Cfrgtkky

In quantum field theory, with the field $phi$ and the momentum $pi$ being operators, their time evolution is governed (in the Heisenberg-picture) by the Heisenberg equation:

begin{align}
dot{phi} = frac{i}{hbar}[ hat{H}, phi] \
dot{pi} = frac{i}{hbar}[ hat{H}, pi]. \
end{align}

Now, in case the Hamiltonian operator $hat{H}=int d^3x ~hat{cal H}$ can be written as an integral over the hamiltonian density $hat{cal H}$, and the fields and the momenta commute at non-equal positions, do the same equations hold as well with the Hamiltonian operator being replaced by it's density? What would the caveats be?

begin{align}
dot{phi} = frac{i}{hbar}[ hat{cal H}, phi] \
dot{pi} = frac{i}{hbar}[ hat{cal H}, pi]. \
end{align}

You have $hat{H} = int d^3x hat{tilde{H}}(x)$. That implies that canonical Relations will be slightly altered.

For a Quantum field Operator $hat{phi}(x',t)$ distributed over space $x'$ and time $t$, you will have a relation like the following:

$[hat{tilde{H}}(x),hat{phi}(x',t)] = frac {partial}{partial t} hat{phi(x',t)} delta(x-x')$.

The Delta function factor ensures not only the commutation of Operators for nonequal space Points; also that after Integration over space, the ordinary commutation Relations are obtained

1. The answer is No. For starters for dimensional reasons. A density carries dimension $L^{-3}$.




2. In the classical (as opposed to the quantum) case, it is tempting to (at least partially) incorporate OP's suggestion for functionals
   $$ F~=~int ! d^3x~f(x), qquad G~=~int ! d^3x~g(x), tag{1} $$
   by changing the definition from the standard field-theoretic canonical Poisson bracket
   $${ F, G} ~:=~int_V ! d^3x ~left(frac{delta F}{delta phi (x)}frac{delta G}{delta pi (x)}-frac{delta F}{delta pi (x)}frac{delta G}{delta phi (x)} right)
   ~=~int_V ! d^3x ~{!{ f(x),g(x)}!} tag{2}$$
   to a same-$x$ Poisson bracket
   $$ {!{ f(x),g(x)}!} ~:=~frac{delta f(x)}{delta phi (x)}frac{delta g(x)}{delta pi (x)}-frac{delta f(x)}{delta pi (x)}frac{delta g(x)}{delta phi (x)}, tag{3} $$
   where $delta f(x)/delta phi (x)$ denote a same-spacetime functional derivative, see e.g. my Phys.SE answer here. In other words, the non-zero fundamental Poisson brackets read
   $${ phi(x),pi(y) } ~=~delta^3(x!-!y)qquadtext{and}qquad
   {!{ phi(x),pi(x) }!} ~=~1,tag{4}$$
   i.e. the same-$x$ Poisson bracket (3) is defined without a Dirac delta distribution. However, in the $x$-local ${!{cdot,cdot}!}$ formalism (3) equality signs typically only hold modulo total spacetime derivative terms.