Is the boost operator conserved (in the context of QFT)? Let $J^{\mu\nu}$ be the generators of Lorentz transformations, so that $J^{ij}$ generates rotations, and $J^{0i}$ generates boosts. The algebra of the Poincaré Group contains
$$
[P^\mu,J^{\alpha\beta}]=i(\eta^{\mu\alpha}P^\beta-\eta^{\mu\beta}P^\alpha)
$$
The $\mu=0$ component of this relation is the commutator of $J^{\alpha\beta}$ with the Hamiltonian. For example, in the case of rotations, it implies
$$
[H,J^{ij}]=0
$$
so that angular momentum is conserved.
On the other hand, in the case of boosts, it implies
$$
[H,J^{0i}]=iP^i
$$
so, apparently, the boost operators don't commute with the Hamiltonian. The only way to make $J^{0i}$ to be conserved is to assume that it is explicitly time-dependent, so that
$$
\dot J^{0i}=i[H,J^{0i}]+\partial_0 J^{0i}
$$
If we take $\partial_0 J^{0i}=P^i$, then we find that $\dot J^{0i}=0$, and this operator is conserved.
Question. Is there any "first principles" reason as to why we should take $\partial_0 J^{0i}=P^i$? Or are the boost operators not conserved in general?

I know that in the framework of canonical quantisation, we take the $J^{\mu\nu}$ operators to be
$$
J^{\mu\nu}=\int\mathrm d\boldsymbol x\ x^\mu \mathcal P^\nu-x^\nu \mathcal P^\mu+\text{spin part}
$$
so that we indeed have $\partial_0 J^{0i}=P^i$. But this clearly relies on the existence of a Lagrangian, and on the assumption that the generators of the Poincaré Group must be those given by Noether's formula, and on all the details of canonical quantisation in general (which IMHO is justified only because of consistency, but there is no "first principles" reason to many of its ingredients). I am interested in the properties of the Poincaré Group without any reference to a Lagrangian or to "quantisation" of a classical system.
 A: Let us suppose that $G$ is a Lie group admitting some faithful unitary strongly continuous representation $U$ on a Hilbert space $\cal H$, so that we can interpret the elements $a$  of the Lie algebra of $G$ in terms of self-adjoint generators of one-parameter unitary groups,
$$U(\exp(s a)) = e^{is A}\:.$$
The operators $A$ are self-adjoint and are commonly defined in a  dense subdomain called Garding space where they are essentially self-adjoint (another interesting domain is the one constructed by Nelson where the exponential in the right-hand side may be developed as in its standard Taylor series).
If $G$ is the largest group of continuous symmetries of the quantum system, one of these one-parameter subgroups should represent the temporal evolution of the system. Let us suppose that it is the group generated by the element $-h$ of the Lie algebra of $G$. We therefore have
$$U(\exp(-t h)) = e^{-it H}$$
I changed the sign since the time evolution is the inverse operation of the time translation. $H$, by definition, is the Hamiltonian observable of the system. Obviously some physical requirements are necessary on $H$, first of all its spectrum must be bounded below etc...I do not stick on them here and I henceforth suppose that $H$ is a well-behaved physical Hamiltonian.
Now we have two possibilities, if $a$ is a generic element of the Lie algebra. One is ($\{\cdot, \cdot\}$ is the standard Lie algebra commutator)
$$\{h, a\}=0\tag{1}\:.$$
As a consequence of Hausdorff - Baker - Campbell equality in $G$, we can exponentiate this identity to a group identity
$$\exp(th) \exp(sa) \exp(-th) = \exp(sa)\quad \forall t,s \in \mathbb R\:.$$
Applying the representation $U$:
$$e^{itH} e^{sA} e^{-itH} =  e^{sA}\quad \forall t,s \in \mathbb R\:.$$
Taking advantage of Stone theorem, taking the strong derivative in $s$ we immediately obtain
$$e^{itH} A e^{-itH} =  A\quad \forall t\in \mathbb R\:.$$
(This identity is completely well-posed even regarding subtleties with domains.)
This identity says nothing but that the Heisenberg evolution of the observable $A$ is constant and thus $A$ is a constant of motion.
What happens if instead 
$$\{h, a\}\neq 0\: \mbox{?}\tag{2}\:.$$
In this case we have 
$$\exp(th) \exp(sa) \exp(-th) = \exp(sa(t))\quad \forall t,s \in \mathbb R\:.\tag{2}$$
where we have used the natural action of $g \in G$ over its Lie algebra, 
$$a \mapsto g^{-1}ag$$
and we have defined the element of the Lie algebra $a(t)$
$$a(t) := \exp(th) a\exp(-th) \quad \forall t\in \mathbb R\:.\tag{3}$$
Fixing a basis $a_1, \ldots, a_n$ of the Lie algebra,
$$a = \sum_j c_j a_j \quad \mbox{for some reals $c_j$}$$
and thus 
$$a(t) = \sum_j c_j(t) a_j \quad \mbox{for some real valued functions $c_j= c_j(t)$}$$
 the found identity can be rephrased to
$$\exp(th) \exp(sa) \exp(-th) = \exp(\sum_{j=1}^n sa(t))\quad \forall t,s \in A\:.$$
Exploiting again the theorem by Stone 
we conclude that for an observable constructed out of the self-adjoint generators $A_j$ (corresponding to the basis of the $a_j$),
$$A = \sum_j c_j A_j \quad \mbox{for some reals $c_j$}$$
it holds 
$$e^{itH} A e^{-itH} =  \sum^n_{j=1}c_j(t) A_j\quad \forall t\in \mathbb R\:.\tag{4}$$ 
(This identity is valid on Garding space and can be extended into a true identity between self-adjoint opertators taking the closures of both sides).
