Normalized equation for particle in a ring

In summary, the normalization constant for the particle in a ring is found by integrating the probability density function, which is defined in terms of theta, over the range of theta. This is because the particle moves along a ring, not a straight line, so the elemental length is rdtheta.
  • #1
Titan97
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Homework Statement


Normalized equation for particle in a ring, where V=0 on a ring of radius 'a' and infinite everywhere else.

Homework Equations

The Attempt at a Solution


Replcing x by rθ,
$$-\frac{\hbar^2}{2I}\frac{\partial^2\psi}{\partial\theta^2}=E\psi$$
By guess, I found out that ##Ae^{ik\theta}## is an eigenfunction of the hamiltonian.
To find ##A##, I used:
$$\int_0^{2\pi}|\psi|^2rd\theta=1$$
But in the solution, they used
$$\int_0^{2\pi}|\psi|^2d\theta=1$$So the normalization constants were different. Which one is correct?
 
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  • #2
Titan97 said:

Homework Statement


Normalized equation for particle in a ring, where V=0 on a ring of radius 'a' and infinite everywhere else.

Homework Equations

The Attempt at a Solution


Replcing x by rθ,
$$-\frac{\hbar^2}{2I}\frac{\partial^2\psi}{\partial\theta^2}=E\psi$$
By guess, I found out that ##Ae^{ik\theta}## is an eigenfunction of the hamiltonian.
To find ##A##, I used:
$$\int_0^{2\pi}|\psi|^2rd\theta=1$$
But in the solution, they used
$$\int_0^{2\pi}|\psi|^2d\theta=1$$So the normalization constants were different. Which one is correct?

You are not integrating a function of ##r, \theta## over two dimensions. It's a simple one-dimensional integral for ##\theta##.
 
  • #3
the particle moves along a ring. so the elemental length should be rdtheta
 
  • #4
Titan97 said:
the particle moves along a ring. so the elemental length should be rdtheta

You've defined your ##\psi## in terms of ##\theta##. ##|\psi|^2## is the probability density function in terms of ##\theta##, not in terms of ##l## where ##l## is the arc length along the ring.

Normalisation means that the total proability is 1, so the simple integral of ##|\psi(\theta)|^2 d\theta## is required.

If you transformed ##\psi## into a function of arc length, ##l##, then you would integrate ##|\psi(l)|^2 dl##.

To illustrate this, a constant pdf for ##\theta## would be ##p(\theta) = \frac{1}{2 \pi}##, whereas a constant pdf for ##l## would be ##p(l) = \frac{1}{2 \pi a}##
 
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What is the normalized equation for a particle in a ring?

The normalized equation for a particle in a ring is a mathematical equation that describes the behavior of a particle confined to move in a circular ring. It takes into account the effects of the ring's boundaries on the particle's motion.

What are the variables in the normalized equation for a particle in a ring?

The variables in the normalized equation for a particle in a ring include the radius of the ring, the particle's mass, the angular frequency of the particle's motion, and the potential energy of the particle.

How is the normalized equation for a particle in a ring different from other equations?

The normalized equation for a particle in a ring takes into account the circular boundary of the ring, while other equations may describe the behavior of particles in different geometries or without boundaries.

What is the significance of the normalized equation for a particle in a ring in physics?

The normalized equation for a particle in a ring is significant in physics as it helps us understand the behavior of particles in circular systems, such as atoms and molecules. It also allows us to make predictions and calculations about the particle's motion in a ring.

How is the normalized equation for a particle in a ring derived?

The normalized equation for a particle in a ring is derived using mathematical principles and techniques, such as the Schrödinger equation and boundary conditions. It is a result of applying quantum mechanics to a circular system with a finite boundary.

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