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I am trying to understand pure and mixed states better. If I have N quantum particles in an isolated system. The many-particle state is a superposition of the product of single-particle states by the appropriate statistics (bosons, fermions, or distinguishable). Would this state be still considered pure since there is no interaction with the environment?

Does that mean isolated systems or microcanonical ensembles are always pure?

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In the case of an isolated system (more precisely, a system that is not entangled with another system), whether a pure or a mixed state is assigned depends only on the state of knowledge of the observer. In principle, the system will be in a pure state. However, if the observer is in any way uncertain about which pure state the system is in, the state is mixed. If you are doing statistical mechanics, that means you have incomplete knowledge about the state of the system and the state will probably be mixed.

Let's take your example of the microcanonical ensemble. This describes a situation where the observer has knowledge of some macroscopic constraints, namely that the system is isolated and that it has total energy $E$, but knows nothing else. Then the correct procedure is to assign equal probability to all microscopic states $|\psi_i\rangle$ that have energy $E$ (equal a priori probabilities). The state of the system relative to the observer is therefore $$ \rho = \sum_{i=1}^N \frac{1}{N} |\psi_i\rangle\langle\psi_i|, $$ where I assumed that there are $N$ states which all have the same energy expectation value $E$. This state is clearly a mixed one. Nevertheless, the system is actually in just one of these (pure) states. The appearance of a mixed state in the description merely reflects the classical uncertainty that the observer has about the system.

With entanglement the situation becomes more complicated. If a system $A$ interacts strongly with another system $B$, the total state will be entangled in general. Obviously this cannot occur in an isolated system, but I will describe what happens here to give some contrast. An entangled state cannot be written as a simple tensor product: $$ |\psi_{AB}\rangle \neq |\phi_A\rangle\otimes|\chi_B\rangle. $$ Instead, an entangled state takes the form $$ |\psi_{AB}\rangle = \sum\limits_i \lambda_i |\phi_{A,i}\rangle\otimes|\chi_{B,i}\rangle. $$ The state of system $A$ alone is obtained by tracing out system $B$: $$ \rho_A = \mathrm{Tr}_B|\psi_{AB}\rangle\langle\psi_{AB}| = \sum\limits_i |\lambda_i|^2 |\phi_{A,i}\rangle\langle\phi_{A,i}|, $$ using the orthonormality of the $B$ states: $\langle\chi_{B,i}|\chi_{B,j}\rangle = \delta_{i,j}.$

So here we necessarily obtain a mixed state due to the entanglement between systems $A$ and $B$. It is important to note that the uncertainty represented by the mixed state is not dependent on the observer's state of knowledge. Rather, it corresponds to the unavoidable quantum uncertainty embodied by the commutation relations.

EDIT: However, as Peter Shor points out, unless you have access to system $B$, the outcomes of your measurements on the mixed state $\rho_A$ are identical to those you would obtain from the pure state $$ |\psi_A\rangle = \sum\limits_i\lambda_i |\phi_{A,i}\rangle. $$ You can only tell if $A$ is entangled by comparing measurement outcomes on $A$ with the measurement outcomes on $B$, in which case you will see Bell inequality-violating correlations that would not be present otherwise.

So in short, the state of an isolated system is not necessarily mixed, unlike the state of a system that interacts with and is thereby entangled with another system. However, when the observer's knowledge of the system is incomplete, which is the case when statistical mechanics is applicable, the state of an isolated system will indeed be mixed. The only exception to this last sentence that I can think of right now is at zero temperature with no ground state degeneracy, in which case a statistical mechanics system may still be in a unique pure state: the ground state.

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Thanks Mark! A further clarification. I want to understand the difference between classical uncertainty and quantum uncertainty in the following isolated case. Lets say I have N bosons in a specific energy state. The multi-particle state is a permutation of single-particle product states. This implies there is degeneracy. I think this is inherent uncertainty so any of these states although a linear superposition of basis product states is still a pure state. Am I right? So classical uncertainty comes into play only when I don't know the expected energy of a state but only the total energy? –  Sankaran Jan 15 '13 at 16:46
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One point that should be made is that if you don't have access to the system $B$ that system $A$ is entangled with, there is no way to tell a pure state on $A$ from a mixed state on $A$. –  Peter Shor Jan 15 '13 at 16:51
    
@Sankaran You do indeed describe a system in a pure state. Any linear superposition of basis kets is again a pure state. However, there is no degeneracy in the usual sense, because all of the states that appear in the sum are really the same state. You have to put in the sum 'by hand', as it were, in order to enforce the permutation symmetry that is broken when you choose an arbitrary ordering of your indistinguishable particles (which you must do in order to write down the state). –  Mark Mitchison Jan 15 '13 at 17:15
    
(contd.) This is why the Fock space 'occupation number' representation is generally preferable in a many-body system; it makes the permutation (anti)symmetry manifest. I am not quite sure what you mean by the last sentence of you comment. I would phrase it thus: The classical uncertainty comes into play when you are ignorant about the state (or equivalently, ignorant about the microscopic details of how the system was prepared), but you know the total energy. –  Mark Mitchison Jan 15 '13 at 17:16
    
@PeterShor Good point, thanks. Will add a comment on that. –  Mark Mitchison Jan 15 '13 at 17:17
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