It is a well-known effect that generic localized operators spread out under evolution by lattice Hamiltonians. For example , $$e^{iHt} \sigma^x _j e^{-iHt}$$ will in general not be supported only at the lattice site $$j$$.
I am not sure how this spreading happens when the Hamiltonian is a field-theoretic Hamiltonian on a continuum. Take for instance the field operator at a space pint $$x$$ evolved to a time $$t$$,
$$\phi(x,t) = \phi(x,0) + it [H,\phi(x,0)] - \frac{t^2}{2} [H, [H, \phi(x,0)]] + \ldots$$
Since the Hamiltonian can be written as a space integral of the Hamiltonian density, it seems that $$[H,\phi(x,0)]$$ and all higher-order commutators are still supported at $$x$$ implying that there is no spreading.
This line of reasoning is obviously wrong, but I cannot find the error. Intuitively I suppose that the operator should spread because of the spatial derivates present in the Hamiltonian. But to me, it seems that $$[\frac{\partial \phi(x)}{\partial x}, \phi(x)]$$ should still be supported at $$x$$.
• The Hamiltonian involves derivatives of the field operators, and therefore so do the commutators. As an example, take $H$ to be the Hamiltonian of a free scalar field, and consider $\dot \phi(x,t)\propto [H,\dot\phi(x,t)]\propto (\nabla^2+m^2)\phi(x,t)$. Even if $\phi$ were a classical field, the equation $\dot \phi(x,t)\propto (\nabla^2+m^2)\phi(x,t)$ would imply spreading. Saying that $\nabla^2\phi(x)$ is supported only at $x$ is misleading, because the definition of $\nabla$ requires considering a neighborhood of $x$. Does this address the question? – Chiral Anomaly Sep 17 at 1:42