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I'm currently interested in writing Clifford gates in the Pauli basis i.e. $C = \sum_{P}\alpha_P P$, where we are summing over all pauli strings of fixed length. For example:

  • $H =\frac{1}{\sqrt{2}}(X+Z)$
  • $CX = \frac{1}{2}(II+XI+IZ-XZ)$
  • $HS = \frac{1}{2}(I+iX+iZ+iY)$ (up to a phase)

The two properties I suspect are true and want to prove are:

  1. All non-zero coeffecients have the same magnitude (possibly up to $\pm 1, \pm i$)
  2. The number of non-zero coeffecients is a power of 2.

My thought process so far is $C =\sum_{P}\alpha_P P = CCC^\dagger = \sum_{P}\alpha_P CPC^\dagger = \sum_{P}\alpha_P P'$ where $P' = CPC^\dagger$. This tells us the coefficients in the orbit of $P$ under conjugation by $C$ must be the same up to $\pm 1$. However, this doesn't give any reason why different orbits couldn't have different magnitudes. A different approach to talk about the relation between orbits would be for a pauli $P$ and $Q = CPC^\dagger$ then

$PC = CQ$

$P \sum \alpha_{P'} {P'} =\sum \alpha_{P'} P' Q $

$P\sum \alpha_{P'} P' = \sum\alpha_{P'} {P'} Q $

So for any pauli $P'$ we must have, $\alpha_{PP'} = (\pm 1,\pm i)\alpha_{P'Q}$ depending on how we need to massage the paulis to form the standard pauli basis. This has allowed to me to construct all the previously mention examples of Cliffords but the process does not feel systematic enough to form a proof.

Is there any standard proofs of these two properties (if they are true)?

Ethan Davies
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1 Answers1

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In this paper, the authors show that any $N$ qubit Clifford can be decomposed in the form $$ G=E_0\prod_{n=1}^k\frac{I+iE_n}{\sqrt{2}} $$ where the $E_n$ are $N$-qubit Pauli matrices. Moreover, the $E_n$ are Pauli matrices $X_{x_n}Z_{z_n}$ where the $v_n=x_nz_n$ are linearly independent (if I've correctly understood that bit of the paper). This must mean that there are no terms in $\{E_n\}_{n=1}^k$ which multiply to give identity. Hence the product $$ \frac{I+iE_n}{\sqrt{2}} $$ must comprise $2^k$ distinct terms, each with an amplitude $1/\sqrt{2^k}$ up to a phase $\pm1,\pm i$.

Update: while the $\{E_n\}_{n=2}^k$ are linearly independent, there is one term $E_1$ which may be linearly dependent on the others. This may affect the conclusion, but it's a very limited set of cases to consider.

DaftWullie
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