QM limit of QFT in Schwartz

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  • Thread starter dm4b
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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 the Klein-Gordon Equation:

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

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?

I realize you can expand in terms of $$p^2/m$$ and make use of $$\nabla^2e^{ipx}=-p^2e^{-ipx}$$ to pull the terms out, but I'm really interested in how he uses the KG equation to achieve the same result.

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

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  • #2
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I figured out the answer to this, if an admin would like to delete the OP. Didn't see a way to do that myself
 
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  • #3
Demystifier
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QM limit of QFT, in general, is quite an interesting and nontrivial topic, so just because you solved a technical difficulty is not a reason to not continue the discussion.

For instance, there is no position operator in the relativistic theory, but there is in the non-relativistic limit. How about that? I have my own answer, but there is no general consensus on that.
 
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