Hi everyone,
I’m currently studying quantum information and I use Qiskit as my main package for quick simulations. I was wondering how this kind of circuit could be physically implemented and I discovered your environment. I wanted to try to convert a Qiskit circuit into a Perceval circuit in order to see how logic gates like H gate or CNOT gate were realized.
If I understand correctly, you are making a CNOT gate based on a linear optical component. In previous readings I thought I had read that this could decrease the “efficiency” of this kind of gate. If I’m not mistaken, does anyone know this efficiency? (0.5, 0.25 maybe?) Moreover, is there a difference in efficiency between post-processed CNOT and heralded CNOT ?
If I’m wrong, does anyone have an explanation of how CNOT gates like this work?
Many thanks in advance,
clementp
Hi clementp,
Your intuition is correct, there is a trade off between the type and amount of CNOT gates you use and the efficiency of the circuit.
However, the important thing to note is that when we say efficiency, we are really talking about the success probability of the circuit.
CNOT gates in photonics are impossible to perform deterministically, they act probabilistically. So what this means in practice is that we take a number of samples and we post-process these samples, discarding the measurements where the CNOT gate has failed.
So, for the post-processed CNOT, we have a success probability of 1/9, for this CNOT we check that no photons have entered the auxillary modes and that we have one photon per pair of qubit modes. This CNOT can only be used if it is the last CNOT gate on given qubit modes, because we have to check that the photons are in the correct modes, the output of one of these gates cannot be inputted into another CNOT.
The heralded CNOT gate has a success probability of 2/27. For this gate we must detect photons in 2 of the auxillary modes to confirm that the gate was successful, these gates can be used in series with any other gate.
So for the success probability of a given circuit you will multiply the success probabilities of all of the CNOT gates in the circuit.
I hope this answers your question.
Cheers,
Patrick
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