Conjuntos universais de portas para SU (3)?


Na computação quântica, estamos frequentemente interessados ​​nos casos em que o grupo de operadores unitários especiais, G, para algum sistema d-dimensional fornece ao grupo todo SU (d) exatamente ou mesmo apenas uma aproximação fornecida por uma cobertura densa de SU (d).

Um grupo de ordem finita, como o grupo Clifford para um sistema d-dimensional C (d), não dará uma cobertura densa. Um grupo de ordem infinita não dará uma cobertura densa se o grupo for abeliano. No entanto, minha intuição grosseira é que um número infinito de operações de mudança de portões e bases do grupo Clifford deve ser suficiente para fornecer uma cobertura densa.

Formalmente, minha pergunta é:

Eu tenho um grupo G que é um subgrupo de SU (d). G tem ordem infinita e C (d) é um subgrupo de G. Todos esses G fornecem uma cobertura densa de SU (d).

Observe que estou particularmente interessado no caso quando d> 2.

Considero o grupo Clifford como definido aqui:

Você pode formular uma definição matemática do grupo Clifford? Eu achei difícil de extrair do papel sem lê-lo em detalhe

@Squark: para N2 arbitrário, considere o subgrupo GU(N) gerado por um operador que "desloca" os vetores de base padrão em ciclicamente, um operadorC N Z = d i a g ( 1 , ω , ω 2 , , ω N - 1 )XCNZ=diag(1,ω,ω2,,ωN1) para e um operador Y = e π i ( N - 1 ) ( N + 1 )ω=exp(2πi/N). (O escalar na frente deYestá em negociação paraN>2; paraN=2,as matrizesX,Y,Zserão as matrizes de spin Pauli usuais.) Então o grupo Clifford é o conjunto de operadores emU(N)que preservaGsob conjugação. Y=eπi(N1)(N+1)/NZXYN>2N=2X,Y,ZU(N)G
Niel de Beaudrap



Esta não é uma resposta completa, mas talvez ajude a responder à pergunta.

Como tem ordem infinita, mas C ( d ) não é, G contém necessariamente uma porta de grupo que não seja de Clifford. No entanto, G tem C ( d ) como um subgrupo. Mas para d = 2, o grupo Clifford mais qualquer outro portão que não esteja no grupo Clifford é aproximadamente universal (veja, por exemplo, o Teorema 1 aqui ). Portanto, todos esses G fornecem uma cobertura densa em S U ( 2 n ) .GC(d)GGC(d)d=2GSU(2n)

No caso em que , parece possível provar que você ainda possui uma cobertura densa ao longo das seguintes linhas (usando a notação do artigo vinculado na pergunta):d>2

  1. Como todos os portões em são unitários, todos os seus autovalores são raízes da unidade, que por simplicidade vou parametrizar por ângulos reais 0 θ i < 2 π .G0θi<2π
  2. As G has infinite order, either G contains gates for which at least one value θk is an irrational multiple of π or contains an arbitrarily good approximation to such an irrational multiple of π. Let us designate one such gate g.
  3. Then there exists an n such that gn is arbitrarily close to, but not equal to the identity.
  4. Since gn is unitary it can be written as exp(iH).
  5. Since the Pauli group as defined in quant-ph/9802007 forms a basis for d×d matrices, you can write H=j,k=0d1αjkXdjZdk, where αjkC and |αjk|ϵ for any ϵ>0 (by [3]), with at least one such αab not equal to zero.
  6. We can then choose C an element from the Clifford group which maps XdjZdk to Zd under conjugation. Thus CgnC=exp(iCHC)=exp(i(αabZd+(j,k)(a,b)αjkXdjZdk)), where α is just a permutation of α and αab=α01.
  7. Note that Zd satisfies Zd(XduZdv)=ωu(XduZdv)Zd. Let us define g=ZdCgnCZd=exp(i(αabZd+(j,k)(a,b)ωjαjkXdjZdk)).
  8. By the Baker-Cambel-Hausdorff theorem, since all α have been made arbitrarily close to the identity, we can evaluate the product of g=g1×...×gd to first order as exp(i(d×(kα0kZk)+(=1dωd)×j0kαjkXdjZdk)). Summing over all routes of unity, for d>1 yields g=exp(i(d×(kbα0kZk)). This is basically a decoupling sequence which decouples the non-diagonal elements.
  9. As only diagonal matrices remain in the exponential, g must be diagonal. Further due to the restrictions on α it necessarily has eigenvalues which are non-zero but proportional to ϵ.
  10. By varying ϵ and repeating the above process it should be possible to generate d linearly independent gates:, such that their product results in a diagonal gate with with irrational and incommensurate phases or an arbitrarily close approximation to one.
  11. By the reference given in Mark Howard's answer this, together with the Clifford group, should suffice for approximate universality.

