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Starting from the lagrangian (linear sigma model without symmetry breaking, here $N$ is the nucleon doublet and $\tau_a$ are pauli matrices)

$L=\bar Ni\gamma^\mu \partial_\mu N+ \frac{1}{2} \partial_\mu\sigma\partial^\mu\sigma+\frac{1}{2}\partial_\mu\pi_a\partial^\mu\pi_a+g\bar N(\sigma+i\gamma_5\pi_a \tau_a)N$

we can construct conserved currents using Noether's Theorem applied to $SU(2)_L\otimes SU(2)_R$ symmetry: we get three currents for every $SU(2)$. By adding and subtracting them, we obtain vector and axial currents.
We could have obtained vector charges quickly by observing that they are just isospin charges, so nucleons behave as an $SU(2)$ doublet (fundamental representation), pions as a triplet (adjoint representation) and sigma as a singlet (so basically it does not transform):

$V_a=-i\int d^3x \,\,[iN^\dagger\frac{\tau_a}{2}N+\dot\pi_b(-i\epsilon_{abc})\pi_c]$

But if I wanted to do the same with axial charges, what Lie algebra/representation must I use for pions and sigma?
I mean, axial charges are

$A_a=-i\int d^3x \,\,[iN^\dagger\frac{\tau_a}{2}\gamma_5N+i(\sigma\dot\pi_a-\dot\sigma\pi_a)]$

and I would like to reproduce the second term using a representation of Lie algebra generators of axial symmetry which act on $\sigma$ and $\pi$, but I don't know the algebra (I think it is $SU(2)$), neither the representation to use.
I tried to reproduce that form using the three matrices

$T^1=\begin{bmatrix} 0&-i&0&0\\i&0&0&0\\0&0&0&0\\0&0&0&0 \end{bmatrix}\quad T^2=\begin{bmatrix} 0&0&-i&0\\0&0&0&0\\i&0&0&0\\0&0&0&0 \end{bmatrix}\quad T^3=\begin{bmatrix} 0&0&0&-i\\0&0&0&0\\0&0&0&0\\i&0&0&0 \end{bmatrix}$

which should act on the vector $(\sigma,\pi_1,\pi_2,\pi_3)$, but I calculated their commutator and they don't form an algebra, so I think I'm getting wrong somewhere in my reasoning.

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In the linear sigma model, the chiral action on the pion fields can be implemented on the following matrix combination of the fields:

$$U(2) \ni \Sigma = \sigma + i \tau^a \pi_a $$

An element $ (U_L = exp(\frac{i}{2}\theta^{(L)}_a \tau^a), U_R = exp(\frac{i}{2}\theta^{(R)}_a \tau^a)) \in SU(2)_L \otimes SU(2)_R $ acts on \Sigma as follows:

$$\Sigma \rightarrow \Sigma' = U_L \Sigma U_R^{\dagger}$$

The kinetic term of the Lagrangian in the matrix representation is given by:

$$L_{kin} = \frac{1}{2} \partial_{\mu}\Sigma \partial^{\mu}\Sigma^{\dagger}$$.

This term is manifestly invariant under all transformations. The interaction term has also a manifestly invariant form:

$$L_{int} = \bar{N}_L \Sigma N_R+ \bar{N}_R \Sigma^{\dagger} N_L$$.

where $N_{L,R} = (1\pm \gamma_5)N $. Thus the whole Lagrangian is invariant under the chiral transformations.

The vector transformation is generated by the subgroup characterized by:

$$\theta^{(L)} = \theta^{(R)} = \theta^{(V)}$$

The axial transformation is generated by the subset characterized by:

$$\theta^{(L)} = -\theta^{(R)} = \theta^{(A)}$$

Substituting in the transformation equations of $\Sigma$ and keeping only the linear terms (this is sufficient for the application of the Noether's theorem), we obtain:

-Vector transformation:

$$ \pi_a' = \pi_a +\epsilon_{abc}\theta^{(V)}_b \pi_c $$

$$ \sigma' = \sigma$$

-Axial transformation:

$$ \pi_a' = \pi_a +\theta^{(A)}_a \sigma $$

$$ \sigma' = \sigma + \theta^{(A)}_a \pi_a$$

Now it is not hard to see that these transformations generate the correct contributions of the pionic fields to the currents.

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  • $\begingroup$ Thank you, it works but there must be a minus sign in axial transformation law of $\sigma$. Anyway, is there a way to put it in a form like $Q_a=-i\int d^3x\,\,(\pi_i T^a_{ij} \phi_j)$, where here $\pi_i=\dot\phi_i$ is the conjugate momenta? I can't figure out the $\phi_i$ and the $T^a_{ij}$, which should form a representation of $SU(2)$ $\endgroup$ – gian_25 Mar 9 '14 at 14:20
  • $\begingroup$ @gian_25 The matrix representation is just the 4-dimensional fundamental representation of $SO(4) = SU(2) \times SU(2)$, denoting the indices of the 4-space corresponding to $(\sigma, \pi_1, \pi_2, \pi_3)$ by $0,1,2,3$. Then$ (T^{V}_i)_{jk} = i \epsilon_{ijk}, (T^{(A)}_i)_{jk} = i (\delta_{ji}\delta_{k0}+ \delta_{ki}\delta_{i0})$. Please observe that the commutator of two axial generators is a vector generator. $\endgroup$ – David Bar Moshe Mar 9 '14 at 15:04
  • $\begingroup$ Ok, I think there is still a minus missing and $T^{(A)}_i$ become the same matrices I have written before, but now I understand what their commutators are. Thank you for the answer! $\endgroup$ – gian_25 Mar 9 '14 at 16:01

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