# Klein Gordon Hamiltonian

#### shoehorn

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

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

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

$$\int d^3x'\,\pi(x')\int d^3y\delta^{(3)}(x-y)\delta^{(3)}(x'-y) = \int d^3x'\,\pi(x')\delta^{(3)}(x-x')$$

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

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#### George Jones

Staff Emeritus
Gold Member
It might be a bit better to do the $x'$ 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.

#### Marco_84

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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