I see here in Olivia DeMatteo's notes, she states:

When we consider the action of the entire Clifford group on a single non-identity Pauli, it maps that Pauli to each of the $d^2 − 1$ other possible Paulis an equal number of times. Since we have $|C_n|/|P_n|$ possible Cliffords, we get mapped to each Pauli $\frac{|Cn|/|Pn|}{d^2−1}$ times.

This is also stated in this paper

[...] conjugation under the Clifford group maps each nonidentity Pauli element to every other nonidentity Pauli element with equal frequency.

I don't see a proof for this claim anywhere. Does anyone know of a proof for it? I think the proof for the cardinality of the Clifford group can be used to prove it, but I'm sure there is a simpler way to prove it.

  • $\begingroup$ Note that (in this part of his notes but not in other parts of his notes) DeMatteo is thinking of the Pauli group mod global phase. So this Pauli group is really an elementary abelian group. The quotient group $ C_n/P_n $ is the full automorphism group of this elementary abelian group. $ C_n/P_n $ acts transitively on all $ |P_n|-1 $ nonidentity Paulis (recall $ d^2 $ is size of the Pauli group mod phases). So by orbit stabilizer theorem the stabilizer $ P \to P $ has exactly size $ |C_n/P_n|/(|P_n|-1) $. Maps $ P \to Q $ are a coset of the stabilizer $ P \to P $ so same size for all $ Q $. $\endgroup$ Commented Mar 24, 2023 at 16:29

1 Answer 1


Conjugating a non-identity Pauli operator $P$ by Clifford operators yields all non-identity Pauli operators, including $P$, with equal frequency.

Proof Let $G_n$ and $C_n$ denote the $n$-qubit Pauli and Clifford groups, respectively. For $P, Q\in G_n$, let $C_{P\to Q}$ denote the set of Clifford operators that map $P$ to $Q$ under conjugation, i.e.

$$ C_{P\to Q}=\{U\in C_n\,|\,UPU^\dagger=Q\}. $$

For any $P\in G_n$, $C_{P\to P}$ is a subgroup of $C_n$. Indeed, for any $U\in C_{P\to P}$ its inverse$^1$ $U^\dagger\in C_{P\to P}$ and for any $U, V\in C_{P\to P}$ their product $UV\in C_{P\to P}$.

Let $P,Q\in G_n$ and suppose that neither $P$ nor $Q$ is the identity$^2$. Then there exists$^3$ $V\in C_{P\to Q}$. It is easy to see that

$$ C_{P\to Q} = VC_{P\to P} $$

where $VC_{P\to P}$ is the left coset

$$ VC_{P\to P} := \{VU\,|\,U\in C_{P\to P}\} $$

of the subgroup $C_{P\to P}$ in $C_n$. By group theory, all cosets have the same size, so

$$ |C_{P\to Q}| = |C_{P\to P}| $$

for all $P,Q\in G_n$. Conclusion follows from the observation that $|C_{P\to Q}|$ is precisely the number of Cliffords that map $P$ to $Q$ under conjugation.$\square$

$^1$Strictly speaking, we do not need to check for closure under inverses. Explanation why is an exercise for the reader.
$^2$ The assumption is important, because $C_{I\to I}=C_n$ and $C_{I\to R} = C_{R\to I} =\emptyset$ for all non-identity $R\in G_n$.
$^3$ Proof that such $V$ exists is another exercise for the reader.

  • $\begingroup$ Just pointing out that you and the DeMatteo notes linked above are both working with the Pauli group mod global phase. Otherwise Paulis like $ X $ and $ iX $ would not be conjugate since $ X $ has order $ 2 $ and $ iX $ has order $ 4 $. $\endgroup$ Commented Mar 24, 2023 at 15:41

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