# Plane waves in QFT

Suppose we work in the metric $(-1,+1)$.

How do we describe an incoming particle with a plane wave; $\exp(-\mathrm ikx)$ or $\exp(+\mathrm ikx)$? What's the difference? Does it change if we work in the other flat metric $(+1,-1)$?

It seems it's pretty arbitrary and not standard?

More generally, how to determine direction of propagation of a plane wave written in exponential form?

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It seems Peskin and Schroeder uses $$exp(-ikx)$$ for incoming momenta. – Mauchy May 7 '13 at 11:53
I believe that it's a convention based on the fact that most people are right-handed, so it 'seems' more natural to have something coming from the left and going to the right. Since a plane wave if something like $e^{i (\omega t - kx)}$, and it goes right if k>0, you have the answer above. – Hydro Guy May 7 '13 at 13:31
How do you figure out the direction in which the particle is moving? Apply the momentum operator on the wavefunction of interest and obtain the momentum eigenvalue. Then look at the sign of the momentum, If it is positive, then you have right mover, and left mover if it is negative. So it is not conventional. – Prathyush May 7 '13 at 14:37

First of all, we don't usually talk about the direction of propagation of a plane wave in QFT. Plane waves are said to exist at all spacetime coordinates with a certain internal momentum, k. And, in reference to some of the comments, in QFT, we don't normally operate with wavefunctions. We promote wavefunctions to operators and act on states. But in this case, we don't need a state.

Usually, a plane-wave solution can be expressed as:

$$a_ke^{-ikx}~~OR~~a_k^\dagger e^{ikx}$$ In the mostly minus notation, $(+,-,-,-)$, where $a_k$ and $a_k^\dagger$ are annihilation and creation operators of a particle respectively. In the mostly plus notation, it is the same as if we switch the minus signs in the exponents (it's slightly more complicated, but that's the gist). The solutions can be written as:

$$a_ke^{ikx}~~OR~~a_k^\dagger e^{-ikx}$$

These mean the same thing as long as the convention is stated. As long as you're consistent, the physics come out to be exactly the same.
As for determining the direction of propagation, that statement doesn't make a lot of sense in this context. Because this solution is valid everywhere, the plane-wave exists everywhere, so it doesn't really move. Furthermore, it's tough to specify the direction a particle is moving because in QFT, there can be no single particle systems with momentum eigenstates. This is because QFT is a relativistic version of QM and at the energy levels it applies to, even in the ground state, particles are able to spontaneously pop into existence and muss up your expectations.

However, you can tell the direction something is moving through time. Since the time component in the exponential is always positive or always negative, you can say that particles with a creation operator are moving forward in time and particles with an annihilation operator are moving backward in time (that is, they can be thought of as anti-particles).

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