# Particle on a ring (components in the postion basis)

• Lambda96
In summary: Bigr)$$In summary, the conversation discusses two tasks, b and g, related to calculating the state of a particle on a ring. The individual steps and calculations for task b are provided, but the individual is stuck and unsure how to proceed. For task g, the individual is also stuck and seeking help. They continue to work through the problem by changing a summation index and reducing the sum to one term. The conversation ends with the individual thanking the other person for their help. Lambda96 Homework Statement see screenshots Relevant Equations none Hi, I have problems with the task part b and g To solve the task, we have received the following information Task b First, I wrote down what the state ##\psi## looks like$$\psi=\frac{1}{\sqrt{N}} \sum\limits_{k}^{} \psi_k\psi=\frac{1}{\sqrt{N}} \sum\limits_{k}^{} \frac{1}{\sqrt{N}} \sum\limits_{j}^{} e^{ikaj} \vec{e}_j$$Then I to calculate ##\psi^j=\braket{\vec{e}_j|\psi}##.$$\braket{\vec{e}_j|\psi}=\vec{e}_j^{\dagger} \cdot \frac{1}{\sqrt{N}} \sum\limits_{k}^{} \frac{1}{\sqrt{N}} \sum\limits_{j}^{} e^{ikaj} \vec{e}_j\braket{\vec{e}_j|\psi}=\vec{e}_j^{\dagger} \cdot \frac{1}{N} \sum\limits_{k}^{} \sum\limits_{j}^{} e^{ikaj} \vec{e}_j$$Now I unfortunately do not know how to proceed further. But I don't understand, if all momentums are equally probable, why the particle should be 100% at location N and not at other locations like 1 and 2 and so on. What makes the point N so special that the particle should be there in contrast to the other points? To solve task g, we have received the following information Task g If I understood the task correctly, then the wave function is collapsed, to the eigenvector of the momentum operator, more precisely to ##\psi_0##. The wave function has with 100% the eigenvalue of ##\psi_0## after the uncertainty principle, the uncertainty would have to become extremely large concerning the position, which means that the particle can be everywhere on the ring and thus the probability for each position is equally large. Last edited: Lambda96 said:$$\braket{\vec{e}_j|\psi}=\vec{e}_j^{\dagger} \cdot \frac{1}{\sqrt{N}} \sum\limits_{k}^{} \frac{1}{\sqrt{N}} \sum\limits_{j}^{} e^{ikaj} \vec{e}_j$$In the second sum on the right, you should change the summation index ##j## to some other symbol so that the summation index is not confused with the ##j## in ##\vec{e}_j^{\dagger}##. For example, you could write$$\braket{\vec{e}_j|\psi}=\vec{e}_j^{\dagger} \cdot \frac{1}{\sqrt{N}} \sum\limits_{k}^{} \frac{1}{\sqrt{N}} \sum\limits_{l}^{} e^{ikal} \vec{e}_l$$Lambda96 Thanks TSny for your help , I have now changed the index of the summation from ##j## to ##l## and have now calculated the following.$$\braket{\vec{e}_j|\psi}=\vec{e}_j^{\dagger} \cdot \frac{1}{\sqrt{N}} \sum\limits_{k}^{} \frac{1}{\sqrt{N}} \sum\limits_{l}^{} e^{ikal} \vec{e}_l\braket{\vec{e}_j|\psi}=\frac{1}{N} \sum\limits_{k}^{} \sum\limits_{l}^{} e^{ikal} \vec{e}_j^{\dagger} \cdot \vec{e}_l\braket{\vec{e}_j|\psi}=\frac{1}{N} \sum\limits_{k}^{} \sum\limits_{l}^{} e^{ikal} \delta_{jl}\braket{\vec{e}_j|\psi}=\frac{1}{N} \sum\limits_{k}^{} e^{ika} \delta_{j1}+e^{2ika} \delta_{j2}+ \ldots +e^{Nika} \delta_{jN}\braket{\vec{e}_j|\psi}=\frac{1}{N} \Bigl( e^{ia} \delta_{j1}+e^{2ia} \delta_{j2}+ \ldots +e^{Nia} \delta_{jN}+e^{2a} \delta_{j1}+e^{4ia} \delta_{j2}+ \ldots +e^{2Nia} \delta_{jN}+ \ldots + e^{Nia} \delta_{j1}+e^{2Nia} \delta_{j2}+ \ldots +e^{N^2ika} \delta_{jN} \Bigr)\braket{\vec{e}_j|\psi}=\frac{1}{N} \Bigl(\Bigl(e^{ia}+e^{2a} +\ldots+e^{Nia} \Bigr) \delta_{j1}+\Bigl( e^{2ia}+e^{4ia}+ \ldots +e^{2Nia} \Bigr)\delta_{j2}+ \ldots + \Bigl( e^{Nia}+e^{2Nia} + \ldots +e^{N^2ika} \Bigr)\delta_{jN}\Bigr)$$Now I'm stuck The individual entries in the brackets look almost like states ##\vec{\psi}_k##, i.e.$$\braket{\vec{e}_j|\psi}=\frac{1}{\sqrt{N}} \Bigl(\frac{1}{\sqrt{N}} \Bigl(e^{ia}+e^{2a} +\ldots+e^{Nia} \Bigr) \delta_{j1}+\frac{1}{\sqrt{N}} \Bigl( e^{2ia}+e^{4ia}+ \ldots +e^{2Nia} \Bigr)\delta_{j2}+ \ldots + \frac{1}{\sqrt{N}} \Bigl( e^{Nia}+e^{2Nia} + \ldots +e^{N^2ika} \Bigr)\delta_{jN}\Bigr)\braket{\vec{e}_j|\psi}=\frac{1}{\sqrt{N}} \Bigl(\vec{\psi}_1 \delta_{j1}+\vec{\psi}_2 \delta_{j2}+ \ldots + \vec{\psi}_N \delta_{jN}\Bigr)$$Now does it mean that if, for example, ##\vec{e}_j=\vec{e}_2##, that only the state ##\braket{\vec{e}_2|\psi}=\frac{1}{\sqrt{N}} \Bigl(\vec{\psi}_2 \delta_{22}\Bigr)## remains and the particle is at location 2? Lambda96 said:$$\braket{\vec{e}_j|\psi}=\frac{1}{N} \sum\limits_{k}^{} \sum\limits_{l}^{} e^{ikal} \delta_{jl}$$Good. Consider the sum$$ \sum\limits_{l}^{} e^{ikal} \delta_{jl}$$Since ##\delta_{jl}## equals zero for any ##l \neq j##, all of the terms in the summation are zero except for one term. So, the sum reduces to one term that can be written in terms of ##k##, ##a##, and ##j##. Lambda96 and vanhees71 Thanks TSny for your help Then I can write the term as follows$$\braket{\vec{e}_j|\psi}=\frac{1}{N}\sum\limits_{k}^{} \sum\limits_{l}^{}e^{ikal} \delta_{jl}\braket{\vec{e}_j|\psi}=\frac{1}{N}\sum\limits_{k}^{} e^{ikaj}\braket{\vec{e}_j|\psi}=\frac{1}{N}\Bigl( e^{iaj}+e^{2iaj}+e^{3iaj}+ \ldots +e^{Niaj} \Bigr)$$Lambda96 said:$$\braket{\vec{e}_j|\psi}=\frac{1}{N}\sum\limits_{k}^{} \sum\limits_{l}^{}e^{ikal} \delta_{jl}\braket{\vec{e}_j|\psi}=\frac{1}{N}\sum\limits_{k}^{} e^{ikaj}\braket{\vec{e}_j|\psi}=\frac{1}{N}\Bigl( e^{iaj}+e^{2iaj}+e^{3iaj}+ \ldots +e^{Niaj} \Bigr)$$Good, except the sum over ##k## does not go from ##k = 1## to ##k = N##. Recall that the values of ##k## are ##k = 2\pi n /L## for ##n = 0, 1, 2, ... , N-1##. Lambda96 and SammyS Thanks again for your help TSny , also thanks for the hint with the index ##k## Could I then write the second line as follows?$$\braket{\vec{e}_j|\psi}=\frac{1}{N}\sum\limits_{k}^{} e^{ikaj}\braket{\vec{e}_j|\psi}=\frac{1}{N}\sum\limits_{n=0}^{N-1} e^{\frac{2i \pi n a j}{L}}$$Lambda96 said: Thanks again for your help TSny , also thanks for the hint with the index ##k## Could I then write the second line as follows?$$\braket{\vec{e}_j|\psi}=\frac{1}{N}\sum\limits_{k}^{} e^{ikaj}\braket{\vec{e}_j|\psi}=\frac{1}{N}\sum\limits_{n=0}^{N-1} e^{\frac{2i \pi n a j}{L}}
Yes, that looks right.

