3
$\begingroup$

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?

$\endgroup$

2 Answers 2

3
$\begingroup$

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

$\endgroup$
7
  • $\begingroup$ 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, 2019 at 10:54
  • $\begingroup$ @Upstart Assuming the subscripts are actually 5 and 6, then yes. $\endgroup$
    – DaftWullie
    Nov 26, 2019 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, 2019 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, 2019 at 12:29
  • $\begingroup$ Should I ask it as a separate question? $\endgroup$
    – Upstart
    Nov 26, 2019 at 12:59
0
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.