# Fourier mode expansion of quantum fields

My question pertains to understanding the notion of Fourier modes in the context of quantum fields. For simplicity I have stuck to free scalar fields.

As I understand it, to solve the (homogeneous) Klein-Gordon (KG) equation $$(\Box +m^{2})\hat{\phi}(t,\mathbf{x})=0$$ one expands the field operator $\hat{\phi}(t,\mathbf{x})$ in terms of its corresponding Fourier modes $$\hat{\phi}(t,\mathbf{x})=\int\frac{d^{3}k}{(2\pi)^{3}}\hat{\tilde{\phi}}(t,\mathbf{k})e^{i\mathbf{k}\cdot\mathbf{x}}$$ and as such, in order for this to be a solution to the KG equation, the Fourier modes $\hat{\tilde{\phi}}(t,\mathbf{k})$ must satisfy $$\partial_{t}^{2}\hat{\tilde{\phi}}(t,\mathbf{k})+\omega^{2}(\mathbf{k})\hat{\tilde{\phi}}(t,\mathbf{k})=0$$ where $\omega(\mathbf{k})=\sqrt{\mathbf{k}^{2}+m^{2}}$ is the oscillation (angular) frequency of the modes.

What I'm unsure about is, when one expresses the solution in terms of creation and annihilation operators, such that $$\hat{\phi}(t,\mathbf{x})=\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{\sqrt{2\omega(\mathbf{k})}}\left(\hat{a}(\mathbf{k})e^{-i\omega t+i\mathbf{k}\cdot\mathbf{x}}+\hat{a}^{\dagger}(\mathbf{k})e^{i\omega t-i\mathbf{k}\cdot\mathbf{x}}\right)$$ what are the Fourier (or frequency) modes?

Are they the terms $\hat{a}(t,\mathbf{k})=\hat{a}(\mathbf{k})e^{-i\omega t}$ and $\hat{a}^{\dagger}(t,\mathbf{k})=\hat{a}^{\dagger}(\mathbf{k})e^{i\omega t}$, oscillating at (angular) frequency $\omega(\mathbf{k})$?! If this is the case, should each a combination (for each particular value of $\mathbf{k}$) be interpreted as describing a harmonic oscillator, oscillating at (angular) frequency $\omega(\mathbf{k})$?!

• just check your last full equation, I think it should be $2\pi$ to the power 3/2, not 3, sorry for being pedantic – user108787 Aug 5 '16 at 11:04
• @count_to_10 I've just checked in Peskin & Schroeder, and according to them it should be $2\pi$ to the power 3 (and not 3/2). I think it's just a choice though, one can either choose it to be symmetric such that both the Fouier transform and its inverse have (inverse) fractional powers (3/2) $2\pi$ associated with them, or one can just choose it such that one of them has $1/(2\pi)^{3}$ associated with it and the other has no factors of $2\pi$ associated with it. – Will Aug 5 '16 at 11:17
• this might help physics.stackexchange.com/questions/127141/… especially the last answer, which relates to your question, and the people involved get into a question of intrepretation..... – user108787 Aug 5 '16 at 11:42
• @count_to_10 Thanks for the link. From what I understand from reading the last answer, it is the linear combination of $\hat{a}(t,\mathbf{k})=\hat{a}(\mathbf{k})e^{-i\omega t}$ and $\hat{a}^{\dagger}(t,\mathbf{k})=\hat{a}^{\dagger}(\mathbf{k})e^{i\omega t}$ that (at least heuristically) are the modes of the field $\hat{\phi}$ (one for each value of $\mathbf{k}$), each oscillating at frequency $\omega(\mathbf{k})$?! – Will Aug 5 '16 at 12:00
• I self study, so in my naive way I thought that only numbers were important in physics, but from the comments on this site, I realise words and their meaning are just as important : ) Best of luck getting a definitive answer. – user108787 Aug 5 '16 at 12:09