Is it always possible to diagonalize a Clifford gate $ g $ using a gate $ V $ from the third level $\mathcal{C}^{(3)}$ of the Clifford hierarchy? In other words can every Clifford gate be written as $$ g=VDV^{-1} $$ where $ V $ is from $\mathcal{C}^{(3)}$ and $ D $ is diagonal?

Update: Just rehashing the argument from @AdamZalcman mostly for my own benefit. Any matrix which can be expressed $ VDV^{-1} $ for $ D $ diagonal and $ V \in \mathcal{C}^{(3)} $ is of the form $ aI+bVZV^{-1} $. When $ b=0 $ then all scalar matrices can be obtained this way. When $ a=0 $ then all $ VZV^{-1} $ for $ V \in \mathcal{C}^{(3)} $ can be obtained this way. $ VZV^{-1} $ has projective order 2 (squares to a scalar matrix) and it is in the Clifford group (since $ V \in \mathcal{C}^{(3)} $ and $ Z $ is Pauli) so it must be either a non identity Pauli or a Hadamard type matrix, up to global phase. The final case to consider is $ a,b $ both nonzero. The only Clifford gates that can be written as a nontrivial combination of $ 1 $ and a Pauli are the square roots of Pauli gates. However no Clifford gates can be written as nontrivial combination of $ 1 $ and a Hadamard (such a matrix would have Pauli coordinates exactly three of which are nonzero, no Clifford gate has such coordinates see second column in table below). Thus we have exhausted all Clifford gates which can be written as $ aI+bVZV^{-1} $ for $ V \in \mathcal{C}^{(3)} $ and we finally conclude that no order 3 single qubit Clifford gate can be written as $ VDV^{-1} $ for $ V \in \mathcal{C}^{(3)} $.

Note that throughout the argument we an remove the up to global phase/ up to scalar ambiguity by specializing to the case of determinant 1 Cliffords. Then this shows determinant 1 Clifford gates with (projective) order 3 cannot be written as $ VDV^{-1} $ for $ D $ diagonal and $ V \in \mathcal{C}^{(3)} $.


1 Answer 1


TL;DR: No, this isn't always possible. In fact, a third of the gates in the single-qubit Clifford group cannot be diagonalized by a matrix in $\mathcal{C}^{(3)}$.

First note that the Pauli matrices $\{I, X, Y, Z\}$ form a basis of the complex vector space of all $2\times 2$ complex matrices. Moreover, $\{I, Z\}$ is a basis of the subspace consisting of the diagonal complex matrices. Consequently, there are $a,b\in\mathbb{C}$ such that $D=aI+bZ$. Now, suppose that conjugation by $V\in\mathcal{C}^{(3)}$ sends $Z$ to the Clifford $C\in\mathcal{C}^{(2)}$. Then $g=aI+bC$. Moreover, conjugation preserves the trace, so $C$ is a traceless Clifford. By the listing in the Appendix, the only traceless single-qubit Cliffords are non-identity Paulis and the Hadamards.

Case $1$: $C$ is a Pauli

If $C=P\in\{X, Y, Z\}$ then $$ g=aI+bP\tag1 $$ so by the expansions of Cliffords in the Pauli basis in the Appendix, we see that $g$ is either the identity, a Pauli matrix or a square root of a Pauli matrix.

Case $2$: $C$ is a Hadamard

If on the other hand, $C$ is a Hadamard, then $g$ is a linear combination of the identity and two Paulis $$ g=aI+\frac{b}{\sqrt2}P\pm\frac{b}{\sqrt2}Q\tag2 $$ where $P,Q$ are two distinct non-identity Paulis. By inspecting the expansions of single-qubit Cliffords in the Pauli basis listed in the Appendix, we see that $(2)$ implies that $g$ is the identity or a Hadamard.

Finally, since $(1)$ and $(2)$ exhaust the possible expansions of $g$ in the Pauli basis, we conclude that none of the eight order-$3$ single-qubit Cliffords can be diagonalized by a matrix from the third level of the Clifford hierarchy.

Appendix: Listing of $\mathcal{C}^{(2)}$

The following list exhausts all twenty four single-qubit Cliffords: $$ \begin{array}{c|c|c|c|c} \text{Name}&\text{Expansion in Pauli basis}&\text{Count}&\text{Order}&\text{Trace}\\ \hline \text{Identity}&I&1&1&2\\ \text{Paulis}&P&3&2&0\\ \text{Square roots of Paulis}&\frac{1}{\sqrt2}(I\pm iP)&6&4&\sqrt2\\ \text{Hadamards}&\frac{1}{\sqrt2}(P\pm Q)&6&2&0\\ \text{Order-3 Cliffords}&\frac12(I\pm iP\pm i Q\pm iR)&8&3&1 \end{array} $$ where $P,Q,R\in\{X,Y,Z\}$ are distinct non-identity Paulis.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.