10

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 $G = \langle -X, Y, Z\rangle$ we see that $S$ is Clifford. On the other hand, $TXT^\dagger = \begin{pmatrix} & e^{-i\pi/4}\\ e^{i\pi/4}&\end{pmatrix} \notin G$, ...


8

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 significantly more spacetime volume than Clifford gates. For reference, see the introduction sections here and here. The natural expectation would be that no quantum ...


8

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 dimension $p$ as the unitary normaliser of the generalised Pauli group (e.g. Gottesman, Nielsen & Chuang). As such, it is clearly not a finite group as the ...


8

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 of the Bloch sphere. It is easy to see, that only rotations about the $X,Y,Z$ axis with angles $\theta=\pi/2,\pi,3\pi/2,2\pi$ preserve the octahedron. This ...


7

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. I think one can use an argumentation based on stabiliser / Clifford extent and the results in the BBCCGH paper from 2019. Similarly, one can also argue for ...


7

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 operation on one of the qubits) will be a matrix equivalent to a product of Pauli operations. If you check this for each $X_q$ and $Z_q$ for each qubit $q$, the ...


7

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 $|1\rangle$ state and apply a $Z$ gate to it. PS "inverse identity gate" is a really bad name for it. The identity operation is its own inverse.


6

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 $k = \sqrt{-1} Z$ respectively, thus you only need to write $H$ and $P$ the linear combinations: $H = -\frac{\sqrt{-1} }{\sqrt{2}}(i+k)$ and $P = \frac{1+\sqrt{-1}}{2}(1-k)$ However,...


6

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_{\substack{x_r,x_c \in \{0,1\}^k \\ x_b \in \{0,1\}^{n-k}}}\!\!\!\!\! i^{p(x_b,x_c,x_r)} (-1)^{q(x_b,x_c,x_r)} \bigl\lvert T_1[x_r;x_b] \bigr\rangle\!\bigl\langle T_2[...


6

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 result in a $n$-bit non-entangled output. The $CCNOT$ gate however is of course entangling, really all controlled gates are entangling, which means there exist ...


6

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 you really care about is making your operations in such a way as to keep the fault-tolerant threshold as small as possible. If you calculate a fault-tolerant ...


5

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 counterexample take Hadamard for $C$. Then $$ U=T^\dagger H T = \frac{1}{\sqrt{2}}\begin{bmatrix} 1 & e^{i\pi/4} \\ e^{-i\pi/4} & -1 \end{bmatrix} $$ is not ...


5

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 vertices of a $d$-dimensional regular polygon. One may think of a point as a discrete location of a particle, then the shift operator $X$ shifts the particle ...


5

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 one: $d\to 1$ (or the same thing starting with $0$, depending on notation). As per the "boost" operator $Z$, I have usually seen those referred to as "clock ...


5

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 commutative diagram \begin{align} <X,Y,Z> & \hookrightarrow & U(2)\\ \downarrow & & \downarrow\\ <[X],[Z]> & \hookrightarrow & PU(...


5

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, you need to add a second term, which could be either $-ZI$ or $-IZ$. Either way, you can easily see how to make a product $-ZI$ out of your stabilizers.


5

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 subgroup acts by assumption as $$ U (\sigma_1 \otimes \dots \otimes \sigma_n) U^\dagger = \tilde U (\sigma_1\otimes\dots\otimes\sigma_{n-1})\tilde U^\dagger \...


5

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 follow the notation $\pm X/2$ for the $\pm\frac{\pi}{2}$ rotation around the $X$ axis as used in the table B.6 on page 101 in Julian Kelly's PhD thesis. We can make ...


5

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: Stabilizer Generators: X_ -> +_X Z_ -> +_Z _X -> +X_ _Z -> +Z_ Unitary Matrix: [+1 , , , ] [ , , +1 , ] [ , +1 , , ] [ , ...


5

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 and Clifford groups, respectively. For $P, Q\in G_n$, let $C_{P\to Q}$ denote the set of Clifford operators that map $P$ to $Q$ under conjugation, i.e. $$ C_{P\...


5

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 bulk sampling: Note the Y axis; that's a log-log plot. There are some simulators that are faster in specialized circumstances, e.g. for piecewise separable ...


4

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 which are stabilised by operators with two or more factors of $\pm i Y$. This is only enough to simulate real stabiliser circuits, which is what you're noticing....


4

There is a very closely related representation of the tableau representation of Aaronson (and Gottesman), which works not only for qubits but for qudits of arbitrary finite dimension, which works particularly well for purely Clifford circuits (i.e. at most one terminal measurement). In this alternative representation, one has tableaus describing how ...


4

The diffusion operator is a multi controlled not operation (modulo some hadamards). It's not a Clifford operation. Also any useful oracle you'd use with Grover's algorithm won't be Clifford operations either, since any Clifford oracle accepts 0%, 50%, or 100% of all inputs which makes search trivial.


4

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 $H$ given by: $$ \begin{gather} S=\sum_{j=0}^{p-1}\omega^{j(j+1)2^{-1}}|j\rangle\langle j| \\ H = \frac{1}{\sqrt{p}}\sum_{j=0}^{p-1}\sum_{k=0}^{p-1}\omega^{jk}|j\...


4

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 Table III of https://arxiv.org/abs/2011.03494 , quantum chemistry algorithms looking at properties of the FeMoCo molecule use half a billion Toffoli gates. ...


4

No, it's not possible. For example, being able to directly measure $X+Y$ would allow you to prepare T states and thereby perform T gates, which are not stabilizer operations. If the fact that $X$ and $Y$ don't commute bothers you, then note that preparing $|+\rangle^{\otimes n}$ and then measuring $\sum_{k=0}^{n-1} Z_k$ and getting a result of $n-2$ would ...


4

TL/DR The dimension of the stabilizer $\mathcal{S}$ is $2^{n}$ because there are exactly $n$ generators to make the dimension of the stabilizer's eigenspace exactly $1$. Then, each element in $\mathcal{S}$ is a unique combination of the generators for a total of $2^{n}$ combinations. Long version It's easy to show that the stabilizer $\mathcal{S}$ forms a ...


3

One way that you might do it is as follows. Consider the unitary sequence $V=HTHT$. Because it's unitary, we can write it in the form $$ V=e^{i\gamma}e^{-i\theta\vec{n}\cdot\vec{\sigma}}=e^{i\gamma}(\cos\theta I-i\sin\theta \ \vec{n}\cdot\vec{\sigma}) $$ where $\vec{n}\cdot\vec{n}=1$. If you work through the details, you'll find that $$ \cos\theta=\cos^2\...


3

You might be interested in controlled version of $-I$. Despite the fact that you can neglect global phase in case of non-controlled gates, you cannot do so in case of controlled version. The controled gate $-I$ is described by matrix \begin{pmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -...


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