Answer to Question #114870 in Quantum Mechanics for Basia

Question

For the eigenstate with n=o,1, and 2 ,compute the probability that the coordinate of a linear harmonic oscillator in its ground state has value greater than oscillator than the amplitude of a classical oscillator of the same energy level

Answer 1

In quantum case $E_n=\Big(n+\frac{1}{2}\Big)\hbar\omega$

In classic case $E=kx^2/2=m\omega^2A^2/2$

Let's find the amplitude $A$ of oscillations for a classical oscillator with energy equal to the energy of a quantum oscillator in the quantum state $n$.

We obtain $E_n=m\omega^2A^2_n/2 \rightsquigarrow A_n=\sqrt{(2n+1)\frac{\hbar}{m\omega}}$

In quantum case

$\psi_n(x)=\frac{1}{\sqrt{2^n n!}}\cdot (\frac{m\omega }{\pi \hbar })^{1/4}\cdot e^{-\frac{m\omega x^2}{2\hbar}}\cdot H_n(\sqrt{\frac{m\omega }{\hbar }} x)$

where $H_n(y)$ are Hermite polynomials, $H_0(y)=1$ .

We want to compute the probability that the coordinate of a linear harmonic oscillator in its ground state (i.e. $n=0$) has value greater than ''classic zone'' $x\in(-A,A)$, i.e.

$$\int\limits_{A_0}^\infty |\psi_0(x)|^2\,dx=\sqrt{\frac{m\omega}{\pi \hbar}}\int\limits_{A_0}^\infty e^{-m\omega x^2/\hbar}\,dx =$$ $$=\sqrt{\frac{m\omega}{\pi \hbar}} \cdot\frac{1}{2}\sqrt{\frac{\pi \hbar}{m \omega}}\operatorname{erfc}(1) \approx 0.0787$$

Or approximately $7.87 \%$