In quantum computing we are often interested in cases where group of special unitary operators, G, for some d-dimensional system gives either the whole group SU(d) exactly or even just an approximation provided by a dense cover of SU(d).

A group of finite order, such as the Clifford group for a d-dimensional system C(d), will not give a dense cover. A group of infinite order will not give a dense cover if the group is Abelian. However, my rough intuition is that an infinite number of gates and basis changing operations of the Clifford group should suffice to provide a dense cover.

Formally, my question is:

I have a group G that is a subgroup of SU(d). G has infinite order and C(d) is a subgroup of G. Do all such G provide a dense cover of SU(d).

Note that I am particular interested in the case when d>2.

I take the Clifford group to be as defined here: http://arxiv.org/abs/quant-ph/9802007

  • $\begingroup$ Can you formulate a mathematical definition of the Clifford group? I found it difficult to extract from the paper without reading it in detail $\endgroup$
    – Vanessa
    Jan 14, 2012 at 18:02
  • $\begingroup$ @Squark: for $N\geqslant 2$ arbitrary, consider the subgroup $G \subseteq \mathbf U(N)$ generated by an operator $X$ which "shifts" the standard basis vectors on $\mathbb C^N$ cyclically, an operator $Z = \mathrm{diag}(1, \omega, \omega^2, \ldots, \omega^{N-1})$ for $\omega = \exp(2\pi i/N)$, and an operator $Y = \mathrm e^{\pi i (N-1)(N+1)/N} ZX$. (The scalar in front of $Y$ is up for negotiation for $N > 2$; for $N = 2$ the matrices $X,Y,Z$ will be the usual Pauli spin matrices.) Then the Clifford group is the set of operators in $\mathbf U(N)$ which preserves $G$ under conjugation. $\endgroup$ Jan 23, 2012 at 18:14

5 Answers 5


This is not a complete answer, but perhaps it goes some way towards answering the question.

Since $G$ has infinite order but $C(d)$ is not, then $G$ necessarily contains a non-Clifford group gate. However $G$ has $C(d)$ as a subgroup. But for $d=2$ the Clifford group plus any other gate not in the Clifford group is approximately universal (see e.g. Theorem 1 here). Therefore all such $G$ provide dense cover on $SU(2^n)$.

For the case where $d>2$ it seems like it may be possible to prove that you still get dense cover along the following lines (using the notation of the paper linked to in the question):