The physical content of the found identity is that, 
even if the representative $A$ of the Lie algebra of observables  is not a constant of motion, its "à la Heisenberg" evolution is however described in terms of a linear combination of the generators and the time dependence affects the numerical coefficients only. 
This is a highly non-trivial result. Actually, the result can be turned into a statement concerning existence of  constants of motion parametrically depending on time. This is the standard procedure usually adopted in QFT, especially for the boost generator.
Assuming  that (4) is valid, one define the observable parametrically depending on time in Schroedinger picture (and once again the dependence on time just appears in the coefficients)
$$A_S(t) := \sum^n_{j=1}c_j(-t) A_j$$
With this definition
$$A_S(0) := A$$
and, from (4), where $U_t:= e^{-itH}$
$$A_H(t) := U^*_t A_S(t) U_t := A_S(0) \quad \forall t \in \mathbb R$$
which, formally, on some domain could be re-written 
$$ \partial_t A_H(t) + i[H, A_H(t)]=0\:.$$
If $G$ is the Poincare' group,  the boost generators $K_j$ are treated that way. One defines the time-parametrized boost operators $K_{Sj}(t)$ in the Schroedinger picture which always take a form like this where $j=1,2,3$,
$$K_{Sj}(t) = K_j - t Z_j$$
where $Z_j$ is a certain constant linear combination of the generators of the Lie algebra and $K$ is the standard boost generator you get by the commutation relations. In QFT it has a contribution due to the spin and an orbital part. $K$ is important because it is related with the relativistic position operator as it can be understood if performing the non relativistic limit (i.e. replacing the Poincaré group with Galilei's group)
$$K_{Sj}(t) = mX_j - t P_j\:,$$
where $m$ is the mass of the system and $X_j$ the position of its center of mass.
A: You are correct that, in the abstract Poincaré algebra, the "Hamiltonian" $P^0$ does not commute with the boosts $K^i = J^{0i}$. But, crucially, "conservation" has no meaning in the abstract setting. Conservation of a quantity is a statement about a dynamical system, either in the Lagrangian or in the Hamiltonian formalism. In particular, while Noether's theorem requires the symmetry to be an off-shell symmetry of the Lagrangian, the conservation of the current holds only upon use of the equations of motion, i.e. on-shell. Also, the Poincaré algebra as an algebra knows nothing about the special role time plays for the idea of conservation. It's just an algebra.
Therefore, there is no way at all to expect a "first principles" reasoning that $J^{0i}$ should be conserved in all dynamical systems carrying a representation of the Poincaré algebra, but in those that are Poincaré invariant, we necessarily need to have
$$ \mathrm{i}[H,J^{0i}] + \partial_0 J^{0i} = 0 \tag{1}$$
by exactly the reasoning you have presented.
Formally, to consider the time-dependence of $J^{0i}$ and eq. (1) properly within a Hamiltonian framework, we would have to define an extended phase space by adding the time $t$ and its conjugate momentum $p_t$ to the phase space and defining the extended Hamiltonian $\tilde{H} = p_t + H(t,x,p)$. One can show that a quantity $f(t,x,p)$ is a constant of motion of the original system if and only if
$$ \{f,\tilde{H}\} = 0$$
for this extended system. Note that $\{f,\tilde{H}\} = \{f,p_0\} + \{f,H\}$, so this is eq. (1) in disguise. For a reference doing this extension in detail (and also discussing the quantization of this setting), see e.g. "Time-dependent mechanical symmetries and extended Hamiltonian systems" by Kuwabara.
We can, however, argue rather straightforwardly that eq. (1) is not an unreasonable condition considering the physical meaning of $J^{0i}$ and $P^i$:
On Minkowski space, an infinitesimal boost in the x-direction is given by $x\mapsto x+\epsilon t$ for infinitesimal $\epsilon$, that is, $J^{0x} x = t$. Taking the derivative w.r.t. $t$, this immediately becomes $\partial_0 (J^{0x} x) = 1$, so the infinitesimal transformation generated by $\partial_0 J^{0x}$ is the translation $x\mapsto x+\epsilon$, which is exactly the transformation the momentum $P^x$ generates. That is, the formula you write for $J^{0i}$ at the end of your question that is allegedly dependent on all the details of quantization of QFT...isn't! Even classically, we can say that on functions on Minkowski space, we have the representation
$$ J^{0i} = x^0 \partial^i - x^i\partial^0$$
for the generators of boosts, and $\partial_0 J^{0i} = \partial^i = P^i$.
