# Could Time-Evolution be antiunitary?

There are serveral Arguments for Time-Evolution to be unitary, for example, time-evolution should preserve the norm of each given state (because elseways the probabillity Interpretation would not work), it should be bijective (because elseways one would lose information).

This however does not fix wether time-evolution is unitary or antiunitary, since a antiunitary Operator could as well we bijective, and preserve the norm. Are there additional reasons why time-evolution must be unitary?

I tried to come up with an argument, which is that for the limit of no time passed, the time-evolution should become unity, which obviously is not antiunitary. Is that the argument? Are there other reasons why time-evolution should be unitary? Or are there infact formulations of quantum-mechanics / QFTs with antiunitary time-evolutions?

Edit: I don't asume the Schrödinger equation to hold. Instead, I'm interested in what postulates have to be made to have the Schrödinger equation be the outcome. Postulates so far are only that states (represented by a vector in a complex-valued hilbertspace) give a probability interpretation for observables (represented by Operators acting on that Hilbert space). Additionally I want the superposition principle to hold, and time-evolution should (because of the probability interpretation) not change the norm of a state vector.

• For zero time ($t=0$) the time evolution is the identity, a unitary operator. As $U(t)$ should be continuous it has to remain a unitary operator. Another argument is the fact that the product of two anti-unitaries is a unitary. However, we have $U(t) = U(t/2)^2$ for a stationary process, so ... – Fabian Aug 27 '19 at 9:40
• @Fabian This was another thought of mine as well. I would just have liked to have this thoughts confirmed. – Quantumwhisp Aug 27 '19 at 9:53
• Possible duplicate: physics.stackexchange.com/q/333323/50583 and its linked questions. Additionally, in order to properly answer this question you should state what your starting point here is - e.g. if you start from the Schrödinger equation as a given then the answer is trivial, but if you start from some other idea of what "quantum mechanics" is, it might not be. – ACuriousMind Aug 27 '19 at 16:46
• The answer to this question is relevant (with $A$ the Hamiltonian). – Fabian Aug 28 '19 at 10:27
• @Fabian Your first comment constitutes a pretty good answer in my understanding. – Dvij D.C. Aug 29 '19 at 8:37