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Background from Is there a closure property for the entire Clifford hierarchy? : The Clifford hierarchy (on $n$ qubits), is a collection of nested subsets $\mathcal C^{(1)} \subset \mathcal C^{(2)} \subset \mathcal C^{(3)} \subset \cdots \subset \mathbf U(2^n)$ of the unitary operators on $n$ qubits, defined recursively as follows:

  • $ \mathcal C^{(1)} $ is the Pauli group on $n$ qubits, i.e., the set of ($n$-fold tensor products of) Pauli operators with real or imaginary scalar factors: tensor products of $\{ \mathbf 1, X, Y, Z \}$ with a scalar factor of $\pm 1$ or $\pm i$.

  • For $k > 1$, we recursively define $\mathcal C^{(k)}$ as

    $$ \mathcal C^{(k)} = \Bigl\{ U \in \mathbf U(2^n) \mathrel{\Big\vert} \forall P \in \mathcal C^{(1)} : U P U^\dagger \in \mathcal C^{(k-1)} \Bigr\}$$ — that is, the set of unitaries that, when used to conjugate a Pauli operator, yields an operator at one level lower in the hierarchy.

So, for example, the Clifford group is the second level of the Clifford hierarchy, $\mathcal C^{(2)}$, as any $U \in \mathcal C^{(2)}$ maps a Pauli operator to another Pauli operator (i.e., an element of $\mathcal C^{(1)}$) by conjugation.

The gates $ X,Z $ are in the Pauli group, the controlled gates $ CX,CZ $ are in the Clifford group. In general $ C^{r}X, C^{r}Z $ are in the $ r+1 $ level of the Clifford hierarchy. Similarly, the phase gate $ S $ is in the Clifford group and $ C^{r}S $ is in the $ r+2 $ level of the Clifford hierarchy.

Is it true that if $ U $ is a Clifford gate then controlled $ U $ is in the third level of the Clifford hierarchy? And in general is it true that the $ r $-controlled version of a Clifford gate will always be in the $ r+2 $ level of the Clifford hierarchy?

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I want to guess no. The main reason it usually works is $C-V X_0 C-V^{\dagger} = X_0(C-V)^2 V^\dagger $, which is helpful in getting lower down the hierarchy. The case that throws a spanner in the works is $C-HS$, since $HS$ is a twelfth route of unity, we can't escape the factor of three and then $C-HS\, X\, C-HS^\dagger$ isn't Clifford despite $HS$ being Clifford.

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  • $\begingroup$ that is a great identity, where did you learn it? I love that so nice that the squaring just takes you down a level of the hierarchy (at least for diagonal gates and lots of other gates) but I see it doesn't work for facet gate. Is there a similar identity for qutrit gates? $\endgroup$ Commented Sep 5 at 1:41
  • $\begingroup$ Most of my research is around blind quantum computation, so conjugating Cliffords is a big part of it! The more general version of the formulae in a blind setting is here (arxiv.org/pdf/2404.07052). For the qutrit case, it would depend on how you define a control gate in that world but if its CV |a>|psi> =|a> V^a |psi>, then we would have CV X^b CV^dag |a> = |a+b> V^-a V^(a+b) |psi>. Working everything out in ternary will give some expressions for it like CV^(f(a,b)) V^(g(a,b)). $\endgroup$ Commented Sep 6 at 9:53
  • $\begingroup$ The more annoying one I found, was sqrt(H) = (1/sqrt(2)) (I+H) exists no where in the Clifford Hierarchy! Controlled versions and square roots only behave nicely in the X,Y,Z directions. $\endgroup$ Commented Sep 6 at 9:55
  • $\begingroup$ Is the $C-HS$ gate you describe in any level of the Clifford hierarchy? $\endgroup$ Commented Sep 26 at 3:05
  • $\begingroup$ It shouldn't be. Say C-HS is in the n'th level of the hierarchy, C-(HS)^2 must be in the n-1 level due to the great identity and the the hierarchy is closed under left and right multiplication by Cliffords. C-(HS)^4 is n-2, C-(HS)^8 is n-3, C-(HS)^16 = C-(HS)^4 is n-4. but we know this gate to be in n-2 th level. It's not in the hierarchy but the terms it can become are finite and they repeat in cycles. $\endgroup$ Commented Sep 26 at 8:06

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