Hamiltonian density is time-independent, so how does energy get transported in QFT? In a QFT on $d+1$-dimensional Minkowski spacetime, the Hamiltonian is:
$$\tag{1} \hat{H} = \int d^{d}\vec{x} \hat{\mathcal{H}}(\vec{x}), $$
where $\hat{\mathcal{H}}(\vec{x})$ is the Hamiltonian density operator at position $\vec{x}$. In the Heisenberg picture, the Hamiltonian density itself evolves as
\begin{align}\tag{2}
   \partial_t \hat{\mathcal{H}}(\vec{x}) & = \frac{i}{\hbar}[\hat{H},\hat{\mathcal{H}}(\vec{x})] \\
    & = 0
\end{align}
where the last line follows since $$\tag{3}[\hat{\mathcal{H}}(\vec{x}),\hat{\mathcal{H}}(\vec{y})]=0$$ for all $\vec{x},\vec{y}$.
So $\hat{\mathcal{H}(\vec{x})}$ is constant over time. But this surely can't be true: it should be possible for the energy at a point (more generally, in a region) to change over time.
Where did I go wrong?
 A: The Hamiltonian density $\hat{\cal H}_H(\vec{x},t)$ in the Heisenberg picture could in principle have explicit time dependence
$$\frac{d}{dt}\hat{\cal H}_H(\vec{x},t) ~=~\frac{1}{i\hbar}[\hat{\cal H}_H(\vec{x},t),\hat{H}_H(t)]+\left(\frac{\partial \hat{\cal H}_S(\vec{x},t)}{\partial t}\right)_H.\tag{A}$$
The commutator $$[\hat{\cal H}_H(\vec{x},t),\hat{\cal H}_H(\vec{y},t)]\tag{B}$$ vanishes for spacelike separated operators because of locality, but there could in principle be contact terms proportional to (derivatives of) a Dirac delta distribution $\delta^d(\vec{x}-\vec{y})$.
A: For a vibrating string we have energy
$$
H= \int_{-\infty}^{\infty} \left\{\frac 12 \rho \dot y^2+ \frac 12 T{y'}^2 \right\}dx
$$
and the local energy conservation law is
$$
\frac{\partial}{\partial t}\left\{\frac 12 \rho \dot y^2+ \frac 12 T{y'}^2\right\} + \frac{\partial}{\partial x}\{-T\dot y y'\}=0,
$$
So -- even classically --- you can see from the $\partial_x$ term that the Poisson bracket of two hamiltonian densities involves a $\delta'(x-x')(Ty'\dot y)$, which is, in some sense, zero at $x=x'$  but still gives a non-zero contribution after integration by parts. This is what @Qmechanic is saying.
