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0 votes
1 answer
178 views

Why does isospin symmetry survive chiral symmetry breaking?

In QCD there is a breaking of chiral symmetry such that, for the lightest quarks only, we have $$\langle \overline{u} u \rangle = \langle \overline{d} d \rangle = \mathrm{const.}$$ where the constant ...
Kris's user avatar
  • 841
0 votes
1 answer
70 views

Do we have an analytic calculation to derive $\frac{F^2}{4}\,\text{Tr}\left\{\partial_\mu U\partial^\mu U^{\dagger}\right\}$ from the QCD Lagrangian?

I have studied the quark condensate and chiral perturbation theory. However, I am not sure where the "kinetic term" of the pion $$\frac{F^2}{4}\operatorname{Tr}\left\{\partial_\mu U\partial^...
StupiXPerson's user avatar
3 votes
2 answers
160 views

How do we known that $\langle \bar{\psi}_i \psi_j\rangle=(250 MeV)^3\delta_{ij}$?

I have started to read the phenomenology of QCD in low energy regime. I understand that, from the QCD renormalization group equation, the QCD becomes nonperturbative theory when energy scale is below $...
StupiXPerson's user avatar
9 votes
2 answers
789 views

How is the pion related to spontaneous symmetry breaking in QCD?

In chapter 19 of An Introduction to Quantum Field Theory by Peskin & Schroeder, they discuss spontaneous symmetry breaking (SSB) at low energies in massless (or nearly massless) QCD, given by $$\...
Flynn Linton's user avatar
2 votes
0 answers
68 views

Goldstone bosons in 2 and 3 quark flavor symmetries [closed]

In my (undergraduate) advanced elementary particles class last semester, we learnt that for a 2 quark (u/d) model the symmetry of the Lagrangian is (and breaks as) $$ U(2)_L \otimes U(2)_R = SU(2)_L \...
Yaezir's user avatar
  • 23
4 votes
1 answer
298 views

Global symmetries QCD goldstone bosons

Beside the local $SU(3)$-Color-symmetrie The QCD Lagrangian also has global symmetries: $$L_{QCD}=\sum_{f,c}\bar{q_{fc}}(i\gamma^\mu D_\mu - m ) q_{fc} - \frac{1}{4}F^a_{\mu \nu} F^{a \mu \nu} $$ $SU(...
taxus1's user avatar
  • 63
0 votes
1 answer
185 views

How does the Nambu-Goldstone mode explain the absence of parity doubling?

I've been doing some reading about chiral symmetry breaking since it was not touched in my particle physics course I found these slides As explained in the above link, if we take $|\psi \rangle$ as ...
Monopole's user avatar
  • 3,454
3 votes
0 answers
265 views

Why is the chiral condensate a negative quantity?

The chiral condensate serves as an order parameter for the chiral phase transition. Thus, it is a finite quantity in one phase and vanishes in the other phase. It is given as a vacuum expectation ...
Bernd's user avatar
  • 457
3 votes
0 answers
69 views

Restoration of symmetry explicitly broken by anomaly

What is the meaning of the restoration at finite temperature of a symmetry that is "broken" by the presence of an anomaly. If the symmetry is not there why is it restored at finite ...
Kvothe's user avatar
  • 861
0 votes
0 answers
76 views

Are the arrangements of quarks in hadron ground-state wavefunctions rotationally symmetric?

The Hamiltonian of quantum chromodynamics (like the rest of the Standard Model) is rotationally symmetric. My question is whether these space symmetries are spontaneously broken in the ground state of ...
tparker's user avatar
  • 48.3k
3 votes
1 answer
170 views

$U(1)_A$ effects on the baryons?

We know that the axial $U(1)_A$ is anomalous thus not a global symmetry. Therefore there is no direct associated pseudo goldstone boson for $U(1)_A$. This makes the $\eta'$ much more massive than the ...
ann marie cœur's user avatar
1 vote
1 answer
324 views

Quark condensate and spontaneous symmetry breaking?

It is known the quark condensate $<\bar{\psi}^{i}_L\psi^j_R>=\sigma \delta^{ij}$($i,j$ are flavour indices ) breaks the symmetry group $SU(N_f)_L\times SU(N_f)_R$. Because it is only invariant ...
Sven2009's user avatar
  • 995
0 votes
1 answer
181 views

Chiral symmetry in massless QCD

The QCD Lagrangian for two flavors is: $-\frac{1}{4} G\tilde{G}+i\bar{u}\displaystyle{\not} D u+i\bar{d} \displaystyle{\not} D d-m_u\bar{u} u-m_d\bar{d}d$ or alternaively $-\frac{1}{4} G\tilde{G}+i\...
user avatar
2 votes
1 answer
526 views

Peskin's treatment of Pions as Goldstone Bosons

After restoring the mass terms in the Lagrangian \begin{align} \mathcal{L}=\bar{u} i \not D u+\bar{d i} \not D d-m_{u} \bar{u} u-m_{d} \bar{d} d, \end{align} one obtains equations of motion for the ...
user4580791's user avatar
3 votes
0 answers
360 views

Pion mass in theta vacuum

Does the mass of the charged pion depend on the QCD vacuum angle? I've seen it said---e.g., in these TASI lectures---that when the quark mass matrix is real and there is a nonzero QCD vacuum angle $\...
plusplusplus's user avatar

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