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

Using the plane wave ansatz $$\phi(x) = e^{ik_\mu x^\mu}$$ the solution to the Klein-Gordon equation $(\Box + m^2) \phi(x) =0$ can be written as a sum of solutions, since the equation is linear and the superposition principle holds, as $$\phi(x) = \sum_{{k}} \left( Ae^{ik_\mu x^\mu} + Be^{-ik_\mu x^\mu} \right).$$ How does one find the coefficients? More exactly, why does it turn out they are the annihilation and creation operators with the factor $1/\sqrt{2E}$?

The various books and sources I've checked just confused me even more. Peskin and Schroeder just plug in the integral equation (Fourier modes) by analogy with the harmonic oscillator solution. Schwartz gives a very strange reason that the energy factor is just for convenience. In Srednicki the author writes it as $f(k)$ without an explicit form. In Mandl and Shaw, they just state the equation without any justification.

My best guess is that those come from the quantization process, but how does one do it in this case explicitly?

Answer

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.