Chiral Lagrangian symmetry

LAHLH
Messages
405
Reaction score
2
Hi,

If I have the Lagrangian L=i\chi^{\dagger\alpha i}\bar{\sigma}^{\mu}(D_{\mu})_{\alpha}^{\beta}\chi_{\beta i}+i\xi^{\dagger}_{\bar{i}\alpha}\bar{\sigma}^{\mu}(\bar{D}_{\mu})^{\alpha}_{\beta}\xi^{\beta i}-1/4 F^{a\mu\nu}F_{\mu\nu}^{a} where \alpha,\beta are colour indices, and i=1,2 is a flavour index (the Lagrangian is for two massless quarks, approximating u,d quarks only), and spinor indices are supressed. chi and xi are both LH Weyl fiels. See Srednicki ch83 for more details, available free online.

Then it's obvious that this Lagrangian has global flavour symmetry \chi_{\alpha i}\to L_{i}^{j}\chi_{\alpha j}, ,\xi^{\alpha\bar{i}}\to (R*)^{\bar{i}}_{\bar{j}} \xi ^{\alpha\bar{j}}, where L and R* are constant unitary matrices and the c.c. of R just a notational convention. So we have U(2)_L \times U(2)_R sym.

Then I can see that if we set L=R*=e^{i\alpha}I , equivalent to \Psi\to e^{-i\alpha\gamma_5}\Psi in terms of Dirac field then there is an anomaly in this axial U(1) sym, so I presume we just exclude this? then left over is the non-anomlous symmetry. Srednicki says this is SU(2)_L \times SU(2)_R \times U(1)_V, why is this the case? how has excluding this anomlous axial U(1) symmetry reduced U(2)_L\times U(2)_R TO SU(2)_L\times SU(2)_R\times U(1)_V?

thanks for any pointers
 
Physics news on Phys.org
U(2) = SU(2) x U(1), maybe mod Z2. It's been a little while. You have the U(1)_L and U(1)_R. Then you define U(1)_V = U(1)_L + U(1)_R and U(1)_A = U(1)_L - U(1)_R.
 
chrispb said:
U(2) = SU(2) x U(1), maybe mod Z2. It's been a little while. You have the U(1)_L and U(1)_R. Then you define U(1)_V = U(1)_L + U(1)_R and U(1)_A = U(1)_L - U(1)_R.

I see, thank you. Don't suppose you know a good reference to read up about this in particular, my group theory is a bit rusty at the moment..
 
I learned most of my group theory from Dresselhaus and Tinkham's Group Theory books, Georgi's Lie Algebras in Particle Physics (available online for free, though not as related to this issue in particular) and Fecko's Differential Geometry and Lie Groups for Physicists. I especially like the last book.
 
I was wondering if anyone could help me understand in more detail how SU(2)XSU(2) breaks into axial and vector parts? Also, am I correct in thinking since SU(2) has 3 generators su(2)xsu(2) has 9? I believe three of these are axial generators? since when they are broken you get three pseudogoldstone pions? So 6 vec gens? how does one see all this, if true?
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...
Back
Top