What does the square root of Laplacian mean?

Question

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.

Answer 1

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

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