# Computing the evolution of expectation values of the Pauli observable

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?

$$\langle X\rangle=\text{Tr}(X\rho)$$. 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 $$\langle X\rangle\mapsto n_X\langle X\rangle+n_Y\langle Y\rangle+n_Z\langle Z\rangle$$
• 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. Commented Sep 7, 2023 at 15:30
• 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 $a=0$. Commented Sep 7, 2023 at 15:40