7

You can also check their equivalence using Operator, a class from the Qiskit's quantum_info module as follows. from qiskit import * from qiskit.quantum_info import Operator import numpy as np Build and convert each circuit to an operator.( used your name convention for the circuits ) The code below builds and converts the first circuit qrz to op1 qrz = ...


5

The answer is no. Define X=[[0,1,0],[0,0,1],[1,0,0]] Z=[[1,0,0],[0,w,0],[0,0,w^2]], w^3=1 Then the Pauli group is generated by X and Z and is of order 27. With H being your matrix, you can check that H'XH and H'ZH are not in the group. Calculations like this are easy to do in gap The dim=3 counterpart of the Hadamard gate is the 3 dimensional Fourier ...


4

Yes, they should be. You can check their statevector to confirm that they indeed generate the same state. First, you can generate some arbitrary initial state and use the following two methods to show that they generate the same statevector. For instance: init_circ = QuantumCircuit(3) init_circ.ry(2.5,0) init_circ.ry(1.2, 1) init_circ.ry(0.5, 2) print(...


4

Yes. In general, a trivial construction of controlled-$U$, given a circuit for $U$ is to make every circuit element controlled off the control qubit. However, this introduces more overhead than necessary. If you have a part of your circuit that looks like $VWV^\dagger$, it is only necessary to make the $W$ controlled because the $VV^\dagger$ cancel in the ...


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

Based on the paper Elementary gates of ternary quantum logic circuit, the extensions of the Z gate are as follows $$ \begin{gather} Z^{[0]} = \text{diag}\{-1, I_2\} = \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \\ Z^{[1]} = \text{diag}\{1, -1, 1\} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 &...


3

Swapping 4-level qudits is equivalent to swapping pairs of qubits. Because you can encode a 4-level qudit into a pair of qubits. Similarly, swapping 8-level qudits will be equivalent to swapping triplets of qubits. The swap gate is convenient enough that this should hold regardless of how you map the 4-level qudit into the qubits (e.g. whether you map qudit |...


3

I believe Neilsen and Chuang were the first to use this particular notation. Previous work had referred to $S$ and $T$ as $\sigma_z^{1/2}$ and $\sigma_z^{1/4}$, respectively (Boykin et al. 1999). The use of $S$ may have been inspired by Deutsch's "S-matrix" (Deutsch 1989), though this was really a root-of-NOT gate. The use of $T$ may have been ...


2

Error correcting codes work the same on entangled qubits as any other qubit. All Alice and Bob have to do is separately encode their qubit into a Shor code. They each run an encoding circuit, apply noise, then run a decoding circuit and apply the corrections it inferred.


2

As you say, the difference is in the global phase. Let me explain using the first of your examples, $$ \left[\begin{array}{cc} \frac{1-i}{2} & \frac{1+i}{2} \end{array}\right]. $$ Mathematically, this is the same as $$ \left[\begin{array}{cc} \frac{e^{-i\pi/4}}{\sqrt{2}} & \frac{e^{i\pi/4}}{\sqrt{2}} \end{array}\right]=\left[\begin{array}{cc} e^{-i\...


2

Partial answer After discussing this with someone and going back through the API documentation on dynamical decoupling, it was clear on how these numbers were chosen. They corresponds to the gate duration time on the specific device in use. Which makes sense as why the specified durations in the questions are different. I didn't read the documentation ...


2

This is the same as Gate time or the time needed to execute that certain gate. If you look at the calibration data description of a specific device through the IBMQ website, and you will see that there is an option to see the CNOT gate time. For example, if you select your backend to be ibmq_santiago then you can see something like: Note the $CNOT_{q0,q1}$ ...


2

The recovery operator $R_{xz}$ is the same one as you would apply when doing quantum state teleporation $R'^\dagger_{xz}$ will be some combination of Clifford gates and Pauli gates, the latter classically conditioned on the Bell basis measurement outcome. I think that a single qubit ($n=1$) will be sufficient to demonstrate whats going on here. Let the ...


1

From a brief skim of the paper, the variance (second moment) is defined at the start of section 3, and worked with extensively throughout that section. Higher moments are defined at the start of section 4.


1

I was also flummoxed by the apparent puzzle why the value of $\mathbb{E}[\langle z|C|0^n\rangle|^2]$ is $2/2^n$ instead of $1/2^n$, but in my opinion I think the confusion arises from the symbol $\mathbb{E}$ (expectation value) whose meaning is rather ambiguous. Expectation value over what? Case 1: Fix a bit-string $z$. We ask for the value of $\langle z|C|0^...


1

If you don't want to write your own function to do this then one way to do this is through qiskit pauli_measurement. For example: from qiskit import QuantumCircuit, QuantumRegister, ClassicalRegister from qiskit.quantum_info import Pauli from qiskit.aqua.operators.legacy import pauli_measurement qr = QuantumRegister(4) cr = ClassicalRegister(4) qc = ...


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