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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?

Sanchayan Dutta
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Upstart
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2 Answers2

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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, $$ \text{Tr}_A|\psi\rangle\langle\psi|=\left(\text{Tr}_1|\phi\rangle\langle\phi|\right)\otimes \left(\text{Tr}_4|\phi\rangle\langle\phi|\right)=\frac{1}{4}I\otimes I $$

DaftWullie
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  • 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 $\dfrac{I\otimes I\otimes I}{8}$? – Upstart Nov 26 '19 at 10:54
  • @Upstart Assuming the subscripts are actually 5 and 6, then yes. – DaftWullie Nov 26 '19 at 11:52
  • 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 $2, 3, 5,6$ – Upstart Nov 26 '19 at 12:22
  • @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. – DaftWullie Nov 26 '19 at 12:29
  • Should I ask it as a separate question? – Upstart Nov 26 '19 at 12:59
  • @Upstart If you're still unclear what the answer should be, then yes. On the other hand, if it is simply that what you really wanted to see was the mechanical calculation of the reduced density matrix that you could apply as a general method, and that my shortcut for this specific case is not insightful enough, you might just clarify the current question and remove the "tick" from my my answer (I won't be offended, even if some on SE view this as bad form) – DaftWullie Nov 26 '19 at 13:54
  • No sir I actually understood your answer. I have asked another question https://quantumcomputing.stackexchange.com/questions/8953/density-matrix-inferences – Upstart Nov 26 '19 at 14:05
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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).

user185597
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