All Questions
7
questions
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\...
2
votes
1
answer
531
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 ...
1
vote
2
answers
240
views
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 ...
1
vote
1
answer
279
views
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 $$SU(3)_R\ \...
3
votes
1
answer
483
views
Relevance of the condensate $\langle 0 | \bar q q | 0 \rangle$ to SSB of chiral QCD symmetry
The QCD lagrangian with two massless flavours of quarks takes the form $$\mathcal L = \sum_{i=u,d} i \bar q_i \gamma_{\mu} D^{\mu} q_i.$$ Defining operators to project out the left and right handed ...
14
votes
1
answer
3k
views
Physical meaning of the chiral condensate in QCD
Considering the QCD Lagrangian in the chiral limit, where all the quarks masses are set to zero. Then the Lagrangian has the following chiral symmetry:
$$
SU(L)_{V} \times SU(L)_{A} \times U(1)_{V} \...
2
votes
1
answer
239
views
Chiral current VEV below the QCD scale
Let's have pure QCD. I know that after spontaneous symmetry breaking quark bilinear form are replaced by their averaged values:
$$
\bar{q}_{i}q_{j} \to \langle \bar{q}_{i}q_{j}\rangle \approx \Lambda_{...