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Let’s consider the two-qubit state |Ψ⟩ =(1/2)|00⟩ + i(√3/4)|01⟩ +(3/4)|10⟩. a) Find the expectation values for the values of both qubits separately.

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You can find the single-qubit expectation values by tracing out the degrees of freedom of the other qubit. Mathematically, this corresponds to $\rho_A = \text{Tr}_B(|{\psi}_{AB}\rangle\langle{\psi_{AB}}|)$, if for a two-qubit system you want to know what happens to qubit $A$ only. The diagonal of the reduced density matrix gives you the new probabilities, so that $\text{Tr}(Z\rho_A)$ yields the expectation value of qubit $A$.

Another way to calculate this is to partition the global measurement outcomes, here the state $|{00}\rangle$ with probability $p_{00} = \frac{1}{4}$, state $|{01}\rangle$ with probability $p_{01} = \frac{3}{16}$ and state $|{10}\rangle$ with probability $p_{10} = \frac{9}{16}$, and add the probabilities of the states that have your qubit of interest in a certain state, ignoring whatever state the other qubit is in. For example, the probability that qubit 1 is in the 0-state is given by $p_{00}+p_{01}$, while the probability it is in the 1-state is given by $p_{10}$. In this case, the expectation value of qubit $A$ is $p_{00}+p_{01} - p_{10}.$

The two methods are equivalent.

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