Question

The purpose of this problem is to compare the time dependencies for systems in a superposition of two energy eigenstates in an infinite square well to those in a simple harmonic oscillator.
Consider two systems (an infinite square well and a simple harmonic oscillator) that have the same value for their ground state energy E_ground.

1) What is E₃, the energy of the 2nd excited state (the third lowest energy) of the infinite square well system in terms of E_ground? E₃ =

2) What is E₃, the energy of the 2nd excited state (the third lowest energy) of the simple harmonic oscillator system in terms of E_ground? E₃ =

3) Now suppose the wave function for the infinite square well system is a superposition of two energy eigenstates, namely its ground state and its second excited state. Assuming E_ground = 20 eV, what is t₁, the minimum time it takes for the probability density ρρ(x,t₁) of the infinite square well system to return to its original value (ρρ(x,0)) at t = 0?

t₁ =

4) Now suppose the wave function for the simple harmonic oscillator system is a superposition of two energy eigenstates, namely its ground state and its second excited state.
Assuming E_ground = 20 eV, what is t₁, the minimum time it takes for the probability density ρρ(x,t₁) of the simple harmonic oscillator system to return to its original value (ρρ(x,0)) at t = 0?
t₁ =

5) Now, both the infinite square well system and the simple harmonic oscillator system still have the same value for their ground state energies, E_ground, but the wavefunctions for both systems are described as a superposition of two energy eigenstates, namely their ground state and their third excited state.
What is R(SHO/Square), the ratio of t₁(SHO), the minimum time it takes for the probability density ρρ(x,t_1,SHO) of the simple harmonic oscillator system to return to its original value (ρρ(x,0)) to t₁(Square), the minimum time it takes for the probability density ρρ(x,t_1,Square) of the infinite square well system to return to its original value (ρρ(x,0)) ? i.e., R(SHO/Square) = t₁(SHO) / t₁(Square).

R(SHO/Square) =

Answer 1