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I have a quantum state described by $N$ qubits, and I don't know anything about this quantum state except the expectation value of the single-qubit Pauli observables of the $i$-th qubit ($\langle X_i \rangle$, $\langle Y_i \rangle$ and $\langle Z_i \rangle$). Then I apply a single-qubit quantum gate $U$ to the $i$-th qubit. I know the $2\times 2$ matrix describing this quantum gate.

Is it possible to determine $\langle X_i \rangle^f$, $\langle Y_i \rangle^f$ and $\langle Z_i \rangle^f$ after the application of $U$ by knowing $\langle X_i \rangle$, $\langle Y_i \rangle$, $\langle Z_i \rangle$ and the elements of $U$?

Could it be useful to find a way to do that?

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$\langle X\rangle=\text{Tr}(X\rho)$. So, if we update by a single qubit unitary, $\rho\mapsto U\rho U^\dagger$, we have that $$ \langle X\rangle\mapsto \text{Tr}(XU\rho U^\dagger)=\text{Tr}(U^\dagger XU\rho). $$ Now, $$ U^\dagger XU=n_XX+n_YY+n_ZZ. $$ (Basically, the Pauli matrices with identity form a basis, but since $U^\dagger XU$ is traceless, it has no identity component.) You should be able to calculate the values $n_X,n_Y,n_Z$. This means that $$ \langle X\rangle\mapsto n_X\langle X\rangle+n_Y\langle Y\rangle+n_Z\langle Z\rangle $$

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  • $\begingroup$ Thank you, I think this is the answer I was looking for. Con you please further explain the equivalence of the two traces and the part of the pauli basis (why $U^{\dagger}XU=n_XX+n_YY+n_ZZ$)? Thanks again. $\endgroup$ Commented Sep 7, 2023 at 15:30
  • $\begingroup$ trace is invariant under permutations, so if you have a product of operations inside the trace, you can pick up an operator from one end of the product and stick it at the other end. $\endgroup$ Commented Sep 7, 2023 at 15:38
  • $\begingroup$ Any $2\times 2$ matrix can be written as $aI+n_XX+n_YY+n_ZZ$ (just write out the matrix, and see that you can independently solve for all 4 matrix elements). However, $\text{Tr}(U^\dagger XU)=\text{Tr}(XUU^\dagger)=\text{Tr}(X)=0$, while $\text{Tr}(aI+n_XX+n_YY+n_ZZ)=2a$ so $a=0$. $\endgroup$ Commented Sep 7, 2023 at 15:40

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