# 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?

• This sounds very surprising to me too. Perhaps the energy density is indeed a constant for free theories, and that energy redistribution is a feature of interacting theories? Nov 25, 2022 at 4:51

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 $$\delta^d(\vec{x}-\vec{y})$$.
• Surely $[\mathcal{H}(\vec{x}), \mathcal{H}(\vec{x})]= 0$. Where do the $\delta^d(\vec{x}-\vec{y})$ terms come from?? Nov 25, 2022 at 17:36
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.
• Is this really what @Qmechanic is saying? I can't make any connection between what you and s/he said. Since $[\mathcal{H}(\vec{x}),\mathcal{H}(\vec{x})]=0$, I can't see where the delta function comes from - perhaps you could edit your answer to address this? Dec 2, 2022 at 17:11
• He is saying that $[H(x),H(x')]\propto \delta'(x-x')$ and, as $\delta'(x)=- \delta'(-x)$, we can say that $\delta'(0)=0$. Usualy distributions cannot be evalauted at a point but any approximant to $\delta'(x)$ will be zero at $x=0$ In the case I give that $\delta'$ is what gives the energy flux. The $\delta'(x-x')$ comes from the commutator of $y'$ and $\dot y$ since $[\dot y(x), y(x')]\propto \delta(x-x')$ Dec 2, 2022 at 20:01