## Related questions with answers

For the harmonic oscillator potential energy $U = \frac{1}{2}{kx^2},$ the ground-state wave function is $\psi(x)=A e^{-(\sqrt{m \kappa} / 2 \hbar) x^{2}}$ and its energy is $\frac{1}{2} \hbar \sqrt{\kappa / m}$

(a) Find the classical turning points for a particle with this energy.

(b) The Schrodinger equation says that $\psi(x)$ and its second derivative should be of the opposite sign when E > U and of the same sign when E< U. These two regions are divided by the classical turning points. Verify the relationship between $\psi(x)$ and its second derivative for the ground-state oscillator wave function.

Solution

Verified$\textbf{Given Information: }$

$\begin{align*} E &= \dfrac{1}{2} m \omega _0^2 A^2 \\ \omega_0 &= \sqrt{\dfrac{k}{m}} \\ \Psi (x) &= Ae^{\frac{\sqrt{mk}}{2\hbar}x^2} \\ \end{align*}$

## Create a free account to view solutions

## Create a free account to view solutions

## Recommended textbook solutions

#### Physics for Scientists and Engineers: A Strategic Approach with Modern Physics

4th Edition•ISBN: 9780133942651 (8 more)Randall D. Knight#### Mathematical Methods in the Physical Sciences

3rd Edition•ISBN: 9780471198260 (1 more)Mary L. Boas#### Modern Physics for Scientists and Engineers

2nd Edition•ISBN: 9780805303087 (1 more)Randy Harris#### Fundamentals of Physics

10th Edition•ISBN: 9781118230718 (3 more)David Halliday, Jearl Walker, Robert Resnick## More related questions

1/4

1/7