A A question about the mode expansion of a free scalar field

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1. Aug 4, 2016

"Don't panic!"

In the canonical quantisation of a free scalar field $\phi$ one typical constructs a mode expansion of the corresponding field operator $\hat{\phi}$ as a solution to the Klein-Gordon equation, $$\hat{\phi}(t,\mathbf{x})=\int\frac{d^{3}k}{(2\pi)^{3}}\frac{1}{\sqrt{\omega(\mathbf{k})}}\left(\hat{a}(\mathbf{k})e^{-ik\cdot x}+\hat{a}^{\dagger}(\mathbf{k})e^{ik\cdot x}\right)$$ where $k\cdot x:=k_{\mu}x^{\mu}=k_{0}t-\mathbf{k}\cdot\mathbf{x}$.

What I'm unsure about is what exactly is the term "mode" referring to? Is it the whole component $\hat{a}(\mathbf{k})e^{-ik\cdot x}$, or is it simply $\hat{a}(\mathbf{k})$?

Furthermore, one has that the frequency $\omega(\mathbf{k})$ satisfies the equation $$\omega^{2}(\mathbf{k})=\mathbf{k}^{2}+m^{2}$$ but is $\omega(\mathbf{k})$ the frequency of each mode or of the field $\hat{\phi}$ itself? I think it's the frequency of each mode (with momentum $\mathbf{k}$, but then, given my first question, I am unsure whether this is the frequency corresponding to $\hat{a}(\mathbf{k})e^{-ik\cdot x}$, or is it simply $\hat{a}(\mathbf{k})$?!

2. Aug 4, 2016

Jilang

Isn't it just the different k?

3. Aug 4, 2016

"Don't panic!"

Yes, but I was wondering specifically whether the operators $\hat{a}(\mathbf{k})$ and $\hat{a}^{\dagger}(\mathbf{k})$ are referred to as the "frequency modes" of the scalar field $\hat{\phi}$, i.e. one set for each value of $\mathbf{k}$, such that each set (corresponding to each value of $\mathbf{k}$) for a particular value of creation/annihilation operators correspond to an independent harmonic oscillator, oscillating at a frequency $\omega(\mathbf{k})$?!