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In the introductory text to quantum field theory by Peskin & Schroeder, they state that in analogy to the simple harmonic oscillator in quantum mechanics, the free scalar field can be expressed as:

$$\phi(\vec{x}) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2 \omega_p}} \left( a_p e^{i \vec{p}\cdot\vec{x}} + a^{\dagger}_p e^{-i \vec{p}\cdot\vec{x}} \right) \tag{2.25}$$

In quantum mechanics $\phi$ would be written as:

$$\phi = \frac{1}{\sqrt{2 \omega_p}} \left( a + a^{\dagger} \right)$$

I can see the similarities between the two expressions, as well as the fact that one may expand the free Klein-Gordon field as:

$$\phi(\vec{x},t) = \int \frac{d^3 p}{(2\pi)^3} e^{i \vec{p}\cdot\vec{x}} \phi(\vec{p},t).\tag{2.20b} $$

However I don't get how to reach the final expression given above, especially the exponential with negative sign in the second term. It's probably just a small thing that I am not seeing, but I would be thankful if somebody could give me a hint.

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I'll adopt the abbreviation $kx:=k_0x^0-\mathbf{k}\cdot\mathbf{x}$. The Klein-Gordon equation $(\square +m^2)\phi=0$ can be solved by a Fourier transform. Writing $\phi(x)=\int d^4ke^{-ikx}\tilde{\phi}(k)$ we get $(k^2-m^2)\tilde{\phi}(k)=0$, i.e. $\tilde{\phi}(k)=\tilde{\varphi}(k)\delta(k^2-m^2)$ for some function $\tilde{\varphi}(k)$. Using$$\delta(k^2-m^2)=\dfrac{\delta(k_0-\omega_\mathbf{k})+\delta(k_0+\omega_\mathbf{k})}{2\omega_\mathbf{k}}$$gives$$\phi(x)=\int\dfrac{d^3k}{(2\pi)^32\omega_\mathbf{k}}(a_+(k)e^{-ikx}+a_-^\ast(k)e^{ikx})$$with$$a_+(k):=(2pi)^3\tilde{\varphi}(\omega_\mathbf{k},\,\mathbf{k}),\,a_-(k):=(2pi)^3\tilde{\varphi}^\ast(-\omega_\mathbf{k},\,\mathbf{k}).$$For real $\phi$, $a_-=a_+^\dagger$, so defining $a:=a_+$ we're done. (You have an erroneous $\sqrt{}$ sign in the Lorentz-invariant integration operator.) Note in particular your $e^{i\mathbf{k}\cdot\mathbf{x}}$ coefficient of $a_k$ is $e^{i(k_0x^0-kx)}$, but I've absorbed the $e^{ik_0x^0}$ factor into my definition of $a_k$ to make the above result's Lorentz-invariance manifest.

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    $\begingroup$ Thank you very much for your answer, this is very clear! Are you sure regarding the erroneous $\sqrt{}$? In Peskin it seems to be there (eq. 2.25, page 21), with $\omega_p = \sqrt{p^2 + m^2}$. Or maybe I misunderstand what you mean? $\endgroup$ – Jxx Aug 24 '18 at 22:08
  • $\begingroup$ I have one more question regarding the result you give for $\tilde{phi}(k)$ in your first paragraph: is the integral really there or is it just equal some function times the delta function? $\endgroup$ – Jxx Aug 24 '18 at 22:40
  • $\begingroup$ @Julien (i) I'm referring to your first equation having $\dfrac{1}{\sqrt{2\omega_\mathbf{k}}}$. The frequency is of course already a square root. My answer's power counting is based on pp.47/8 here. (ii) Ah, that was a mistake I've now fixed. There's a funny story there: while I originally wrote the answer I accidentally missed an integral sign or two, but I obviously overdid putting it into places. $\endgroup$ – J.G. Aug 24 '18 at 22:42
  • $\begingroup$ Thanks for your answer! That's strange, in Peskin and in David Tong's notes there is a square root indeed, although I can see that there isn't in your link and I hear your argument. I don't have a link to Peskin, but the lectures of D. Tong (which you can find here: damtp.cam.ac.uk/user/tong/qft/qft.pdf ) are based on Peskin I believe (see eq. 2.18 page 23). $\endgroup$ – Jxx Aug 24 '18 at 22:49
  • $\begingroup$ Well, we definitely don't need the extra square root. As has been explained here, it's just the $1/k_0$ factor from the Dirac delta in my answer above. $\endgroup$ – J.G. Aug 24 '18 at 22:55

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