1
$\begingroup$

What is the logical gate speed of a photonic quantum computer? says

In a simple world the speed of a photonic quantum computer would just be the speed at which it’s possible to make small (fixed sized) entangled states. GHz rates for such are plausible and correspond to the much slower MHz code-cycle rates of a superconducting machine.

But it doesn't mention just how much MHz the Superconducting clock rates are.

Which classical processor would be on par with the QPU mentioned in, for example, How to factor 2048 bit RSA integers in 8 hours using 20 million noisy qubits when it comes to clock speed? Would it be the Pentium II (233 MHz to 450 MHz)? The i486 (16 MHz to 100 MHz)?

$\endgroup$

1 Answer 1

1
$\begingroup$

Very much depends on the machine. I'm not a hardware engineer but my rules of thumb are:

Unitary Layer of two qubit gates followed by layer of single qubit gates: 10ns to 100ns.

Dissipative Layer of measure and/or reset: 100ns to 1000ns.

Surface code cycle: 300ns to 3000ns

Reaction time: dominated by the decoder not the hardware; hopefully below 10us

The non-clifford reaction cycle time is the closest thing to a true "logical" clock rate, because all Cliffords can be precomputed just in time via gate teleportation. Cliffords don't take seconds, they take qubit*seconds

$\endgroup$
2
  • $\begingroup$ At our most pessimistic, is it fair to say that our 8 million physical qubit machine is ~100 kHZ: en.wikipedia.org/wiki/ENIAC given the ~10us reaction time? $\endgroup$ Commented Jul 8 at 21:13
  • 1
    $\begingroup$ @VictoryOmole The architectures are so different that it's not really meaningful to compare those numbers. In the ENIAC a clock cycle was a load+add+save. The reaction cycle is more akin to the gate delay from one logic gate to the next. You would need to build an adder out of those gates, and add up the delays, in order to do a more apples-to-apples comparison. $\endgroup$ Commented Jul 8 at 22:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.