A step in derivation of Schrodinger eq. from scalar QFT

I've read the related questions and understand the following derivation except for one key step that involves a minus sign. (From Schwartz ch.2):

$$i \langle 0| \partial_t \phi(\vec x,t) | \psi\rangle = \biggl\langle 0 \Bigm| \int \frac{d^3p}{(2\pi)^3} \frac{\sqrt{\vec p^2 + m^2}}{\sqrt{2\omega_p}} (a_p e^{-ipx} - a_p^{\dagger} e^{ipx})\Bigm|\psi\biggl\rangle \tag{2.85}$$ $$= \left\langle 0\Bigm| \sqrt{m^2- \vec \nabla^2} \phi_0(\vec x,t) \Bigm|\psi\right\rangle\tag{*}$$ which gives us $i\partial_t \psi(x) \approx \left( m - \frac{\vec \nabla^2}{2m} \right) \psi(x)$.

My question is the step from (2.85) to (*). The quantum field, defined

$$\phi_0(\vec x,t) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} (a_p e^{-ipx} + a_p^{\dagger} e^{ipx})\tag{2.81}$$

contains a plus sign, not a minus sign as in (2.85). I agree that the F.T. of $\sqrt{\vec p^2 + m^2}$ is $\sqrt{m^2 - \vec \nabla^2}$, but doesn't that minus sign ruin the equality?

• The only contribution should be from the $a_p$ term, so the sign of the $a_p^\dagger$ term makes no difference. – knzhou Dec 25 '17 at 22:54
• @knzhou Could you explain? (2.85) is equivalent to saying $\int \frac{d^3p}{(2\pi)^3} \frac{\sqrt{\vec p^2 + m^2}}{\sqrt{2\omega_p}} (-a_p^{\dagger} )e^{ipx} = \sqrt{m^2 - \vec \nabla^2} \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} (+a_p^{\dagger}) e^{ipx}$ which isn't clear to me – Dwagg Dec 25 '17 at 23:12
• But both sides are sandwiched with $\langle 0|$ and $| \psi \rangle$, so this equation is actually $0 = 0$ and it's fine. It's like how you can have $e^0 = \cos(0)$ even though $e^x \neq \cos(x)$ in general. – knzhou Dec 25 '17 at 23:16