# Does the Heisenberg equation for fields and canonical momentums hold as well for the hamiltonian density operator instead of the Hamiltonian operator?

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}

## 2 Answers

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 $$\{ \phi(x),\pi(y) \} ~=~\delta^3(x\!-\!y)\qquad\text{and}\qquad \{\!\{ \phi(x),\pi(x) \}\!\} ~=~1,\tag{4}$$ 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.