Why isn't this complete? If you flesh out the details in your vague steps (step 10 in particular), it seems like it might work.
Peter Shor

@PeterShor: For exactly that reason: I haven't fleshed out all of the steps. I think it should work, but I acknowledge it is not rigorous. I'll see if I can flesh out 10.
Joe Fitzsimons

Nice. This seems like a good approach.

I'm giving the bounty to this answer because I think the chances are that a proof along these lines will answer the question. The other answers are very useful as well.
Peter Shor

@PeterShor: Thanks! I was feeling a bit guilty that my first answer was incorrect.
Joe Fitzsimons


I believe the answer to original question is probably yes, but unfortunately, I can't say that definitively. I can help answer Peter's extended question, however.

In math/0001038, by Nebe, Rains, and Sloane, they show that the Clifford group is a maximal finite subgroup of U(2^n). Solovay has also shown this in unpublished work that "uses essentially the classification of finite simple groups." The Nebe et al. paper also shows that the qudit Clifford group is a maximal finite subgroup for prime p, also using the classification of finite groups. This means that the Clifford group plus any gate is an infinite group, which makes one of the assumptions of the original question redundant.

Now, both Rains and Solovay told me that the next step, showing that an infinite group containing the Clifford group is universal, is relatively straightforward. However, I don't know how that step actually works. And more importantly for the original question, I don't know if they were only considering the qubit case or also the qudit case.

Actually, I might add that I don't understand the Nebe, Rains, and Sloane proof either, but would like to.


It's not clear to me whether you're asking about SU(3) or SU(3n) acting on a tensor product of qudits. I'll assume you're asking about SU(3). It's not clear to me (despite what I said in a previous version of my answer) that the statement for SU(3) implies the statement for SU(3n).

As long as the set of gates doesn't lie in a subgroup of SU(3), it will generate a dense cover of SU(3). So you need to check whether any of the infinite subgroups of SU(3) contains the Clifford group. I am fairly sure they don't, but I can't say for sure. Here is a math overflow question giving all the Lie subgroups of SU(3).

I read the third last sentence of the question as saying that the Clifford group was a subgroup of the particular group G that Earl is considering. Hence my answer below, but perhaps I have misunderstood or misread something.
Joe Fitzsimons

The difficulty with your answer is that it your reference seems only to talk about SU(2), while the OP is asking about SU(3) and the analogous group to the Clifford group in SU(3) (and also qudits of dimension d>3). Your reference answers his question for d=2. What we need is that the theorem from your reference also holds in SU(3); namely, that there are no subgroups containing the SU(3) Clifford group.
Peter Shor

Ah, I see. I'll delete my answer. From the context of the notes I linked to it sounded like the theorem applied in arbitrary dimensions, not just the case where d=2. However, upon digging up the source that appears not to be the case. Thanks for pointing out the error.
Joe Fitzsimons

Ultimately, I will be interested in SU(3n). However, because this is entailed by universality in SU(3) + the Clifford group, this is how I phrased the question to keep it simple. I also had a quick look at the reference Joe provided and could only see results for d=2.

Also, I will follow Peters suggestion and check the Lie subgroups on the math overflow reference, though it might take me a while to get through all of it!


I thought I should update this thread before the site is frozen forever.

Daniel's answer is on the right lines. This "next step" that he mentions appears in Nebe, Rains and Sloane's later book, "Self-Dual Codes and Invariant Theory".

The answer to this question is therefore "Yes" - and it follows directly from Corollary 6.8.2 in Nebe, Rains and Sloane's book.

I am grateful to Vadym Kliuchnikov who pointed this out to me while I was visiting Waterloo.

I should clarify that "Yes" is the direct answer to Earl's formal question above and this is shown by Corollary 6.8.2 in the book.
Dan Browne


I think the following paper may contain the relevant constructions for proving qudit universality

In particular, the comment at the end of section 4 says that Controlled-phase CZ, Fourier transform F, and a diagonal gate D with irrational and incommensurate phases gives approximate universality. (This is a sufficient condition on D but I'm pretty sure it is not a necessary condition.)

If your G is of the correct form (and diagonal gates would seem a natural choice) then the result applies

An alternative approach would be to create the ancilla states required for implementation of the qudit Toffoli, or directly using G along with Cliffords to implement the Toffoli. It's hard to say whether this is possible without knowing more about G.

Welcome to the site, Mark!
Joe Fitzsimons

Hi Mark. Thanks for your answer. Though I am interested in the most general case, I am particularly interested in a case where I know I have an infinite number of gates because it is generated by a gate with phases that are irrational multiples of π. However, the "irrational" gate is not diagonal in the computational basis, and so I can't apply the results you cite.
Ao utilizar nosso site, você reconhece que leu e compreendeu nossa Política de Cookies e nossa Política de Privacidade.
Licensed under cc by-sa 3.0 with attribution required.