Solving a circuit implementing a two-level unitary operation

Question

The circuit below implements the following two-level unitary transformation:

$\tilde{U}$ is a unitary matrix: $\tilde{U} = \left[\begin{matrix} a & c \\ b & d \end{matrix}\right]$

where $a, b, c, d$ are any complex numbers.

As we can see, $U$ acts non-trivially only on the states $\lvert 000 \rangle, \lvert 111 \rangle$.

How would you solve the circuit for the input state $\lvert 000 \rangle$ or $\lvert 111 \rangle$? My problem is figuring out how to deal with the state $\tilde{U}A$ in last two CNOT gates.

EDIT: to clarify what I want:

• I start with $\lvert \psi_{0} \rangle = \lvert 0, 0, 0 \rangle$
• after the first CNOT I get $\lvert \psi_{1} \rangle = \lvert 0, 0, 1 \rangle$
• after the second CNOT I get $\lvert \psi_{2} \rangle = \lvert 0, 1, 1 \rangle$
• ...

How would you write $\lvert \psi_{3} \rangle$, $\lvert \psi_{4} \rangle$, $\lvert \psi_{5} \rangle$? For this specific circuit, is it even possible to write the full steps like that?

Answer 1

$$ | \psi_3 \rangle = a | 0 1 1 \rangle + b | 1 1 1 \rangle\\ $$

Because the 1 on B and C criterion is met.

$$ | \psi_4 \rangle = a | 0 0 1 \rangle + b | 1 1 1 \rangle\\ $$

Because only the first term meets the criterion for the controls so it is the only part affected to flip the B index.

$$ | \psi_5 \rangle = a | 0 0 0 \rangle + b | 1 1 1 \rangle\\ $$

Because only the first term meets the criterion for the controls so it is the only part affected to flip the C index.