Mode Expansion in Klein-Gordon QFT I have a confusion regarding the mode expansion of the Klein-Gordon field theory. I am following Peskin and Schroeder. My questions are about how we formally get to the expansion of the KG QFT in terms of harmonic oscillator modes (I am happy with how we do this in the classical case which is a straightforwards Fourier transform).
Peskin and Schroeder claim (eq 2.25) that the equation:
$$\phi(\vec{x}) = \int \frac{d^3\vec{k}}{(2\pi)^3} \frac{1}{\sqrt{2\omega_\vec{k}}}(a_\vec{k}e^{i\vec{k}\cdot\vec{x}}+a^\dagger_{\vec{k}}e^{-i\vec{k}\cdot\vec{x}})$$
Follows purely by analogy with the solution to the harmonic oscillator where $x = \frac{1}{\sqrt{2\omega}}(a+a^\dagger)$. Is there a way to see this not by analogy, but by actually FT-ing the quantum field?
They then proceed (eq 2.27) to claim the following. I think this follows simply by changing variables $\vec{k}\mapsto -\vec{k}$ in the second integral (but am not 100% convinced):
$$\phi(\vec{x}) = \int \frac{d^3\vec{k}}{(2\pi)^3} \frac{1}{\sqrt{2\omega_\vec{k}}}(a_\vec{k}+a^\dagger_{-\vec{k}})e^{i\vec{k}\cdot\vec{x}}$$
which looks rather more like how I would try to do a FT of the field due to only having positive frequencies. However I don't see how you can tell you need that specific form of the Fourier coefficients.
An answer relating what you'd do in the classical case (FT field, solve harmonic oscillator for each mode and sub back into mode expansion) to how you do this in the classical case would be welcome (e.g. the quantum case has operators in places the classical case has variables and daggers where the classical case has complex conjugates. This isn't surprising, but I don't see how to rigorously obtain these results.)
 A: From the classical field theory, we can write the field (in flat spacetime) as a fourier transform of mode functions:
$$
\hat \phi (\vec x, t)=\int\frac{\mathrm{d}^n\vec k}{(2\pi)^{3/2}}e^{- \mathrm{i} \vec k \cdot \vec x}\hat \phi_{\vec k}(t)\qquad\qquad\qquad\qquad\qquad(1)
$$
(Note that the Fourier Transform is an integral over all of real space, even in the classical theory)
When you do canonical quantization, you promote the field to operator status, then impose the equal time commutation relation. The mode functions are also then are promoted to operators and inherit their commutation relations.
Now we define the following time-dependent operators in terms of the mode functions:
$$
\hat a^-_{\vec k}(t)=\sqrt{\frac{\omega_{k}}{2}}\left(\hat \phi_{\vec k}(t)+\frac{\mathrm{i}\hat \pi_{\vec k}(t)}{\omega_{k}}\right),
\qquad
\hat a^+_{\vec k}(t)=\sqrt{\frac{\omega_{k}}{2}}\left(\hat \phi_{\vec {-k}}(t)-\frac{\mathrm{i}\hat \pi_{\vec{-k}}(t)}{\omega_{k}}\right)
\qquad\qquad (2)
$$
From the field equations of motion, we can find the equations of motion for these operators. If we solve those equations, we find:
$$
\hat a^{\pm}_{\vec k}(t)=\hat a^{\pm}_{\vec k}e^{{\pm}\mathrm{i}\omega_{k}t}
$$
If we then invert the relation (2) we can write
$$
\hat \phi_{\vec k}=\frac{1}{\sqrt{2\omega_{k}}}\left(\hat a^-_{\vec k}(t)+\hat a^+_{-\vec k}(t)\right)\quad\qquad\qquad(3a)\\
\qquad=\frac{1}{\sqrt{2\omega_{k}}}\left(\hat a^-_{\vec k}e^{-\mathrm{i}\omega_{k}t}+\hat a^+_{-\vec k}e^{\mathrm{i}\omega_{k}t}\right)\qquad\quad(3b)
$$
Plugging (3b) into (1) gives the Equation you were asking about from Peskin and Schoeder.
In summary, the role of the Fourier coefficients as the creation/annihilation follows directly from the classical field theory+ canonical Quantization.
Reference:  V. F. Mukhanov and S. Winitzki, Introduction to Quantum Fields in Classical Backgrounds (See Chapter 3, 4 and 6, in particular Section 4.1)
A: Rule number 1: we quantize fields after we transition to the canonical picture. So, we start with the Lagrangian for the Klein-Gordon field ($c=1$, $\hbar=1$):
$$L = \int \mathrm{d}^3 x  \left[\frac{1}{2}\dot\phi^2(\mathbf{x}) - \frac{1}{2}[\nabla\phi(\mathbf{x})]^2 - \frac{m^2}{2}\phi^2(\mathbf{x})\right]$$
We immediately transform into the classical canonical picture, getting our canonically conjugate momentum to $\phi(\mathbf{x})$ from the definition
\begin{align}
   \pi(\mathbf{x}) & \equiv \frac{\delta L}{\delta \dot\phi(\mathbf{x})} \\
   & = \dot\phi(\mathbf{x}).
\end{align}
From there we can transition to our classical Hamiltonian
$$
H = \int \mathrm{d}^3 x\left[\frac{1}{2}\pi^2(\mathbf{x}) + \frac{1}{2}[\nabla\phi(\mathbf{x})]^2 + \frac{m^2}{2}\phi^2(\mathbf{x}) \right]
$$
Now that we have our non-commuting variables, we can quantize by imposing the equal time commutation relations,
\begin{align}
   [\phi(\mathbf{x}),\phi(\mathbf{y})] & = [\pi(\mathbf{x}),\pi(\mathbf{y})] = 0 \\
   [\phi(\mathbf{x}),\pi(\mathbf{y})] & = i \delta(\mathbf{x}-\mathbf{y}).
\end{align}
Now, we want to change from real space to mode space. There are multiple ways to do this that have varying degrees of complexity associated with them.


