Chiral Lagrangian symmetry

1. Dec 21, 2011

LAHLH

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

2. Dec 21, 2011

chrispb

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.

3. Dec 22, 2011

LAHLH

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..

4. Dec 22, 2011

chrispb

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.

5. Jan 6, 2012

LAHLH

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?