1. Oct 3, 2012

### HomogenousCow

I am confused about mixed quantum states, if the only observable states are pure eigenstates, since we have to measure to observe, what is the physical meaning of a mixed state?

2. Oct 3, 2012

### cgk

Letting alone the measurement only on pure states'' issue: Mixed states are usually used in two situations: (i) to represent statistical ensembles, where we do not know the microscopic details of the involved states, and (ii) to represent open subsystems of larger quantum systems.

On (i): Say you have an ideal gas with a given set of thermodynamic properties (N, T, V, etc). You will typically not be able or willing to write a pure state for the entire gas, because there are too many particles and the actual coordinates of the individual atoms/molecules do not really matter. In this case you can set up a ensemble state (which is a mixed state) which averages over all realizations of the macroscopic properties you have (e.g., N, T, V) under the side condition that you do not impose anything in the ensemble which you do not know (that is, take the ensemble which maximizes the (information-)Entropy of the ensemble). After doing that, you can derive all processes which only involve the thermodynamic quantities using the ensemble alone.

(ii) If you are only interested in a sub-system of a larger system, then you can trace out the environment degrees of freedom and obtain an embedded subsystem, involving only the degrees of freedom you are interested in. This embedded subsystem will normally *not* be in a pure state, even if the full system it is created from (by reduction) was in a pure state.

3. Oct 3, 2012

### Dickfore

Think of a beam of atoms coming from an oven. Their spin is unpolarized, meaning there is a 50-50 chance of getting spin up or spin down in any direction for a set of measurements performed on a set of atoms.

Could you construct a state vector with the afforementioned property?