Undergrad Unitary Time Evolution: Explaining Open Quantum Systems

[jamie.j1989]

Hi, I am a bit confused about unitary time evolution, I understand that a closed quantum system can be explained by unitary time evolution which ensures that the probability of all possible outcomes is always 1. But for an open quantum system we can't in general explain it with a unitary time evolution. This is where I get confused, we explain the evolution of an open quantum systems with the density matrix $\rho(t)$, which must always satisfy $Tr[\rho(t)]=1$, so my question is, how is this not unitary time evolution? Thanks.

 

[ShayanJ]

The density matrix may refer to the state of the system+environment which we expect to have a unitary evolution. But when we trace out the environmental degrees of freedom, we are left with a density matrix which describes what we should expect if we have don't have access to the environmental degrees the freedom. So $\rho_{total}$ has a unitary evolution but $\rho_{system}=Tr_{environment}(\rho_{total})$ doesn't!

 

[Zafa Pi]

More often than not open system measurements are described by the Copenhagen/Collapse Interpretation. Collapse is not invertible let alone unitary.

 

[A. Neumaier]

Collapse preserves the condition $\mbox{Tr}~ \rho=1$. This is a much weaker condition than nitarity, which says that $\rho(t)=U(t)\rho(0)U(t)^*$ with a unitary $U(t)$.

 

[Zafa Pi]

Collapse preserves the condition $\mbox{Tr}~ \rho=1$. This is a much weaker condition than nitarity, which says that $\rho(t)=U(t)\rho(0)U(t)^*$ with a unitary $U(t)$.

Indeed, ρ becomes the projection onto the collapsed eigenvector.