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In Matthew Schwartz's QFT text, he derives the Schrodinger Equation in the low-energy limit. I got lost on one of the steps.

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:

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

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?

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

Qmechanic
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EthanT
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    For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein. – Qmechanic Feb 10 '19 at 06:13
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    Possible duplicate: https://physics.stackexchange.com/q/244994/2451 – Qmechanic Feb 10 '19 at 06:14
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    expand the square root in powers of $p^2/m^2$ and make use of $\nabla^2e^{ip\cdot x}=-p^2e^{ip\cdot x}$; this allows you to pull the $\nabla$'s out of the momentum integral, which can then be evaluated using what you already know, and then sum up the series to rewrite as a square root operator – Wakabaloola Feb 10 '19 at 06:25
  • @Wakabaloola That looks different than what the author was suggesting, but sure seems to do the trick, regardless. I should have caught that myself, I suppose. Thanks! Do you also happen to know the connection to the Klein-Gordon equation above and how one would use that to derive that term? – EthanT Feb 10 '19 at 16:20
  • if i interpret your question correctly: yes, once you’ve determined that d/dt acts as a square root operator on that correlator you can act with it twice – Wakabaloola Feb 10 '19 at 16:26
  • @Wakabaloola That's the part where I believe I am lost Why exactly would d/dt act as a square root operator? – EthanT Feb 10 '19 at 16:30
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    in detail: if you’re happy with the derivation of (2.85) now you can hit both sides of the resulting equation with an additional id/dt and proceed precisely as before. (note also that derivatives commute) -what do you find? – Wakabaloola Feb 10 '19 at 16:38
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    (what i meant above is: what you’ve already shown tells you that id/dt acting on the correlator is the same as the square root derivative operator acting on the correlator) – Wakabaloola Feb 10 '19 at 16:42
  • I knew this was going to turn out to be something that should have been obvious! Thanks @Wakabaloola , that did the trick. I get it now :) – EthanT Feb 10 '19 at 17:02

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