# Tag Info

0

OP is basically asking (v4) the following. How the equation $$\int\! \mathrm{d}^4x~ \left(i{\cal A}(x) + i\sum_r J_r(x) \langle F^r\left[x,\Phi\right] \rangle_J\right) ~=~0, \tag{4.5}$$ or equivalently, $$\int\! \mathrm{d}^4x~ \left(i{\cal A}(x) + i\sum_r J_r(x) F^r\left[x,\frac{\delta}{i \delta J}\right]\right)Z[J] ~=~0, \tag{4.5'}$$ ...

4

The derivation in the given reference indeed seems confused and inconsistent. The crucial error seems to me that $$S[\phi + \epsilon\delta\phi] = S[\phi]$$ is just not true for an infinitesimal symmetry. The definition of a symmetry is that $S[\phi']=S[\phi]$ (modulo boundary terms) for the finite transformation $\phi\mapsto \phi'$. Writing this ...

0

There is a simple proof that the cancellation is impossible (at least unless you are willing to add to the classical a term proportional to $\hbar$), I am reformulating an answer by @Qmechanic in a simpler language: The anomaly, or the measure variation which is the typical source of the anomaly, contributes a term which is independent of $\hbar$, while any ...

4

Comments to the question (v2): Traditionally, the classical action $S$ sits in the Boltzmann factor $\exp\left[\frac{i}{\hbar} S\right]$ behind an inverse power of $\hbar$ in the path integral, while the path integral measure is independent of $\hbar$. In the conventional way of counting, we say that the Jacobian $J$ from the path integral measure is a ...

Top 50 recent answers are included