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I ran across this statement in a professor's notes and I think it's just a typo, but I wanted to take the opportunity to check my understanding. So in his notes it says:

even if we have complete knowledge of quantum systems, they still can be in the[a] mixed state

As far as I understand, a mixed state is simply a classical mixture of quantum states. If we inherit the definition of maximum information from the von neumann entropy, then we should define maximum information to correspond to an entropy of ln(1) = 0. But since the density matrix is not a rank 1 projector in a mixed state, this condition is not met. Is there some other notion of maximum information that I am not aware of, do I have some misunderstanding, or is this just a typo?

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Your reasoning is correct about Von Neumann entropy, so your professor 's statement seems strange. Now, all is in the exact meaning of the words "complete knowledge". For instance, if you look at an entangled pure system, you have a complete knowledge of the (whole) system, but you have an incomplete knowledge of the sub-systems... – Trimok Dec 7 '13 at 13:28
Yeah I think you're right, thanks. That's almost certainly what he meant. – Bobak Hashemi Dec 7 '13 at 13:47
up vote 2 down vote accepted

The other thing that I can think of, is when you are not interested in some parts of your system(i.e. environment), so you trace it out. Now if the environment is not separable from the rest of the system, which is usually the case; what you are left with(the reduced state) is a mixed state.

Note that in this case:

$$\rho_{AB}\ne \text{Tr}_B(\rho_{AB})\otimes \text{Tr}_A(\rho_{AB})$$

What might be also interesting, is that the reverse of this procedure also works. Namely, the purification (which means that every mixed state acting on finite dimensional Hilbert spaces can be viewed as the reduced state of some pure state) holds.

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yes I agree he must have meant reduced states, that's a fair way to put it I suppose, maybe not the best for pedagogy. – Bobak Hashemi Dec 7 '13 at 13:44

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