Quantisation of Lorentz charge A conserved charge can be derived from the Noether current corresponding to the Lorentz symmetry:
$$
Q_i = \int d^3x (x_iT_{00}-tT_{0i})
$$
where $T_{\mu\nu}$ is the usual stress-energy tensor. Suppose the Lagrangian is the free field Lagrangian $$ \mathcal{L}=\frac{1}{2}(\partial\phi)^2-\frac{1}{2}m^2\phi^2$$
How to show explicitly that the quantised charge $\hat{Q}_i$ satisfies the Heisenberg equation
$$
\frac{d\hat{Q}_i}{d t }~=~\frac{1}{i\hbar}[\hat{Q}_i,\hat{H}]+
\frac{\partial\hat{Q}_i}{\partial t }~?
$$
I tried to perform the calculation by substituting the mode expansion but got  stuck with a huge mess of mixed derivatives and can't get to the result I want. I am self-studying QFT so I have no one to ask.
 A: You can write the charge operator and the Hamiltonian in terms of the ladder operators. The charge operator can also be written in term of those as it should be a combination of $\phi$ and $\pi$ (momentum conjugate). From there on it should be fine, as you have an equation containing just $a_p$ and $a_p^\dagger$. Also one usually assumes that $Q_i$ doesn't depend explicitly on time, so $\frac{\partial Q_i}{\partial t} = 0$. 
A: I suggest that you start by using the classical equations of motion for the the $\phi$ field to see what happens there. The quantum calculation will work in the same because  the quantum commutators work analogously to the classical Poisson brackets. By the way --- I think it would help conceptually to call the   conserved quantity $E X_i$ rather than $Q_i$. It is, after all, the total energy $E\equiv \int d^3 x\, T^{00}$ times the position in space of the "center of mass" (i.e the energy centroid) at time $t=0$.  It is not a "charge" like the electric charge. For general time you will then see that
$$
X_i(t) = X_i+t P_i/E
$$
where ${P_i}\equiv  \int d^3x\, T^{0i}$ is the total momentum.
