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I am trying to work through Peskin and Schröder and am a little stuck in Chapter 4 [section 4.2 p. 83 below eq. (4.13)], when he first treats interacting fields. The subject is the quartic interaction in Klein-Gordon Theory. They claim that:

"At any fixed time, we can of course expand the field (in the interacting theory) as before (in the free theory) in terms of ladder operators."

I don't see why this should be possible in general. Their argument for the ladder operators and the expansion in plane waves in the case of the free theory was that from the Klein-Gordon equation we get Fourier modes independently satisfying harmonic oscillator equations. However as far as I can see, once I add an interacting term I don't get these equations anymore.

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You are always free to define $$a_{\boldsymbol k}\equiv \int\mathrm d^3 x\ \mathrm e^{ikx}(\omega_{\boldsymbol k}\phi(x)+i\pi(x)) \tag{1}$$ where $\pi=\dot\phi(x)$. If you take the time derivative of this definition, you get $$\dot a_{\boldsymbol k}= \int\mathrm d^3 x\ \mathrm e^{ikx}(\partial^2+m^2)\phi(x) \tag{2}$$ which is non-zero for an interacting field. Therefore, $a=a(t)$, where $t$ is the time slice you chose in $(1)$; in other words, our definition of $a$ is not in general independent of the time slice, so $a$ depends parametrically on the value of $t$ at that slice.

Now, inverting the Fourier transform in $(1)$, you get $$\phi(x)=\int\frac{\mathrm d^3 k}{(2\pi)^32\omega_{\boldsymbol k}}\ \mathrm e^{-ikx}a_{\boldsymbol k}(t)+\text{h.c.}$$

which is essentially P&S's statement. Note that, in general, this statement is mostly devoid of any practical meaning; its just a trivial consequence of the inversion theorem of the Fourier transform.

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    $\begingroup$ Thanks a lot. So as I understand it now, the ladder operators have changed. If I now want to impose the commutation relations on the ladder operators can I do it at all times simulatoneously. Or do I have to pick one time say t1, impose the commutation relations at t1 and than find the commutation relations at all other times by acting with the time evolution operator? $\endgroup$ Commented Feb 7, 2017 at 17:01

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