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

transformation matrix

quantum circuit

$\tilde{U}$ 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?

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  • $\begingroup$ what's the difference between white circle and black circle? $\endgroup$ – user1271772 Oct 19 '18 at 18:07
  • $\begingroup$ The black circle represent a control which should be 1 to be applied. A white one would require a 0. $\endgroup$ – cnada Oct 19 '18 at 18:28
  • $\begingroup$ You already said what U does through the full matrix/ circuit. Do you want to say what is after each gate individually? $\endgroup$ – AHusain Oct 19 '18 at 19:04
  • $\begingroup$ @AHusain precisely, suppose you do not have the matrix representation, you only have the circuit, therefore you need to solve it. $\endgroup$ – Davide_sd Oct 19 '18 at 19:37
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    $\begingroup$ So you have written down the matrices for each of the three types of gates in the circuit and are having trouble with the multiplication step $U=U_1 U_2 U_3 U_2 U_1$? You can use computer algebra to do that if you're having trouble doing a bunch of 8 by 8 matrix multiplications. $\endgroup$ – AHusain Oct 19 '18 at 19:56
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$$ | \psi_3 \rangle = a | 0 1 1 \rangle + b | 1 1 1 \rangle\\ $$

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.

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