Causality and anti-particles

Question

How can I quantitatively and qualitatively understand the fact that there is a relevance between the existence of anti-particles and causality?

Answer 1

This is a consequence of the fact that there is no positive frequency function which is zero outside the light cone. If you have a particle in relativity, its dynamics require that it goes faster than light, and to restore causality, it must go back in time. This is explained in my answer here: Is anti-matter matter going backwards in time? .

If you have a quantum particle with positive energy, the propagation function $G(x-y)$ is the amplitude to go from x to y. This propagation is said to be causal if the propagator is zero unless x is to the future of y, so that in a time-space decomposition, $G(t,r)$ is zero for t<0. In this case, the Fourier transform $G(\omega,k)$ cannot vanish for all $\omega<0$, because it is impossible for a nonzero function and its Fourier transform to both be exactly zero in a half-plane.

To see this, the condition of vanishing of $G(t,r)$ for t<0 implies the analyticity of the Fourier transform for $\omega$ with a negative imaginary part, since in this region, the Fourier transform of G becomes a sum of decaying exponentials. An analytic function can't be zero in a region without being zero everywhere, so the Fourier transform of a future directed function is not strictly positive energy.

Because of this, there is no relativistic particle formalism in which the particles both have positive energies and causal propagation. You can either deal with fields, in which case the particle notion is non-local, or you can deal with particles, but then they go back in time.

The back-in-time formalism is using the standard non-causal Feynman propagator, which is

$$ G(\omega,k) = {i\over \omega^2 - k^2 - m^2 - i\epsilon}$$

up to numerator modifications for higher spin, with the $i\epsilon$ pole prescription. This has two poles in $\omega$ for any $k$, and the pole prescription pushes one pole to have slightly positive imaginary part and the other pole to the slightly negative imaginary part. There are singularities in both directions in imaginary $\omega$ direction, which means that the propagation is non-causal.

The part that goes forward in time is the positive energy part; the part that goes back in time is the negative energy part.

Answer 2

This is mainly an issue of the complex Klein Gordon field (There's no such requirement for the Dirac field for instance) It's most easily shown with the self propagator of the complex Klein Gordon field using plane waves in the x-direction:

The Klein Gordon equation is.

$\frac{\partial^2 \psi}{\partial t^2} ~~=~~ \left(\frac{\partial^2}{\partial x^2} - m^2 \right)~\psi$

A straightforward generator of time evolution would be.

$\frac{\partial \psi}{\partial t} ~~=~~\pm i \sqrt{-\frac{\partial^2}{\partial x^2} + m^2}~\psi$

With a (-) sign for particles and a (+) sign for anti-particles. However, this generator is non local since it corresponds to an infinite series of derivatives. In fact it's equal to a convolution with a Bessel K function.

$\frac{\partial \psi}{\partial t} ~~=~~\pm i \sqrt{-\frac{\partial^2}{\partial x^2} + m^2}~\psi ~~=~~ \frac{m}{x}K_1(mx)~*~\psi$

source: wikipedia

This means instantaneous propagation since $\partial\psi/\partial t$ depends on non local values of $\psi$. This problem then enters the general time evolution operator for arbitrary $t$.

$\psi(t) ~~=~~ \exp\left\{ \pm i \sqrt{-\frac{\partial^2}{\partial x^2} + m^2}\right\}~\psi$

However, now comes the trick:

The sum of the particle and anti particle propagators is local (within the light cone)

$\psi(t) ~~=~~ \frac12\Big(\exp\left\{ + i ...\right\} +\exp\left\{ - i ...\right\}\Big)\psi ~~=~~ \cos\left\{ \sqrt{-\frac{\partial^2}{\partial x^2} + m^2}\right\}~\psi$

because the Taylor series expansion of the cosine only contains even powers of the argument there is no more square root operator.

It has to be said that the part outside the light cone is small to begin with, in the order of the Compton radius of the particle. But it also shrinks further as propagation progresses. For an electron it's about $10^{-13}m$ at the start of the propagation but only $10^{-20}m$ after a lightmicron (the time in which light propagates 1 $\mu m$). It shrinks further away linear with time.

This issue does not occur for the time evolution operator of the correct equation for the electron : The Dirac equation. This equation is linear and $\partial\psi/\partial t$ does not contain the above square root.

Hans.