# 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.)

• possible duplicate: A question on using Fourier decomposition to solve the Klein Gordon equation. – AccidentalFourierTransform Dec 10 '18 at 14:46
• @AccidentalFourierTransform I don't think that question is a duplicate as it asks why you only FT over 3-space and are doing the part that is exactly analogous to the classical bit I can do. I want to know why in the quantum case there need to be negative frequencies and operator valued Fourier co-efficients (at least, I think that's what I want to know.) – jacob1729 Dec 10 '18 at 14:54
• I believe it has to do with the field being real, in contrast to the complex scalar case, where you can see you have two types of operators. This just depends on the model you are dealing with. – ohneVal Dec 10 '18 at 17:01
• possible duplicate: physics.stackexchange.com/questions/372425/… – knzhou Dec 10 '18 at 17:07
• The reason you need negative frequencies is simply because the Klein-Gordan equation constrains $(\omega^2 - k^2 - m^2) \phi(\omega, \mathbf{k}) = 0$, which has both positive and negative $\omega$ as solutions. Contrast this to the Schrodinger equation, $(\omega - k^2/2m) \psi(\omega, \mathbf{k}) = 0$. – knzhou Dec 10 '18 at 17:09