Explicit form of the translation operator generators in the Poincare group? Let $P_0$ be the generator for temporal translation and $P_1, P_2, P_3$ be for spatial translations. Let $p_μ$ be the momentum operator in the $x_μ=x^μ$ direction. I watched a lecture where the guy said that $P_μ=p_μ=i\eta^{μ \nu}\partial_\nu$ (this is all in planck units for simplicity and $\eta^{μ\nu}$ is the $(+,-,-,-)$ Minkowski metric) and for $μ=0, P=H$ where $H$ is the Hamiltonian operator (which makes sense to me because of the Schrodinger equation). But, with a little wikipedia research, I found that the spatial translation operator is $T=\exp(-ix^μp_μ)$ and the temporal translation operator is $T=\exp(iHt)$. So shouldn't the translation generators of the Poincare group then be $$P_μ=i\eta^{μ\nu}x_μp_{\nu}~?$$ What am I missing here?
 A: Let $|x\rangle$ be an element of the position basis. Define the translation operator $T(a)$ as
$$
|x+a\rangle\equiv T(a)|x\rangle \tag{1}
$$
for any vector $a^\mu$. As $T$ is unitary, it can be written as
$$
T(a)=\mathbb I-ia_\mu P^\mu \tag{2}
$$
for some hermitian$^1$ operators $P^\mu$. If we expand the l.h.s. $(1)$ to linear order in $a$ (just a Taylor expansion) we get
$$
|x+a\rangle=|x\rangle+a^\mu\partial_\mu|x\rangle+\mathcal O(a^2) \tag{3}
$$
On the other hand, if we expand the r.h.s. of $(1)$ to linear order in $a$, we get
$$
T(a)|x\rangle=|x\rangle-ia_\mu P^\mu|x\rangle \tag{4}
$$
which, by comparison with $(3)$ means that
$$
P_\mu=i\partial_\mu \tag{5}
$$
Therefore, the momentum operator is $P_\mu=i\partial_\mu$ (in the position basis), and the translation operator is
$$
T(a)=\mathrm e^{-ia_\mu P^\mu}=\mathrm e^{a^\mu \partial_\mu} \tag{6}
$$
So, what does $T(x)=\mathrm e^{-ix^\mu P_\mu}$ mean? Well, this is just the translation operator $(6)$, in the specific case $a^\mu=x^\mu$, i.e., we are displacing by the vector $x$. Note that this is just a specific case of $(6)$, which is sometimes used in QFT; for example, if $\hat\Phi(x)$ is a field (operator), then you can relate its value at any $x$ with the value at $x=0$ through $\hat\Phi(x)=T(x)\hat\Phi(0)T^\dagger(x)$. In some cases, this is useful, for example to prove that the vev $^2$ of any field is just a constant (i.e., position independent). Anyway, the fact that we can write $T(x)$ doesn't change the form of the generator of $T$, which means that the momentum operator is always given by $(5)$.

$^1$ self-adjoint, I know...
$^2$ vev = vacuum expectation value = $\langle\text{VAC}|\hat\Phi(x)|\text{VAC}\rangle$
