# The expectation values for the values of both qubits [closed]

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 $$p_{00}+p_{01} - p_{10}.$$