*

*Fourier transform: pros - momentum eigenfunctions; cons - our operators aren't Hermitian anymore.

*Harteley transform: pros - operators are Hermitian, no constraints on real/imaginary parts; cons - derivatives aren't quite as simple anymore.

*Sine/Cosine transform: pros - operators are Hermitian; cons - you're basically inspecting the guts of the Fourier transform, making a lot of extra work. Sometimes it's needed, though.


Because it's the lingua franca, we'll do the Fourier transform here. Perform the substitution $\phi(\mathbf{x}) = \int \mathrm{d}^3 k \phi(\mathbf{k}) \frac{e^{i\mathbf{k}\cdot\mathbf{x}}}{(2\pi)^{3/2}}$, and similar for $\pi$, you'll find that
\begin{align}
  H & = \int \mathrm{d}^3 k\left[\frac{1}{2}\pi(-\mathbf{k}) \pi(\mathbf{k}) + \frac{\mathbf{k}^2 + m^2}{2}\phi(-\mathbf{k}) \phi(\mathbf{k}) \right], \\
  [\phi(\mathbf{k}),\phi(\mathbf{k}')] & = [\pi(\mathbf{k}),\pi(\mathbf{k}')] = 0, \text{ and} \\
   [\phi(\mathbf{k}),\pi(\mathbf{k}')] & = i \delta(\mathbf{k}+\mathbf{k}').
\end{align}
To get the above into a more familiar form, you'll have to use the fact that $\phi^\dagger(\mathbf{x}) = \phi(\mathbf{x})\Rightarrow \phi^\dagger(\mathbf{k}) = \phi(-\mathbf{k}).$ Leading to
\begin{align}
  H & = \int \mathrm{d}^3 k\left[\frac{1}{2}\pi^\dagger(\mathbf{k}) \pi(\mathbf{k}) + \frac{\mathbf{k}^2 + m^2}{2}\phi^\dagger(\mathbf{k}) \phi(\mathbf{k}) \right], \\
  [\phi(\mathbf{k}),\phi(\mathbf{k}')] & = [\pi(\mathbf{k}),\pi(\mathbf{k}')] = 0, \text{ and} \\
   [\phi(\mathbf{k}),\pi^\dagger(\mathbf{k}')] & = i \delta(\mathbf{k}-\mathbf{k}').
\end{align}
This implies that $\phi$ is canonically conjugate to $\pi^\dagger$ not $\pi$. This causes a conundrum: how do we build our ladder operators? For the simple harmonic oscillator we had $a = \sqrt{\frac{m\omega}{2}} \left(x + \frac{i}{m\omega} p\right)$. Which do we use
\begin{align}
   a(\mathbf{k}) & = \sqrt{\frac{\omega}{2}} \left(\phi(\mathbf{k}) + \frac{i}{\omega} \pi^\dagger(\mathbf{k})\right) \text{ or} \tag1\\
   a(\mathbf{k}) & = \sqrt{\frac{\omega}{2}} \left(\phi(\mathbf{k}) + \frac{i}{\omega} \pi(\mathbf{k})\right)? \tag2
\end{align}
The answer is: it depends on what you care about, because those are both valid in one of the mode pictures available to us.
To see why, it's useful to dig in to the guts. The Fourier transform can be written in terms of the cosine and sine transforms to give
$$
\phi(\mathbf{k}) = \phi_+(\mathbf{k}) + i \phi_-(\mathbf{k}),
$$
Note that the cosine transform produces an even function, and the sine transform an odd one, and that's why I chose the subscript plus and minuses. Interestingly, the constraints on these functions affect their commutation relations. The nontrivial ones are:
$$
[\phi_{\pm}(\mathbf{k}),\pi_{\pm}(\mathbf{k}')] = \frac{i}{2}\delta(\mathbf{k}-\mathbf{k}').
$$
The Harteley transform is then $\phi_H(\mathbf{k}) = \phi_+(\mathbf{k}) +  \phi_-(\mathbf{k}).$
Now, without a doubt we can define ladder operators for the sine and cosine transforms, since they're Hermitian
\begin{align}
  a_\pm(\mathbf{k}) & = \sqrt{\frac{\omega}{2}} \left(\phi_\pm(\mathbf{k}) + \frac{i}{\omega} \pi_\pm(\mathbf{k})\right) \text{ with} \\
 [a_\pm(\mathbf{k}),a_\pm(\mathbf{k}')] & = \frac{1}{2}\delta(\mathbf{k}-\mathbf{k}').
\end{align}
Note that each $a_\pm$ is even or odd just like the operators it's built from.
Similarly, the ladder operator for the Harteley mode works just fine $a_H(\mathbf{k}) = \sqrt{\frac{\omega}{2}} \left(\phi_H(\mathbf{k}) + \frac{i}{\omega} \pi_H(\mathbf{k})\right) \Rightarrow [a_H(\mathbf{k}),a_H(\mathbf{k}')] =\delta(\mathbf{k}-\mathbf{k}').$
The reason for defining the ladder operators on the standing waves is because it allows us to inspect what (1) and (2) are doing (you can do a change of variables in the integral to show that (1) is what Peskin & Schroder are using). In terms of the standing mode ladder operators 
$$a_P(\mathbf{k}) = a_+(\mathbf{k}) + i a_-^\dagger(\mathbf{k}).$$
In other words, this operator cannot be rightly called either a creation or annihilation operator - it does both. If you really wanted to annihilate a traveling wave, you'd build your operator as
$$a_T(\mathbf{k}) = a_+(\mathbf{k}) \pm i a_-(\mathbf{k}),$$
where, here the $\pm$ here is intended to convey uncertainty over the correct choice (the plus choice is identical to (2), by the way).
Why choose (1), then? I haven't worked it out, myself, but two possibilities occur to me:


