Vector/axial current and vector/axial transformation

Question

As is well known, in particle physics, the vector current is defined as a current of the form $$J_V^\mu = \bar{\psi} \gamma^\mu \psi $$ and the axial current as $$J_A^\mu = \bar{\psi} \gamma^5 \gamma^\mu \psi $$ These currents are important for establishing, for example, the charged weak current as one of the V-A form.

Now, in Nuclear Physics theories that relate the interactions of elementary particles with the nucleus (chiral theories, Skyrme model), which are commonly based on group theory, it is established that a vector transformation is $$U \rightarrow MUM^\dagger, \quad M,U \in SU(2) $$ and an axial transformation is $$U \rightarrow MUM, \quad M,U \in SU(2) $$ By similarity, a charged current associated with the nucleus is also of the V-A form.

Therefore, my question is the following: what is the relationship between these transformations and the currents, aside from the name? What has led to the idea that they are related?

Answer 1

They are not just related, they are identical, joined at the hip. I'll illustrate by the original legendary 2-flavor chiral model, with the chiral group $\text{SU}(2)_L \times \text{SU}(2)_R$. This group is spontaneously broken down to $\text{SU}(2)_V$, by the QCD vacuum: it is realized ''nonlinearly'', in the Nambu–Goldstone mode, i.e., the $Q_V$ annihilate the vacuum, but the $Q_A$ do not! This is visualized nicely through a geometrical argument based on the fact that the Lie algebra of $\text{SU}(2)_L\times\text{SU}(2)_R$ is isomorphic to that of SO(4).

Your currents as given are malformed into singlets. They need isospin triplet matrices τ to make sense into the current algebra underlying the chiral picture: $$\begin{cases} q_L \mapsto q_L'= M_L q_L = \exp{\left(- i {\boldsymbol{\theta}}_L \cdot \tfrac{\boldsymbol{\tau}}{2} \right)} q_L \\ q_R \mapsto q_R'= M_R q_R = \exp{\left(- i \boldsymbol{ \theta}_R \cdot \tfrac{\boldsymbol{\tau}}{2} \right)} q_R \end{cases},$$ where τ denote the three Pauli matrices in the flavor space and θ_L, θ_R are the corresponding rotation angles for the L and R groups, respectively.

The corresponding symmetry group $\text{SU}(2)_L\times\text{SU}(2)_R$ is the chiral group, controlled by the six conserved currents $L_\mu^i = \bar q_L \gamma_\mu \tfrac{\tau^i}{2} q_L$, $R_\mu^i = \bar q_R \gamma_\mu \tfrac{\tau^i}{2} q_R$, which can equally well be expressed in terms of the vector and axial-vector currents you should have written, $V_\mu^i = L_\mu^i + R_\mu^i$ , $A_\mu^i = R_\mu^i - L_\mu^i$.

The corresponding conserved charges generate the algebra of the chiral group, $$\left[ Q_{I}^i, Q_{I}^j \right] = i \epsilon^{ijk} Q_I^k \qquad \qquad \left[ Q_{L}^i, Q_{R}^j \right] = 0,$$ with $I=L,R$, or, equivalently, $$\left[ Q_{V}^i, Q_{V}^j \right] = i \epsilon^{ijk} Q_V^k, \qquad \left[ Q_{A}^i, Q_{A}^j \right] = i \epsilon^{ijk} Q_V^k, \qquad \left[ Q_{V}^i, Q_{A}^j \right] = i \epsilon^{ijk} Q_A^k.$$

Application of these commutation relations to hadronic reactions dominated current algebra calculations in the early seventies of the last century.

So the chiral field $U$ transforms differently on the left and the right, $$M_L U M_R^\dagger=\exp(i\boldsymbol{\theta}_A\cdot \boldsymbol{\tau}/2 -i\boldsymbol{\theta}_V\cdot \boldsymbol{\tau}/2 )~ U ~\exp(i\boldsymbol{\theta}_A\cdot \boldsymbol{\tau}/2 +i\boldsymbol{\theta}_V\cdot \boldsymbol{\tau}/2 ).$$

θ_A= 0 , then, manifestly yields your vector transformation, while θ_V =0 the axial one.