Implementation of filter operation

6

1

If I want to implement the measurement operation corresponding to filtering, i.e. $$ M_1=\left(\begin{array}{cc}1 & 0 \\ 0 & \alpha \end{array}\right)\qquad M_2=\left(\begin{array}{cc}0 & 0 \\ 0 & \sqrt{1-\alpha^2} \end{array}\right), $$ how would I do that?

DaftWullie

Posted 2019-08-06T10:59:57.667

Reputation: 35 722

Answers

4

These measreuements describe a non-projective measurement. We typically convert these into projective measurements by introducing ancilla qubits.

In this case, define a unitary $U$ such that $$ U|0\rangle=\alpha|0\rangle+\sqrt{1-\alpha^2}|1\rangle. $$ Take the qubit that we want to measure, and introduce an ancilla in the state $|0\rangle$. Apply controlled-$U$ controlled from your qubit to be measured, and targeting the ancilla. Finally, perform a standard, $Z$, measurement on the ancilla qubit. Answers 0 and 1 correspond to implementing $M_1$ and $M_2$ respectively.

To see this explicitly, consider the possible inputs of $|0\rangle$ and $1\rangle$. Everything else will follow by linearity. $$ |0\rangle|0\rangle\mapsto |0\rangle|0\rangle\qquad |1\rangle|0\rangle\mapsto |1\rangle(\alpha|0\rangle+\sqrt{1-\alpha^2}|1\rangle). $$ So, input $|0\rangle$ always returns $|0\rangle$ (good since $M_1|0\rangle=|0\rangle$ and $M_2|0\rangle=0$), while $|1\rangle$ returns either $M_1|1\rangle$ or $M_2\rangle$ depending on the measurement result.

DaftWullie

Posted 2019-08-06T10:59:57.667

Reputation: 35 722