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Suppose I share two Bell states among two participants Alice and Bob in the following manner : $$ |\psi\rangle=\left(\dfrac{|0\rangle_1|0\rangle_2+ |1\rangle_1|1\rangle_2}{\sqrt{2}}\right)\left(\dfrac{|0\rangle_3|0\rangle_4+ |1\rangle_3|1\rangle_4}{\sqrt{2}}\right) $$ Now suppose Alice has qubits $(1,4)$ and Bob has $(2,3)$. I want to find out the density matrices corresponding to Alice, Bob, and combined.

For the first case should I calculate $|\psi\rangle\langle\psi|$, what should be done, in case there was only one Bell pair shared between Alice and Bob, I would have done $$ \rho_A = \mathrm{Trace}_B(\rho)$$ can this be generalized when there are more than one Bell pair shared in the sense that I have shared? Can somebody help?

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Yes, the overall density matrix shared between Alice and Bob is $|\psi\rangle\langle\psi|$. To get the desnity matrix of either Alice or Bob, you should calculate $$ \text{Tr}_B|\psi\rangle\langle\psi|\qquad\text{Tr}_A|\psi\rangle\langle\psi| $$ respectively.

However, in this particular case, the calculation is much simply. Let $|\phi\rangle$ be the Bell pair such that $$ |\psi\rangle=|\phi_{12}\rangle|\phi_{34}\rangle. $$ Because there's a separable partition between (1,2) and (3,4), this is not changed by the partial trace. Thus $$ \text{Tr}_B|\psi\rangle\langle\psi|=\left(\text{Tr}_2|\phi\rangle\langle\phi|\right)\otimes \left(\text{Tr}_3|\phi\rangle\langle\phi|\right). $$

You imply that you know how to do the partial trace for a single Bell state. The answer is $I/2$. So, we have $$ \text{Tr}_B|\psi\rangle\langle\psi|=\frac{1}{4}I\otimes I, $$ the maximally mixed state of two qubits. Similarly, $$ \text{Tr}_A|\psi\rangle\langle\psi|=\left(\text{Tr}_1|\phi\rangle\langle\phi|\right)\otimes \left(\text{Tr}_4|\phi\rangle\langle\phi|\right)=\frac{1}{4}I\otimes I $$

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  • $\begingroup$ What if we had another bell state $\dfrac{|0\rangle_3|0\rangle_4+|1\rangle_3|1\rangle_4}{\sqrt{2}}$ with the previous ones. That should change the combined density matrix, but the individual matrices would be $\dfrac{I\otimes I\otimes I}{8}$? $\endgroup$ – Upstart Nov 26 '19 at 10:54
  • $\begingroup$ @Upstart Assuming the subscripts are actually 5 and 6, then yes. $\endgroup$ – DaftWullie Nov 26 '19 at 11:52
  • $\begingroup$ But, when we have the third state $\dfrac{|0\rangle_5|0\rangle_6+ |1\rangle_5|1\rangle_6}{\sqrt{2}}$ with the $5$th and $6th$ qubit with say Charlie, then if we wanted $\rho_A$, then that would imply partial trace over $2, 3, 5,6$ $\endgroup$ – Upstart Nov 26 '19 at 12:22
  • $\begingroup$ @Upstart True. This is why we don't do follow-up questions in comments - there's not enough space to be explicit enough about the assumptions. I was assuming you meant that Alice would have one of 5 or 6 and Bob would have the other. $\endgroup$ – DaftWullie Nov 26 '19 at 12:29
  • $\begingroup$ Should I ask it as a separate question? $\endgroup$ – Upstart Nov 26 '19 at 12:59
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One can perhaps guess the answer without full calculation. Noting that "tracing" intuitively means losing information, then, if you A is maximally entangled with B, then you lose information about B (or A) then you end up with no information about A. That is basically how qubits lose information to the environment (here B is like an environment for A).

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