Elements of the (qubit) Clifford Hierarchy are unitary matrices. For a good definition of the Clifford Hierarchy see: Is there a closure property for the entire Clifford hierarchy?

While a complete structure theorem for the Clifford Hierarchy, $\mathcal{CH}$, is still lacking, can we prove that if $U\in \mathcal{CH}$ then $U^{\dagger}\in \mathcal{CH}$?


$U$ is semi-Clifford

Semi-Clifford means that $U = C_L D C_R$ where $C_L, C_R$ are Clifford and $D$ is a diagonal gate in $\mathcal{CH}$.

Then, $U^\dagger = C_R^\dagger D^\dagger C_L^\dagger$. By [1] the diagonal gate in level k form a group so the inverse of $ D $ is also in level $ k $. Each level of the Clifford Hierarchy is closed under left and right multiplication by Clifford gates. So in this case $U^\dagger$ is even in the same level of $\mathcal{CH}$ as $U$.

$U$ is generalized semi-Clifford

Generalized semi-Clifford means that $U = C_L P D C_R$ where $C_L, C_R$ are Clifford, $D$ is a diagonal gate in $\mathcal{CH}$, and $P$ is a permutation (on states) in $\mathcal{CH}$. This was proven in [2].

In this case, $U^\dagger = C_R^\dagger D^\dagger P^\dagger C_L^\dagger = C_R^\dagger P^\dagger (P D^\dagger P^\dagger) C_L^\dagger$. And since the term in parentheses is a diagonal matrix with the same $2^k$ root-of-unity entries as $D$, it must be in $\mathcal{CH}$ (since $D$ is in $\mathcal{CH}$). We can also see that $U^\dagger$ must be generalized semi-Clifford and $U^\dagger$ is in $\mathcal{CH}$ iff $P^\dagger$ is in $\mathcal{CH}$. To complete the proof of this case I need to show that $P^\dagger$ is in $\mathcal{CH}$ which I currently do not know how to do.

General $U$

It is not known if all elements in $\mathcal{CH}$ are generalized semi-Clifford. Nevertheless, we may be able to prove (or disprove) this conjecture for general $U \in \mathcal{CH}$.
  • $\begingroup$ +1 great question! I just got back from thailand so I've been away from QCSE and arXiv so first of all congrats on posting your preprint (reference [2] in this question) it looks great! Anyway, I think that asking for closure under inverses is the most natural follow up to quantumcomputing.stackexchange.com/questions/26499/… where it is shown that the Clifford hierarchy is not closed under products. Wish I'd thought to ask this! I'll put a bounty on this soon if it doesn't get any action $\endgroup$ Commented Feb 8, 2023 at 18:25
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    $\begingroup$ For the semi-clifford case you use that the Clifford group is closed under inverses and the diagonal gates in the $ k $ th level of the hierarchy are closed under inverses, but I feel like you need to mention a final step which is that the $ k $ level of the Clifford hierarchy is closed under left and right multiplication by Clifford gates e.g. proposition 3 of arxiv.org/abs/0712.2084. Since $ D^\dagger $ is in the $ k $ level and $ C_R^\dagger, C_L^\dagger $ are Clifford then the product $U^\dagger = C_R^\dagger D^\dagger C_L^\dagger$ is also in the $ k $ level of the hierarchy. $\endgroup$ Commented Feb 8, 2023 at 18:38
  • $\begingroup$ Or are you making some stronger claim that the entire group generated by the level k diagonal gates together with the Clifford gates actually is contained in level k? $\endgroup$ Commented Feb 8, 2023 at 18:42
  • $\begingroup$ You're correct. I'm making use of the proof that left and right multiplication by Clifford gates preserve the level in $\mathcal{CH}$. For clarity I should definitely mention that. Thanks for pointing it out. $\endgroup$ Commented Feb 10, 2023 at 4:56
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    $\begingroup$ I think I found an order 3 permutation in the third level of CH. Let CCX(c,c,t) and CX(c,t) denote the controlled-controlled not and the controlled not gates, respectively. And let c and t denote the qubits that the controls and targets act upon. Then, I claim that CCX(1,2,3)CX(3,4)CX(4,3) is an order three permutation at the third level in CH. $\endgroup$ Commented Feb 14, 2023 at 21:40

1 Answer 1


This partial answer breaks the question of the closure of the levels of the Clifford hierarchy under inversion into two parts: the easy question of closure under complex conjugation and the harder question of closure under transposition. It also offers a proof of the former and an informal argument for the latter.

