Peskin and Schroeder confusion on promoting Classical Klein-Gordon equation to quantum field equation I am reading "An introduction to quantum field theory" by Peskin and Schroeder and I am confused. I appreciate your help.
Here's the context to my question: In chapter 2, the book introduces quantum field theory by first talking about classical field theory and Klein-Gordon equation. The book expands a typical classical field in momentum space as,
\begin{equation}
\phi(x, t)=\int\frac{d^3p}{(2\pi)^3}e^{ipx}\phi(p, t).
\end{equation}
The classical Klein-Gordon equation in momentum space becomes (equation (2.21)),
\begin{equation}
\left[\frac{\partial^2}{\partial t^2} + (|p|^2 + m^2)\right]\phi(p, t) = 0, \tag{2.21}
\end{equation}
which is the equation of motion of a (classical) simple harmonic oscillator with frequency $\omega_p=\sqrt{|p|^2 + m^2 }$. The book then claims that the solution to quantum harmonic oscillator is well known (equation (2.23)),
\begin{equation}
\phi = \frac{1}{\sqrt{2\omega}}(a + a^{\dagger}).\tag{2.23}
\end{equation}
Where $a$ is the ladder operator. If I understand the book correctly, it finally claims that if we use the solutions of quantum harmonic oscillator as the solution to Klein-Gordon equation in momentum space, i.e. $\phi(p,t) = \frac{1}{\sqrt{2\omega_p}}(a_p + a_p^\dagger)$, we obtain the solution to the quantum Klein-Gordon equation. Such solution satisfies the commutation relation automatically so I can see where the author is going with this idea. The solution to quantum Klein-Gordon equation is then (equation (2.25)),
\begin{equation}
\phi(x) = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_pe^{ipx} + a^\dagger_pe^{-ipx})\tag{2.25}
\end{equation}
Here's my question: Claiming equation (2.23) as the solution to the equation (2.21) is a bit non-rigorous, don't you think? Put it this way, how can we prove that,
\begin{equation}
\left[\frac{\partial^2}{\partial t^2} + (|p|^2 + m^2)\right]\frac{1}{\sqrt{2\omega_p}}(a_p + a_p^{\dagger}) = 0 ???
\end{equation}
The disconnect I feel here stems from the fact that although expanding $\phi$ as ladder operators is useful in finding the eigenstate (and eigenvalue) to Hamiltonian of harmonic oscillator, equation (2.21) is just a differential equation that, at first glance, isn't about finding eigenstate. Operation-wise, finding eigenstate is so much different from solving differential equation that I think a mathematical proof is needed to before we can claim that the solution to quantum harmonic oscillator is also a solution to the equation of motion of the classical harmonic oscillator. I would like to ask kindly if you can prove a proof to my final equation to fill in the gap in knowledge I current have while reading the book.
 A: Please first see this reference about solving the quantum simple harmonic oscillator with raising and lowering operators.
Next, realize that we are dealing with a free field theory, and the Hamiltonian (which governs the dynamics) is the free field Hamiltonian:
$$
H = \sum_\vec p \omega_p a^\dagger_{\vec p }a_{\vec p}\;,
$$
which is clearly a sum over individual simple harmonic oscillator Hamiltonians (one for each $\vec p$ value). (To put is another way the free field is just a bunch of uncoupled simple harmonic oscillators.)
In quantum mechanics, the time-dependence of an operator in the Hamiltonian picture (or, once we add interactions, we will call this the interaction picture) is given by:
$$
a_{\vec p}(t) = e^{-iHt}a_{\vec p}e^{iHt}
$$
Taking the time derivative we find, as usual:
$$
\dot a_{\vec p} = -i[H, a_{\vec p}(t)] = i\omega_{\vec p}a_{\vec p}(t)\;.
$$
Take a second time derivative to find:
$$
\ddot a_{\vec p} = -\omega_p^2 a_{\vec p}\;.
$$
Or, using $\omega_{\vec p}^2 = |\vec p|^2 + m^2$ and rearranging:
$$
\ddot a_{\vec p} + (|\vec p|^2 + m^2)a_{\vec p} = 0\;,
$$
just like we want.
I am quite sure you can work out the analogous relation for $a^\dagger_{\vec p}$ on your own.
