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Provided that $\mathsf{MIP}^*=\mathsf{RE}$ there can be Bell inequalities that have violations achievable only for infinite dimensional quantum systems (vide discussions in Post1 and Post2). Does this implies that this kind of behaviour, once found, will not be simulable by a finite dimensional quantum computer? What are implications for the Church-Turing thesis or its extended quantum version?

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    $\begingroup$ Why limit to the extended Church-Turing thesis? Why not ask for implications for the Church-Turing thesis itself? $\endgroup$
    – Mark Spinelli
    Commented Aug 20, 2022 at 16:20
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    $\begingroup$ You don't need $\mathsf{MIP}^*=\mathsf{RE}$ to see that there are quantum behaviours requiring infinite-dimensional quantum systems (i.e. Hilbert spaces) that cannot be modelled with finite-dimensional systems; this is a consequence of the separation between the $C_q$ and $C_{qs}$ models arxiv.org/abs/1804.05116 $\endgroup$
    – Condo
    Commented Aug 21, 2022 at 10:42
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    $\begingroup$ I would also recommend reading Scott Aaronson's blog post on the result scottaaronson.blog/?p=4512 there are also many good questions and discussions in the comments. $\endgroup$
    – Condo
    Commented Aug 22, 2022 at 14:10
  • $\begingroup$ @Condo thanks for the reference, it is a very interesting one and indeed makes the reference to MIP*=RE unnecessary. Essentially this means that not all physical processes can be captured by finite dimensional Hilbert space dynamics and therefore not all processes can be simulated by quantum computation with qubits? I didn't knew that this was well-known. $\endgroup$
    – R.W
    Commented Aug 25, 2022 at 14:36
  • $\begingroup$ @R.W I think it could depend on your definition of "physical process." Essentially, there are these various models for quantum behaviours, which manifest as these various correlation sets $C_q, C_{qs}$, and $C_{qc}$, but we don't know which one "nature" really abides by. In fact, I've heard it suggested that "physical" correlations should be a closed set, but neither $C_q$ nor $C_{qs}$ has this property. $\endgroup$
    – Condo
    Commented Aug 25, 2022 at 15:37

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