shoehorn
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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'\, \int d^3y \frac{\delta\phi(x)}{\delta\phi(y)}\frac{\delta\mathcal{H}(x')}{\delta\pi(y)}<br /> = \int d^3x' \int d^3y\, \delta^{(3)}(x-y)\pi(x')\delta^{(3)}(x'-y)
=\int d^3x'\,\pi(x')\int d^3y\delta^{(3)}(x-y)\delta^{(3)}(x'-y)
The question I have is whether or not I can integrate the dirac distributions as follows:
\int d^3x'\,\pi(x')\int d^3y\delta^{(3)}(x-y)\delta^{(3)}(x'-y)<br /> = \int d^3x'\,\pi(x')\delta^{(3)}(x-x')
so as to give me \dot{\phi}(x) = \pi(x)? Have I got this right?
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'\, \int d^3y \frac{\delta\phi(x)}{\delta\phi(y)}\frac{\delta\mathcal{H}(x')}{\delta\pi(y)}<br /> = \int d^3x' \int d^3y\, \delta^{(3)}(x-y)\pi(x')\delta^{(3)}(x'-y)
=\int d^3x'\,\pi(x')\int d^3y\delta^{(3)}(x-y)\delta^{(3)}(x'-y)
The question I have is whether or not I can integrate the dirac distributions as follows:
\int d^3x'\,\pi(x')\int d^3y\delta^{(3)}(x-y)\delta^{(3)}(x'-y)<br /> = \int d^3x'\,\pi(x')\delta^{(3)}(x-x')
so as to give me \dot{\phi}(x) = \pi(x)? Have I got this right?
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