Fourier decomposition of the field into plane waves [QFT] I am going through the text Quantum Field Theory and the Standard Model and have a question regarding a formula used in chapter 2 on the general solution to the differential equation: $\left(\partial_{t}^{2}+\vec{p} \cdot \vec{p}\right) a_{p}(t)=0$. The author states that the answer is just $\phi(x, t)=\int \frac{d^{3} p}{(2 \pi)^{3}}\left(a_{p} e^{-i p x}+a_{p}^{\star} e^{i p x}\right)$, a Fourier Decomposition into plane waves, however I do not have the slightest clue where this formula was obtained from. I want to add that $p x=\omega_{p} x_{0}-\vec{p} \cdot \vec{x} .$
 A: You're probably looking at a free scalar theory whose lagrangian is given by $$\mathcal L = \frac12\partial_\mu\phi\partial^\mu\phi - \frac12\phi^2.$$  Varying the action leads to the classical equations of motion $$(\partial_\mu\partial^\mu+m^2)\phi=0.$$  If you Fourier decompose the field $$\phi(x) = \int \frac{d^3p}{(2\pi)^3}\left( a_{\vec p}e^{-ip\cdot x} + a^*_{\vec p}e^{ip\cdot x} \right)_{p^0=\sqrt{\vec{p}^2+m^2}},$$ you can see that the above automatically solves the equations of motion.  Notice importantly how we've taken $p^0=\sqrt{\vec p^2+m^2}$, which your book seems not have have explicitly mentioned.  As mentioned in the comments, at this classical level the $a_{\vec p}$ and $a_{\vec p}^*$ are just Fourier coefficients.  Also, the way I've written them, they're not dependent on time in any way.  However, one could imagine stripping off the time dependence from the Fourier modes and putting those into the $a$'s; one could define $$a_{\vec p}(t)\equiv a_{\vec p}e^{-ip^0t}$$ and thus $$\phi(x) = \int \frac{d^3p}{(2\pi)^3}\left( a_{\vec p}(t)e^{i\vec p\cdot \vec x} + a^*_{\vec p}(t)e^{-i\vec p\cdot \vec x} \right).$$  Remember always that we're taking $p^0=\sqrt{\vec p^2+m^2}$ implicitly.  Then one should find that $$(\partial_t\partial_t+(\vec p\cdot\vec p+m^2))a_{\vec p}(t) = 0.$$
A: So the aim is to solve the equation $(\Box+m^2)\phi(\vec{x},t)=0$ for $\phi$ ($m=0$ in Schwartz's example, but we'll leave it in). Note that (as Schwartz writes) $$(\Box+m^2)\phi(\vec{x},t)=(\partial_t^2-\vec{\nabla}^2+m^2)\phi(\vec{x},t)=0\tag*{(1)}$$ Something you will get used to is flipping between differential operators in position space and their form in Fourier space. The Fourier conjugates of $\Box$ and $\Delta=\vec{\nabla}^2$ are namely $-p^2$ and $-\vec{p}^2$, respectively. To see how this works, consider the field expressed in terms of its $\vec{p}$-space Fourier conjugate, which we will call $a_\vec{p}(t)$: $$\phi(\vec{x},t)=\int\frac{d^3\vec{p}}{(2\pi)^3}a_\vec{p}(t)e^{i\vec{p}\cdot\vec{x}}.$$ Insering this into eq. 1 then gives
\begin{align*}
(\Box+m^2)\phi(\vec{x},t)&=\int\frac{d^3\vec{p}}{(2\pi)^3}(\partial_t^2-\vec{\nabla}^2+m^2)a_\vec{p}(t)e^{i\vec{p}\cdot\vec{x}}\\
&=\int\frac{d^3\vec{p}}{(2\pi)^3}(e^{i\vec{p}\cdot\vec{x}}\partial_t^2a_\vec{p}(t)-a_\vec{p}(t)\vec{\nabla}^2e^{i\vec{p}\cdot\vec{x}}+m^2a_\vec{p}(t)e^{i\vec{p}\cdot\vec{x}})\\
&=\int\frac{d^3\vec{p}}{(2\pi)^3}(\partial_t^2+\vec{p}^2+m^2)a_\vec{p}(t)e^{i\vec{p}\cdot\vec{x}}\\
&=0
\end{align*}
So we find that the Fourier coefficients for each $\vec{p}$ are solutions to the conjugate equation to (1): $$(\partial_t^2+\vec{p}^2+m^2)a_\vec{p}(t)=0$$ which should be familiar as the differential equation for simple harmonic motion, solved by $a_\vec{p}(t)=a_\vec{p}e^{-i\omega t}$ where $\omega=\sqrt{\vec{p}^2+m^2}$ and $a_\vec{p}$ is constant in time (but of course depends on $\vec{p}$). The final step is to realize that $\phi^\dagger(\vec{x},t)$ is also a solution to (1) (just take the Hermitian conjugate of both sides), so we arrive at the full general solution $$\phi(\vec{x},t)=\int\frac{d^3\vec{p}}{(2\pi)^3}\left(a_\vec{p}e^{-ipx}+a^\dagger_\vec{p}e^{ipx}\right)$$ with $px=\omega t-\vec{p}\cdot\vec{x}$.
