Given two states $\rho_a$ and $\rho_b$, and knowing that the SWAP gate swaps two qubits, how can one prove that visibilities

$v_1=Tr[\rho_a^2 \rho_b^2] = Tr[S_{AB} S_{BC} S_{CD} \rho_a \otimes \rho_a \otimes \rho_b \otimes \rho_b ]$

$v_2=Tr[(\rho_a \rho_b)^2] = Tr[S_{BC} S_{CD} S_{AB} S_{BC} S_{AB} \rho_a \otimes \rho_a \otimes \rho_b \otimes \rho_b]$

Reference: https://arxiv.org/pdf/1501.03099.pdf


1 Answer 1


Explicitly write out what the trace is supposed to be: $$ \text{Tr}(S_{AB}S_{BC}S_{CD}\rho_a\otimes\rho_a\otimes\rho_b\otimes\rho_b)=\sum_{i,j,k,l}\langle ijkl|S_{AB}S_{BC}S_{CD}\rho_a\otimes\rho_a\otimes\rho_b\otimes\rho_b|ijkl\rangle. $$ Now you can apply the swaps to the basis elements $$ =\sum_{i,j,k,l}\langle jikl|S_{BC}S_{CD}\rho_a\otimes\rho_a\otimes\rho_b\otimes\rho_b|ijkl\rangle $$ and keep going... $$ =\sum_{i,j,k,l}\langle jkli|\rho_a\otimes\rho_a\otimes\rho_b\otimes\rho_b|ijkl\rangle $$ So, this is the same thing as $$ \sum_{ijkl}\langle j|\rho_a|i\rangle\langle i|\rho_b|l\rangle\langle l|\rho_b|k\rangle\langle k|\rho_a|j\rangle. $$ It's worth a comment about how I chose the ordering of those terms. I just started with the first index ($j$). Because it's closing index was $i$, the next term I wrote down started with $i$, and so on. Now, knowing the result I want, I'll use the commutativity of multiplication to choose a slightly different order: $$ \sum_{ijkl}\langle k|\rho_a|j\rangle\langle j|\rho_a|i\rangle\langle i|\rho_b|l\rangle\langle l|\rho_b|k\rangle. $$ Next, notice that there's lots of completeness relations appearing here: $\sum_j|j\rangle\langle j|=I$. Thus, this is the same as $$ \sum_k\langle k|\rho_aI\rho_aI\rho_bI\rho_B|k\rangle=\text{Tr}(\rho_a^2\rho_b^2). $$

The other one will work just the same.


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