Quantum noob here. Apologies if my question is trivial.

According to google, IBM has a quantum computer with 433 qubits.

Based on this (How many logical qubits are needed to run Shor's algorithm efficiently on large integers ($n > 2^{1024}$)?) post, one should be able to run Shor's algorithm for numbers with 100+ digits on such a quantum computer.

Why do current large-scale quantum computers like IBM's fail to run Shor's algorithm?

Is it the difference between physical and logical qubits?


1 Answer 1


Yes, there's a big difference between physical and logical qubits. The (physical) qubits that make up a device such as the IBM one suffer from noise. Basically, errors that mean the operations you're performing aren't exactly the ones you wanted. Even if small, these build up rapidly to make even very small calculations unviable.

The way that we beat this is to use error correcting codes. You encode multiple physical qubits into on logical qubit and, if you do everything right, there's a threshold error rate for which you gain a benefit. The challenge is that the sorts of error rates on these real devices are very similar to the threshold values. This means that you need to throw lots of resources (i.e. many physical qubits for one logical qubit) at the problem to try and get some benefit. 433 physical qubits might not even give you much benefit for even a single logical qubit. In practice, we're probably talking needing millions of physical qubits to run the sort of algorithm you're talking about. (See here for a fairly recent estimate of resource requirements.)

  • $\begingroup$ Dear @DaftWullie, thank you for the explanation! So basically, even if the number of qubits is large in a quantum computer, it is not clear, how much logical qubits they encode. Also, I will check out the article you sent. $\endgroup$
    – Adelhart
    Commented Nov 3, 2023 at 11:27
  • $\begingroup$ Yes, there is still a choice to be made about how you do the error correction, and therefore how many physical qubits encode a single logical qubit. $\endgroup$
    – DaftWullie
    Commented Nov 3, 2023 at 13:41

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