How Does U(x) Relate to Pion Dynamics in Chiral Symmetry Breaking?

  • Context: Graduate 
  • Thread starter Thread starter LAHLH
  • Start date Start date
  • Tags Tags
    Chiral
Click For Summary
SUMMARY

The discussion focuses on the relationship between the composite field \(\chi_{\alpha i} \xi^{\alpha \bar{j}}\) and pion dynamics in the context of chiral symmetry breaking within the framework of SU(2)_L × SU(2)_R. The expectation value \(\langle 0| \chi_{\alpha i} \xi^{\alpha \bar{j}} |0\rangle = -v^3 \delta^{\bar{j}}_i\) is established as a mechanism for breaking axial symmetries, while the low-energy effective Lagrangian for the three pions is constructed using the unitary matrix \(U(x) = \exp\left[2i\pi^a(x)T^a/f_\pi\right]\). The discussion highlights the significance of universality classes in statistical mechanics, indicating that Lagrangians invariant under the same symmetries will converge to the same infrared fixed point, relevant to pion physics.

PREREQUISITES
  • Understanding of chiral symmetry and its breaking in particle physics.
  • Familiarity with composite fields and their role in quantum field theory.
  • Knowledge of effective Lagrangians and their construction in low-energy physics.
  • Basic concepts of renormalization group flow and universality classes.
NEXT STEPS
  • Study the construction of effective Lagrangians in quantum field theory.
  • Explore the role of Goldstone bosons in symmetry breaking, specifically in the context of pions.
  • Investigate the implications of renormalization group flow in particle physics.
  • Examine the mathematical framework of composite fields and their applications in quantum field theory.
USEFUL FOR

Particle physicists, theoretical physicists, and graduate students studying quantum field theory, particularly those interested in chiral symmetry and pion dynamics.

LAHLH
Messages
405
Reaction score
2
Hi,

So to break the axial part of SU(2)_L \times SU(2)_R we use the composite field \chi_{\alpha i} \xi^{\alpha \bar{j}} (since this is Lorentz scalar, colour singlet, and has the right transformation under the flavour symmetries, only breaking the axial part etc). We assume that \langle 0| \chi_{\alpha i} \xi^{\alpha \bar{j}} |0\rangle=-v^3 \delta^{\bar{j}}_i. (I understand roughly the mathematics of this operator does the job of breaking axial but not vector generators, but from my reading of symmetry breaking I don't really understand how we just *choose* a field in some arbitrary way like this; I thought the field had to be field involved in your Lagrangian and then you would find a continuous family of ground states of it etc)

Then my text talks about constructing a low energy effective lagrangian for the three pions (psuedo goldstone bosons) by letting |U\rangle be a low energy state for which the expectation value of \chi_{\alpha i} \xi^{\alpha \bar{j}} varies slowly in flavour space as a function of spacetime:
\langle U|\chi_{\alpha i} \xi^{\alpha \bar{j}}|U\rangle=-v^3 U^{\bar{j}}_i(x)

with U(x) a spacetime dependent unitary matrix that can be written U(x)=exp\left[2i\pi^a(x)T^a/f_\pi\right] (\pi are pion fields, T generators, f is pion decay const)

Then why do we think of U(x) as an effective low energy field? why does specifiying it's Lagrangian to be most gen consistent with symmetries, say anything about the pions? what is really going on here? I don't see why we are forming a Lagrangian from this object that was just on the RHS of an expectation value of this composite field.
 
Physics news on Phys.org
I'm studying this right now as well and had the same question. I think the answer has to do with the concept of "universality class" from statistical mechanics. Basically, I think the idea is that two Lagrangians that are invariant under the same symmetries will flow to the same fixed point in the IR under the renormalization group. In this case, the IR is pion physics.
 
U is a field that - when expanded in the pion fields - contains the free kinetic energy terms + higher interaction terms which are - due to power counting - non-renormalizable. You can see that by looking at power counting of \varphi^n theories which are 'generated' by expanding U.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
4K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 38 ·
2
Replies
38
Views
5K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 8 ·
Replies
8
Views
6K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K