Suppose we have two qubits in the state (a|0⟩+b|1⟩)|0⟩, to which we apply the following circuit.

(a) Is the resulting state entangled?

Yes

No
(b) Specify the state of each qubit after we applied the circuit.

what circuit?

To determine whether the resulting state is entangled, we need to first apply the circuit to the initial state (a|0⟩+b|1⟩)|0⟩ and see if it can be expressed as a product state.

The circuit description is missing from the question, so we cannot provide a specific answer without knowing the gates and their sequence involved. However, let's assume that the circuit consists of some gates acting on both qubits.

To find the resulting state of each qubit after applying the circuit, we need to apply the gates individually to each qubit while keeping track of the transformations.

Let's suppose the first qubit undergoes a sequence of gates with the starting state (a|0⟩+b|1⟩). We can apply each gate's matrix operation to the initial state to get the final state of the first qubit.

Similarly, assuming the second qubit undergoes a sequence of gates with the starting state |0⟩, we can apply the matrix operations of each gate in the correct order to get the final state of the second qubit.

After finding the final state of each qubit, we can write down the resulting state of the two-qubit system as the tensor product of the final states of the individual qubits.

If this resulting state can be expressed as a product state (a⨂b)|00⟩, then the two qubits are not entangled. However, if the resulting state cannot be factored as a product state, then the two qubits are entangled.

In summary, to answer the question:

(a) To determine if the resulting state is entangled, apply the circuit to the given initial state and check if the resulting state can be expressed as a product state.

(b) Analyze the circuit and apply the gates in the correct order to find the final state of each qubit. Then, express the resulting state of the two-qubit system as the tensor product of the final states of the individual qubits.