# Density matrix of a product of Bell states

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?

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

• 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

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).