Lambda96

You’re very welcome.

Lambda96

## What is a particle on a ring in quantum mechanics?

A particle on a ring is a fundamental problem in quantum mechanics that describes a particle constrained to move on a circular path with a fixed radius. This system is often used to model electrons in a circular orbit or to understand rotational motion in molecules.

## How is the Hamiltonian for a particle on a ring formulated?

The Hamiltonian for a particle on a ring is given by $$\hat{H} = -\frac{\hbar^2}{2I} \frac{\partial^2}{\partial \phi^2}$$, where $$\hbar$$ is the reduced Planck's constant, $$I$$ is the moment of inertia of the particle, and $$\phi$$ is the angular coordinate around the ring. This Hamiltonian takes into account only the kinetic energy of the particle since the potential energy is constant on the ring.

## What are the eigenfunctions and eigenvalues of the Hamiltonian for a particle on a ring?

The eigenfunctions of the Hamiltonian for a particle on a ring are the angular momentum eigenstates given by $$\psi_m(\phi) = \frac{1}{\sqrt{2\pi}} e^{im\phi}$$, where $$m$$ is an integer (representing the quantum number). The corresponding eigenvalues, which represent the energy levels, are $$E_m = \frac{\hbar^2 m^2}{2I}$$.

## How is the wavefunction of a particle on a ring expressed in the position basis?

In the position basis, the wavefunction of a particle on a ring is expressed as $$\psi(\phi)$$, where $$\phi$$ is the angular coordinate. The general solution can be written as a linear combination of the eigenfunctions: $$\psi(\phi) = \sum_{m=-\infty}^{\infty} c_m e^{im\phi}$$, where $$c_m$$ are the coefficients determined by the initial conditions or boundary conditions of the problem.

## What is the significance of angular momentum quantization in the particle on a ring model?

Angular momentum quantization in the particle on a ring model signifies that the angular momentum of the particle can only take on discrete values, which are integer multiples of $$\hbar$$. This quantization is a direct consequence of the periodic boundary conditions and the wave nature of particles in quantum mechanics. It leads to discrete energy levels and has important implications in understanding the rotational spectra of molecules and other quantum systems with circular symmetry.

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