A Help Understanding a Quantum Circuit Identity

CMJ96
Messages
50
Reaction score
0
Hello I have the following quantum circuit identity for converting a controlled U gate (4x4 matrix) into a series of CNOT gates and single qubit gates
$$ U= AXA^{\dagger}X$$
where A is a unitary matrix.
Here is a picture of the mentioned identity.
hCYpzW8.png

Can someone help me understand conceptually what is going on here? How do you actually define A?
 

Attachments

  • hCYpzW8.png
    hCYpzW8.png
    6.5 KB · Views: 1,007
Physics news on Phys.org
CMJ96 said:
Can someone help me understand conceptually what is going on here?
The idea is to use a CNOT gate and single qubit operations instead of trying to implement the controlled U. Single qubit operations are simpler to do in real experiments, so that the focus is to implement a CNOT gate and try to use it as much as possible. As you see, the price to pay is that you have to perform two CNOTs and two single-qubit operations instead of a single operation.

CMJ96 said:
How do you actually define A?
Depends on what you want U to achieve. The idea is to find the A that allows you to end up in the same state as with the controlled U gate.
 
So when A is introduced, would it be another 4x4 unitary matrix (assuming the control U gate is a 4x4 matrix)?
 
CMJ96 said:
So when A is introduced, would it be another 4x4 unitary matrix (assuming the control U gate is a 4x4 matrix)?
From ##U= AXA^{\dagger}X##, you see that ##U## and ##A## have the same dimension. You can look at the action of ##U## and ##A## on the lower qubit only, in which case they are 2x2 matrices, but if you want the full controlled gate, which has to be a two-qubit operator, then you have to write them as a 4x4 matrices.
 
  • Like
Likes CMJ96
Hi I've gone away and had a think about this, and now I feel I understand what is happening pretty well, however I'm still struggling with applying it.
$$
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & \frac{\sqrt{3}}{2} & \frac{-1}{2} \\
0 & 0 & \frac{1}{2} & \frac{\sqrt{3}}{2}
\end{bmatrix}
$$
Is it appropriate to be using the aforementioned circuit identity to write this controlled U as single qubit gates and CNOT's? I have been trying for a while now to figure out what unitary matrix fits into the equation for A and can't quite get it
 
Last edited:
CMJ96 said:
Hi I've gone away and had a think about this, and now I feel I understand what is happening pretty well, however I'm still struggling with applying it.
$$
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & \frac{\sqrt{3}}{2} & \frac{-1}{2} \\
0 & 0 & \frac{1}{2} & \frac{\sqrt{3}}{2}
\end{bmatrix}
$$
What basis are you using for the matrix representation? What matrix is that supposed to be? What is the U you are trying to apply to the second qubit?
 
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
Is it possible, and fruitful, to use certain conceptual and technical tools from effective field theory (coarse-graining/integrating-out, power-counting, matching, RG) to think about the relationship between the fundamental (quantum) and the emergent (classical), both to account for the quasi-autonomy of the classical level and to quantify residual quantum corrections? By “emergent,” I mean the following: after integrating out fast/irrelevant quantum degrees of freedom (high-energy modes...

Similar threads

Back
Top