Question

The variational method can be used to solve for the ground state wavefunction and energy of a harmonic oscillator. Using a trail wavefunction of , where the function is defined between $-\pi /2\lambda < x< \pi /2\lambda$ . The Hamiltonian operator for a 1D harmonic oscillator is $H = -h/4\pi \mu *d^2/dx^2+kx^2/2$ Solving for the wavefunction gives $E(\lambda )=h^2\lambda ^2/4\pi \mu +0.161k/\lambda ^2$

Find $\lambda$ that gives the lowest energy and compare from the trial function to the exact value, $E_{0}=h\omega /4\pi$ where $\omega$$=\sqrt{(k/\mu )}$

coS

Answer 1

41卜 3 0.322 k IT 4 12 4IT 47T