The issue here is that there two *distinct* actions of the translation group on fields present in your computations.

**Definitions of the group actions.**

The two group actions to which I'm referring are as follows. For conceptual simplicity, let $\phi$ denote an operator-valued field defined on $\mathbb R$; generalization to higher dimensions is straightforward. Let $\mathcal H$ denote the Hilbert space of the theory, then we assume (at least in theories for which we want to talk about translation-invariance) that there is a unitary representation $\hat U$ of the translation group of $\mathbb R$ acting on the Hilbert space.

This unitary representation then induces an action $\rho_1$ of the translation group acting on fields as follows:
\begin{align}
(\rho_1(a)\hat \phi)(x) = \hat U(a)\hat\phi(x)\hat U(a)^{-1}. \tag{1}
\end{align}
On the other hand, we can define a second action of the translation group acting on fields as follows:
\begin{align}
(\rho_2(a)\hat \phi)(x) = \hat \phi(x-a) \tag{2}
\end{align}

**Infinitesimal generators.**

Each of the group actions above possesses an infinitesimal generator.

To determine what that is for $\rho_1$, we write $\hat U(a) = e^{-ia\hat P}$, so that $\hat P$ is the infinitesimal generator of $\hat U$, and we notice that if we expand the right hand side of $(1)$ in $a$ we have
\begin{align}
\hat U(a)\hat\phi(x)\hat U(a)^{-1}
&= (\hat I - ia\hat P)\hat \phi(x) (\hat I + ia\hat P) + O(a^2) \\
&= \hat \phi(x) + ia\hat\phi(x)\hat P - ia\hat\phi(x) \hat P + O(a^2) \\
&=\hat \phi(x) -ia\Big(\hat P\hat \phi(x)-\hat \phi(x) \hat P\Big) + O(a^2) \\
&= \hat\phi(x) -ia[\hat P,\hat \phi(x)] + O(a^2)
\end{align}
inspecting the term that is first order in $a$, we see immediately that the operator
\begin{align}
\hat \phi(x) \mapsto [\hat P,\hat \phi(x)]
\end{align}
is the infinitesimal generator of the first group action $\rho_1$. It turns out by the way that this operator has a special name: the adjoint operator, and it is often denoted $\mathrm{ad}_{\hat P}$. So all in all, we see that $\mathrm{ad}_{\hat P}$ is the infinitesimal generator of $\rho_1$ since we have shown that
\begin{align}
(\rho_1(a)\hat \phi)(x) = \big(\hat I -ia \,\mathrm{ad}_{\hat P} + O(a^2)\big)\hat\phi(x)
\end{align}
As an aside, this is all intimiately related to the so-called Hadamard Lemma for the Baker-Campbell-Hausdorff formula.

To determine the infinitesimal generator for $\rho_2$, we expand the right hand side of $(2)$ in $a$ using Taylor's formula to obtain
\begin{align}
\hat\phi(x+a)
&= \hat \phi(x) - a \partial\phi(x) + O(a^2) \\
&= \left(1-ia(-i\partial) + O(a^2)\right)\hat\phi(x) \\
\end{align}
so that the infinitesimal generator of the group action $\rho_2$ is $-i\partial$. We can summarize these results as follows. Let's call the infinitesimal generator of $\rho_1$ $P_1$ and the infinitesimal generator of $\rho_2$ $P_2$, then we have shown that
\begin{align}
P_1 = \mathrm{ad}_{\hat P}, \qquad P_2 = -i\partial
\end{align}
Notice, in particular, that these are not the same mathematical objects.

**Fields that transform in special ways under translations.**

Although the group actions $\rho_1$ and $\rho_2$ are distinct and have distinct generators, it is sometimes the case in field theory that one considers fields $\hat \phi$ that transform as follows:
\begin{align}
\hat U(a)\hat\phi(x) \hat U(a)^{-1} = \hat\phi(x-a).
\end{align}
Notice that the left hand side of this is just the action of $\rho_1$ on $\hat\phi$, and the right hand side is the action of $\rho_2$ on $\hat\phi$, so for this special class of fields, the two group actions agree! In this special case, it also follows that the infinitesimal generators of $\rho_1$ and $\rho_2$ agree on these special fields;
\begin{align}
P_1 \hat\phi(x) = P_2\hat\phi(x),
\end{align}
or more explicitly
\begin{align}
\mathrm{ad}_{\hat P}\hat\phi(x) = -i\partial\hat\phi(x).
\end{align}

This post imported from StackExchange Physics at 2014-06-21 21:34 (UCT), posted by SE-user joshphysics