All Questions
Tagged with quantum-chromodynamics symmetry-breaking
43
questions
1
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2
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240
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Sufficient and Necessary Conditions for Chiral Symmetry Breaking
In their 2005 paper, the authors write (just below eq. 3.19)
we see that a non-zero value of $F_0$ is a necessary and sufficient criterion for spontaneous chiral symmetry breaking. On the other ...
- 2,648
2
votes
1
answer
260
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Why is the approximate $\rm U(2)\times U(2)$ global symmetry of QCD that has a special importance?
I was looking at Peskin and Schroeder (Section 19.3, page $667-668$). They talk about $\rm U(2)\times U(2)$ symmetry for the QCD Lagrangian in the limit of massless $u$ and $d$ quarks. However, this ...
- 26.8k
1
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1
answer
279
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Group structure of QCD‘s chiral symmetry (breaking)
With $3$ flavors of massless quarks, the QCD Lagrangian is invariant under flavor transformations$$SU(3)_V\ \otimes\ SU(3)_A\ \otimes\ U(1)_V\ \otimes\ U(1)_A.$$
Now, this is equivalent to $$SU(3)_R\ \...
- 2,648
1
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2
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379
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Derivation of Casher-Banks relation
Consider two-point function $\langle \bar{\psi}\psi\rangle$ in a model with massive fermions $\psi$ and gauge field:
$$
\langle \bar{\psi}\psi\rangle =\frac{1}{V}\sum_{n} \frac{1}{\lambda_{n} +im},
$$...
- 8,901
4
votes
1
answer
248
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Simple explanation of the QCD VEV in terms of instantons
I've heard that instantons in QCD generate quark bilinear condensate $\langle \bar{q}_{L}q_{R}\rangle$ which is responsible for spontaneous symmetry breaking. Is there any clear and simple way to ...
- 8,901
2
votes
0
answers
86
views
No global monopoles in QCD
If a global symmetry gets both spontaneously and explicitly broken, the explicit symmetry breaking pattern is crucial for understanding the formation of topological defects.
For example, in the axion ...
- 1,783
1
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0
answers
46
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Do hierarchical condensates yield instantaneous or sequential symmetry breaking?
I am wondering whether the formation of a hierarchical vacuum condensate yields an instantaneous or sequential symmetry breaking in a cosmological phase transition. Let me illustrate this question ...
- 1,783
1
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2
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1k
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Breaking of $SU(3)$ symmetry by bi-fundamental representation
Are there any general theorems which fix the possible symmetry breaking patterns of Lie groups (such as $SU(3)$) by vacuum expectation values of fields in specific representations (such as the quark ...
- 1,783
2
votes
1
answer
141
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gauge-invariant 6-quark order parameter
In this Review paper in p.1462, bottom left: Rev.Mod.Phys.80:1455-1515,2008 -- Color superconductivity in dense quark matter
It says that "There is an associated gauge-invariant 6-quark order ...
- 4,336
0
votes
1
answer
147
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chiral symmetry condensate and 2SC, CFL breaking C, P and T symmetry?
Because we know that chiral symmetry condensate causes the chiral symmetry breaking, and it produces Goldstone modes of pseudo-scalars, so I believe that chiral symmetry breaking also breaks the T ...
- 4,336
8
votes
1
answer
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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-...
- 1,114
3
votes
1
answer
403
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Scalar pseudo-Goldstones from hypothetical $SU(3)_V$ symmetry breaking in QCD
The eight light pseudoscalar mesons of QCD are the pseudo-Goldstone bosons of the spontaneously broken chiral (axial) $SU(3)_A$ quark flavor symmetry.
If we consider the hypothetical case of also ...
- 1,783
1
vote
0
answers
195
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Why does the $U(2n)$ flavor symmetry break down to a $U(1)$ group and an $SU(2n)$ group?
I am studying quantum field theory using Srednicki's textbook. Problem 83.1 is:
Suppose that the color group is $G_C=SO(3)$ rather than $SU(3)$, and that each quark flavor is represented by a Dirac ...
- 1,653
9
votes
1
answer
739
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Why do we have a non-zero quark vacuum condensate even though the QCD coupling goes to zero in the deep infrared?
It is well-known that QCD has a Landau pole at $\Lambda_{\rm QCD}\sim 200$ MeV, which means that the perturbative QCD coupling becomes strong at this scale. Conventionally, this is claimed to be the ...
- 1,783
4
votes
1
answer
2k
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Question about the linear sigma-model
Suppose the linear sigma-model lagrangian:
$$
L = \bar{N}(i\gamma_{\mu}\partial^{\mu}-g_A \phi)N + |\partial_{\mu}\phi|^{2} - V(|\phi|) - c\sigma ,
$$
where
$$
N = \begin{pmatrix} p \\ n\end{pmatrix}, ...
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