All Questions
Tagged with quantum-chromodynamics symmetry-breaking
43
questions
8
votes
1
answer
2k
views
Peccei-Quinn-symmetry and effective Lagrangian for the Axion field
To solve the strong CP-problem Peccei and Quinn suggested the use of a new $U(1)$-symmetry called the PQ-symmetry. For this symmetry they constructed an effective Lagrangian involving the Nambu-...
0
votes
1
answer
180
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 ...
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
$$\frac{F^2}{4}\operatorname{Tr}\left\{\partial_\mu U\partial^...
3
votes
2
answers
161
views
How do we known that $\langle \bar{\psi}_i \psi_j\rangle=(250 MeV)^3\delta_{ij}$?
I have started to read the phenomenology of QCD in low energy regime. I understand that, from the QCD renormalization group equation, the QCD becomes nonperturbative theory when energy scale is below $...
9
votes
2
answers
796
views
How is the pion related to spontaneous symmetry breaking in QCD?
In chapter 19 of An Introduction to Quantum Field Theory by Peskin & Schroeder, they discuss spontaneous symmetry breaking (SSB) at low energies in massless (or nearly massless) QCD, given by
$$\...
2
votes
0
answers
69
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)
$$
U(2)_L \otimes U(2)_R = SU(2)_L \...
4
votes
1
answer
302
views
Global symmetries QCD goldstone bosons
Beside the local $SU(3)$-Color-symmetrie The QCD Lagrangian also has global symmetries:
$$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(...
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 ...
3
votes
0
answers
267
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 ...
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 ...
1
vote
0
answers
410
views
How to find the explicit value of the fermion vacuum expectation value?
In the derivation for Goldstone modes, a pdf I found online (scipp.ucsc.edu/~dine/ph222/goldstone_lecture.pdf) says it's believed the above-mentioned expectation value $\langle \bar{ \psi} \psi \...
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 ...
2
votes
3
answers
886
views
The status of $SU(3)_C$ symmetry in the Standard Model
In the Standard Model of Particle physics the $SU(2)_{EW}$ symmetry and the $SU(2)$ isospin symmetry are broken. What about $SU(3)_C$? Is it broken too?
if YES, what breaks the symmetry?
If NO, what ...
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 ...
1
vote
1
answer
327
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 ...