# Is QFT time symmetric, and how is it implemented?

In electromagnetism, while the Maxwell equations are time symmetric, there is a choice to restrict solutions specifically to retarded potentials, imposing a time direction on the equations. And in QFT, as far as I'm aware, there is a restriction of what constitute an acceptable field to the small groups with a forward in time pointing momentum (in Weinberg at least).

On the other hand, the propagator usually used in QFT is the Feynman propagator, which is time symmetric. And the axiomatisations of QFT do not seem to include any specific axiom regarding this. Wightman just requires operators to commute outside the lightcone, but doesn't seem to differentiate the two lightcones. Local QFT requires spacetime to be time oriented, but I am not quite sure in what axiom this is actually used, as the category of regions is over the double lightcones.

Is QFT defined to be time symmetric or is there a direction of time assumed? Is that some implicit assumption, or does it derive from the axioms of QFT? Or else what is to prevent to have two different fields from having momentums in different directions?

If I understand your question correctly...

As you surely know, Boltzmann have derived H-theorem about increasement of entropy from his equation (which sets thermodynamical arrow of time), which was derived from time-reversible classical mechanics. Historically this has given rise to criticism of this theorem. Now we know, that the full description of many-particle distribution function is given by BBGKY chain of equations; however, when we assume that higher order particle correlations are negligible, we may break the chain and, after some simplifications, get the Boltzmann equation. This gives us an answer how irreversibility enters in Boltzmann equation (and results, hence, in Boltzmann theorem): by neglecting many-particle correlations we refuse the fluctuations which may return the system back.

Analogical thing happens in general quantum field theory. As we know, each QFT has to be unitary for correct description of physical processes. Independently on theory details (for example, whether the theory has $T$-symmetry or no), by using only unitarity requirement for $S-$matrix, which gives the amplitudes of each process, we may extract from the unitarity condition $SS^{\dagger} = 1$ the quantum Boltzmann H-theorem: $$\frac{d}{dt}\int da \log\left(\frac{P_{a}}{c_{a}} \right)P_{a} \geqslant 0$$ Here $P_{a}da$ is the probability of finding the system in the state $a$, and $c_{a} = \left( \frac{V}{(2 \pi)^{3}}\right)^{N_{c}}$ is the volume of the phase space of $N_{a}$ particles in state $a$.

As for the case of classical Boltzmann H-theorem derivation, we may ask where irreversibility enters the case. The answer is that for derivation of QFT version of Boltzmann H-theorem we need de-facto assume following. If the process start from the some definite state, in general it evolves in the superposition of the states (or, for the density matrix formulation of QM states, the density matrix becomes nondiagonal). For derivation of H-theorem we need to assume that the final state is definite (the density matrix is diagonal). This is equivalent to the measurement procedure, which, as we know, for QM is irreversible. The direction of time is given by the procedure of measurement.

• Doesn't assuming the final state to be diagonal (i.e. result after a measurement) directly contradict the assumption of unitary dynamics? (I.e. you would have to show that the dynamics actually somehow select such states) – ACuriousMind Feb 26 '16 at 17:36
• @ACuriousMind : in fact, yes, since it provides decoherence. – Name YYY Feb 27 '16 at 9:45

No, QFT is not generically time symmetric.

This is seen by observing the strong interaction experimentally violates CP symmetry, which is theoretically then reflected in the Standard Model by the possibility of a phase in the quark and/or neutrino mass matrices. Since CPT must be a symmetry by the CPT theorem, this directly implies T asymmetry, since $[H,CPT] = CP[H,T] + [H,CP]T = [H,CP]T$ if $[H,T] = 0$, and so $[H,CPT]=0$ cannot hold for $[H,CP]\neq 0$.

• My question was not really about T-symmetry, as it concerns fields that are still T-symmetric (such as free scalar or EM fields), more about how to enforce that particles move only in one time direction. – Slereah Feb 26 '16 at 16:02