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I'm following the Chapter 5 (specifically, section 5.4) of Quigg's book Gauge Theories of the Strong, Weak and Electromagnetic Interactions and am confused with the following:

Studying a sigma model with meson fields, he states that it is useful to 'decompose' the gauge group $SU(2)_L \times SU(2)_R$, with elements

$$ G_L=\exp{\left( i\pmb{\alpha}_L \cdot \frac{\pmb{\tau}}{2}\right) }; \;\;\;\;\ G_R=\exp{\left( i\pmb{\alpha}_R \cdot \frac{\pmb{\tau}}{2}\right) } $$

'in terms of $SU(2)_V \times SU(2)_A$', with elements

$$ G_I=\exp{\left( i\pmb{\alpha} \cdot \frac{\pmb{\tau}}{2}\right) }; \;\;\;\;\ G_5=\exp{\left( i\gamma_5\pmb{\alpha}_5 \cdot \frac{\pmb{\tau}}{2}\right) }. $$

where $\pmb{\alpha}=\pmb{\alpha_R}+\pmb{\alpha_L}$ and $\pmb{\alpha}_5=\pmb{\alpha}_R-\pmb{\alpha}_L$.

I want to show that, after one vacuum is favored, the $SU(2)_V$ symmetry breaks, to which (I think) I need the explicit form of said decomposition.

Since the arguments of the exponentials do not commute (because the Pauli-matrices are weighted differently by the gauge parameters) I am not seeing how to write anything with this goal in mind.

How can I do that? Or, if it is not needed, how can I show what I want?

More specifically, he defines a 2x2 mesonic spinless matrix field which transforms as

$$ \Sigma \to \Sigma'=G_L \Sigma G_R^{\dagger}. $$

He then proceeds to write the field in terms of 4 scalars in the sigma(tau)-matrices basis:

$$ \Sigma=\sigma+i\pmb{\pi}\cdot\pmb{\tau}. $$

The claim is that the vacuum state

$$ \left<\sigma\right>_0=v, \;\;\;\;\;\;\; \left<\pi\right>_0=\left(\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}\right) $$

(here $v$ is the VEV of the interaction) is unchanged by isospin rotations but breaks $SU(2)_A$ symmetry.

That is what I want to show.

GaloisFan
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    Not that it addresses your question, but decomposition of the current algebra does not result in a fictitious quoted $SU(2)_V\times SU(2)_A$ : there is no such group. In fact, the axial generators do not close into a Lie algebra, and so cannot exponentiate to a fictitious group $G_5$... Moreover the vector generators do not commute with the axial ones--they transform them! I cannot imagine Chris wrote such in a book... I'm betting your quotes are a bluff. – Cosmas Zachos Jul 18 '18 at 01:07
  • @CosmasZachos With the quotation marks what I (in a silly way) mean is 'in my informal interpretation of what he wrote'. What he indeed say is that it is useful to recast $G_L$ and $G_R$ in terms of the given $G_I$ and $G_5$. I am admittedly somewhat ignorant with respect to dealing with product symmetry groups and apologize if that makes the question confusing. – GaloisFan Jul 18 '18 at 01:35
  • Apologies for frowning at you. The *error, in fact, comes with the book, between eqns (5.4.7a) and (5.4.7b) , where he calls the $G_5$ chiral rotation $SU(2)_A$, an absurdity unworthy of a textbook. You can see the $G_5$s do not close to any group, as composing two of them will net CBH terms lacking a $\gamma_5$! Simply ignore the unfortunate and subversively confusing remark. It would* mess you up subsequently, as these 3 generators are precisely the ones spontaneously broken, and correspond to the pion pseudogoldstons. – Cosmas Zachos Jul 18 '18 at 14:44
  • Related, and also. – Cosmas Zachos Jul 18 '18 at 15:05
  • @CosmasZachos No problem! I updated the question, hopefully it is more clear now. – GaloisFan Jul 18 '18 at 17:15

1 Answers1

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Quite simply, under an $SU(2)_V$ (isospin) rotation, $\pmb{\alpha}_5=0$, so that $\pmb{\alpha_L}=\pmb{\alpha}/2=\pmb{\alpha_R}$, and $$ G_L=\exp{\left( i\pmb{\alpha}_L \cdot \frac{\pmb{\tau}}{2}\right) }=G_R. $$ Consequently, $$ \langle\Sigma\rangle=v 1\!\!1 \to v G_L 1\!\!1 G_L^{\dagger}= v 1\!\!1. $$ The vacuum is invariant under isospin.

By contrast, under the three axial rotations (there is no such thing as an $SU(2)_A$, and its promulgators deserve censure and ridicule for confusing impressionable minds), $\pmb{\alpha}= 0$ so that $\pmb{\alpha}_5 /2=-\pmb{\alpha}_L=\pmb{\alpha}_R$, so that $G_R=G_L^\dagger$, $$ \langle\Sigma\rangle=v 1\!\!1 \to v G_L 1\!\!1 G_L= v G_5^\dagger \neq v 1\!\!1. $$ That means that axial transformations have shifted the vacuum, so the vacuum is not unique and invariant.

  • Note I desist from assigning a group to the $G_5$s, since they cannot close to a group--in sharp contrast to the $G_I$s. You multiply $\Sigma$ on the left and the right by the same transformation matrix. But $G_5(\alpha)G_5(\beta)\neq G_5(\beta)G_5(\alpha)$ for a succession of such operations.(If you track down the details, you actually shift the pions by a constant--the essence of the celebrated Nambu-Goldstone realization.)

  • There is no chimera like $SU(2)_A$, and damn your misleading text which should know better. (These three generators are in what is technically known as a cosets space, and they correspond one-to-one with the massless pions of that model, the goldstons. I can see M G-M, the coinventor of that model, screaming at the mathematical solecism... But, trust me, it is not just math....)

Cosmas Zachos
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