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How can one argue that the $S$-gate is Clifford while $T$-gate is not?

By definition, conjugation by a Clifford gate preserves the Pauli group $G = \langle X, Y, Z\rangle$. It is easy to check that $SXS^\dagger = Y, SYS^\dagger = -X$ and $SZS^\dagger = Z$. Since $G = \...
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Why are non-Clifford gates more complex than Clifford gates?

Yes, you are correct. Non-Clifford gates cannot be transversely implemented, instead implementation generally requires distilling magic states or Toffoli states. In practice this requires ...
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9 votes

How do I check if a gate represented by Unitary $U$ is a Clifford Gate?

Here's a simple strategy based on the idea that Clifford operations conjugate Pauli products into other Pauli products. If $U$ is a Clifford operation, then $U P U^\dagger$ (where $P$ is a Pauli ...
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Is the Clifford group finite?

Sometimes, there is a bit of confusion around the Clifford group in the field ... and it's a matter of definition. A lot of people define the Clifford group $\mathrm{Cl}_n(p)$ of $n$ qudits of prime ...
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How do I check if a gate represented by Unitary $U$ is a Clifford Gate?

Following Dehaene and de Moor (Theorem 6 in particular), every Clifford unitary can be represented (up to a global scalar factor) by an expression of the form $$ U = 2^{-k/2} \!\!\!\!\!\!\sum_{\...
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8 votes

How can one argue that the $S$-gate is Clifford while $T$-gate is not?

Geometrically: For a single qubit, we have the Bloch sphere and the stabiliser states span an octahedron inside it. Unitaries act in the adjoint representation as $SO(3)$, i.e. they induce rotations ...
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Is the SWAP gate a Clifford Gate? How would I express it using the Clifford Gate generators?

It's well known that you can make a swap out of three CNOTs. For reference, Stim's gate documentation includes H+S+CX decompositions of a lot of Clifford gates including the swap: ...
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Is the Clifford group a semidirect product?

$ \newcommand{\F}{\mathbb{F}} \newcommand{\C}{\mathbb{C}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\Sp}{\mathrm{Sp}} \newcommand{\Cl}{\mathrm{Cl}} \newcommand{\P}{\mathcal{P}} $Here's the short version:...
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Is there a closure property for the entire Clifford hierarchy?

It is actually possible to show that there is a simple, single-qubit operator (identified in discussion with John van de Wetering), which is a product of elements of $\mathcal C^{(3)}$ but which does ...
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Are almost-Clifford circuits almost easy to simulate?

Short answer: Yes, this should be possible. However, the details have to be filled out. The key is to relate this to magic monotones. There has been some development since the 2016 Bravyi-Gosset paper....
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How do I create an inverse identity gate?

As a general rule, you wouldn't bother constructing this: it is just a global phase that has no observable consequence. If you really insist on doing this, introduce an ancilla qubit in the $|1\rangle$...
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Why do we care about the number of $T$ gates in a quantum circuit?

If you are trying to implement a fault-tolerant quantum computation, you need to implement unitary gates that act on logical qubits. You typically have a finite set of these gates available, and what ...
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Does conjugation by a Clifford send each non-identity Pauli to every other non-identity Pauli with equal frequency?

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 ...
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Definition of the Pauli group and the Clifford group

The difference in definitions is from either taking the unitary group or the projective unitary group. That accounts for the constant prefactors of $\pm i$ that are missing. In lieu of a tikz ...
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What is the set of generators for the qutrit Clifford group?

From the paper Normal form for single-qutrit Clifford+T operators and synthesis of single-qutrit gates, the Clifford group in $p>2$ dimensions acting on a sigle qudit is generated by $S$ and $H$ ...
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How is a Toffoli gate built without using T gates?

