Question

The normalized ground-state wave function of a one-dimensional quantum harmonic oscillator with force constant \(K\) and mass \(m\) is \(\psi_0(x) = \left(\frac{\alpha}{\pi}\right)^{1/4} e^{-\alpha x^2/2}\), where \(\alpha = \frac{m\omega_0}{\hbar}\) and \(\omega_0^2 = \frac{K}{m}\). Which one of the following is the probability of finding the particle outside the classically allowed region?
(The classically allowed region is where the total energy is greater than the potential energy.) 
 

• A. \(\dfrac{2}{\sqrt{\pi}} \int_1^{\infty} y^2 e^{-y^2} dy\)
• B. \(\dfrac{2}{\sqrt{\pi}} \int_1^{\infty} e^{-y^2} dy\)
• C. \(0.5\)
• D. \(0\)

Hint:

In the harmonic oscillator ground state, the probability of finding the particle outside the classical region is small but nonzero — this reflects quantum tunneling.

Answer 1

The Correct Option is B

Step 1: Determine classical turning point. 
For ground state (\(n = 0\)), energy \(E_0 = \frac{1}{2}\hbar \omega_0\). Potential energy \(V(x) = \frac{1}{2}Kx^2 = \frac{1}{2}m\omega_0^2 x^2\). At the classical limit, \(V(x_c) = E_0 \Rightarrow x_c = \sqrt{\frac{\hbar}{m\omega_0}} = \frac{1}{\sqrt{\alpha}}\). 
 

Step 2: Probability outside this region. 
\[ P = 2 \int_{x_c}^{\infty} |\psi_0(x)|^2 dx = 2 \int_{x_c}^{\infty} \left(\frac{\alpha}{\pi}\right)^{1/2} e^{-\alpha x^2} dx \] Let \(y = \sqrt{\alpha} x \Rightarrow dy = \sqrt{\alpha} dx\): \[ P = \frac{2}{\sqrt{\pi}} \int_1^{\infty} e^{-y^2} dy \]

Step 3: Conclusion. 
Hence, the probability outside the classically allowed region is \(\dfrac{2}{\sqrt{\pi}} \int_1^{\infty} e^{-y^2} dy\).