Math Problem Statement

The “mass on a spring” is a common example of simple harmonic motion. The quantum harmonic oscillator is a quantum mechanical analog of this scenario, in which a particle is placed in the following potential: V(x) = 1/2mw^2*x^2.

The corresponding solutions to the Schrodinger equation have the following form: ψ_n(x) = (1/sqrt(2^nn!))(((mw)/(πh))^(1/4))e^(-(mwx^2)/2h)H_n(sqrt(mw/h)x), n = 0,1,2,3,... where H_n(x) = (-1)^ne^(x^2)(d^n/dx^n)(e^(-x^2)) are the physicists' Hermite polynomials.

Use the Schrodinger equation to determine expressions for the first three energy levels of the quantum harmonic oscillator in terms of w and h. Use these to determine a general expression for the energy of any state, n.

Solution

The problem you're describing deals with the quantum harmonic oscillator, which is one of the most fundamental systems in quantum mechanics. The potential given is V(x)=12mω2x2V(x) = \frac{1}{2} m \omega^2 x^2, representing a harmonic oscillator with mass mm and angular frequency ω\omega. We aim to determine the energy levels of this system using the time-independent Schrödinger equation.

Step 1: The Schrödinger Equation

The time-independent Schrödinger equation for a system is given by: H^ψ(x)=Eψ(x)\hat{H} \psi(x) = E \psi(x) where H^\hat{H} is the Hamiltonian operator. For the quantum harmonic oscillator, the Hamiltonian is the sum of the kinetic and potential energy operators: H^=22md2dx2+12mω2x2\hat{H} = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + \frac{1}{2} m \omega^2 x^2 Here, \hbar is the reduced Planck constant, and ω\omega is the angular frequency of the oscillator.

Step 2: Energy Levels

By solving the Schrödinger equation for this potential, we find that the energy eigenvalues EnE_n are quantized and given by the formula: En=(n+12)ωwheren=0,1,2,3,E_n = \left( n + \frac{1}{2} \right) \hbar \omega \quad \text{where} \quad n = 0, 1, 2, 3, \dots This is the general expression for the energy of the quantum harmonic oscillator. The term 12ω\frac{1}{2} \hbar \omega represents the zero-point energy, which is the energy the system has even in its ground state (n=0n = 0).

Step 3: First Three Energy Levels

To determine the first three energy levels explicitly, substitute n=0n = 0, n=1n = 1, and n=2n = 2 into the general formula:

  1. Ground state (n = 0): E0=(0+12)ω=12ωE_0 = \left( 0 + \frac{1}{2} \right) \hbar \omega = \frac{1}{2} \hbar \omega

  2. First excited state (n = 1): E1=(1+12)ω=32ωE_1 = \left( 1 + \frac{1}{2} \right) \hbar \omega = \frac{3}{2} \hbar \omega

  3. Second excited state (n = 2): E2=(2+12)ω=52ωE_2 = \left( 2 + \frac{1}{2} \right) \hbar \omega = \frac{5}{2} \hbar \omega

Step 4: General Expression for Energy Levels

The general expression for the energy levels of the quantum harmonic oscillator is: En=(n+12)ωE_n = \left( n + \frac{1}{2} \right) \hbar \omega This formula applies to all n=0,1,2,3,n = 0, 1, 2, 3, \dots, where nn is the quantum number representing the energy level.

Summary:

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Math Problem Analysis

Mathematical Concepts

Quantum Mechanics
Schrodinger Equation
Simple Harmonic Motion
Wavefunctions
Hermite Polynomials

Formulas

V(x) = 1/2*m*w^2*x^2
ψ_n(x) = (1/sqrt(2^n*n!))*(((m*w)/(π*h))^(1/4))*e^(-(m*w*x^2)/2h)*H_n*(sqrt(m*w/h)*x)
E_n = (n + 1/2) * ħ * ω

Theorems

Solution of the Time-Independent Schrodinger Equation for the Quantum Harmonic Oscillator
Hermite Polynomials

Suitable Grade Level

Undergraduate Physics (Advanced)