# Harmonic oscillator – probability of being outside classical region

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.15.

The classical harmonic oscillator has an energy of ${E=\frac{1}{2}kx_{0}^{2}}$ where ${k}$ is the spring constant and ${x_{0}}$ is the maximum displacement from the equilibrium position. In terms of the frequency of oscillation, this is ${E=\frac{1}{2}m\omega^{2}x_{0}^{2}}$, so the mass oscillates between ${x_{0}=-\sqrt{2E/m\omega^{2}}}$ and ${x_{0}=\sqrt{2E/m\omega^{2}}}$. For a quantum oscillator, we can work out the probability that the particle is found outside the classical region. In the ground state, we have

$\displaystyle \psi_{0}(x)=\left(\frac{m\omega}{\pi\hbar}\right)^{1/4c}e^{-m\omega x^{2}/2\hbar} \ \ \ \ \ (1)$

The probability that the particle is found between two points ${a}$ and ${b}$ is

$\displaystyle P_{ab}=\int_{a}^{b}\psi_{0}^{2}(x)dx \ \ \ \ \ (2)$

so the probability that the particle is in the classical region is

 $\displaystyle P_{classical}$ $\displaystyle =$ $\displaystyle \int_{-\sqrt{2E/m\omega^{2}}}^{\sqrt{2E/m\omega^{2}}}\psi_{0}^{2}(x)dx\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{m\omega}{\pi\hbar}\right)^{1/2}\int_{-\sqrt{2E/m\omega^{2}}}^{\sqrt{2E/m\omega^{2}}}e^{-m\omega x^{2}/\hbar}dx \ \ \ \ \ (4)$

In the ground state, ${E=\hbar\omega/2}$ so this is

$\displaystyle P_{classical}=\left(\frac{m\omega}{\pi\hbar}\right)^{1/2}\int_{-\sqrt{\hbar/m\omega}}^{\sqrt{\hbar/m\omega}}e^{-m\omega x^{2}/\hbar}dx \ \ \ \ \ (5)$

This is easier to deal with if we introduce a substitute variable

$\displaystyle \xi\equiv\sqrt{\frac{m\omega}{\hbar}}x \ \ \ \ \ (6)$

Then the integral transforms to

$\displaystyle P_{classical}=\frac{1}{\sqrt{\pi}}\int_{-1}^{1}e^{-\xi^{2}}d\xi \ \ \ \ \ (7)$

This integral is the error function, so we get

 $\displaystyle P_{classical}$ $\displaystyle =$ $\displaystyle \mathrm{erf}(1)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0.8427 \ \ \ \ \ (9)$

The probability of being outside the classical region is then

$\displaystyle 1-P_{classical}=0.1573 \ \ \ \ \ (10)$

## 2 thoughts on “Harmonic oscillator – probability of being outside classical region”

1. growescience

I’m
pretty sure the answer as given is correct, since I used Maple to work out the integral explicitly (I didn’t use tables).