Question

The wavefunction of a particle of mass `m` in a unidimensional potential field `U(x)=kx^(2)//2` has in the ground state the form `Psi(x)=Ae^(-alphax^(2))`, where `A` is a normalization factor and `alpha` is positive constant. Making use of a Schrodinger equation, find the constant `alpha` and the energy `E` of the particle in this state.

Answer 1

The Schrodinger equation is `(d^(2)Psi)/(dx^(2))+(2m)/( ħ^(2))(E-(1)/(2)kx^(2))Psi=0`
we are given `Psi=Ae^(-alphax^(2)//2)`
Then `Psi'= -alphaxAe^(-alphax^(2)//2)`
Substituting we find that following equation must hold
`[(alpha^(2)x^(2)-prop)+(2m)/( ħ^(2))(E-(1)/(2)kx^(2))]Psi=0`
since `Psi!=0`, the bracket must vanish identicall. This means that the cofficient of `x^(2)` as well the term independent of `x` must vanish. we get
`prop^(2)=(mk)/( ħ^(2))` and `prop=(2mE)/( ħ^(2))`
Putting `k//m= omega^(2)`, this leads to `prop=(m omega)/( ħ^(2))` and `E=( ħ^(2)omega)/(2)`