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I want to implement 3 level system on a qubit quantum computer.

Currently I feel I can use 2 qubits and encode the states of a 3-level system as such

|0>=|00>, |1>=|01>, |2>=|10>

The co-efficient of the |11> will always be zero and we will ignore it. However the gates that would be implemented on such states will be 3x3 matrices, but the system of 2 qubits is of dimension 4x4.

How to implement unitary gates ($U$) on such systems ?

One method I thought is using a direct sum $U_{3 \text{ x }3} \oplus 0_{1 \text{ x } 1}$, but the resultant matrix does not stay unitary.

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1 Answer 1

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Given a unitary $3 \times 3$ matrix $V$ that performs your desired transformation in the $|00\rangle, |01\rangle,$ and $|10\rangle$ states, the following matrix performs that transformation and leaves $|11\rangle$ intact.

$$ U = \begin{bmatrix} V & 0\\ 0 & 1 \end{bmatrix} $$

Then,

$$ \begin{align} U^*U &= \begin{bmatrix} v_{1,1}^* & v_{2,1}^* & v_{3,1}^* & 0 \\ v_{1,2}^* & v_{2,2}^* & v_{2,3}^* & 0 \\ v_{1,3}^* & v_{2,3}^* & v_{3,3}^* & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} v_{1,1} & v_{1,2} & v_{1,3} & 0 \\ v_{2, 1} & v_{2,2} & v_{2,3} & 0 \\ v_{3,1} & v_{2,3} & v_{3,3} & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} V^* & 0 \\ 0 & 1 \end{bmatrix}\begin{bmatrix} V & 0 \\ 0 & 1 \end{bmatrix} \\ &= \begin{bmatrix} V^*V & 0 \\ 0 & 1 \end{bmatrix} = I \end{align} $$ Where the last follows since we are given that $V$ is unitary. You can see the same is true for $UU^*$, so we have that $U$ is unitary.

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