- #1

- 359

- 4

First he mentions that

$$ \Psi (x) = <x| \Psi>,\tag{2.83}$$

which satisfies

$$i\partial _t\Psi(x)=i\partial_t< 0|\phi (\vec{x},t)|\Psi>=i<0|\partial_t\phi(\vec{x},t)|\Psi>.\tag{2.84}$$

That was all fine and good, but he lost me on the next part, going from the first line (i) to the second (ii).

(i)$$i<0|\partial_t\phi(\vec{x},t)|\Psi>=<0|\int \frac{d^3p}{(2\pi )^3} \frac{\sqrt{\vec{p}^2+m^2}}{\sqrt{2\omega _p}}(a_pe^{-ipx}-a_{p}^{\dagger}e^{ipx})|\Psi>$$

(ii)$$=<0|\sqrt{m^2-\vec{\nabla}^2}\phi_0(\vec{x},t)|\Psi>.\tag{2.85}$$

He apparently uses the Klein-Gordon Equation:

$$\partial _{t}^{2}\phi_0=(\vec{\nabla} ^2-m^2)\phi_0$$

to get the following term

$$\sqrt{m^2-\vec{\nabla}^2}$$

in equation (ii) above, but not quite sure how. Can anyone help me out?

I realize you can expand in terms of $$p^2/m$$ and make use of $$\nabla^2e^{ipx}=-p^2e^{-ipx}$$ to pull the terms out, but I'm really interested in how he uses the KG equation to achieve the same result.

This is on page 24 for those that have the text.