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For the harmonic oscillator potential energy U=12kx2,U = \frac{1}{2}{kx^2}, the ground-state wave function is ψ(x)=Ae(mκ/2)x2\psi(x)=A e^{-(\sqrt{m \kappa} / 2 \hbar) x^{2}} and its energy is 12κ/m\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 ψ(x)\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 ψ(x)\psi(x) and its second derivative for the ground-state oscillator wave function.

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Answered 2 years ago
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Given Information: \textbf{Given Information: }

E=12mω02A2ω0=kmΨ(x)=Aemk2x2\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*}

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