# How is this Fourier transform done?

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?

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 $$$$(\square + m^2) \phi(x, t) = 0$$$$ Then, note that by separation of variables, we can find the following basis of solutions, labeled by $$k$$ and by the sign of $$\omega$$: $$$$\phi_{k, \pm}(x, t) = N_k e^{i (k x \pm \omega(k) t)}$$$$ where $$N_k$$ is a normalization constant, and where $$\omega(k)$$ is a function of $$k$$ that is fixed by the equations of motion to satisfy $$$$\omega(k) = \sqrt{k^2 + m^2}$$$$ This is the main reason I don't like to think of your first expression as a Fourier transform. If we were doing a Fourier transform, then $$\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 $$$$N_k = \frac{1}{2\omega(k)}$$$$ This is derived in any good QFT book or lecture notes, such as Section 2 of David Tong's QFT notes.
• Since $$\phi(x, t) = \phi^\dagger(x, t)$$, we can deduce that $$$$a_{k, -} = a^\dagger_{-k, +}$$$$ Let's define $$a_k \equiv a_{k, +}$$. Then, $$$$\phi(x, t) = \int \frac{d k}{(2\pi)(2 \omega(k))} \left(a_{k} e^{i( k x - \omega(k) t)} + a^\dagger_{-k} e^{i (k x + \omega(k) t)}\right)$$$$ Note that for a complex scalar field, you wouldn't be able to relate $$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$$ in the second term, we get to the mode expansion $$$$\phi(x, t) = \int \frac{d k}{(2\pi)(2 \omega(k))} \left(a_{k} e^{i (k x - \omega(k) t)} + a^\dagger_{k} e^{-i (k x - \omega(k) t)}\right)$$$$

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

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.