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?

  • $\begingroup$ 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? $\endgroup$
    – Ryder Rude
    Nov 25, 2022 at 4:51

2 Answers 2


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 $\delta^d(\vec{x}-\vec{y})$.

  • $\begingroup$ Surely $[\mathcal{H}(\vec{x}), \mathcal{H}(\vec{x})]= 0$. Where do the $\delta^d(\vec{x}-\vec{y})$ terms come from?? $\endgroup$ 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.

  • $\begingroup$ 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? $\endgroup$ Dec 2, 2022 at 17:11
  • $\begingroup$ 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 $[\dot y(x), y(x')]\propto \delta(x-x')$ $\endgroup$
    – mike stone
    Dec 2, 2022 at 20:01

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