- #1

TroyElliott

- 59

- 3

- TL;DR Summary
- I need help understanding a term that appears in the derivation of the time dependent conjugate momentum of the Klein-Gordon field

When deriving ##\Pi(\vec{x},t)## for the Klein-Gordon equation (i.e. plugging ##\Pi(\vec{x},t)## into the Heisenberg equation of motion for the scalar field Hamiltonian), we come across a term that is the following

##\int_{-\infty}^{\infty}d^{3}y \nabla_{y}(\delta(\vec{x}-\vec{y}))\nabla\phi(y)##

We are then told "using integration by parts we get the following"

##\int_{-\infty}^{\infty}d^{3}y \nabla_{y}(\delta(\vec{x}-\vec{y}))\nabla\phi(y) = -\int_{-\infty}^{\infty}d^{3}y \delta(\vec{x}-\vec{y})\nabla^{2}\phi(y).##

For this to be true, the the boundary term ##\delta(\vec{x}-\vec{y})\nabla\phi(y)## evaluated at ##\infty## and at ##-\infty##, must be zero. Are we assuming that the gradient of the the field goes to zero at infinity? And what about the delta function term when it becomes ##\delta(\vec{x}-\infty)##? This blows up at the the boundary at infinity. Any thoughts on why this boundary term should be zero is much appreciated!

Thanks!

##\int_{-\infty}^{\infty}d^{3}y \nabla_{y}(\delta(\vec{x}-\vec{y}))\nabla\phi(y)##

We are then told "using integration by parts we get the following"

##\int_{-\infty}^{\infty}d^{3}y \nabla_{y}(\delta(\vec{x}-\vec{y}))\nabla\phi(y) = -\int_{-\infty}^{\infty}d^{3}y \delta(\vec{x}-\vec{y})\nabla^{2}\phi(y).##

For this to be true, the the boundary term ##\delta(\vec{x}-\vec{y})\nabla\phi(y)## evaluated at ##\infty## and at ##-\infty##, must be zero. Are we assuming that the gradient of the the field goes to zero at infinity? And what about the delta function term when it becomes ##\delta(\vec{x}-\infty)##? This blows up at the the boundary at infinity. Any thoughts on why this boundary term should be zero is much appreciated!

Thanks!