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I am studying QFT with the Peskin and Schroeder textbook, and I am new to this area of physics. I'm struggling with the solution of the Klein-Gordon equation using Fourier integral as a continuum set of oscillators:

\begin{equation} \phi(\mathbf{x}) = \int \frac{d^3 p}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega_\mathbf{p}}} \left(a_\mathbf{p} e^{i \mathbf{p} \cdot \mathbf{x}} + a^\dagger_\mathbf{p} e^{-i \mathbf{p} \cdot \mathbf{x}}\right). \end{equation}

The question is: why doesn't the field $\phi$ depend on time? My answer is that this solution is applicable only in the Schrödinger picture, where the field is a function of spatial coordinates, not time.

The postscriptum question: Can we obtain this solution without knowledge of the harmonic oscillator? Specifically, I mean that perhaps it can be obtained without the procedure of canonical quantization, can't it?

I would appreciate any assistance: from simple reasoning to a purely formal derivation of formulas.

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It does. The bold x notation represents the spacetime 4-vector components, so depends on time and space. The solution is not at all due to second quantization: you’re putting the cart before the horse. That solution is just the solution of the classical Klein Gordon equation, in which we then do second quantization to make it an operator.

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  • $\begingroup$ And if it helps, I’ve made a video here (m.youtube.com/…) explaining how to get the solution $\endgroup$
    – user310742
    Commented May 27, 2023 at 21:23
  • $\begingroup$ I've seen in «Notations and conventions» that bold letters are for spatial coordinates and ordinary letters are for $D = 3+1$ spacetime. $\endgroup$
    – Andrey
    Commented May 27, 2023 at 21:39
  • $\begingroup$ Well, then in that case what you have is the solution to the KG equation without considering time evolution. Similar to how you can have a wave function without considering time evolution. But still, the full solution does depend on t, though. You can check this by plugging e^{-ikt}e^{ikx} and you’ll see it solves the equation. $\endgroup$
    – user310742
    Commented May 27, 2023 at 22:30

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