 Solution

A scalar field has $\delta {\varphi }_a=0$ for a pure change of spacetime argument. With $\delta x^k={ϵ}^k$ and $\delta x^0=0$, its variation at a fixed argument is $\bar{\delta }{\varphi }_a=-{ϵ}^k{\partial }_k{\varphi }_a$. The supplied Noether charge becomesSince $\delta Q/\delta {\pi }_a=-{ϵ}^k{\partial }_k{\varphi }_a$, solution and antisymmetry of the Poisson bracket giveCanonical quantization replaces the Poisson bracket by a commutator divided by $i\hslash$:With this sign convention one may use $U=e^{-i\hat{Q}/\hslash }$ and ${\hat{\varphi }}_a^{\prime }=U^{†}{\hat{\varphi }}_{a}U$. Because $Q=-{ϵ}^k{P}_k$, the negative shift of the field argument follows with the required sign.