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 2 design.

I apologize in advance if the question is too basic.

  • $\begingroup$ 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
    Aug 22 '18 at 8:45
  • 1
    $\begingroup$ @Mithrandir24601 Thanks for the comment. I added a link to the website that links to the pdf. $\endgroup$
    – Blackwidow
    Aug 22 '18 at 16:33

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. Let S be the set of 1-qubit Clifford gates. Then you need to show that $$ \sum_{s\in S}s\otimes s|00\rangle\langle 00|s\otimes s\propto\mathbb{I} $$ The critical thing here is that there’s 2 copies of the state that we’re averaging over.


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