The $T$ gate as well as all possible single qubit rotations are non-entangling operations. That means if we have a circuit composed of single bit rotations, any non-entangled $n$-bit input, it will ...
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Proof for Cardinality of the Clifford Group

What the author wrote is completely correct, they did not make a mistake. The subgroup of Cliffords fixing $X_n$ and $Z_n$ is indeed isomorphic to $C_{n-1}$ as a group, this is simply because this ...
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Getting non-Clifford after performing several Clifford gates in qiskit

The matrix $$ M = \frac{1}{\sqrt{2}}\begin{bmatrix}-i & 1\\-1 & i\end{bmatrix} $$ resembles $$ X/2 = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & -i\\-i & 1\end{bmatrix}\tag1 $$ where we ...
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In the Clifford group, is the center of $ \overline{\text{Cl}_n} \equiv\text{Cl}_n/U(1)$ trivial?

Yes, the center of $\overline{\text{Cl}_n}$ is trivial. Summary The proof consists of two parts. First, we establish that if $[U]\in Z(\overline{\text{Cl}_n})$ is an element of the center then $U$ is ...
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Prove that adding any non Clifford gate to the Clifford group yields a universal gate set

There is at least one other way to prove this I'm aware of. The argument uses the concept of a unitary 2-design and how this restricts the representation theory of a group. To avoid pathological cases,...
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Finite subgroup of $U(4)$ containing a non-Clifford gate and all local Cliffords

This partial answer places additional restrictions on $U$. Constructing unitaries with infinite order By KAK decomposition, $U$ can be written as $$ U=(A_1\otimes A_0)e^{i\alpha X\otimes X + i\beta Y\...
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Finite subgroup of $U(4)$ containing a non-Clifford gate and all local Cliffords

No. There is no way to add a non Clifford gate to the local Clifford group $ Cl_1^{\otimes 2} $ and get a finite group. Definitions: A subgroup $ G $ of $ GL_n(\mathbb{C}) $ is reducible if we can ...
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If $C$ is a Clifford circuit, is there necessarily a Clifford circuit $C'$ such that $CT=TC'$?

No. There always exists a unique unitary $U$ such that $CT=TU$. Namely, $U = T^\dagger C T$. The question is whether $U$ is Clifford. It turns out that this is not guaranteed. For a simple ...
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Definition of the Pauli group and the Clifford group

Note that the second definition actually doesn't make more sense in the context of the stabiliser formalism, as neither of $\pm i Y$ have a +1 eigenspace. That means that you can only describe states ...
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What are boost and shift operators and why are they called so?

The orthonormal basis $|j\rangle$ of the $d$ dimensional finite Hilbert space corresponds to a configuration space of equally spaced clockwise ordered $d$ points on a circle $S^1$ or equivalently, the ...
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What are boost and shift operators and why are they called so?

The shift operator takes his name from the fact that it shifts the position of its input, as in, it sends $1\to2$, $2\to3$ etc, with the last computational basis element being sent back to the first ...
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Isomorphism between the Clifford group and the quaternions

The quaternions are represented faithfully in two dimensions by the unit matrix and the Pauli matrices multiplied by the imaginary unit: $i = \sqrt{-1} X$, $j = \sqrt{-1} Y$ and $k = \sqrt{-1} Z$ ...
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A question from Aaronson 2004 paper

The terms $II$ and $ZZ$ do not uniquely specify the state $|11\rangle$ because you could equally have the state $|00\rangle$. Indeed, you should not include the identity term in your stabilizer. Thus, ...
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Review paper on depth, qubits and $T$ gates number on Clifford+T decomposition for various "typical" algorithms

Copying over from "Are circuits with more than 1000 gates common?". Note that a Toffoli gate is roughly as expensive as 2T gates or 4T gates, depending on your architecture. According to ...
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Clifford gates are transversal What exactly does this transversal mean? What is the difference between non-Clifford gates and Clifford gates?

Transversal and Clifford are not as closely linked as your question would seem to imply. Transversal gates are those for which an error-correcting code can achieve the transformation on a logical ...
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