Lemma 1 (Every level of the Clifford hierarchy is closed under complex conjugation)
For every $k\geqslant 1$, we have $$U\in\mathcal{C}^{(k)}\iff\overline{U}\in\mathcal{C}^{(k)}.\tag1$$ Proof. Equivalence $(1)$ is obviously true for every $U$ in the Pauli group $\mathcal{C}^{(1)}$. Assume that $(1)$ holds for $k$ and let $U\in\mathcal{C}^{(k+1)}$ and $P\in\mathcal{C}^{(1)}$. By definition $UPU^\dagger=V\in\mathcal{C}^{(k)}$. But then $$ \overline{U}P\overline{U}^\dagger=\pm \overline{U}\overline{P}\overline{U}^\dagger=\pm\overline{(UPU^\dagger)}=\pm\overline{V}\in\mathcal{C}^{(k)}\tag2 $$ so $\overline{U}\in\mathcal{C}^{(k+1)}$. $\square$

This immediately gives us

Corollary 2 (Symmetric subset of every level is closed under inversion)
If $U=e^{i\theta}U^T$ for some $\theta\in[0,2\pi)$, then $$U\in\mathcal{C}^{(k)}\iff U^\dagger\in\mathcal{C}^{(k)}\tag3$$ for every $k\geqslant 1$.

This provides a simple alternative route to closure under inversion for diagonal gates and extends it to gates such as $\sqrt{\text{iSWAP}}$ and various other swap gates. In fact, $(3)$ applies to every gate whose Pauli expansion consists of terms for which the parity of the number of $Y$ operators is the same (either all even or all odd). Unfortunately, this set does not include all permutations.

Conjecture 3 (Every level is closed under transposition)
For every $k\geqslant 1$, we have $$U\in\mathcal{C}^{(k)}\iff U^T\in\mathcal{C}^{(k)}\tag4.$$

Informal argument. Clifford hierarchy arises as the set of gates that admit fault-tolerant implementation using gate teleportation, see page $3$ in "Quantum teleportation is a universal computational primitive". The protocol is summarized in figure $2$ of the paper

enter image description here

The protocol consists of two stages. In the first stage, we prepare $n$ Bell pairs $|\Phi^n\rangle$ and apply the desired gate $U\in\mathcal{C}^{(k)}$ to one half of the Bell pairs obtaining $$ |\Psi^n_U\rangle=(I\otimes U)|\Psi^n\rangle.\tag5 $$ In the second stage, we perform Bell measurements $B$ on the input state $|\alpha\rangle$ and the other half of the state created in $(5)$ and finally apply corrections conditioned on the measurement outcomes using $R'^\dagger_{xz}\in\mathcal{C}^{(k-1)}$.

Now, it is easy to check that $$ |\Psi^n_U\rangle=(U^T\otimes I)|\Psi^n\rangle.\tag6 $$ Thus, two changes are needed in the protocol above to effect $U^T$ rather than $U$. First, we can apply $U$ on the middle qubit rather than the bottom qubit (or alternatively apply $U^T$ to the bottom qubit). Second, replace the $R'_{xz}=UR_{xz}U^\dagger$ correction with the $R''_{xz}=U^TR_{xz}\overline{U}$ correction.

We know that $R'_{xz}\in\mathcal{C}^{(k-1)}$, but we don't know whether $R''_{xz}\in\mathcal{C}^{(k-1)}$. However, one of the changes to the protocol necessary to effect $U^T$ rather than $U$ is operationally very simple: apply $U$ on a different qubit. One might then suspect that the associated update to the correction isn't too complicated and in particular, the correction stays in $\mathcal{C}^{(k-1)}$. I don't know how to prove this at the moment, though. $\square$

Lemma 1 and Conjecture 3 combine to give us

Corollary 4 (Every level is closed under inversion)
For every $k\geqslant 1$, we have $$U\in\mathcal{C}^{(k)}\iff U^\dagger\in\mathcal{C}^{(k)}\tag7.$$

  • $\begingroup$ oh dang sorry the bounty expired, I would have totally awarded it to you, this answer looks really fun! $\endgroup$ Commented Feb 18, 2023 at 1:13
  • $\begingroup$ This is a significant advance! You're close to proving not only that $\mathcal{CH}$ is closed under the inverse operation, but the stronger claim that $\mathcal{C}^{(k)}$ is closed under the inverse operation! This is certainly worth the award. Though please update if you make progress on conjecture 3 which, I agree, sounds very reasonable. $\endgroup$ Commented Feb 18, 2023 at 20:35

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