I started reading about Randomized Benchmarking (this paper, arxiv version) and came across "unitary 2 design."

After some googling, I found that the Clifford group being a unitary 2 design is a specific case of "Quantum t-design."

I read the wikipedia page and a few other references (this one for example, non pdf link to the website that links to the pdf).

I would like to have some intuitive understanding of the difference between different t designs and what makes the Clifford group a 2 design.

  • $\begingroup$ Welcome to Quantum Computing SE! It might(?) be worth mentioning that the Clifford group actually forms a 3-design. Also, would you happen to have a non-pdf link (i.e. a link to a webpage that links to the pdf) of the 'Unitary t-designs' reference? $\endgroup$
    – Mithrandir24601
    Commented Aug 22, 2018 at 8:45
  • 1
    $\begingroup$ @Mithrandir24601 Thanks for the comment. I added a link to the website that links to the pdf. $\endgroup$
    – Blackwidow
    Commented Aug 22, 2018 at 16:33

1 Answer 1


The $t$ in $t$-design is essentially a measure of how good a job the set of gates does in terms of randomising a state (the larger t, the more random, with properly random requiring the infinite limit). Often, you want to compute the average of some function over all possible pure input states, which is equivalent to fixing the input state and averaging over all possible unitaries. However, averaging over all possible unitaries is a pain, and is unnecessary if the function you want to compute is simple enough. If the function you want is a polynomial of degree t or less in terms of the coefficients of the input state, it is sufficient to average over a set of gates that comprise a t-design.

Another way of thinking about this is, instead of a degree t polynomial, you can talk about calculating a linear function of t copies of the input state. This is more like you would do in an actual experiment.

As for what makes the Clifford group a 2-design, I guess you just have to sit down and do the maths. There's a proof in section A.1 of this paper. For the special case of the Clifford group on a single qubit, let S be the set of 1-qubit Clifford gates. Then you need to show that $$ \sum_{s\in S}s\otimes s|00\rangle\langle 00|s\otimes s\propto\mathbb{I}+\text{SWAP} $$ The critical thing here is that there’s 2 copies of the state that we’re averaging over.

  • $\begingroup$ I may be confused about unitaries $t$-designs and quantum states $t$-designs, but shouldn't the sum be proportional to $\mathbb{I}+\mathrm{SWAP}$, as mentioned in this answer? The reason I think this shouldn't be proportional to the identity is that it means that the density matrix is the maximally mixed state. However, if I'm not mistaken, performing a SWAP test on the LHS results in $|0\rangle$ with probability $1$ but performing it on the RHS yields it with probability $\frac34$. Am I missing something here? $\endgroup$
    – Tristan Nemoz
    Commented Dec 19, 2022 at 9:50
  • 1
    $\begingroup$ Yes, you're right. Obviously if I take a convex combination of symmetric states, the output is symmetric, and hence cannot be the identity matrix. $\endgroup$
    – DaftWullie
    Commented Dec 19, 2022 at 12:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.