This is a question which wants us to find the variation of the action for the free scalar field under the field transformation
$\phi(x) \mapsto \phi’(x) = \phi(x) – a(x)\partial_\mu\phi(x)$
Deduce the Ward identities generated by translation invariance.
So the solution goes:
$x \mapsto x’^\mu = x^\mu + a^\mu$
$ \phi(x) \mapsto \phi’(x’) = \phi(x)$
$a^\mu = \epsilon \delta^{\mu\rho}$
$\phi’(x) = \phi(x-a) = \phi(x) – a^\mu\partial_\mu\phi(x) = \phi(x) – \epsilon\partial_\rho\phi(x)$
I am struggling to understand the last term on the RHS with $\epsilon\partial_\rho\phi(x)$.
How do we get from $a^\mu\partial_\mu\phi(x)$ to $\epsilon\partial_\rho\phi(x)$.
Any comments would be greatly appreciated.
Best Answer
It is just simple: substituting $a^\mu = \epsilon \delta^{\mu\rho}$, $a^{\mu}\partial_\mu \phi(x) = \epsilon \delta^{\mu\rho}\partial_\mu \phi(x) = \epsilon \partial^\rho \phi(x)$. Recall that we are using Einstein convention for the $\mu$-contraction with the Kronecker delta $\delta^{\mu\rho}$: \begin{align} \sum_{\mu} \delta^{\mu\rho} \partial_\mu = \sum_{\mu} \big(\text{1 if $\mu=\rho$, 0 otherwise}\big) \partial_\mu = \partial_\mu \,. \end{align}
By the way, I would like to say that the index notation here is somewhat sloppy, treating the index $\rho$ as a fixed number or otherwise a "dangling pointer." I suppose writing down equations just by $a^\mu$ (assumed infinitesimal) will be fine.
[Additional comments: there's no such thing as "$\delta^{\mu\rho}$", if you are not assuming Euclidean spacetime and have the inverse Euclidean metric $g^{\mu\nu} = \delta^{\mu\nu}$. Only $\delta^\mu{}_\nu$ is the "true" Kronecker delta: the identity map on tangent spaces. I recommend $a^\mu = \delta^\mu{}_\rho \epsilon^\rho$ or, allowing some sloppiness, at least $a^\mu = \epsilon\delta^\mu{}_\rho$.]