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Given a Hermitian operator $\hat{\phi}(x,t)$, why we can write it in terms of Fourier transformations as $$\hat{\phi}(x,t)=\int^\infty_{-\infty}\frac{dk}{(2\pi)(2\omega)}[\hat{a}(k)e^{ikx-i\omega t}+\hat{a}^*(k)e^{-ikx+i\omega t}]$$

The part I do not understand is how can we have two previously undefined operators $\hat{a}$ and $\hat{a}^*$ in the fourier transform expression. I should note that because \dag is undefined here, I am using * above $\hat{a}$. But in the notes I am told that in this case $\hat{a}^*$ is equal to $\hat{a}$^\dag.

My understanding of fourier transform is, given this operator $\hat{\phi}(x,t)$ we could write it as $$\int^\infty_{-\infty}\int^\infty_{-\infty}dkd\omega \exp[-ikx+i\omega t]\hat{\bar{\phi}}(k,\omega)$$ where $\hat{\bar{\phi}}$ is defined as$$\hat{\bar{\phi}}(k,\omega)=\frac{1}{(2\pi)^2}\int^\infty_{-\infty}\int^\infty_{-\infty}dxdt \exp[ikx-i\omega t]\phi(x,t)$$ But how is this linked to the expression given at the start?

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The first expression isn't a Fourier transform of a general field $\phi(x, t)$. I prefer to think of it as a mode expansion of a solution of the Klein-Gordon equation. (I explain why I say this a little later on in the answer.)

First, note that we are talking about the subset of possible field configurations $\phi(x, t)$ that satisfy the equations of motion \begin{equation} (\square + m^2) \phi(x, t) = 0 \end{equation} Then, note that by separation of variables, we can find the following basis of solutions, labeled by $k$ and by the sign of $\omega$: \begin{equation} \phi_{k, \pm}(x, t) = N_k e^{i (k x \pm \omega(k) t)} \end{equation} where $N_k$ is a normalization constant, and where $\omega(k)$ is a function of $k$ that is fixed by the equations of motion to satisfy \begin{equation} \omega(k) = \sqrt{k^2 + m^2} \end{equation} This is the main reason I don't like to think of your first expression as a Fourier transform. If we were doing a Fourier transform, then $\omega, k$ would be independent coordinates in the frequency domain, and $\phi(x, t)$ would be a general (off-shell) function of space and time and would not satisfy the Klein-Gordon equation. However, here we see that $\omega$ is a function of $k$, which follows from $(\square+m^2)\phi=0$.

Anyway, given the mode functions $\phi_{k, \pm}$, we can write a general solution to the Klein-Gordon equation by expanding $\phi(x, t)$ in terms of these mode functions (note: this expansion of the general solution in terms of a particular basis of solutions is where we introduce the $a_k$ objects; in classical field theory, these would be complex numbers, in quantum field theory, they are operators) \begin{eqnarray} \phi(x, t) &=& \sum_{s = \{+1, -1\} } \int \frac{d k}{2\pi} N_k a_{k, s} e^{i(k x - s \omega(k) t)} \\ &=& \int \frac{d k}{2\pi} N_k \left(a_{k, +} e^{i (k x - \omega(k) t)} + a_{k, -} e^{i (k x + \omega(k) t)}\right) \end{eqnarray} Now we do a few tricks.

  • For convenience, we fix $N_k$ so that the integral has a Lorentz invariant measure. This amounts to choosing \begin{equation} N_k = \frac{1}{2\omega(k)} \end{equation} This is derived in any good QFT book or lecture notes, such as Section 2 of David Tong's QFT notes.
  • Since $\phi(x, t) = \phi^\dagger(x, t)$, we can deduce that \begin{equation} a_{k, -} = a^\dagger_{-k, +} \end{equation} Let's define $a_k \equiv a_{k, +}$. Then, \begin{equation} \phi(x, t) = \int \frac{d k}{(2\pi)(2 \omega(k))} \left(a_{k} e^{i( k x - \omega(k) t)} + a^\dagger_{-k} e^{i (k x + \omega(k) t)}\right) \end{equation} Note that for a complex scalar field, you wouldn't be able to relate $a_{k, +}$ and $a_{k, -}$. Often the notation people use in that case is to introduce $b$ and $c$ operators, with $b_k \equiv a_{k, +}$ and $c_k \equiv a^\dagger_{k, -}$. These represent creation and annihilation operators for particles and anti-particles. However, we will stick to using $a$ for a real scalar field here.
  • Finally, changing the variable of integration from $k$ to $-k$ in the second term, we get to the mode expansion \begin{equation} \phi(x, t) = \int \frac{d k}{(2\pi)(2 \omega(k))} \left(a_{k} e^{i (k x - \omega(k) t)} + a^\dagger_{k} e^{-i (k x - \omega(k) t)}\right) \end{equation}

Having said all that, you could get the mode expansion from the inverse Fourier transform \begin{equation} \phi(x, t) = \int \frac{d \omega}{2\pi}\frac{d k}{2\pi} \tilde{\phi}(\omega, k) e^{-i (\omega t + k x)} \end{equation} with the momentum space representation \begin{equation} \tilde{\phi}(\omega, k) = 2\pi a_k \delta(\omega^2 - k^2 - m^2) = \frac{ a_k}{2 \omega(k)} \left[ 2\pi \delta(\omega-\omega(k)) + 2\pi \delta(\omega+\omega(k)) \right] \end{equation}

The delta function $\delta(\omega^2-k^2-m^2)$ enforces that the field obeys the Klein Gordon equation (that it is on shell). However, the way I derived this was by reverse engineering the form of $\tilde{\phi}(\omega, k)$ to reproduce the mode expansion; working in momentum space doesn't give you new information here.

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