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)

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  • Pero  On Monday, 9 March 2015 at 15:03

    The probability should be twice that shown, due to the way the erf and normal distribution tables work.

    • growescience  On Tuesday, 10 March 2015 at 16:28

      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).

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