Does equality of operators on vacuum imply equality of operators on the whole space? Consider a field operator $\Phi(x)$ which generates states from the vacuum such as
$$ \tag 1 | x \rangle = \Phi(x) | 0 \rangle.$$
Consider also how a translation is implemented on such a state:
$$ \tag 2 | x + y \rangle = e^{iyP} | x \rangle,$$
where $P^\mu$ are the generators of translations.
Now consider the following chain of identies, in which only (1) and (2) are used:
$$ \tag 3 \Phi(x+y) | 0 \rangle
= | x + y \rangle
= e^{iyP} | x \rangle
= e^{iyP} \Phi(x) | 0 \rangle. $$
From this, I would be tempted to conclude that the transformation rule for $\Phi(x)$ is
$$ \tag 4 \Phi(x+y) = e^{iyP} \Phi(x),$$
while we know that the correct transformation rule is
$$ \tag 5 \Phi(x+y) = e^{iyP} \Phi(x) e^{-iyP}.$$
What is wrong with the above reasoning?
 A: $\newcommand{\ket}[1]{\lvert #1 \rangle} \newcommand{\iex}[1]{\mathrm{e}^{\mathrm{i}#1}} \newcommand{\miex}[1]{\mathrm{e}^{-\mathrm{i}#1}}$Your $(3)$ does not imply $(4)$, since equality of operators on one state - the vacuum in this case - does not imply the equality of the operators on the whole space.
To get the correct transformation, consider an arbitrary state $\ket{\psi}$ transforming under a translation by $y$ as $\ket{\psi} \mapsto \iex{yP}\ket{\psi}$. Since $\Phi(x)\ket{\psi}$ is also a state, we must have $\Phi(x)\ket{\psi} \mapsto \iex{yP}(\Phi(x)\ket{\psi})$. But we also know that the transformation works on the individual parts as
$$ \Phi(x)\ket{\psi} \mapsto \Phi(x+y)\iex{yP}\ket{\psi}$$
and since the momentum operator and the field do not necessarily commute, we see that, certainly, $\Phi(x+y) = \iex{yP}\Phi(x)\miex{yP}$ is a prescription that yields the desired transformation rule for all states $\ket{\psi}$.
Note: This is not a wrong prescription in case of the vacuum, since the vacuum is translation invariant, so $(4)$ and $(5)$ have the same effect in that case.
