There is a relation in the textbook, "Quantum Field Theory and the Standard Model, Schwartz"

$$\left \langle 0\left | \sqrt{m^2-\vec{\bigtriangledown }^2}\phi _0(\vec{x},t) \right |\psi \right \rangle=\left \langle 0\left | \int \frac{d^3p}{(2 \pi)^3} \frac{\sqrt{\vec{p}^2+m^2}}{\sqrt{2\omega _p}}\left ( a_pe^{-ipx}-a_p^\dagger e^{ipx} \right )\right |\psi \right \rangle, \tag{2.85}$$

where $$\phi _0(\vec{x},t)=\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega _p}}\left ( a_pe^{-ipx}+a^\dagger _pe^{ipx} \right ).\tag{2.78}$$

I don't know why there is a minus sign appearing in $a_pe^{-ipx}-a_p^\dagger e^{ipx}$ instead of a plus sign.

  • $\begingroup$ It looks like he's defining $\sqrt {m^2 -\nabla ^2}$ to be, in Fourier transform, $\pm E_\mathbf p$ for positive and negative energy eigenfunctions respectively. $\endgroup$
    – pppqqq
    Nov 2 '16 at 13:36

The square root of a differential operator indicates that the Fourier factors of that operators are taken as square roots. In this case,

$$\text{FT}(\nabla^2 \varphi) \propto p^2 \widetilde \varphi$$

$$\text{FT}(\sqrt{\nabla^2} \varphi) \propto \sqrt{p^2} \widetilde \varphi$$

The operator will then be equal to something like

$$\sqrt{\nabla^2} \varphi \propto \int d^3p \ \sqrt{p^2} \widetilde \varphi e^{ipx}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.