# Why does the fact that all quantum systems are open mean that no quantum state can be pure

I am teaching myself about open quantum systems and I am confused by the following statement on the wikipedia page about open quantum systems:

"The fact that every quantum system has some degree of openness also means that no quantum system can ever be in a pure state. A pure state is unitary equivalent to a zero-temperature ground state, forbidden by the third law of thermodynamics."

I was wondering if someone could perhaps elaborate on this statement? If I could see some mathematical justification for this I feel I would be able to better understand what it means. Thanks in advance for any help.

This is essentially saying that as all systems interact with their environment to some degree, there will always be some degree of uncertainty about the state of the system and some degree of entanglement between the system and the environment

• If I could see some mathematical justification for this I feel I would be able to better understand what it means. - I like your answer, but perhaps it is worth to add one or two lines of math (e.g. that due to the interaction, the overall state is likely no product vector, thus the reduced density matrix is mixed, i.e. non-pure)... Commented Apr 13, 2023 at 15:58
• So are we essentially saying that the state of the subsystem will always be entangled with the state of the environment due to the interaction? I understand that entangled systems lead to subsystems that aren't pure states, but it feels like this shouldn't necessarily be the case that all quantum numbers will be effected by the interaction with the environment Commented Apr 14, 2023 at 15:07
• For example, as far as I'm aware, the spin of a neutrino cannot change by interaction with the environment, because the neutrino doesn't interact with E.M fields. If I am correct in my thinking, then as long as we have a neutrino in our subsystem we can express the spin of the neutrino as a pure state on the Hilbert space spanned by the neutrino's spin Commented Apr 14, 2023 at 15:10
• (sorry for the longwinded response,) Commented Apr 14, 2023 at 15:10

I am not sure if this is sufficient, but an example is the Coulomb-only hydrogen atom, which we all know well. The energy eigenstates (with $$E_n$$), are pure states, and their only possible time evolution is:

$$\psi_n(t) = \psi_n(0)e^{-iE_n/\hbar}$$

which means $$||\psi_n||^2$$ never changes, and $$n$$ never changes.

Of course, the Coulomb hamiltonian has only a static electric field. When the EM field is treated as a dynamic field, there are interaction terms. The $$\psi_n$$ are no-longer exact solutions of the Hamiltonian, but since $$\alpha \ll 1$$, we can treat them approximately as such in order to compute transition probabilities and energy shifts.

Per your question, you can consider the Coulomb atom as open because it sits in a bath of dynamic EM field, and thus the state can spontaneously change.

Perhaps generalizing this idea to many atoms in a box, at some non-zero temperature, such that no atom is guaranteed to be in the ground state may elucidate the principle.