  1. As all gates in $G$ are unitary, all of their eigenvalues are roots of unity, which for simplicity I will parameterize by real angles $0 \leq \theta_i < 2\pi$.
  2. As $G$ has infinite order, either $G$ contains gates for which at least one value $\theta_k$ is an irrational multiple of $\pi$ or contains an arbitrarily good approximation to such an irrational multiple of $\pi$. Let us designate one such gate $g$.
  3. Then there exists an $n$ such that $g^n$ is arbitrarily close to, but not equal to the identity.
  4. Since $g^n$ is unitary it can be written as $\exp(-iH)$.
  5. Since the Pauli group as defined in quant-ph/9802007 forms a basis for $d \times d$ matrices, you can write $H = \sum_{j,k = 0}^{d-1}\alpha_{jk} X_d^j Z_d^k$, where $\alpha_{jk} \in \mathbb{C}$ and $|\alpha_{jk}|\leq \epsilon$ for any $\epsilon > 0$ (by [3]), with at least one such $\alpha_{ab}$ not equal to zero.
  6. We can then choose $C$ an element from the Clifford group which maps $X_d^j Z_d^k$ to $Z_d$ under conjugation. Thus $C g^n C^\dagger = \exp(-iCHC^\dagger) = \exp(-i(\alpha_{ab}Z_d + \sum_{(j, k) \neq (a,b)} \alpha'_{jk} X_d^j Z_d^k))$, where $\alpha'$ is just a permutation of $\alpha$ and $\alpha_{ab} = \alpha'_{01}$.
  7. Note that $Z_d$ satisfies $Z_d (X_d^u Z_d^v) = \omega^{u} (X_d^u Z_d^v) Z_d$. Let us define $g_\ell = Z_d^{-\ell} C g^n C^\dagger Z_d^{\ell} = \exp(-i(\alpha_{ab}Z_d + \sum_{(j, k) \neq (a,b)} \omega^{j\ell} \alpha'_{jk} X_d^j Z_d^k))$.
  8. By the Baker-Cambel-Hausdorff theorem, since all $\alpha$ have been made arbitrarily close to the identity, we can evaluate the product of $g' = g_1 \times ... \times g_d$ to first order as $\exp(-i(d\times(\sum_{k} \alpha_{0k} Z^k) + (\sum_{\ell = 1}^d \omega^d)\times\sum_{j \neq 0}\sum_k \alpha_{jk}X_d^j Z_d^k))$. Summing over all routes of unity, for $d>1$ yields $g' = \exp(-i(d\times(\sum_{k\neq b} \alpha_{0k} Z^k))$. 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 $\alpha'$ it necessarily has eigenvalues which are non-zero but proportional to $\epsilon$.
  10. By varying $\epsilon$ and repeating the above process it should be possible to generate $d$ linearly independent gates: $g'_1 ... g'_d$, 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.
  • $\begingroup$ 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. $\endgroup$ Jan 17, 2012 at 23:02
  • $\begingroup$ @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. $\endgroup$ Jan 18, 2012 at 5:56
  • $\begingroup$ Nice. This seems like a good approach. $\endgroup$
    – Earl
    Jan 19, 2012 at 11:26
  • $\begingroup$ 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. $\endgroup$ Jan 21, 2012 at 6:44
  • $\begingroup$ @PeterShor: Thanks! I was feeling a bit guilty that my first answer was incorrect. $\endgroup$ Jan 21, 2012 at 10:57

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.

  • $\begingroup$ For $ SU(p^n) $, $ p $ prime, it is true that any infinite subgroup containing the qupit Clifford group is universal. This follows from the fact that the qupit Clifford group is a 2-design, together with corollary 3.5 of arxiv.org/pdf/1609.05780.pdf which states that an infinite subgroup of $ SU(N)$ is universal if and only if its centralizer in the adjoint rep is 1 dimensional (equivalently it is Ad-irreducible, and note that every group unitary 2-design is Ad-irreducible). Indeed any subgroup of $ SU(N) $ which is infinite and contains a 2-design must be universal $\endgroup$ Jun 21, 2022 at 11:21

It's not clear to me whether you're asking about SU(3) or SU(3$^{n}$) 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(3$^n$).

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).

  • $\begingroup$ 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. $\endgroup$ Jan 13, 2012 at 15:50
  • $\begingroup$ 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. $\endgroup$ Jan 13, 2012 at 19:06
  • $\begingroup$ 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. $\endgroup$ Jan 13, 2012 at 19:51
  • $\begingroup$ Ultimately, I will be interested in $SU(3^{n})$. 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 $d=2$. $\endgroup$
    – Earl
    Jan 14, 2012 at 15:27
  • $\begingroup$ 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! $\endgroup$
    – Earl
    Jan 14, 2012 at 15:29

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.

  • $\begingroup$ 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. $\endgroup$
    – Dan Browne
    May 4, 2012 at 2:06
  • $\begingroup$ Corollary 6.8.2 only deals with $ SU(p^n) $ for $ p $ prime. It does not apply, for example, to $ SU(6) $. $\endgroup$ Jun 21, 2022 at 10:31

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 $G$.

  • $\begingroup$ Welcome to the site, Mark! $\endgroup$ Jan 13, 2012 at 10:01
  • $\begingroup$ 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 $\pi$. However, the "irrational" gate is not diagonal in the computational basis, and so I can't apply the results you cite. $\endgroup$
    – Earl
    Jan 14, 2012 at 15:24

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