- #1

Frank Castle

- 580

- 23

*pure state*and can be represented by a vector (ket), ##\lvert\psi\rangle## in some Hilbert space. I also get that if we do not have complete information about the system, then we are not able to specify the exact state of the system and must therefore resort to an ensemble of pure states. There is a probability associated with each pure state in the ensemble which quantifies the probability that our system is in that specific state. Such systems are said to be in a

*mixed state*and are described in terms of a density matrix which contains all the information about the possible pure states that it could be in and their associated probabilities.

What confuses me is what happens when one brings the concept of superposition into the picture. The Schrodinger equation is linear and as such any linear combination of states is a solution to it. How does this differ from the mixed state case?

Is it simply that one can project a pure state ##\lvert\psi\rangle## onto a particular eigenbasis of an operator (representing one of its observables), and as such, is represented as a linear combination of these eigenbasis vectors. Now, as we have complete information of the system, the coefficients of this expansion correspond to the probabilities that the state with be in a given eigenstate of the the associated operator.

Physically, this corresponds to the fact that before we have made a measurement of the system, we cannot not a priori definitively say that the state is in one eigenstate or another of the given observable. Hence, we must consider the state to be in a superposition of all the possible eigenstates simultaneously, and once a measurement has been made, the system will "collapse" into a given eigenstate.

If this is correct, then I can see how the mixed state formalism differs from the pure state formalism, but otherwise I'm left quite confused.