*

*the commutator of $[a_P,a_P^\dagger]$ looks more like the SHO $[a,a^\dagger]=1$, or

*the number density can be more compactly written in terms of $a_P$ than $a_T$.

A: Eqs. (4.49)-(4.56) here provide an alternative approach, mostly discussing it in terms of the classical version. Writing a solution of $(\square +m^2)\phi=0$ as $\phi(x)=\int d^4ke^{-ikx}\tilde{\phi}(k)$ gives$$\int d^4ke^{-ikx}(m^2-k^2)\tilde{\phi}(k)=0$$(the linked source has a typo, showing $d^4x$ instead). Since $(m^2-k^2)\tilde{\phi}(k)=0$, we can write$$\tilde{\phi}(k)=\delta(k^2-m^2)\tilde{\varphi}(k)$$for some $\tilde{\varphi}$. Define $\omega_\mathbf{k}:=\sqrt{k^2+m^2}$ so$$\delta(k^2-m^2)=\frac{1}{2\omega_\mathbf{k}}\sum_\pm\delta(k_0\pm\omega_\mathbf{k})\\\implies\phi(x)=\int d^3\tilde{\mathbf{k}}(a_+(k)e^{-ikx}+a_-^\ast(k)e^{ikx}),\,d^3\tilde{\mathbf{k}}:=\frac{d^3\mathbf{k}}{(2\pi)^32\omega_\mathbf{k}}.$$(The version you asked for square-roots the $2\omega_\mathbf{k}$ factor, but that's achievable by slightly modifying the CCRs viz. Eq. (5.38).) For real $\phi$, $a:=a_+=a_-$ so $\phi(x)=\int d^3\tilde{\mathbf{k}}(a(k)e^{-ikx}+a^\ast(k)e^{ikx})$. Eqs. (5.33)-(5.37) (ibid.) briefly discuss the quantized version so, placing hats on $\phi,\,a_\pm$ (which the linked source doesn't do),$$\hat{\phi}^\dagger=\hat{\phi}\implies\hat{a}:=\hat{a}_+=\hat{a}_-.$$
