How Does the Klein-Gordon Hamiltonian Define the Evolution of Field Variables?

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Suppose that we take the Klein-Gordon Hamiltonian to be of the form

H = \int d^3x \, \mathcal{H}(x) = \frac{1}{2}\int d^3x\, (\pi^2(x) + (\nabla\phi(x))^2- m^2\phi^2(x))

If we want to compute, say, the evolution equation for \phi(x) we use the Poisson bracket:

\dot{\phi}(x) = \{\phi(x),H\} = \int d^3x'\,\{\phi(x),\mathcal{H}(x')\}

So, if we recall that the definition of the Poisson bracket for some functionals F[\phi,\pi;x), G[\phi,\pi;x) is

\{F,G\} \equiv \int d^3y\,\left( \frac{\delta F}{\delta \phi(y)}\frac{\delta G}{\delta\pi(y)} - \frac{\delta F}{\delta\pi(y)}\frac{\delta G}{\delta\phi(y)}\right)

we then have

\dot{\phi}(x) = \int d^3x&#039;\, \int d^3y \frac{\delta\phi(x)}{\delta\phi(y)}\frac{\delta\mathcal{H}(x&#039;)}{\delta\pi(y)}<br /> = \int d^3x&#039; \int d^3y\, \delta^{(3)}(x-y)\pi(x&#039;)\delta^{(3)}(x&#039;-y)
=\int d^3x&#039;\,\pi(x&#039;)\int d^3y\delta^{(3)}(x-y)\delta^{(3)}(x&#039;-y)

The question I have is whether or not I can integrate the dirac distributions as follows:

\int d^3x&#039;\,\pi(x&#039;)\int d^3y\delta^{(3)}(x-y)\delta^{(3)}(x&#039;-y)<br /> = \int d^3x&#039;\,\pi(x&#039;)\delta^{(3)}(x-x&#039;)

so as to give me \dot{\phi}(x) = \pi(x)? Have I got this right?
 
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It might be a bit better to do the x&#039; integration first. Then, you don't have to use a delta function to evaluate a delta. In the standard theory of distributions, products of distributions aren't defined.
 
There Are no problems with the two deltas product cause they are evaluated at different points. (x-y) and (x'-y). The theory of distributions see Schwarz (a fields medal) says that the problem is doing this inner product (d(x)*d(x),f(x)) where d stand for the delta compact support distributions. and these integral is obviously f(0)*infinite or some not well defined value.
You got it right shoeron: the time derivative of field (not only scalars field) is its conjugate momentum. It is usefull to remebere in fields theory
 
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