# Free expansion of interacting electron gas is an irreversible process

Consider a number of electron in vacuum, confined in a bounded region. These electrons have interaction described by QED. At $$t=0$$, we allowed the system to evolve. Without doing any measurement on the system, we know that the whole process is reversible because this is an unitary evolution.

However we also know that there is no chance the system going back to the original state due to repulsive force, so the process is not time reversible. Certainly the process is not related to charge or parity, therefore the process is irreversible. Contradiction.

So what is wrong in this picture, have I miss something?

• Won't the electrons finally distributed on the boundary surface just like negative charge on a metal body? To be gas like, shouldn't there only be near field force between the particles like electrons in a chargeless metal? – jw_ Mar 22 '20 at 1:45
• @jw_ What I mean is a collection of electron only. I don't know the term "gas" may be misleading. – Ken.Wong Mar 22 '20 at 22:24

• I am not sure this is solving the question. It seems to me that the solution is indeed irreversible as well. Take $x_1(t)$ and $x_2(t)$ being the trajectories of the two particles in a 1-D setting, and integrate them forwards in time until $t_0$. Then $x_1(t_0-t')$ will still bring you trivially to $x_1(0) once$t'=t_0$. Isn't that the definition of reversibility of the solution? – AtmosphericPrisonEscape Mar 20 '20 at 21:38 • the solution is not time reversible, and that is fine. time reversibility is not guaranteed to be a symmetry of solutions to the Maxwell's equations. – NewUser Mar 20 '20 at 21:48 "However we also know that there is no chance the system going back to the original state due to repulsive force" This is untrue. If we could track the position of each $$e^{-}$$ (assuming we are thinking of them as point particles for simplicity), then for any given spatial configuration of the charges, reversing the direction of $$v$$ of each charge would actually lead to backward evolution in time. One can think of a simple analogous situation with two point particles of same sign of charge. Let us say both are at rest initially (time = 0) separated by some distance $$d$$. Since the force is repulsive, at some later time t, the distance will be some $$d' > d$$. Now, what happens if we reverse the directions of the velocities for the charge configuration at time $$t$$? Backward evolution will happen, and after another time interval $$t$$ has elapsed, we will be back to the original configuration (separation = d, both charges at rest). This is indeed time reversed motion So, to summarize. If we can break up the dynamics of composite system (electron gas) into dynamics of simpler units (electrons) interacting microscopically in a Time reversal invariant manner, AND, we are actually able to keep track of the motion of each one of the interacting units, the resultant motion will always be T-reversal invariant. Emergence of macroscopic irreversibility from reversibility on microscopic level always is a result of coarse graining. I am not going to elaborate any further, but reading about the Boltzmann H-Theorem will help. It addresses precisely the question of how macroscopic irreversibility emerges from microscopically reversible motion • The assumptions of H-theorem is molecular chaos, it seems to me that does not hold in the above setting because each interaction will entangle the electrons, which shouldn't give independent behavior. – Ken.Wong Mar 21 '20 at 10:08 • Ok, let me rephrase what I said above then. If you can actually sum up the entire perturbative expansion for the unitary time evolution, then the motion will be automatically reversible in time. That means a very simple thing, that is, if$|\psi_t> = U_t |\psi_o>$then$|\psi_o^T> = U_t |\psi_t^T>$, where the$T\$ superscript denotes time reversed states. Also, the situation you describe (electron gas) will probably not have any macroscopic entanglement, and you will have uncorrelated quasiparticles, to which H-theorem will apply. – insomniac Mar 21 '20 at 11:35