Are there examples of lossless, yet time irreversible systems? I would be pleased to know if there are any in Quantum Mechanics.
Which meaning of "irreversible"?
If it is thermodynamically irreversible, then virtually every system is irreversible (second law) and all systems are lossless (first law.) Actually, it all depends on the degrees of freedom you chose for the system. Quantum mechanical thermodynamics is no different, although it isn't obvious.
I might be wrong, but my gut thinking would be that since the Schrodinger equation can easily be run backwards in time, then if what you mean by "lossless" is that it also involves no increase in entropy, then my guess would be that what you're describing is a system that is both time-symmetric and also thermodynamically reversible, and therefore it cannot be a reversible reaction.
What prompted this, can I ask? Do you have a reason to believe that such systems exist? I am curious!
@J Erik Baxter I have been currently reading about the properties of scattering matrix for various systems.
What I found is
1) S has to be unitary for a lossless system.
2) S^*=S^-1 for time reversible system.
3) S=S^T for a non-reciprocal system.
So all of these cases are different. What I intend to find is if there is a system which breaks time reversal symmetry yet is lossless.
Which meaning of "irreversible"?
If it is thermodynamically irreversible, then virtually every system is irreversible (second law) and all systems are lossless (first law.) Actually, it all depends on the degrees of freedom you chose for the system. Quantum mechanical thermodynamics is no different, although it isn't obvious.
Chetan Jesing Waghela Thank you for your response, it is very helpful. Can you please clarify though what you mean by 'non-reciprocal'?
What I would note is that if S is unitary, that means that S^-1 = S^T, so these aren't exactly three independent conditions, in that it's sort of an 'over-determined' system of equations. e.g., If you assume 1), then 2) and 3) become S^* = S, which means that S is Hermitian, and represents a self-adjoint operator. So assuming one of these conditions will reduce the other two into a single condition, providing a constraint on them.
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