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
Eground.
1) What is E3, the energy of the 2nd excited state (the third lowest energy) of the infinite square well system in terms of Eground? 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 Eground? 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 Eground = 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 Eground = 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,
Eground, 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,t1,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,t1,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) =
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