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Here are the steps: \begin{align} \Delta_F(x) &= \int \frac{d^4 k}{(2\pi)^4} \frac{1}{k^2-m^2+i\epsilon}e^{-k x} \end{align} We firstly to a complex integral upon $k^0$ around the upper plane and the lower plane, the contours are $\gamma^+$ and $\gamma^-$ and the associated divergences are $-\omega$ and $\omega$ respectively, with $\omega = \sqrt{|\vec{k}... 0 Here's a simple explanation. We can understand the destruction field operator$\hat{\psi}(x)=\frac1 {\sqrt V} \sum_p{\hat a _p e^{ipx}}\tag{1}$as the expansion in the basis {$e^{ipx}$}, now changing the sign simply means we expand it in another basis {$e^{-ipx}$}. Since the summation is over all negative and positive values of$p$, the two basis are indeed ... 1 I'm not sure in what level of rigor you want to calculate things, but here the functions are not Lebesgue-integrable over$\Bbb{R}^3$, so strictly speaking everything below has to be interpreted as the sense of tempered distributions in$\Bbb{R}^3$. Note that here we interpret the Fourier transform of a vector-valued mapping to mean to the component-wise ... 1 Quantum Mechanic's answer is a nice heuristic, but I think there's a better answer. The mode expansion you listed is an example of a change of basis in second quantization. Specifically, suppose I have a basis of states$\{ | \alpha \rangle \}$and another set of basis states$\{ | \nu \rangle \}$. In first quantization, I would express the former basis in ... 0 A travelling wave moving with velocity$v$takes some functional form$f(x-vt)$. If$v$is positive the wave is moving to the right and if$v$is negative the wave is moving to the left. Since the time evolution for a free field has$da/dt=-i\omega a$, your operators will evolve as $$\psi(x,t)\propto\sum_p a_p e^{ipx-i\omega t}=\sum_p a_p e^{ip(x-\omega t/p)}... 0 I will answer my question as I understand it now. The field \phi(x,t) has to be purely real. This is not stated anywhere but is implied. This was all that was to my confusion. 0 Since we are talking about waves, I suggest you consider what happens in analogous cases- for example, consider a water wave approaching a barrier in which there is a narrow opening. When the wave passes through the opening, it spreads in a circular front. That is a fundamental property of wavelike motion on a surface- a tendency to spread. The linear nature ... 1 I revisited this problem after some time, and even though I think @CR Drost provided a very informative answer, I don't quite think it's "exactly" the solution to my problem. Indeed, after doing the calculation rigorously and following each step carefully, I think the answer is that the fermionic action (at zero temperature) is actually supposed to ... 3 These are called Discrete Fourier Transforms. We define discrete Fourier transform g_p (p=0, 1, 2, \dots, N-1) of a function f is$$ g_p = \frac{1}{\sqrt{N}} \sum_{k=0}^{N-1} f_k \, e^{2\pi ikp/N} $$Inverse Fourier transform is defined in similar way,$$ f_j = \frac{1}{\sqrt{N}} \sum_{p=0}^{N-1} g_p \, e^{-2\pi ijp/N} $$Refer Arfken, 7th Edition, 20.6, ... 0 So the aim is to solve the equation (\Box+m^2)\phi(\vec{x},t)=0 for \phi (m=0 in Schwartz's example, but we'll leave it in). Note that (as Schwartz writes)$$(\Box+m^2)\phi(\vec{x},t)=(\partial_t^2-\vec{\nabla}^2+m^2)\phi(\vec{x},t)=0\tag*{(1)}$$Something you will get used to is flipping between differential operators in position space and their form ... 0 You're probably looking at a free scalar theory whose lagrangian is given by$$\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}... 3 I'll work in a system of units where$\hbar =1$, so that the wave-number$k$and the momentum$p$are equivalent, but you can easily generalise this simple argument. It looks like what you're trying to calculate is the (complex conjugate of the) momentum-space wavefunction for the state$|\psi_E\rangle$,$\tilde{\psi}_E(k)\$. If you know the position-space ...