$#*! 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.