Let $ G $ be a finite group of quantum gates. Is it true that any circuit made using only gates from the finite group $ G $ can be efficiently simulated on a classical computer?

Here by circuit made from $ G $ I mean a circuit in which all gates are from $ G $ and all states are prepared and measured in the computational basis.

The Gottesman-Knill theorem says that any Clifford ( $ G $ equals the Clifford group) circuit can be simulated efficiently on a classical computer. So the answer is yes for at least some choices of a finite group of gates $ G $. I would imagine that if $ G $ is a finite abelian group then the theory of computation is very simple and can also be simulated efficiently on a classical computer.

It is worth noting that classical reversible computation (on $ n $ bits say) is $ G $ circuits with $ G=S_{2^n} $ the symmetric group on the $ 2^n $ bit strings.

Every group of size $ |G| $ is a subgroup of a symmetric group on $ |G| $ letters. So we can take $ log_2(|G|) $ bits and certainly find $ G $ is a subgroup of the symmetric group on $ 2^{log_2(|G|)} $ bit strings. The way to realize a group as a permutation group is just to act it on itself by left multiplication


anyway I have no idea is this can be done efficiently but certainly this seems like a way to represent circuits using a finite group of gates as just reversible classical circuits.

Again my question is about efficient classical simulation: if $ G $ is a finite group of quantum gates can every circuit in which all gates are from the group $ G $ and all states are prepared and measured in the computational basis be simulated efficiently on a classical computer?

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    $\begingroup$ Following the proof of GK theorem, it seems that it cannot be trivially extended to every finite group. On the other hand, I do not see simple counter-examples as well. $\endgroup$ Mar 6, 2022 at 17:34
  • $\begingroup$ I suppose that the Clifford case has a lot more structure and a lot more going for it than you've picked out in your question. There are two groups: the Clifford group and (an Abelian subgroup of) the Pauli group. There are three very useful things here: (i) the initial state is easily decomposed in terms of the Pauli subgroup, (ii) since the subgroup is Abelian, we can just deal with $n$ individual terms instead of a product of $n$ terms (which would have $2^n$ components), and (iii) the action of the Clifford group on the Pauli group is easily calculated $\endgroup$
    – DaftWullie
    Mar 7, 2022 at 8:02
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    $\begingroup$ One difficulty that we have with this question is that I don't believe the gap between quantum universality and quantum simulability is well understood. After all, that would instantly answer your question - a finite group of quantum gates cannot be approximately universal. So, if the answer to your question were negative, we'd be looking at a group which is neither universal nor classically simulable. $\endgroup$
    – DaftWullie
    Mar 8, 2022 at 7:52
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    $\begingroup$ This paper is potentially relevant semanticscholar.org/paper/… $\endgroup$ Apr 7, 2022 at 2:36
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    $\begingroup$ @IanGershonTeixeira I think if you take $G$ to be any polycyclic group then you should be able to efficiently simulate gates in its normalizer $C$ (in some bigger group, for example all unitaries if working in a complex representation). You track the action of a gate in $C$ by tracking its affect on the generators of $G$ : $G \to U^{-1} G U$. This just maps the exponents of the generators to another set of exponents...a generalized tableaux of sorts. $\endgroup$
    – unknown
    Jun 4, 2022 at 16:43

1 Answer 1


Let me give an answer which is not really intended as an answer (in that it doesn't address what I suspect the question is aimed at. For example, it does not cover the case of Clifford gates), and rather aiming to get clarification of the question itself.

Imagine you have a finite group $G$ of size $|G|=N$. Since it is a group, every pair has an action $g_ig_j=g_k$. Since we know this is a group, we must have a proof that it's a group, which means we must be able to calculate the values $k$ for a pair of inputs $(i,j)$. Hence, let us pre-calculate all $N^2$ results $g_ig_j$ and store them in a (sorted) lookup-table. Given that $N$ is finite, the entire calculation must be finite.

Now, if we have a circuit of length $M$ comprising elements of $g$, the entire computation comprises looking up pairs in turn and replacing them with single elements until all we have left is the single element $g$ that represents the entire computation. This only requires $M-1$ lookups in the table. Hence, the simulation time is $O(M)$.

Now, what I assume the question really wants to ask is to let $N$ be a function of $n$, the problem size (i.e. number of qubits) and that, for any natural number $n$, $N$ is finite. This, for example, allows the sizes such as $N\sim 2^n$ (or even crazier), but excludes continuous groups. My answer does not apply to those.

Or, perhaps the specification may be (which could be equivalent): let $G$ be a finite group of gates acting on at most $k$ qubits (e.g. $k=2$), and consider the family of circuits on $n$ qubits comprising application of members of $G$ on arbitrary subsets of qubits. Is the simulation efficient (i.e. polynomial) in $n$?

Both of these possible specifications have an element of scaling with the number of qubits which is not explicitly present in the question specification.

  • $\begingroup$ I agree, this is an important point. The group is not just one, there is a different group for each number of qubits $n$. So a scaling must be specified. If I'm not wrong, the scale you mention, $N\propto 2^n$, applies to Clifford gates. So the task could be stated as: show that $N\propto 2^n$ implies that the output of the circuit, applied to $\left|0\right>$, can be classically calculated in polynomial time. Btw, this is my interest; maybe the user that asked the question is more interested in something else! $\endgroup$ Apr 1, 2022 at 10:41

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