Question

Consider the energies allowed for bound states of a half-harmonic oscillator, having the potential$$U(x)=\left\\{\begin{array}{ll}\frac{1}{2} m \omega_{0}^{2} x^{2} & \text { for } x>0 \\\\\infty & \text { for } x \leq 0\end{array}\right.$$ Using simple arguments based on the characteristics of normalized wave functions, what are the energies allowed for bound states in this potential?

Short answer

Answer: The allowed energy levels for bound states in a half-harmonic oscillator are given by$$ E_{n'} = \hbar \omega_0 \left(2n' + \frac{1}{2}\right)~, n' = 0, 1, 2, ... $$where n' is an integer, ħ is the reduced Planck constant, and ω₀ is the angular frequency.

Step-by-step solution

Understand the potential function for the half-harmonic oscillator

In this exercise, we have a half-harmonic oscillator which means the potential function U(x) is non-infinite only for x>0 and is infinite for x≤0. The potential function for x>0 is given by$$ U(x) = \frac{1}{2}m \omega_0^2 x^2 $$where m is the mass of the particle, ω₀ is the angular frequency, and x is the position coordinate.

Characteristics of a normalized wave function

Normalized wave function ψ(x) must satisfy the following properties: 1. The wave function is continuous and smooth. 2. The wave function is normalized, meaning the probability of finding a particle in all possible positions is equal to 1:$$ \int_{-\infty}^{\infty} |\psi(x)|^2 dx = 1 $$

Analyze the boundary condition at x=0

Since the potential function is infinite for x≤0, the wave function ψ(x) must go to zero as x approaches 0. This ensures that the wave function and its derivatives are continuous and smooth, satisfying the properties of a normalized wave function. The boundary condition at x=0 is$$ \psi(0) = 0 $$

Determine the allowed energy levels

For a full harmonic oscillator, the energy levels are given by$$ E_n = \hbar \omega_0 \left(n + \frac{1}{2}\right) $$where n is an integer (0, 1, 2, ...), and ħ is the reduced Planck constant. However, in the case of a half-harmonic oscillator, the wave function must satisfy the boundary condition at x=0. If the wave function of a full harmonic oscillator satisfies this condition, we can use the same energy levels. The wave functions corresponding to even values of n (n = 0, 2, 4, ...) are symmetric around x=0, and hence they satisfy the given boundary condition. Therefore, only these energy levels are allowed and represented by$$ E_{n'} = \hbar \omega_0 \left(2n' + \frac{1}{2}\right) $$where n' is an integer (0, 1, 2, ...). The allowed energies for bound states in a half-harmonic oscillator are$$ E_{n'} = \hbar \omega_0 \left(2n' + \frac{1}{2}\right)~, n' = 0, 1, 2, ... $$