# theorems for universal set of quantum gates for SU(d)

+ 3 like - 0 dislike
1247 views

It seems that there is a theorem that for prime dimension d, the set of Clifford gates and one non-Clifford gate together forms a universal set of quantum gates for SU(d). It also seems that for a universal set of gates, there is a requirement of a gate using which can provide non-trivial rotation by an irrational multiple of Pi. Does this imply that using any non-Clifford gate for prime d, one can construct such a gate ? Also, it probably does not follow directly that the non-Clifford gate used to get such a non-trivial rotation itself has irrational eigenvalues but it seems that is usually the case- for example, the qubit T gate and qudit versions of it.

Is there any similar theorem as mentioned above, for composite d ? It seems that if one finds a gate using which one can construct a non-trivial rotation by an irrational multiple of Pi, that should be enough to have universal set of gates even for composite d. Is it obvious that such a gate has to be non-Clifford ?

This post imported from StackExchange at 2016-06-20 15:13 (UTC), posted by SE-user user25957
Irrational multiple of pi rotation is a sufficient condition (allow to approximate any angle), not a requirement. Except if you mean that an irrational angle should be easily approximated (otherwise the set is obviously non-universal)

This post imported from StackExchange at 2016-06-20 15:13 (UTC), posted by SE-user Frédéric Grosshans
I think my theorem was not written precisely and probably that's the reason for your comment. According to the theorem, Clifford gates and a Non-Clifford gate give the universal set of gates. I think this is not just approximate universal in the sense that it can be actually used to give a rotation which is an irrational multiple of pi, just like the T gate.

This post imported from StackExchange at 2016-06-20 15:13 (UTC), posted by SE-user user25957
You cannot have exactly all irrational rotations with a discrete set of gates: the number of combinations is countable while the number of irrational angles is uncountable. A set is universal is the set of operations you can prepare is dense in the total set of operations, i.e. you can approximate any gate

This post imported from StackExchange at 2016-06-20 15:13 (UTC), posted by SE-user Frédéric Grosshans

+ 4 like - 0 dislike

I'm not aware of any proof that the Clifford group + any non-Clifford element gives a universal set of quantum gates. The closest related result that I know is that the Clifford group + any non-Clifford element densely generates an infinite group. (This is theorem 6.5 of Nebe et al. http://arxiv.org/abs/math/0001038v2). But this falls short of proving that the Clifford group + any non-Clifford element is universal, because it is possible you could densely generate a continuous group on $n$ qubits that is not all of $U(2^n)$. So I believe that proving this result remains open. Note that in the case of the Clifford group on qubits, one can easily show that the Clifford group + any non-Clifford one-qubit gate is universal, using the representation theory of $SU(2)$. But the more general statement about adding arbitrary elements to the Clifford group is open.

Update: The answer to the first problem is known. In particular, the statement that the Clifford group + any non-Clifford element yields a universal gate set follows from two theorems of Nebe, Rains and Sloane. First, Theorem 6.5 of Nebe et al (http://arxiv.org/abs/math/0001038v2) shows that adding any element to the Clifford group renders it infinite. Second, Corollary 6.8.2 of Nebe et al. (Self-Dual Codes and Invariant Theory (Springer 2006)) states that any infinite group which contains the Clifford group contains all of $SU(d^n)$. Combining these shows that the Clifford group + any non-Clifford element is universal, in any local dimension $d$ (prime or composite). This was previously discussed on stackexchange here: Universal sets of gates for SU(3)?

With regards to rotations by irrational multiples of pi: it's not necessary for your gate set to contain a rotation by an irrational multiple of pi to be universal - for example the gate set $CNOT, H,T$ is universal where $T$ is a Z-rotation by $\pi/8$. All that's required for universality is that the gate set densely generates all rotations. A priori it's possible that even as you take longer and longer products of your gate elements, all the resulting rotations are by rational multiples of pi, but the gate is still universal as the rational rotations are dense in the reals. So I don't believe one can argue that "the gate added to the Clifford group must generate an irrational rotation to be universal". We only know that "the gate added to the Clifford group must densely generate and irrational rotation to be universal".

This post imported from StackExchange at 2016-06-20 15:13 (UTC), posted by SE-user Adam Bouland
answered May 11, 2016 by (40 points)
Hi Adam, thanks for the answer. I found that the proof of the theorem has been discussed in the appendix D of PhysRevX.2.041021. Also, rotation by irrational multiple of pi can be done using T gate in the manner discussed in Boykin et al. / Information Processing Letters 75 (2000) 101–107).

This post imported from StackExchange at 2016-06-20 15:13 (UTC), posted by SE-user user25957
I do believe that "the gate added to the Clifford group must generate an irrational rotation to be universal" may not be true but could you give a counter-example where the converse doesn't work/do you have an example where you add an irrational rotation to the Clifford group and don't get universality ?

This post imported from StackExchange at 2016-06-20 15:13 (UTC), posted by SE-user user25957
I agree the statement "the gate added to the Clifford group must generate an irrational rotation to be universal" could be true - but I don't know how to prove it, nor do I know of a counterexample.

This post imported from StackExchange at 2016-06-20 15:13 (UTC), posted by SE-user Adam Bouland
Hey, I mentioned that reference for rotation by irrational multiple of pi. The proof for the theorem is mentioned in appendix D of PhysRevX.2.041021

This post imported from StackExchange at 2016-06-20 15:13 (UTC), posted by SE-user user25957
For the converse, if I had to guess I'd guess that adding any irrational rotation to the Clifford group actually does generate all unitaries. I just don't know of any proof of this fact.

This post imported from StackExchange at 2016-06-20 15:13 (UTC), posted by SE-user Adam Bouland
 Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead. To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL. Please consult the FAQ for as to how to format your post. This is the answer box; if you want to write a comment instead, please use the 'add comment' button. Live preview (may slow down editor)   Preview Your name to display (optional): Email me at this address if my answer is selected or commented on: Privacy: Your email address will only be used for sending these notifications. Anti-spam verification: If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:p$\hbar$ysic$\varnothing$OverflowThen drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds). To avoid this verification in future, please log in or register.