quantum field theory - Calculation of Feynman invariant amplitude with internal global symmetry indices: trace overand isospin

This is a complete rewriting of the older post, making more clear the problem.

The issue here is to compute the $$|M|^2 =4a^2(\delta_{ad}\delta_{bc} - \frac{1}{2} \delta_{ab}\delta_{cd}) ^2 (u_{1b}^{\mu}\bar{u}_{1 \mu a} u^{\mu}_{1'b}\bar{u}_{1'\mu a})\\ \times(u_{2'd}^{\nu}\bar{u}_{2'\nu c}u_{2d}^{\nu}\bar{u}_{2\nu c})$$ invariant Feynman amplitude, and in fact, the search for a possible, if it exists, way for writing this as a trace both on $SU(2)$ and Lorentz indices.

One of terms of the above amplitude is: $$T_1 = \delta_{ad}\delta_{bc}\delta_{ad}\delta_{bc} \omega = 4 (u_{1b}^{\mu}\bar{u}_{1 \mu a} u^{\mu}_{1'b}\bar{u}_{1'\mu a})(u_{2'a}^{\nu}\bar{u}_{2'\nu b}u_{2a}^{\nu}\bar{u}_{2\nu b}),$$ where $\omega$ is the series of spinors before contracting the deltas and the 4 coefficient comes from the contraction of deltas.

Question: I'm having a problem seeing how could I simplify this expression. Can I simultaneously reform this term to a trace? The possibilities are a trace over:

1. Just spin indices; but what happens then with the internal indices?


2. Internal isospin trace; but what then happens with spin indices?


3. A trace over both indices?

Thank you.

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Details

Assume a Lagrangian with an interaction term of the form $L_{int}=\bar \psi \phi \cdot \tau \psi$ , globally symmetric under the action of $SU(2)$ group. Think that the field $\psi$ describes a nucleon as an isospin SU(2) doublet with entries the proton and neutron and the field $\phi$ an isospin triplet (in the adjoint representation) of SU(2) formed by the three scalar pions.

The problem is to calculate the invariant Feynman amplitude. We have, writing all the indices explicitly:

$$M=a(q^2) (\bar{u}_{1'\mu a}\tau^{ab}_k u^{\mu}_{1b})( u_{2'\nu c}\tau^{cd}_k\bar{u}^{\nu}_{2d}) ,$$ where $\tau_k$ are the $SU(2)$ generators; Pauli matrices and q is the interaction momentum of the propagator. The complex conjugate then is: $$M^{*}=a^{*}(q^2) (\bar{u}_{1\mu a}\tau^{ab}_k u^{\mu}_{1' b}) (u_{2\nu c}\tau^{cd}_k\bar{u}^{\nu}_{2' d}) .$$

This way one should recognise the scalar nature over both $\mu, \nu$ Lorentz indices and over $a,b,c,d, k$ isospin indices. A similar calculation can be found in chapter 46 of Srednicki, without the extra internal symmetry and it's indices. There, it's simple then to recognise the trace over spin indices of the $|M|^2 =MM^{*}$.

For the generators there is the following completeness relation: $\tau_{ab}^k \tau_{cd}^k = 2(\delta_{ad}\delta_{bc} - \frac{1}{2} \delta_{ab}\delta_{cd}) .$

So finally, the amplitude is:

$$|M|^2 =4a^2(\delta_{ad}\delta_{bc} - \frac{1}{2} \delta_{ab}\delta_{cd}) ^2 (u_{1b}^{\mu}\bar{u}_{1 \mu a} u^{\mu}_{1'b}\bar{u}_{1'\mu a})\\ \times(u_{2'd}^{\nu}\bar{u}_{2'\nu c}u_{2d}^{\nu}\bar{u}_{2\nu c})$$

The delta identity gives three terms; the first, let's say $T_1$, is: $$T_1 = \delta_{ad}\delta_{bc}\delta_{ad}\delta_{bc} \omega = 4 (u_{1b}^{\mu}\bar{u}_{1 \mu a} u^{\mu}_{1'b}\bar{u}_{1'\mu a})\\ \times(u_{2'a}^{\nu}\bar{u}_{2'\nu b}u_{2a}^{\nu}\bar{u}_{2\nu b}),$$ where $\omega$ is the series of spinors before contracting the deltas and the 4 coefficient comes from the contraction of deltas.