14
votes
Accepted
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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12
votes
Accepted
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 ...
- 3,352
10
votes
Accepted
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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10
votes
Accepted
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 ...
- 4,487
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 ...
- 33.3k
9
votes
Accepted
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:...
- 4,487
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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8
votes
Accepted
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....
- 4,487
8
votes
Accepted
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 $|1\rangle$...
- 55.7k
8
votes
Accepted
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_{\...
- 11.9k
8
votes
Accepted
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:
...
- 33.3k
8
votes
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 ...
- 55.7k
8
votes
Accepted
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,...
- 4,487
8
votes
Accepted
Is every Clifford gate conjugate to a diagonal Clifford gate?
No, $D$ isn't guaranteed to be Clifford. That would require all entries on the diagonal to differ by multiples of 90 degrees (all be 1, $i$, $-i$, or -1 up to global phase). But that would imply $D$ ...
- 33.3k
7
votes
Accepted
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 ...
- 11.9k
7
votes
Accepted
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$ ...
- 2,865
7
votes
Accepted
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 ...
- 55.7k
7
votes
Accepted
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 ...
- 20.8k
7
votes
What are the relations between the permutation group and the Clifford group?
I guess you're talking about unitaries which preserve computational basis states, i.e. which act as $U|x\rangle = |f(x)\rangle$ where $f:\,\mathbb F_2^n \rightarrow \mathbb F_2^n$ is a reversible ...
- 4,487
6
votes
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 ...
- 3,613
6
votes
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 ...
6
votes
Accepted
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 ...
- 4,487
6
votes
Accepted
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
$$
where we ...
- 20.8k
6
votes
Accepted
Review paper on depth, qubits and $T$ gates number on Clifford+T decomposition for various "typical" algorithms
Copying over from "Are circuits with more than 1000 gates common?". Note that a Toffoli gate is roughly as expensive as 2T gates or 4T gates, depending on your architecture.
According to ...
- 33.3k
6
votes
Accepted
What is the fastest classical simulator for quantum circuits with only Clifford gates?
As far as I know, the current fastest simulator for stabilizer circuits is my simulator Stim (source code on github, paper in Quantum, python package on pypi). This is especially true if you're doing ...
- 33.3k
6
votes
Accepted
In the Clifford group, is the center of $ \overline{\text{Cl}_n} \equiv\text{Cl}_n/U(1)$ trivial?
Yes, the center of $\overline{\text{Cl}_n}$ is trivial.
Summary
The proof consists of two parts. First, we establish that if $[U]\in Z(\overline{\text{Cl}_n})$ is an element of the center then $U$ is ...
- 20.8k
6
votes
Finite subgroup of $U(4)$ containing a non-Clifford gate and all local Cliffords
This partial answer places additional restrictions on $U$.
Constructing unitaries with infinite order
By KAK decomposition, $U$ can be written as
$$
U=(A_1\otimes A_0)e^{i\alpha X\otimes X + i\beta Y\...
- 20.8k
6
votes
Accepted
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 ...
- 2,920
6
votes
Accepted
Is this single qubit gate in the Clifford hierarchy?
I assume that you mean entries in $\mathbb Q[\zeta_{2^k}]$, up to a global phase, right? This is certainly true for $k=1,2$: Clearly, any element of the Pauli group has, up to a global phase, entries ...
- 4,487
6
votes
Accepted
Is the Clifford hierarchy finite?
The third level is certainly finite, as it is a subset of all maps from the Pauli to the Clifford group, and both are finite sets. By induction, every level is thus finite.
It is a much harder problem ...
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