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Suppose that we have two entangled particles A and B with pure state vector

$|\psi\rangle=a|0\rangle_A |1\rangle_B + b|1\rangle_A |0\rangle_B \hspace{1cm}(1)$

When we take the partial trace over the degrees of freedom of one of the two entangled particles, lets say B, we get the reduced density matrix for the A, which is identical to a mixed state density matrix. But is the state of the particle A a real mixed state? i.e. is its state some definite state, $|0\rangle_A$ or $|1\rangle_A$, but we simply don't know which one? Wouldn't that mean that the whole system is in $|0\rangle_A |1\rangle_B$ or $|1\rangle_A |0\rangle_B$, and not in the superposition (1)?

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  • $\begingroup$ Why would tracing over $B$ states affect the coordinates of $|\psi\rangle$ in the $A$ basis? The way I see it, the $|\tilde\psi\rangle$ you get after taking the partial trace will still be a superposition of $|0\rangle_A$ and $|1\rangle_A$. $\endgroup$ – delete000 Mar 6 '13 at 12:45
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    $\begingroup$ For measurements relating only to system A, the state is indistinguishable from a mixed state. Because there is another system, B, which you are making inaccesible via the partial trace, you cannot say whether A is correlated with B or not. So by measurements on A only you cannot tell whether the system is in a pure or mixed state. $\endgroup$ – Michael Mar 6 '13 at 13:12
  • $\begingroup$ @delete000 The two systems are entangled, that's why. :) Try working out the reduced density matrix for A. $\endgroup$ – Michael Mar 6 '13 at 13:14
  • $\begingroup$ @Michael Brown The particle A (or B) is certainly not in a pure state (there is no pure wavefunction for it) and this can be verified experientially (e.g. by measuring spin in different directions). But also in some books I read that it's not correct to suppose that the particle alone is in mixed state. So what is it then? $\endgroup$ – Andreas K. Mar 6 '13 at 13:21
  • $\begingroup$ @Andyk I meant pure or mixed in the full Hilbert space (including system B, on which you are not doing measurements). Of course you're right that by repeated measurements you could rule out a pure state in the reduced Hilbert space of A only. Sorry, I should have been more clear. $\endgroup$ – Michael Mar 6 '13 at 13:23
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But is the state of the particle A a real mixed state?

That is not the right question to ask. The question should be: how does any particular observer describe system A. An observer who doesn't know about the entanglement (cut off from B) will describe A as a mixed state, while an observer who knows about the entanglement with B will describe them as entangled. As per conventional understanding, the whole universe should be in a pure state and the appearance of a mixed state anywhere means that the state is entangled with something we don't know about. So in essence, all mixed states arise because there is some correlation/entanglement/information an observer doesn't know about. There isn't any notion of a "real" mixed state.

is its state some definite state, |0⟩A or |1⟩A, but we simply don't know which one?

No, it is not described by a definite state, or any pure state which is a linear combination of the basis states. It's described by a "mixed state", which is specified by its density matrix -- which tells you with what probability a projection gives $|0\rangle$ or $|1\rangle$. That's all the information you know about a mixed state: its density matrix.

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When you have a system in the state $|\psi\rangle = a|01\rangle+b|10\rangle$ and take the partial trace over the subsystem B, you will find subsystem A in a mixed state as you say. When you measure the state of the subsystem A, not caring about the state of subsystem B, then again, you will find out that it is a mixed state. But you cannot say that the system A is in a mixed state before the measurement. It is in a pure entangled state. Locally, the state of each subsystem appears mixed but that is only because you do not have the complete information about the state. For that, you need the complete composite system, i.e., both subsystems A and B and perform a joint measurement on them both.

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  • $\begingroup$ So A alone isn't in the mixed state: $|0\rangle$ with probability $|a|^2$ or $|1\rangle$ with probability $|b|^2$, and it isn't in the pure state $a|0\rangle+b|1\rangle$. However its (reduced) density matrix is identical to the mixed density matrix. But density matrix is only a calculation tool for measurements and doesn't define the state of A. Right? $\endgroup$ – Andreas K. Mar 7 '13 at 8:01
  • $\begingroup$ So we can only say that A is in entangled state with B and nothing more. $\endgroup$ – Andreas K. Mar 7 '13 at 8:08
  • $\begingroup$ Yes, you get it right. We can't really much about a system if it is entangled with other system. From the local point of view, it looks like it's in a mixed state but that is not the correct description of the state. $\endgroup$ – Ondřej Černotík Mar 7 '13 at 13:39

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