$#*! I unlearned #1: scalar field theory

The Hamiltonian:

$$H = \frac12 \int d^3 x \left[ \pi^2 + (\nabla \phi)^2 + m^2\phi^2 \right]$$

where $\pi(\mathbf{x})$ is the conjugate momentum density of field $\phi(\mathbf{x})$ is quadratic. In particular by going to momentum space with:

$$\begin{align} \phi(\mathbf{x}) &= \int \frac{d^3p}{(2\pi)^3} \sqrt{\frac{1}{2\omega_{\mathbf{p}}}} \left(a_{\mathbf{p}} + a_{-\mathbf{p}}^{\dagger} \right) e^{i \mathbf{p} \cdot \mathbf{x}} \\ \pi(\mathbf{x}) &= \int \frac{d^3p}{(2\pi)^3} (-i) \sqrt{\frac{\omega_{\mathbf{p}}}{2}} \left(a_{\mathbf{p}} - a_{-\mathbf{p}}^{\dagger} \right) e^{i \mathbf{p} \cdot \mathbf{x}} \end{align}$$

$H$ become just sum of uncouple oscillators:

$$H = \int \frac{d^3p}{(2\pi)^3} \omega_{\mathbf{p}} \left( a_{\mathbf{p}}^{\dagger} a_{\mathbf{p}} + \frac12 [a_{\mathbf{p}}, a_{\mathbf{p}}^{\dagger} ] \right)$$

with well-known spectrum and stuff. More important are these commutation relations:

$$[ H, a_{\mathbf{p}} ] = - \omega_{\mathbf{p}} a_{\mathbf{p}} \,,\quad [ H, a_{\mathbf{p}}^{\dagger} ] = \omega_{\mathbf{p}} a_{\mathbf{p}}^{\dagger}$$

The time dependent of $\phi$ and $\pi$ are obtained in the usual way:

$$\phi(x) = \phi(\mathbf{x},t) = e^{iHt} \phi(\mathbf{x}) e^{-iHt} \,.$$

The usual commutation relations hold at any time $t$:

$$[\phi(\mathbf{x},t), \pi(\mathbf{y},t)] = \delta^{3}(\mathbf{x} - \mathbf{y}) \,,\quad [ \phi(\mathbf{x},t), \phi(\mathbf{y},t) ] = [ \pi(\mathbf{x},t), \pi(\mathbf{y},t) ] = 0 \,.$$

Using the Heisenberg EQO $i\frac{\partial Q}{\partial t} = [ Q, H ]$, we get $i \frac{\partial \phi}{\partial t} = i \pi$ and $i \frac{\partial \pi}{\partial t} = i (\nabla^2 - m^2) \phi$, which combine to give:

$$\frac{\partial^2}{\partial t^2} \phi = (\nabla^2 - m^2) \phi \,,$$

the Klein-Gordon equation. Again writing in term of creation and annihilation operators:

$$\begin{align} \phi(x) &= \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2 p^{0}}} \left(a_{\mathbf{p}} e^{-i p \cdot x} + a_{\mathbf{p}}^{\dagger} e^{i p \cdot x} \right) \\ \pi(x) &= \frac{\partial \phi(x)}{\partial t} \end{align}$$

Note that the field operator $\phi$ is Hermitian, so this field is its own anti-particle.

The probability amplitude of creating a field at $y$ and observing it in $x$ is the propagator:

$$D(x-y) = \left\langle 0 \left| \phi(x) \phi(y) \right| 0 \right\rangle = \int \frac{d^3p}{(2\pi)^3} \frac{1}{2 p^0} e^{-ip \cdot (x-y)}$$

Consider a time-like interval: $x^0 - y^0 = t$ and $\mathbf{x} - \mathbf{y} = 0$:

$$D(x-y) = \frac{4\pi}{(2\pi)^3} \int_0^{\infty} dp \frac{p^2}{2\sqrt{p^2+m^2}} e^{-i \sqrt{p^2+m^2} t} = \frac{1}{4\pi^2} \int_{m}^{\infty} dE \sqrt{E^2 - m^2} e^{-i Et} \to e^{-imt}$$

So we get an oscillating propagator. Nothing surprising here

Now consider a purely spacial separation: $x^0 - y^0 = 0$ and $\mathbf{x} - \mathbf{y} = \mathbf{r}$:

$$D(x-y) = \int \frac{d^3p}{(2\pi)^3} \frac{e^{i\mathbf{p} \cdot \mathbf{x}}}{2 p^0} = \frac{-i}{2(2\pi)^2r} \int_{\infty}^{\infty} dp \frac{p e^{ip r}}{\sqrt{p^2 + m^2}}$$

The integrand has two branch point at $\pm m$, so to do this integral, we push the integral up and wrap around the upper branch cut. Let $z = -ipr$, we obtain:

$$D(x-y) =\frac{1}{4\pi^2r^2} \int_{mr}^{\infty} dz \frac{z e^{-z}}{\sqrt{z^2 - (mr)^2}} \to e^{-mr}$$

at large $r$. The propagator seems to violate causality since it is finite for a space-like interval.