Suppose that we take the Klein-Gordon Hamiltonian to be of the form(adsbygoogle = window.adsbygoogle || []).push({});

[tex]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))[/tex]

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

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

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

[tex]\{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)[/tex]

we then have

[tex]\dot{\phi}(x) = \int d^3x'\, \int d^3y \frac{\delta\phi(x)}{\delta\phi(y)}\frac{\delta\mathcal{H}(x')}{\delta\pi(y)}

= \int d^3x' \int d^3y\, \delta^{(3)}(x-y)\pi(x')\delta^{(3)}(x'-y)[/tex]

[tex]=\int d^3x'\,\pi(x')\int d^3y\delta^{(3)}(x-y)\delta^{(3)}(x'-y)[/tex]

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

[tex]\int d^3x'\,\pi(x')\int d^3y\delta^{(3)}(x-y)\delta^{(3)}(x'-y)

= \int d^3x'\,\pi(x')\delta^{(3)}(x-x')[/tex]

so as to give me [itex]\dot{\phi}(x) = \pi(x)[/itex]? Have I got this right?

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# Klein Gordon Hamiltonian

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