2
$\begingroup$

I'm reading this paper while the author states in the eq(A1) that, for a $2n$ qubits maximally entangled state $|\Psi ^+\rangle \langle \Psi ^+|$, we can write it with Pauli operators $P_u\in\left\{ I,X,Y,Z \right\} ^{\otimes n}$ as $$ |\Psi ^+\rangle \langle \Psi ^+|=\frac{1}{4^n}\sum_u{P_u\otimes P_{u}^{T}} \tag 1. $$ I can verify this point for two qubits case. The dentisy matrix of the two qubits maximally entangled state is $$ |\Psi ^+\rangle \langle \Psi ^+| =\frac{1}{\sqrt{2}}\left( |00\rangle +|11\rangle \right) \frac{1}{\sqrt{2}}\left( \langle 00|+\langle 11| \right) =\frac{1}{2}\left( \begin{matrix} 1& 0& 0& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 0\\ 1& 0& 0& 1\\ \end{matrix} \right). \tag 2 $$ It's easy to verify that eq(2) can also be rewritten as the following $$ \frac{1}{4}\left( I\otimes I+X\otimes X+Z\otimes Z+Y\otimes Y^T \right) =\frac{1}{4}\left( I\otimes I+X\otimes X+Z\otimes Z-Y\otimes Y \right) \tag 3 $$ which corresponds to the right-hand side of eq(1).

My question is, how can we rigorously show the general case, i.e. eq(1)?

$\endgroup$

1 Answer 1

2
$\begingroup$

Once you've shown it for $n=1$ it follows for any $n$. This is because you can rearrange the expression for a $2n$-qubit maximally entangled state between a pair of subsystems $A$ and $B$ into a tensor product of $n$-many 2-qubit maximally entangled states on those subsystems: \begin{align} |\Psi^+_{2n}\rangle_{AB}&:=\sum_{i \in \{0,1\}^n} |i\rangle_{A}\otimes |i\rangle_{B} \tag{1} \\&=\sum_{i_1,\dots, i_n \in \{0,1\}} |i_1\dots i_n\rangle_{A}\otimes |i_1\dots i_n\rangle_{B} \tag{2} \\&=\sum_{i_1,\dots, i_n \in \{0,1\}} |i_1\rangle_A |i_1\rangle_B \otimes \cdots \otimes|i_n\rangle_A |i_n\rangle_B \tag{3} \\&= \left( \sum_{i_1 \in \{0,1\}} |i_1\rangle_A |i_1\rangle_B \right)\otimes \cdots\otimes\left( \sum_{i_n \in \{0,1\}} |i_n\rangle_A |i_n\rangle_B \right) \tag{4} \\&= |\Psi^+_2\rangle_{AB}\otimes \cdots \otimes |\Psi^+_2\rangle_{AB}.\tag{5} \end{align} Here, $|\Psi^+_2\rangle$ is the two-qubit maximally entangled state you chose in your initial question. Then, use your expression for this state in the Pauli basis and afterwards put the subsystems back into their original order: \begin{align} |\Psi^+_{2n}\rangle\langle \Psi^+_{2n}|_{AB} &= |\Psi^+_2\rangle \langle \Psi^+_2|_{AB}\otimes \cdots \otimes |\Psi^+_2\rangle \langle \Psi^+_2|_{AB} \tag{6} \\&= \left( \frac{1}{4} \sum_{P_1 \in \{I,X,Y,Z\}} P_1^A\otimes (P_1^B)^T \right) \otimes \cdots \otimes \left( \frac{1}{4} \sum_{P_n \in \{I,X,Y,Z\}} P_n^A\otimes (P_n^B)^T \tag{7} \right) \\&= \frac{1}{4^n} \sum_{P_1,\dots,P_n \in \{I,X,Y,Z\}} P_1^A\otimes (P_1^B)^T \otimes \cdots \otimes P_n^A\otimes (P_n^B)^T \tag{8} \\&= \frac{1}{4^n} \sum_{P_1,\dots,P_n \in \{I,X,Y,Z\}} \left[P_1^A\otimes \cdots P_n^A\right] \otimes \left[P_1^B \otimes \cdots \otimes P_n^B\right]^T \tag{9} \\&= \frac{1}{4^n} \sum_{P \in \{I,X,Y,Z\}^n} P\otimes P^T, \tag{10} \end{align} where the superscripts denote in which subsystem each Pauli is acting (i.e. some of the kronecker products are redundant/unnecessary and there is probably a cleaner notation to make this point).

$\endgroup$

Your Answer

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

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