- #1

eudo

- 29

- 8

[tex]<s^+,p'|j^\mu_{em,s}|s^+,p> = e(p+p')^\mu e^{-i(p-p')\cdot x}[/tex]

where

[tex]|s^+,p>=\sqrt{2E}\hat{a}^\dagger(p)|0>[/tex]

and

[tex]j^\mu_{em,s}=ie(\phi^\dagger\partial^\mu \phi - (\partial^\mu \phi^\dagger)\phi[/tex]

Now, it turns out the solutions for this problem are online here

But I have a question about one of the steps. They expand the [itex]\phi[/itex]'s in terms of [itex]\hat{a}[/itex] and [itex]\hat{b}[/itex] operators and note that the [itex]\hat{a}[/itex]'s and [itex]\hat{b}[/itex]'s commute, so that we can move all the [itex]\hat{b}[/itex]'s to the right where they will give zero when acting on |0>. So we're only left with a term that has the [itex]\hat{a}[/itex] operators.

But it seems to me we still have a term

[tex]

ie\sqrt{4EE'}<0|\hat{a}(p')\int\frac{d^3\boldsymbol{k'}}{(2\pi)^3\sqrt{2\omega'}}\hat{b}(k')e^{-ik'\cdot x}\int \frac{d^3\boldsymbol{k}}{(2\pi)^3\sqrt{2\omega}}ik^\mu \hat{b}^\dagger(k)e^{ik\cdot x}\hat{a}^\dagger(p)|0>

[/tex]

Where did this term go? The [itex]\hat{b}[/itex] and [itex]\hat{b}^\dagger[/itex] operators do not commute.