Question #114870
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
1
Expert's answer
2020-05-11T20:04:14-0400

In quantum case En=(n+12)ωE_n=\Big(n+\frac{1}{2}\Big)\hbar\omega

In classic case E=kx2/2=mω2A2/2E=kx^2/2=m\omega^2A^2/2

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

We obtain En=mω2An2/2An=(2n+1)mωE_n=m\omega^2A^2_n/2 \rightsquigarrow A_n=\sqrt{(2n+1)\frac{\hbar}{m\omega}}


In quantum case

ψn(x)=12nn!(mωπ)1/4emωx22Hn(mωx)\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 Hn(y)H_n(y) are Hermite polynomials, H0(y)=1H_0(y)=1 .


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


A0ψ0(x)2dx=mωπA0emωx2/dx=\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 ==mωπ12πmωerfc(1)0.0787=\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%7.87 \%


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