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 $x$.

  • $\begingroup$ 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? $\endgroup$ – Chiral Anomaly Sep 17 '19 at 1:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.