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Consider the linear adiabatic evolution path taking us from a single qubit Hamiltonian Hinitial=(I+σx)/2{“version”:”1.1″,”math”:”\(H_{initial} = (I+\sigma_x)/2\) “} at time t=0{“version”:”1.1″,”math”:”\(t=0\)”} to a final Hamiltonian Hfinal=(I−σz)/2{“version”:”1.1″,”math”:”\(H_{final} = (I-\sigma_z)/2\)”} at time t=tf{“version”:”1.1″,”math”:”\(t=t_f\)”} such that H(t)=(t/tf)Hfinal+(1−t/tf)Hinitial{“version”:”1.1″,”math”:”\(H(t) = (t/t_f) H_{final} + (1- t/t_f) H_{initial}\)”} for 0≤t≤tf{“version”:”1.1″,”math”:”\(0 \leq t \leq t_f\)”}.  What are the ground states of the qubit at the initial and final times, i.e. at t=0{“version”:”1.1″,”math”:”\(t=0\)”} and t=tf{“version”:”1.1″,”math”:”\(t=t_f\)”}?

In the following circuit, what is the probability for the measurement outcome of Qubit 1 to be in state |1⟩? Here, the states are expressed in the convention |Qubit2⊗Qubit1⟩{“version”:”1.1″,”math”:”\(\vert Qubit_2 \otimes Qubit_1 \rangle\)”}.

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