Shelf in a box, treating the shelf as a weak perturbation

[Futurestar33]

Homework Statement

Shelf in a box, treating the shelf as a weak perturbation. The pertubation is not given what should I use?

Relevant Equations

H= Ho+ H'
E_n = <Y_n|H'|Y_n>

In this problem I am supposed to treat the shelf as a weak perturbation. However it doesn't give us what the perturbed state H' is. At the step V(x) = Vo, but that is all that is given and isn't needed to determine H'.

This isn't in a weak magnetic field so I wouldn't you use H'=qEx and then treat X as an operator.

The other option I would use is H'=lamdaX, (but that is usually given in a problem as well)

 

[DrClaude]

At the step V(x) = Vo, but that is all that is given and isn't needed to determine H'.

What do you mean? This is exactly what is needed to know $H'$.

Start with the Hamiltonian in the absence of the shelf, then figure out how it is different if there is a shelf.

 

[Futurestar33]

Im confused at this then.
When there is no shelf the H= ∇^2Ψ=-K^2Ψ where K= √(2mE)/hbar
General expression being Ψ= Csin(kx)+Dcos(kx) , k=npi/a

When there is a shelf H = ∇^2Ψ = -η^2Ψ where η=√(2mE-Vo)/hbar
General expression being Ψ= Ae^(κx)+Be^(-κx).

I know I am not supposed to plug those in for H'

are you saying I just use En = <Yn|H'| Yn> ==> <Yn| Vo |Yn> ?

 

[Futurestar33]

Ok I got it, I was just thinking that usually it is given or some form of an operator.
but in this case it does just end up up being the above.

 

[DrClaude]

are you saying I just use En = <Yn|H'| Yn> ==> <Yn| Vo |Yn> ?

Yes. Remember that the Hamiltonian is kinetic energy + potential energy. This is not evident in the case of the particle in a box, because in the domain where the wave function is non-zero, the potential is zero.

$H'$ is defined by parts, but calculating $\langle \psi_n |H' | \psi_n \rangle$ is easy since its an integration over a constant potential in a finite region of space.