Related Experiment Video
Updated: Jun 24, 2025

06:53
Scanning SQUID Study of Vortex Manipulation by Local Contact
Published on: February 1, 2017
6.8K
Where is the orbital angular momentum in vortex superposition states?
Optics Express
|June 11, 2024
Summary
We investigated orbital angular momentum (OAM) in coaxial vortex superposition states. Specific singular points emerge, with the center point
Area of Science:
- Optical physics
- Quantum optics
- Light-matter interactions
Background:
- Orbital angular momentum (OAM) is a fundamental property of light.
- Coaxial vortex superposition states involve combining multiple light beams with OAM.
- Understanding OAM distribution is crucial for applications in optical manipulation and communication.
Purpose of the Study:
- To explore the spatial distribution of OAM in coaxial vortex superposition states.
- To identify and characterize singular points within the intensity distribution of these states.
- To demonstrate the conservation of OAM during the coaxial interference process.
Main Methods:
- Applying the independent propagation principle of light for coaxial vortex superposition.
- Analyzing the spatial intensity distribution to locate singular points.
- Drawing analogies with rigid body angular momentum superposition for theoretical insights.
Main Results:
- Identified two types of singular points in the spatial intensity distribution.
- Proposed that the center singular point's topological charge is doubled by the number of superposition components.
- Showed that singular points at overlapping angular regions are shared by superposition components.
- Confirmed that the total OAM equals the sum of OAM in superposition states, demonstrating conservation.
Conclusions:
- The study elucidates the complex OAM distribution in coaxial vortex superposition states.
- Singular points play a key role in determining the topological charge and OAM.
- Orbital angular momentum is conserved throughout the coaxial interference process.
Related Concept Videos
Angular Momentum: Single Particle
6.1K
Angular momentum is directed perpendicular to the plane of the rotation, and its magnitude depends on the choice of the origin. The perpendicular vector joining the linear momentum vector of an object to the origin is called the “lever arm.” If the lever arm and linear momentum are collinear, then the magnitude of the angular momentum is zero. Therefore, in this case, the object rotates about the origin such that it lies on the rim of the circumference defined by the lever arm...
6.1K
Conservation of Angular Momentum
10.2K
A system's total angular momentum remains constant if the net external torque acting on the system is zero. Considering a system that consists of n tiny particles, the angular momentum of any tiny particle may change, but the system's total angular momentum would remain constant. The principle of conservation of angular momentum only considers the net external torque acting on the system. While there are internal forces exerted by different particles within the system that also produce...
10.2K
Angular Momentum
200
Angular momentum characterizes an object's rotational motion and is defined as the moment of its linear momentum about a specified point O. When a particle moves along a curved path in the x-y plane, the scalar formulation calculates the magnitude of its angular momentum, utilizing the moment arm (d), representing the perpendicular distance from point O to the line of action of the linear momentum. Despite being scalar in formulation, angular momentum is inherently a vector quantity. Its...
200
Conservation of Angular Momentum: Application
10.9K
A system's total angular momentum remains constant if the net external torque acting on the system is zero. Examples of such systems include a freely spinning bicycle tire that slows over time due to torque arising from friction, or the slowing of Earth's rotation over millions of years due to frictional forces exerted on tidal deformations. However in the absence of a net external torque, the angular momentum remains conserved. The conservation of angular momentum principle requires a...
10.9K
Angular Momentum about an Arbitrary Axis
195
Imagine a rigid body with a mass denoted as 'm', which has its center of mass at point G and is rotating around an inertial reference frame. The angular momentum at an arbitrary point P can be calculated by taking the cross product of the position vector and linear momentum vector for each individual mass element.
The velocity of a mass element comprises its translational velocity and the relative velocity instigated by the body's rotation. Substituting the velocity equation into...
The velocity of a mass element comprises its translational velocity and the relative velocity instigated by the body's rotation. Substituting the velocity equation into...
195
Angular Momentum: Rigid Body
8.7K
The total angular momentum of a rigid body can be calculated using the summation of the angular momentum of all the tiny particles rotating in the same plane. Considering all the tiny particles rotating in the x-y plane, the direction of angular momentum of all such particles and that of the rigid body would be perpendicular to the plane of the rotation along the z-axis.
This calculation can get complicated when tiny particles within the rigid body are not rotating in the same plane but have...
This calculation can get complicated when tiny particles within the rigid body are not rotating in the same plane but have...
8.7K

