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I am confused about what it is exactly that a reduced density operator describes. To illustrate, I came across the following seemingly paradoxical argument.

Consider a bipartite system $AB$, described by the pure state:

$$| \psi \rangle = \sum_{a,b} \psi(a,b) |ab \rangle$$

Its density matrix is defined as:

$$\rho \equiv | \psi \rangle \langle \psi |$$

With reduced density operators:

$$\rho_A = tr_B(\rho) \qquad \rho_B = tr_A(\rho)$$

We can construct a new operator:

$$\rho_{AB} = \rho_A \otimes \rho_B$$

Which is only equal to $\rho$ if $|\psi\rangle$ is a separable state (no correlations between $A$ and $B$). Another way to say this is:

$$0 = S(\rho) \leq S(\rho_A) + S(\rho_B) = S(\rho_A \otimes \rho_B) = S(\rho_{AB})$$

Because if $A$ and $B$ are entangled, then $\rho$ contains information about these correlations, but $\rho_{AB}$ does not (is this wrong?).

So say $A$ and $B$ are entangled, then $\rho_{AB}$ does not contain that information, but I was under the impression that $\rho_A$ ánd $\rho_B$ do contain (some, all?) of this correlation information; otherwise both $A$ and $B$ should be described by pure states. I realise that $\rho_{AB}$ has no physical meaning in the case of entanglement, but even as a purely mathematical construct I don't see how that information can just vanish.

I guess my question is this: what information exactly does the reduced density operator encompass? And, if any of it is related to the correlations, how can this be reconciled with the argument in the paragraph above?

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The reduced density operator of A contains all of the information about that system you can get by measuring that system alone. It does not contain information about what systems A is entangled with. For that, you would need the density operator of set of all systems with which A is entangled. The information about correlations is also contained in A's Heisenberg picture observables, see http://xxx.lanl.gov/abs/quant-ph/9906007.

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  • $\begingroup$ I don't think I fully understand yet. Does it simply not contain information about what A is entangled with, or does it not even contain information about the fact that A is entangled with something? What about the case where A is entangled with only B, and nothing else? $\endgroup$ – Timsey May 29 '14 at 12:52
  • $\begingroup$ A's reduced density operator does not contain information about what A is entangled with. If A was not entangled with anything it would be in a pure state. But A's reduced density operator only tells you about the probabilities of measurements on A. It doesn't give you the probabilities of measurements on AB so it doesn't tell you that A is entangled specifically with B. $\endgroup$ – alanf May 29 '14 at 12:58
  • $\begingroup$ Alright, so - if I understand correctly - in this case the RDO tells us A is entangled with something, but not with what. Then, in the hypothetical situation where A and B are the only things in existence, would it be correct to conclude that A and B are entangled with eachother? If so, shouldn't this information be encoded in $\rho_{AB}$ as well? $\endgroup$ – Timsey May 29 '14 at 13:27
  • $\begingroup$ If A and B are entangled with one another and with nothing else then their joint state is a pure state not the tensor product of the RDOs of A and B. The RDOs are not supposed to contain that information. They are just a tool for doing calculations. $\endgroup$ – alanf May 29 '14 at 15:12
  • $\begingroup$ Yes, I know that, but that is precisely what seems contradictory to me. I realise the tensor product of the RDOs only makes physical sense if A and B are separable (then $\rho = \rho_{AB}$. If the separate RDOs do not contain information about entanglement, then obviously neither does $\rho_{AB}$. However, how can the RDO of A give the correct answers if A and B are entangled, and the RDO is not supposed to contain any information about entanglement? It seems contradictory that the RDO of A correctly describes A if A is entangled, when it doesn't contain any information about the entanglement. $\endgroup$ – Timsey May 29 '14 at 16:37

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