Preserving Causality for Field Operators

It is important to examine the quantity:

\begin{eqnarray}
\langle 0 \vert \phi(x)\phi(y)\vert 0 \rangle
\end{eqnarray}

Or two field operators at different points in space time, acting on the vacuum. In particular, for a causal theory, we would like all space-like separated operatores to commute, i.e.:

\begin{eqnarray}
[\mathcal{O}(x),\mathcal{O}(y)] &=& 0 \ \mathrm{if} \  \eta_{\mu \nu}x^\mu y^\nu <0
\end{eqnarray}

Or two space-like separated observables cannot effect each other. Let us examine the commutator for our mode-expanded fields:

\begin{eqnarray*}
\phi(x) &=& \sum_k \frac{1}{\sqrt{2L^3 \omega_k}}\left(a_ke^{-ikx}+a^*_k e^{ikx}\right)\\
\left[\phi (x), \phi (y)\right] &=& \sum_{k,p} \frac{1}{2L^3} \frac{1}{\sqrt{\omega_k \omega_p}}\left[\left(a_ke^{-ikx}+a^* e^{ikx}\right)\left(a_pe^{-ipy}+a^*_p e^{ipy}\right)\right]\\
&=& \sum_{k,p} \frac{1}{2L^3} \frac{1}{\sqrt{\omega_k \omega_p}}\left[a_k,a^*_p \right]e^{-ikx+ipy}+\left[a^*_k,a_p \right]e^{+ikx-ipy}\\
&=& \sum_{k,p} \frac{1}{2L^3} \frac{1}{\sqrt{\omega_k \omega_p}}\delta_{k,p}\left(e^{-ikx+ipy}-e^{+ikx-ipy}\right)\\
&=& \sum_{k,p} \frac{1}{2L^3} \frac{1}{\omega_k}\left(e^{-ik(x-y)}-e^{+ik(x-y)}\right)
\end{eqnarray*}

We can know re-write this difference in space time coordinates as:

\begin{eqnarray}
x-y &=& \langle \tau, \vec{\xi} \rangle
\end{eqnarray}

and expand out the four vector contractions:

\begin{eqnarray}
\left[\phi(x),\phi(y) \right]&=& \sum_{k,p} \frac{1}{2L^3} \frac{1}{\omega_k}\left(e^{-i\omega_k \tau}e^{ik\xi}-e^{+i\omega_k\tau}e^{-ik\xi}\right)
\end{eqnarray}

Turning our integral into a sum now, we write:

\begin{eqnarray}
\sum_k \frac{1}{V} \to \int \frac{d^3k}{(2\pi)^3}\\
\left[\phi(x),\phi(y) \right]&=& \int \frac{d^3k}{(2\pi)^3} \frac{1}{2 \omega_k}\left(e^{-i\omega_k \tau}e^{ik\xi}-e^{+i\omega_k\tau}e^{-ik\xi}\right)
\end{eqnarray}

Integration over our angular coordinates is trivial, since we can set $\vec{\xi}$ along the $k_z$ axis and get rid of $\theta$ immediately.

\begin{eqnarray}
\left[\phi(x),\phi(y) \right]&=& \int_0^\infty \int_{1}^{-1} \frac{k^2 dk du }{(2\pi)^2} \frac{1}{2 \omega_k}\left(e^{-i\omega_k \tau}e^{ik\xi u}-e^{+i\omega_k\tau}e^{-ik\xi u}\right)
\end{eqnarray}

Notice we have set $u=\cos(\psi)$, the angle from $k_z$, above. Integrating over $u$ gives us the typical bessel functions:

\begin{eqnarray}
\left[\phi(x),\phi(y) \right]&=& \int_0^\infty \frac{k^2 dk }{(2\pi)^2} \frac{1}{2 \omega_k}\left(\frac{e^{-i\omega_k \tau}e^{ik\xi u}}{ik\xi}-\frac{e^{+i\omega_k\tau}e^{-ik\xi u}}{-ik\xi} \right) \vert^{-1}_{1}\\
\left[\phi(x),\phi(y) \right]&=& \frac{-i}{\xi}\int_0^\infty \frac{k dk }{(2\pi)^2} \frac{1}{2 \omega_k}\left(e^{-i\omega_k \tau}e^{ik\xi u}+e^{+i\omega_k\tau}e^{-ik\xi u}\right) \vert^{-1}_{1}
\end{eqnarray}

Taking a long hard look at the above equation, one might be able to see that we're going to get zero if $\tau=0$ since, the evaluation at endpoints of $u$ will cancel. We can also expand out the endpoints of $u$ to get:

\begin{eqnarray}
\left[\phi(x),\phi(y) \right]&=& \frac{1}{\xi}\int_0^\infty \frac{dk}{2\pi^2} \frac{k}{\omega_k}\sin(k\xi)\sin(\sqrt{k^2+m^2}\tau)
\end{eqnarray}

Now we see immediately that the $\tau \to 0$ limit gives us a vanishing commutator, and the $\xi \to 0$ limit gives us something that is finite, like

\begin{eqnarray}
\left[\phi(x),\phi(y) \right]&=& \int_0^\infty \frac{dk}{2\pi^2} \frac{k^2}{\sqrt{k^2+m^2}}\sin(\sqrt{k^2+m^2}\tau)
\end{eqnarray}

This will give an asymptotic solution of the form $e^{imt}$ which can be shown by stationary phase arguments. And we find, therefore, that the commutator for space-like separation is non-vanishing.

It is important to note that, during this entire discussion, we could have framed the commutator problem as the difference of two propagators, the probability for a particle to be created at $y$ and destroyed at $x$:

\begin{eqnarray}
\langle 0 \vert \phi(x) \phi(y)\vert 0 \rangle &=& D_{xy}
\end{eqnarray}

and therefore our commutator looks like:

\begin{eqnarray}
[\phi(x),\phi(y)] &=& D_{xy}-D_{yx}=0 \ \forall \ \eta_{\mu \nu}x^\mu y^\nu <0
\end{eqnarray}

David Tong, in his online notes, says we can ``wrap words around this''. When the events of creation and annihilation are space-like separated, we can ``re-order'' the events by a proper Lorentz boost.  This means that the two amplitudes for the processes $x\to y$ and $y \to x$ cancel. For a complex scalar field, this turns into a statement about the amplitude of a particle and an anti-particle travelling from $x \to y$, which once again cancel for space-like separation.