7 ms·

What are they doing about error correction? Not that it makes this any less impressive (everything starts somewhere), are they simply ignoring the problem?

vtomole 9y ago
It's a bit too early for quantum computers to do fault-tolerant (error corrected) computation. This is because you need more than one physical qubit to make a logical (error corrected) qubit. You need 7 physical qubits to encode a logical qubit if you use the Steane code [0]. So with 50 qubits, they could theoretically make 7 error corrected qubits. [0]: https://en.wikipedia.org/wiki/Steane_code https://en.wikipedia.org/wiki/Steane_code
- greeneggs 9y ago
  In fact, the smallest quantum error-correcting code using qubits uses only five qubits. [1] So you could fit 10 of these codewords into 50 qubits. But that's not right. At least two more qubits are needed for fault-tolerant error correction. So that means you could fit nine codewords into 50 qubits (since 9x5+2=47 < 50). But that's not right. There exist more efficient codes that put multiple encoded qubits into a single code block. [2] For example, three qubits can be encoded into eight, still with distance three. [3] Six of these could fit into 50 qubits (since 6x8+2=50), giving 18 encoded qubits. But that's not right. The IBM systems are superconducting qubits, with very constrained interactions. Not every qubit can talk directly to every other qubit. So you'd probably want every code block to have its own extra qubits dedicated to error correction. If you need 8+2 qubits per code block, then you could fit five code blocks, for 15 encoded qubits, into 50. Obviously this is really complicated. [1] https://en.wikipedia.org/wiki/Stabilizer_code#Example_of_a_stabilizer_code https://en.wikipedia.org/wiki/Stabilizer_code#Example_of_a_s... [2] http://www.codetables.de/TableIII.php http://www.codetables.de/TableIII.php [3] http://www.codetables.de/QECC.php?q=4&n=8&k=3 http://www.codetables.de/QECC.php?q=4&n=8&k=3
  - jackfoxy 9y ago
    I recently had a conversation with a Microsoft quantum researcher, and this a close approximation to his answer. I just wanted a number. It's complicated.
    - quickthrower2 9y ago
      Sounds like 15 is that number
      - cerved 9y ago
        It's also not 15 at the same time.
agumonkey 9y ago
I just saw a talk by I forgot which bell labs engineer was on the transistor team. He told it took two or more years to get the errors to a useful level. It's easy to think transistor are magic solid devices, they were not.
- raverbashing 9y ago
  But as they are analog devices, there are no errors with transistors There are manufacturing defects, there are inefficient designs, inefficient production methods, poor circuit designs (like using them in common base - not wrong depending on the situation - instead of common emitter)
  - posterboy 9y ago
    every measurement has an error tolerance, starting with the uncertainty principle and including transistor manufaction.
    - raverbashing 9y ago
      Yes, tolerances and margins, but no device is sold as "perfect" quite the contrary, and the smaller the tolerance the more costly it is. So it's an error in the metrological sense but not in the defective sense
mtgx 9y ago
They don't seem to be ignoring it. They are using this idea of "quantum volume" which incentivizes them to take into account both the number of qubits and the error rate to determine how useful their quantum computer is in solving problems: https://www.research.ibm.com/ibm-q/resources/quantum-volume.pdf https://www.research.ibm.com/ibm-q/resources/quantum-volume....

vtomole 9y ago

It's a bit too early for quantum computers to do fault-tolerant (error corrected) computation. This is because you need more than one physical qubit to make a logical (error corrected) qubit. You need 7 physical qubits to encode a logical qubit if you use the Steane code [0]. So with 50 qubits, they could theoretically make 7 error corrected qubits. [0]: https://en.wikipedia.org/wiki/Steane_code https://en.wikipedia.org/wiki/Steane_code

greeneggs 9y ago
In fact, the smallest quantum error-correcting code using qubits uses only five qubits. [1] So you could fit 10 of these codewords into 50 qubits. But that's not right. At least two more qubits are needed for fault-tolerant error correction. So that means you could fit nine codewords into 50 qubits (since 9x5+2=47 < 50). But that's not right. There exist more efficient codes that put multiple encoded qubits into a single code block. [2] For example, three qubits can be encoded into eight, still with distance three. [3] Six of these could fit into 50 qubits (since 6x8+2=50), giving 18 encoded qubits. But that's not right. The IBM systems are superconducting qubits, with very constrained interactions. Not every qubit can talk directly to every other qubit. So you'd probably want every code block to have its own extra qubits dedicated to error correction. If you need 8+2 qubits per code block, then you could fit five code blocks, for 15 encoded qubits, into 50. Obviously this is really complicated. [1] https://en.wikipedia.org/wiki/Stabilizer_code#Example_of_a_stabilizer_code https://en.wikipedia.org/wiki/Stabilizer_code#Example_of_a_s... [2] http://www.codetables.de/TableIII.php http://www.codetables.de/TableIII.php [3] http://www.codetables.de/QECC.php?q=4&n=8&k=3 http://www.codetables.de/QECC.php?q=4&n=8&k=3
- jackfoxy 9y ago
  I recently had a conversation with a Microsoft quantum researcher, and this a close approximation to his answer. I just wanted a number. It's complicated.
  - quickthrower2 9y ago
    Sounds like 15 is that number
    - cerved 9y ago
      It's also not 15 at the same time.

agumonkey 9y ago

I just saw a talk by I forgot which bell labs engineer was on the transistor team. He told it took two or more years to get the errors to a useful level. It's easy to think transistor are magic solid devices, they were not.

raverbashing 9y ago
But as they are analog devices, there are no errors with transistors There are manufacturing defects, there are inefficient designs, inefficient production methods, poor circuit designs (like using them in common base - not wrong depending on the situation - instead of common emitter)
- posterboy 9y ago
  every measurement has an error tolerance, starting with the uncertainty principle and including transistor manufaction.
  - raverbashing 9y ago
    Yes, tolerances and margins, but no device is sold as "perfect" quite the contrary, and the smaller the tolerance the more costly it is. So it's an error in the metrological sense but not in the defective sense

mtgx 9y ago

They don't seem to be ignoring it. They are using this idea of "quantum volume" which incentivizes them to take into account both the number of qubits and the error rate to determine how useful their quantum computer is in solving problems: https://www.research.ibm.com/ibm-q/resources/quantum-volume.pdf https://www.research.ibm.com/ibm-q/resources/quantum-volume....