Why can the Klein-Gordon field be Fourier expanded in terms of ladder operators?

You might feel more comfortable if we run the reasoning 'in reverse' a bit. I'm basically just recounting what all the standard texts do, but without jumping ahead in the interpretation.

• We begin with the canonically quantized Klein-Gordon field $\phi(\mathbf{x})$ and its conjugate momentum $\pi(\mathbf{x})$, so that $[\phi(\mathbf{x}), \pi(\mathbf{y})] \propto i \delta(\mathbf{x}-\mathbf{y})$.
• We know that generally physics is nicer in Fourier space, especially for free fields where plane waves are classical solutions, so we define the Fourier transforms $\phi(\mathbf{p})$ and $\pi(\mathbf{p})$.
• By analogy with the harmonic oscillator, we suspect that the quantity $$a_{\mathbf{p}} \propto \phi(\mathbf{p}) + i \pi(\mathbf{p})/\omega_p$$ will be nice to work with. At this stage, $a_{\mathbf{p}}$ is just some operator with no physical interpretation.
• When we compute the commutation relations, we find $$[a_{\mathbf{p}}^\dagger, a_{\mathbf{q}}] \propto \delta(\mathbf{p}-\mathbf{q})$$ with an annoying energy-dependent proportionality constant, and all other commutators zero.
• From undergraduate quantum mechanics, these commutation relations imply that applying $a_{\mathbf{p}}^\dagger$ gives a ladder of states for each value of $\mathbf{p}$. Then we physically interpret $(a_{\mathbf{p}}^\dagger)^n |0 \rangle$ to contain $n$ particles of momentum $\mathbf{p}$, so that $a_{\mathbf{p}}^\dagger$ really is a creation operator, i.e. it creates particles.
• The only issue is that these states are not normalized; we get an annoying energy-dependent normalization factor. (To see this, try calculating the norm of $a_{\mathbf{p}}^\dagger |0 \rangle$ using the commutation relation above.) Tracing backwards, we find that it goes away if we replace $a_{\mathbf{p}}$ with $\sqrt{\omega_{\mathbf{p}}} a_{\mathbf{p}}$, recovering the standard definition.

When Peskin does this whole thing in one step, you should view it as just a very inspired definition (where we've outlined where the 'inspiration' might come from above), not a statement that needs to be proved. It's a more advanced version of guessing a sinusoid to solve the harmonic oscillator.

Let us start with the ansatz (I'll assume mostly plus metric signature)

\begin{equation} \hat\phi(x) = \int \frac{\mathrm{d}^3 \mathbf{k}}{(2\pi)^{3/2}}\left(\hat A_\mathbf{k} e^{i k\cdot x} + \hat B_\mathbf{k}e^{-ik\cdot x}\right) \end{equation} where $\hat A_\mathbf{k}$ and $\hat B_\mathbf{k}$ are some arbitrary operators and $k^0 = E_k$. We first note that $\phi$ is a real field which means that we must have $B_\mathbf{k} = A_\mathbf{k}^\dagger$. Also a quantum field must obey the equal time commutation relation

\begin{equation} \left[ \hat \phi(\mathbf{x},t),\hat{\dot\phi} (\mathbf{y},t)\right] = i\delta(\mathbf{x-y}). \end{equation} Plugging our ansatz into this relation we get the condition

\begin{equation} 2E_k \left[\hat A_\mathbf{k}, \hat A_\mathbf{k'}^\dagger \right] = \delta(\mathbf{k-k'}) \end{equation} This is just the ladder operator commutation relation with an extra normalization factor. For convenience we can define a rescaled operator $\hat a_\mathbf{k} \equiv \sqrt{2E_k}\hat A_\mathbf{k}$ which has the conventional commutation relation.

Detailed calculation

Taking the time derivative of the field we get

\begin{equation} \hat{\dot \phi}(x) = \int \frac{\mathrm{d}^3 \mathbf{k}}{(2\pi)^{3/2}}\left(-iE_k\hat A_\mathbf{k} e^{-i E_kt + i\mathbf{k\cdot x}} + iE_k\hat A_\mathbf{k}^\dagger e^{i E_kt - i\mathbf{k\cdot x}}\right). \end{equation}

Then the commutator between the field and its time derivative (more generally its conjugate) is

\begin{eqnarray} \left[ \hat \phi(\mathbf{x},t),\hat{\dot\phi} (\mathbf{y},t)\right] = \int \frac{\mathrm{d}^3 \mathbf{k}\mathrm{d}^3 \mathbf{k'}}{(2\pi)^{3}}\left\{iE_{k'}\left[\hat A_\mathbf{k},\hat A_\mathbf{k'}^\dagger\right] e^{-i(E_k-E_{k'})t+i(\mathbf{k\cdot x - k'\cdot y)}} + iE_k\left[\hat A_\mathbf{k'},\hat A_\mathbf{k}^\dagger\right] e^{i(E_k-E_{k'})t-i(\mathbf{k\cdot x - k'\cdot y)}}\right\} = \int \frac{\mathrm{d}^3 \mathbf{k}\mathrm{d}^3 \mathbf{k'}}{(2\pi)^{3}}\left\{iE_{k'}\left[\hat A_\mathbf{k},\hat A_\mathbf{k'}^\dagger\right] e^{-i(E_k-E_{k'})t} + iE_k\left[\hat A_\mathbf{-k'},\hat A_\mathbf{-k}^\dagger\right] e^{i(E_k-E_{k'})t}\right\}e^{i(\mathbf{k\cdot x - k'\cdot y)}} \end{eqnarray} where we have used the fact that operators commute with themselves and changed variables $\mathbf{k\rightarrow -k}$, $\mathbf{k'\rightarrow -k'}$ in the second term. This should be equal to the delta function

\begin{equation} i\delta(\mathbf{x-y}) = i \int \frac{\mathrm{d}^3 \mathbf{k}}{(2\pi)^3}e^{i\mathbf{k\cdot(x-y)}} = i \int \frac{\mathrm{d}^3 \mathbf{k}\mathrm{d}^3 \mathbf{k'}}{(2\pi)^3}\delta(\mathbf{k-k'})e^{i(\mathbf{k\cdot x - k'\cdot y)}} \end{equation} which means we must have

\begin{equation} \left[\hat A_\mathbf{k},\hat A_\mathbf{k'}^\dagger\right] + \left[\hat A_\mathbf{-k'},\hat A_\mathbf{-k}^\dagger\right] = \frac{\delta(\mathbf{k-k'})}{E_k} \end{equation}

to which $\left[\hat A_\mathbf{k},\hat A_\mathbf{k'}^\dagger\right]=\delta(\mathbf{k-k'})/2E_k$ is the solution.