2
$\begingroup$

OPENQASM2.0 has only one two-qubit gate: controlled not. For a teleportation experiment, I need to perform a measurement in the Bell basis. That is, I need a two-qubit gate with matrix representation

$$\begin{bmatrix} 0&1&1&0\\ 0&-1&1&0\\ 1&0&0&1\\ 1&0&0&-1 \end{bmatrix}/\sqrt{2}.$$

To use this library, I need to decompose this gate into a combination of CNOTs and elementary single-qubit gates such as X,Y,Z, etc.

I don't expect the person answering this question to give the decomposition, but hope that they can point me to a helpful resource. I am familiar with linear algebra and have tools such as Mathematica.

$\endgroup$
1
  • $\begingroup$ If you know how to create a Bell pair from any of the four states |00>, |01>, |10>, |11> using H and CNOT gates, then running the circuit backward (H and CNOT are self-inverse) will map any on of the 4 Bell states to the 4 computational basis states. Measure them and this will have implemented the Bell state measurement. $\endgroup$
    – holl
    Dec 13, 2021 at 15:52

1 Answer 1

2
$\begingroup$

The key observation is that swapping the first and third column yields

$$ \frac{1}{\sqrt2}\begin{bmatrix} 0 & 1 & 1 & 0 \\ 0 & -1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & -1 \end{bmatrix} \to \frac{1}{\sqrt2}\begin{bmatrix} 1 & 1 & 0 & 0 \\ 1 & -1 & 0 & 0 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 1 & -1 \end{bmatrix} = I\otimes H\tag1 $$

where $H=\frac{1}{\sqrt2}\begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix}$ is the Hadamard gate. The coefficients $\frac{1}{\sqrt2}$ are needed to ensure unitarity. The column swap corresponds to the permutation matrix

$$ P = \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}\tag2 $$

which maps $|00\rangle$ to $|10\rangle$ and $|10\rangle$ to $|00\rangle$. In other words, it flips the first qubit when the second qubit is in the $|0\rangle$ state. This is the usual controlled-NOT gate with two modifications: the first qubit is the target and the second is the control and the gate acts non-trivially if the control qubit is in the $|0\rangle$ state rather than the usual $|1\rangle$. Thus,

$$ P = (I\otimes X)\circ\text{CNOT}_{2,1}\circ(I\otimes X).\tag3 $$

Putting it all together, we get

$$ \frac{1}{\sqrt2}\begin{bmatrix} 0 & 1 & 1 & 0 \\ 0 & -1 & 1 & 0 \\ 1 & 0 & 0 & 1 \\ 1 & 0 & 0 & -1 \end{bmatrix} = (I\otimes H) \circ (I\otimes X)\circ\text{CNOT}_{2,1}\circ(I\otimes X)\tag4 $$

where the last expression is written in the linear algebraic convention, i.e. the corresponding gates are executed from right to left.

$\endgroup$
2
  • $\begingroup$ This CNOT gate seems to be different from what IBM does; the matrix for their gate has identity for the upper left quadrant. The one needed to use this answer appears to be {{1,0,0,0},{0,0,0,1},{0,0,1,0},{0,1,0,0}} $\endgroup$
    – Anna Naden
    Dec 7, 2021 at 23:47
  • $\begingroup$ It's the same gate, but with control and target qubits swapped. $\endgroup$ Dec 7, 2021 at 23:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.