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Suppose we have a quantum circuit like this. All the gates are known except for one. For any input of q[0] and q[1], we know the corresponding output. I have provided the output state for four different input state which forms a basis . How can we know that which gate is the unknown gate? enter image description here How to solve these types of problems?

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  • $\begingroup$ Please refrain from posting one question several times (quantumcomputing.stackexchange.com/questions/30027/…). Rather edit the former question. $\endgroup$ Feb 4, 2023 at 21:50
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    $\begingroup$ That question was more general one. I couldn't work out the input state in this question. The answer on this question gives me the way of how to solve the question effectively. $\endgroup$
    – wizzywizzy
    Feb 5, 2023 at 12:49

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The effect of a circuit can be explained by matrix-vector multiplications.

If you start with a state $\vec{q}$ and if the gates in your figure are represented by the matrices $A$, $B$, and $U$ (where $U$ is the unknown), then the output will be given by $\vec{q_f} = ABU\vec{q}$.

The $AB$ factor can be replaced by a single matrix called $C$, and with symbolic computation (a.k.a computer algebra) you can combine $U$ and $C$ too.

Let's try this, assuming that your "controlled note" get is just what others usually call "controlled not" or CNOT. Then we have:

$$ A = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix},\tag{1} $$

$$ B = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix},\tag{2} $$

$$ U = \begin{bmatrix} a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p \end{bmatrix},\tag{3} $$

$$ \vec{q} = \begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix},\tag{4} $$

and

$$ \vec{q}_f = \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix}.\tag{5} $$

This would mean that we have:

$$\tag{6} \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{bmatrix} \begin{bmatrix} a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p \end{bmatrix}\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}, $$ $$\tag{7} \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p \end{bmatrix}\begin{bmatrix} 0 \\ 1 \\ 0 \\ 0 \end{bmatrix}, $$

$$ \begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} f \\ b \\ j \\ n \end{bmatrix}.\tag{8} $$

You now have found 4 out of 16 unknown variables. You can do the same for the other 3 scenarios to get the other 4x3 = 12 unkown variables. You can use this matrix-vector multplier to make the computations easier.

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  • $\begingroup$ How do you form the matrix B? I guess it represents Pauli X matrix. But isn't pauli X a 2*2 matrix $\endgroup$
    – wizzywizzy
    Feb 4, 2023 at 15:38
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    $\begingroup$ B is the "left Kronecker product" of the identity matrix and the X gate, so in MATLAB/Octave it would be kron(eye(2),[0 1; 1 0 ]). $\endgroup$ Feb 4, 2023 at 15:39
  • $\begingroup$ So it's like extending X gate to a 4*4 matrix having the same effect? $\endgroup$
    – wizzywizzy
    Feb 4, 2023 at 15:42
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    $\begingroup$ Precisely. The identity matrix is happening to qubit 1, and the X matrix is happening to qubit 2. $\endgroup$ Feb 4, 2023 at 15:43
  • $\begingroup$ Thanks a lot.That clears my confusion $\endgroup$
    – wizzywizzy
    Feb 4, 2023 at 15:46

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