We have designed, fabricated and tested an XY addressable readout architecture for superconducting flux qubits in an integrated circuit designed to enable adiabatic quantum optimization algorithms. The readout architecture for an N-qubit system comprises N hysteretic dc SQUIDs and N rf SQUID latches controlled with wires. The latching elements are coupled to the qubits and provide exceptional readout fidelity. The dc SQUIDs are then coupled to the latches to provide coarse readout. A key advantage of this architecture is that the latches are unaffected by the ac currents generated in the dc SQUIDs during switching. We place a lower bound on the readout fidelity of 99.9999% in an 8 qubit test system.

[soon to be arxiv:0905.0891]

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Thats great you are genies. Soon you will build quantum computer and discover some algorithms for awsome games, which graphics will be milions times better than in current games.

This is a very nice paper.

I had a question: Would it be possible to implement the same active compensation scheme (that you described for tuning away the spread of Ic in the CJJ qubits in a previous paper) to tweak the response of the QFP? I guess it would require an extra 2sqrt(N) lines to make the QFP controllable too, so it depends just how much of a problem this 1-2% variation threshold turns out to be when the system is scaled.

The paper mentions the fact that the back-action on the qubit can be minimised by using the QFP, ie. you are performing a fully quantum-non-demolition measurement. I was thinking about the case of the phase qubit – I don’t think you have this problem as you can ‘freeze’ the qubit state by quickly lowering and raising the barrier. I wonder if you can do this in the flux qubit if you can raise the barrier quickly enough (non-adiabatically)? Even if that is the case theoretically (which I’m not sure it is), it would require signal line and flux response times much faster than the Rabi frequency… hmm. (In the phase qubit this is not a problem as you already need RF lines to apply the microwave drive).

I conclude through my ramblings that the QFP sounds like a good plan!๐

Suz,

With something like 3sqrt(N) lines I think we could do analog compensation for flux offsets (either extrinsic or asymmetry related) in the QFPs. We would have to have the latching event propagate down one of these arrays (implemented by XY control of the compound junctions of the QFPs) in lock-step with application of compensation flux (just another “Y” or “X” line) to the QFPs. Might work. The win would be allowing reduced M between the QFP and qubit, but we don’t gain much in that direction.

The point about the minimal action of the QFP on the qubit wasn’t meant to be about a QND measurement (sorry for the lack of clarity in writing). If the qubit were in a superposition state (ie we hadn’t fully raised the qubit barrier), then raising the barrier of the QFP would be similar to a projective measurement because the coupled QFP / coupler system energy splitting would go to zero (and after the coherence time passes the system would have to be in an appropriate left/left or right/right state). But, this evolution results in us measuring the qubit flux. Doing the same with a dc SQUID does not work since the dc SQUID applies flux to the qubit as its parameters are changed.

Andrew

Haven’t had a chance to glance at the paper yet, but something quirky occurred to me while teaching my QM course this spring that seems to relate to this. I used a rough draft of Ben Schumacher and Michael Westmoreland’s forthcoming undergraduate QM text and they had a problem I assigned where the student shows, essentially, that the two

classicalreadouts for a measurement have to be orthogonal (which purely classical statesusuallyrepresent anyway) in order for the quantum measurements they represent to be distinguishable. Using the same idea, on an exam I had the students figure out the associated requirement if, instead, there were N measurements being made and what happened as N approached infinity. In such a case, it turns out that it is no longer necessary for the readouts to be orthogonal if the quantum states are also not orthogonal. It seems paradoxical, but is what pops out of the math. Any thoughts?The designed mode of operation for this system requires that the system’s Hamiltonian be diagonal in the readout basis prior to the readouts being fired, so the above issues never come up.

It may be possible to operate the readout system by latching the qubit state into the QFP when the qubit’s barrier is low and the qubit’s state is not diagonal in the readout basis. This might give a projective measurement but you’d have to work through the details.

Ah, I see. Yeah, I did the math and you can get a projective measurement under the scheme I mention, but only in the limit as N approaches infinity so it isn’t necessarily practical, but is, in my mind, a counter-intuitive result.

Super nice post. Keep more coming like this๐