The aricles onUniversal quantum computer andQuantum Turing machine describe the same concept.Universal quantum computer should redirect toQuantum Turing machine, just likeUniversal computer redirects toTuring machine.—Precedingunsigned comment added byRobinK (talk •contribs)20:00, 5 July 2009 (UTC)[reply]
- Merged. --Robin (talk)16:42, 9 July 2009 (UTC)[reply]
I think the page on QTMs would be much more useful if it actually contained a definition of what a QTM is.Simphilip (talk)04:47, 1 December 2010 (UTC)[reply]
- I added a very short and somewhat flawed hand-waving definition. Hope that helps.67.198.37.16 (talk)22:02, 15 August 2015 (UTC)[reply]
Computational equivalence
[edit]The lead states of QTMs, citing Yao ‘Quantum circuit complexity’:
- However, the computationally equivalent quantum circuit is a more common model.
However,Revisiting the simulation ofquantum Turing machines byquantum circuits states:
- Yao’s 1995 publication ‘Quantum circuit complexity’ proved that quantum Turing machines and quantum circuits are polynomially equivalent computational models: t≥n steps of a quantum Turing machine running on an input of length n can be simulated by a uniformly generated family of quantum circuits with size quadratic in t, and a polynomial-time uniformly generated family of quantum circuits can be simulated by a quantum Turing machine running in polynomial time.
Given that we care about characterising complexity classes, the looseness in this two-way tie seems to fall rather short of the claim of computational equivalence. Should this be amended? —Charles Stewart(talk)15:04, 15 July 2021 (UTC)[reply]
- according to my understanding, the standard notion of equivalence in complexity theory (and certainly in quantum computing) is "up to polynomial overhead". We care mostly about problems in the class BQP, and under an encoding with at most poly overhead, that class does not change. Moreover, the overhead is needed both to simulate a quantum circuit on a QTM and to simulate a QTM on a circuit - so in general one cannot say which model is the "better" one. See eg.[1] for another work showing "polynomial equivalence" between two models of quantum computation. --Qcomp (talk)21:12, 15 July 2021 (UTC)[reply]
