**Motivation for the Wess-Zumino term**

There are many similar terms with mesons which capture anomalous chiral structure of underlying QCD. How to capture all of them? The answer is given by the theorem, which states following: in general, non-abelian anomaly in 4 dimensions is related Chern-Simons character in 5 dimensions, which is called the Wess-Zumino term. Such theorem also is an elegant way to derive the Wess-Zumino term.

So instead of looking for all terms like $\pi^{0}F_{EM}\tilde{F}_{EM}$ by direct application of anomaly equations it's preferable to gauge the Wess-Zumino term, since it is straightforward. Many anomalous terms arise not from triangle diagram, but, for example, from anomalous pentagonal diagram, which leads to great complication of calculations based on modification of naive Ward identities for given class of diagrams.

How to gauge the Wess-Zumino term? There is an old simple way which is called trial and errors method. Namely, let's start from the Wess-Zumino free term:
$$
N_{c}\Gamma_{WZ} = i\frac{N_{c}}{240 \pi^{2}}\int d^{5}x\epsilon^{ijklm}\text{Tr}\left[L_{i}L_{j}L_{k}L_{l}L_{m}\right],
$$
where $L_{i} \equiv U\partial_{i}U^{-1}$ and $U$ is the goldstone bosons matrix.

Since $U$ here acts in 5-dimensional manifold, there is no direct way to gauge the Wess-Zumino term by simple elongation of derivative. However, we know the explicit form of $U_{EM}(1)$ gauge variation of $U$ field, namely
$$
U(x) \to e^{iQ\epsilon(x)}Ue^{-iQ\epsilon (x)} \Rightarrow \delta U = i\epsilon (x)[Q, U]
$$
Then
$$
\delta L_{i} = i\epsilon (x)[Q, L_{i}] + i(\partial_{i}\epsilon (x))UQU^{-1} - i\partial_{i}\epsilon (x)Q,
$$
and
$$
N_{c}\Delta \Gamma_{WZ} = -\frac{iN_{c}}{48 \pi^{2}}\int d^{5}x\partial_{i}\left[ \partial_{i}\epsilon(x)\epsilon^{ijklm}\text{Tr}\left[ Q(T_{k}T_{l}T_{m} + L_{k}L_{l}L_{m})\right]\right] =
$$
$$
=-\frac{i}{48 \pi^2}\int d^{4}x\epsilon^{\mu \nu \alpha \beta}\partial_{\mu}\epsilon (x) \text{Tr}\left[ QT_{\nu}T_{\alpha}T_{\beta} + L_{\nu}L_{\alpha}L_{\beta}\right] \equiv -\int d^{4}x\partial_{\mu}\epsilon(x)J^{\mu},
$$
where $T_{i} \equiv (\partial_{i}U^{-1})U$ and
$$
J^{\mu} \equiv \frac{iN_{c}}{48 \pi^{2}}\epsilon^{\mu \nu \alpha \beta}\text{Tr}\left[ Q(T_{\nu}T_{\alpha}T_{\beta} + L_{\nu}L_{\alpha}L_{\beta})\right]
$$
For first step, we may modify the Wess-Zumino term by adding
$$
-\int d^{4}xA_{\mu}J^{\mu},
$$
where $A_{\mu}$ is EM 4-potential.

Since $\delta J_{\mu}$ isn't zero, we need to add also the part whose variation coincides term with $\delta J_{\mu}$. Thus we obtain that the Gauged Wess-Zumino term is
$$
\tag 1 \tilde{\Gamma}_{WZ} = N_{c}\Gamma_{WZ} - \int d^{4}xA_{\mu}J^{\mu} +
$$
$$
+\frac{iN_{c}}{24 \pi^{2}}\int d^{4}x\epsilon^{\mu \nu \alpha \beta}(\partial_{\mu}A_{\nu})A_{\alpha}\text{Tr}\left[ Q^{2}(\partial_{\beta}U)U^{-1} + Q^2Q^{-1}\partial_{\beta}U + QUQU^{-1}(\partial_{\beta}U)U^{-1}\right]
$$

For QCD with $u, d$ quarks, $Q = \text{diag}\left(\frac{2}{3}, -\frac{1}{3}, \right)$. By using
$$
U = e^{i\frac{\sqrt{2}\pi_{a}t_{a}}{f_{\pi}}}, \quad \pi_{a}t_{a} = \begin{pmatrix} \frac{\pi^{0}}{\sqrt{2}} & \pi^{-} \\ \pi^{+} & \frac{\pi^{0}}{\sqrt{2}}\end{pmatrix},
$$
we may get $\pi^{0} \to 2\gamma$ vertex from the last summand of $(1)$ and $\gamma \to 3 \pi$ vertex from the second one. Of course, by enlarging the number of quarks to 3, we get more anomalous terms.

This post imported from StackExchange Physics at 2017-05-08 16:15 (UTC), posted by SE-user Name YYY