Suppose I prepare a Bell state $|\beta_{00}\rangle$, and distribute the product state $|\beta_{00}\rangle_{12}|\beta_{00}\rangle_{34}|\beta_{00}\rangle_{56}$ without telling them which state I prepared, with $A$ having qubits $1$ and $4$, $B$ having $2,6$, C having $3,5$.

Now I calculate the individual density matrices for $A, B, C$ which comes out to be $\dfrac{\mathbb{I}\otimes\mathbb{I}}{4}$ for each implying no one has the information about the complete state he/she has since the density matrix is completely mixed. My first question is, is this correct?

Now, say I want to calculate the entropy of the entire system, and then the entropy for each subsystem $A, B, C$. The entropy of each system can be seen to be $2$ by looking at the eigenvalues of the partial density matrices of $A, B, C$. Suppose I now tell them what state I shared, will the entropy of the entire system and the subsystem change? What if $A$ measures his particles $1,4$, then will the entropy change?

  • $\begingroup$ ''implying no one has the information about the complete state he/she has, since the density matrix is completely mixed.'' $\endgroup$ – Upstart Nov 27 '19 at 10:49
  • $\begingroup$ isn't that what you already asked here? $\endgroup$ – glS Nov 27 '19 at 10:54
  • $\begingroup$ But the answer was not what I was looking for $\endgroup$ – Upstart Nov 27 '19 at 10:56

The entropy of a system depends on your state of knowledge about it. If you know that a state is pure, then its entropy is zero. If you "tell $A,B,C$ that the overall shared state is pure", then they know as much and will, therefore, say that the entropy of the overall state is $0$ (note that this is essentially a tautological statement). However, for them measure that the overall state is indeed pure, they need to be able to do nonlocal measurements. If they can only do local measurements, then as far as they are concerned their state is indistinguishable from $\rho^A\otimes\rho^B\otimes \rho^C$.

Stated differently, you can think of the entropy of a system as a property of the way it's being measured. If $A$ measures only their state, they will associate to the measured state an entropy $S(\rho^A)$, and this regardless of whatever it's been done by the other parties. Even if you give $A$ full knowledge about the shared state, $A$ still will only be able to measure the entropy $S(\rho^A)$ with only their share of the state.

  • $\begingroup$ Okay, am I right in deducing that entropy for $A$ is $2$ as it is for others? $\endgroup$ – Upstart Nov 27 '19 at 12:24
  • $\begingroup$ the entropy of $I/d$ is indeed $\log d$ $\endgroup$ – glS Nov 27 '19 at 15:38

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