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Consider Wiesner's quantum money scheme. With today's devices and today's error correction and mitigation schemes, how long can we hold $n$ logical qubits such that they are all (logically) in a product state, with each logical qubit being drawn uniformly at random from one of the BB84 states $\{|0\rangle_L,|1\rangle_L,|+\rangle_L,|-\rangle_L\}$, say, for $n=10$ or $20$ or $30?$

In Wiesner's scheme:

  1. We don't need to act fault-tolerantly on the logical qubits, as there's no two-qubit computation being done on the logical qubits in Wiesner's scheme;
  2. In particular each qubit is prepared and measured in one of the four BB84 basis states - thus assuming easy preparation of $|0\rangle_L$, the only relevant logical single-qubit gates are the Hadamard and the X/NOT gate (no arbitrary phases or T gates are needed); and
  3. Because one whole bill can be written as a product state over all logical qubits, indeed, I can imagine that the $n$ qubits in a bill are prepared and distributed over multiple processors - one processor holding one (logical) qubit, another holding another, etc.
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Today's logical memories are not good enough yet to realize Wiesner's quantum money scheme. Take a look at the screenshot below, taken from a recording of Shraddha Singh's QIP 2023 talk on Real-time quantum error correction beyond break-even. It shows the difference between the lifetime of a logical qubit (logical lifetime) and a bare, unencoded qubit (best physical lifetime). A red arrow indicates that the lifetime of the logical qubit is shorter(!) than the lifetime of a physical qubit. A green arrow indicates an increase in lifetime. There are two green arrows, but they are way too short.

enter image description here

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  • $\begingroup$ Ah I realize now that I misinterpreted your question. I'm unfortunately not familiar with the details of Weisner's scheme. But if you only need preparation and measurement, it's a lot easier than FTQC. In general anything not requiring a universal gate set (such as Clifford gates) is a lot easier than universal FTQC. $\endgroup$
    – Peter-Jan
    Feb 19 at 12:53
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    $\begingroup$ Thanks - what’s the keyword to search for? Quantum memory? Also importantly in Wiesner the qubits are in a product state and never entangled amongst themselves - does that make it even simpler? $\endgroup$ Feb 19 at 13:17
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    $\begingroup$ Yes, quantum memory. And yes that makes it easier as multiple devices could be used. $\endgroup$
    – Peter-Jan
    Feb 19 at 13:24

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