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

  1. Aug 4, 2016 #1
    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. jcsd
  3. Aug 4, 2016 #2
    Isn't it just the different k?
     
  4. Aug 4, 2016 #3
    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})##?!
     
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