + +m_π ^2( U + U^+)] .$$ This is from the second paper, where there is no discussion where that matrix $U$ came from and did we construct it. On the first paper it begins at page 11 by discussing the need of realizing the chiral group non-linearly because $SU(2)_L \times SU(2)_R $ is isomorphic to $SO(4)$ so that we need four dimension for describing rotations but we only have the three “dimensions” of the pionic fields. To bring the four vector $( **π** , σ) $ in a non-linear form the author examines the effect of $SO(4)$ rotation by six angles and expresses all the four-vector terms by $π$. Then is writes: “As a last remark note that the four-dimensional rotations of $(π, σ)$ can be conveniently written using the 2 × 2 matrix notation by introducing the unitary matrix $U$”. Then it uses the $U$ matrix as mentioned before.

Leutwyler in his classic paper (http://arxiv.org/abs/hep-ph/9311274v1) discusses all this: he writes that we can think of the pionic fields as coordinates in the quotient space $G/H$ where $G$ is the Lie group space for which the Hamiltonian or Lagrangian is invariant but $H$ is the subgroup of $G$ so that only the ground state is invariant. He also says that we replace the pionic $π$ field by $U$ because the last has linear transformation properties.

Questions: So the questions are:

  1. How do we construct the matrix $U$?

  2. How can I understand that mathematically we can collect all the pionic $π$ degrees of freedom in a 2X2 unitary matrix that has some relation with the action of $SO(4)$ rotations on the four vector of the pionic fields?

  3. Is this four vector the same as asking how can we treat the pionic fields as coordinates (and understand the geometric argument for the non-linear realization of chiral group)?

  4. Is this something that can always be done? I mean can we always construct a vector space where the fields that don' t satisfy the symmetries of the $H$ group ( the fields of the spontaneously broken symmetry) are coordinates and so construct $U$?

  • $\begingroup$ @Qmechanic Thank you for the edit. Do you believe i should do something to narrow the number of questions? I thought that 3 and 4 questions of the edit you made were analogous, understanding one of them could give answer to the other. $\endgroup$ – Constantine Black Nov 22 '16 at 15:49

Collecting the 4-vector components of Epelbaum to a 2x2 matrix is very analogous to the spinor map of the Lorentz group. In his (3.26), freezing out the σ via the hypersphere $S^3$ constraint (3.24) yields the 2x2 unitary matrix $$ U=1\!\! 1 \sqrt{1-|\varpi/F |^2} +i \frac{\vec{\varpi}}{F}\cdot \vec{\tau} ~. $$

Now, reparameterizing, (with Gürsey (1961) AnnPhys 12 91-117), $$ \vec{\varpi}\equiv \vec{\pi}~ \frac{\sin (|\pi/F|)}{|\pi/F|}, $$ this readily turns to Epelbaum's (3.32), $$ U=1\!\! 1 ~\cos |\pi/F| + i{\hat{\pi}}\cdot \vec{\tau} ~ \sin |\pi/F| =e^{i~\vec{\tau}\cdot \vec{\pi}/F}, $$ the very expression you are expanding in your question, and as he reiterates, $\pi$ and $\varpi$ yield the same S-matrix elements. $\pi$ and $\varpi$ transform identically under the (V) isospin rotations, but $\pi$ shifts slightly differently in subheading orders than $\varpi$ under the broken (A) axial shifts.

However, this is an exceptional case, because it is exceptionally pedagogical! His $G=SO(4) \sim SU(2)\times SU(2)$, $H= SO(3)\sim SU(2)$ and the pion projective coordinates are just the three coordinates of a hypersphere. Never again, for general $N$ flavors: no such helpful SO(4). You stick to $G=SU(N)\times SU(N)$ and $H=SU(N)$ for chiral models, so $N^2 -1$ pseudo scalar goldstons, φ , and the simple group element $U=e^{i~\vec{T}\cdot \vec{\phi}/F}$, which may be expanded in the φs.

For non-chiral Gs, it is even harder, as you have n=dim(G/H) goldstons, but the generic CCWZ construction you find in better QFT texts and WP goes through.

  • $\begingroup$ Thank you Mr Zachos for your response; I just saw your answer. Correct the statements below if they aren't correct. 1) Unitary matrix U can be identified as an object that, having exponential form in one parameterization, is expanded and gives the the general effective Lagrangian- it is a dense collection of the fields in one expression that has linear properties. 2) On the other hand, U can be constructed by geometric arguments of the non-linear realization of the chiral groub via the action of the group of rotations on the pion triplet. Continue... $\endgroup$ – Constantine Black Nov 22 '16 at 15:45
  • $\begingroup$ ...Continue... Obviously there is a gap in my line of thought right here. To collect the pionic degrees of freedom in a four vector; is this something trivial that can always be done? The fields can always be thought of as coordinates in a vector space of the Lie group, or something like that? Forgive my lack of rigor and thank you again. $\endgroup$ – Constantine Black Nov 22 '16 at 15:45
  • $\begingroup$ 1) correct; the exponential of iφ.T always has simple properties--it is a group element. 2) ONLY for SU(2), as a said. It fails, even for SU(3). The geometric arguments work for the exponentials, however: they are the CCWZ framework, subtle and omitted her. But, yes, the goldstons φ furnish a basis of a Lie algebra for chiral models. $\endgroup$ – Cosmas Zachos Nov 22 '16 at 16:07

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