Density matrix of a product of Bell states

Question

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?

Answer 1

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

Answer 2

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).