Problem

 My problem for tomorrow: untangling the Bell's Inequality circuit,

So far:

                                                            

*     *     *

   Been reading the docs from IBM; turns out one always begins a circuit with

a  qubit in the |0> state. Below, the teleportation protocol explained:

                                                                                                                                         

*     *     *

It is the presence of a T-gate ( a pi/4 phase change) which makes this circuit a quantum one, 

impossible to simulate on a classical computer. Hadamard, CNOT and T form a complete set

of quantum gates, which can get us anywhere on the Bloch sphere!

Certainly! Let’s delve into the fascinating world of quantum gates.

The S gate (also known as the P gate or phase gate) is indeed a Clifford gate. Here’s why:

1. Clifford Gates:

   • The Clifford gates are a set of mathematical transformations that normalize the n-qubit Pauli group. These gates play a crucial role in quantum computing and quantum information theory.
   • The Clifford group consists of transformations that map tensor products of Pauli matrices to other tensor products of Pauli matrices through conjugation.
   • The Clifford group is generated by three gates: Hadamard, S (phase), and CNOT.
   • Quantum circuits composed solely of Clifford gates can be efficiently simulated using classical computers due to the Gottesman–Knill theorem1.

2. S Gate:

   • The S gate is a single-qubit gate that induces a π/2 phase (i.e., a quarter turn) and is sometimes referred to as the phase gate.
   • Mathematically, the S gate corresponds to a square root of the Pauli-Z gate:

     $$𝑆=\sqrt{𝑍}$$

     .
   • Since the S gate can be expressed as a tensor product of Pauli matrices, it falls within the Clifford group.
   • In Qiskit, you can apply the S gate to a quantum circuit using the s() method2.

3. Universal Quantum Gates:

   • While the Clifford gates alone are not sufficient for universal quantum computation, augmenting them with certain non-Clifford gates (such as the T gate) creates a universal gate set.
   • The T gate, which induces a π/4 phase, is not a Clifford gate because it cannot be expressed as a tensor product of Pauli matrices.
   • However, when combined with the Clifford gates, the T gate allows for universal quantum computation.

In summary, the S gate is indeed a Clifford gate, making it an essential building block in quantum circuits. If you have any more questions or need further clarification, feel free to ask! 😊12345

Usingthe web

Certainly! Let’s delve into the fascinating world of quantum gates.

The S gate (also known as the P gate or phase gate) is indeed a Clifford gate. Here’s why:

1. Clifford Gates:

   • The Clifford gates are a set of mathematical transformations that normalize the n-qubit Pauli group. These gates play a crucial role in quantum computing and quantum information theory.
   • The Clifford group consists of transformations that map tensor products of Pauli matrices to other tensor products of Pauli matrices through conjugation.
   • The Clifford group is generated by three gates: Hadamard, S (phase), and CNOT.
   • Quantum circuits composed solely of Clifford gates can be efficiently simulated using classical computers due to the Gottesman–Knill theorem1.

2. S Gate:

   • The S gate is a single-qubit gate that induces a π/2 phase (i.e., a quarter turn) and is sometimes referred to as the phase gate.
   • Mathematically, the S gate corresponds to a square root of the Pauli-Z gate:

     $$𝑆=\sqrt{𝑍}$$

     .
   • Since the S gate can be expressed as a tensor product of Pauli matrices, it falls within the Clifford group.
   • In Qiskit, you can apply the S gate to a quantum circuit using the s() method2.

3. Universal Quantum Gates:

   • While the Clifford gates alone are not sufficient for universal quantum computation, augmenting them with certain non-Clifford gates (such as the T gate) creates a universal gate set.
   • The T gate, which induces a π/4 phase, is not a Clifford gate because it cannot be expressed as a tensor product of Pauli matrices.
   • However, when combined with the Clifford gates, the T gate allows for universal quantum computation.

In summary, the S gate is indeed a Clifford gate, making it an essential building block in quantum circuits. If you have any more questions or need further clarification,