Particle in a box - perturbation theory

[odesa]

Homework Statement

Find the ground state energy of a particle restricted to move in one dimension subject to the potential in the attachement using perturbation theory.

Homework Equations

Yo = (2/a)^1/2 sin(nπx/a)

The Attempt at a Solution

I'm not sure how to account for the potential since there are two wells.

My setup right now using braket notation is E = 2<Y|Y> + <Y|Vo|Y> where Y is the wavefunction Yo = (2/a)^1/2 sin(nπx/a)

Can I get some feedback as to if I am headed in the right direction or how to get there if I am not?

Thank you.

 

[Jhero]

Thank you for your question. To find the ground state energy of a particle restricted to move in one dimension subject to the potential in the attachment, we can use perturbation theory.

First, let's define our unperturbed system, which is the particle restricted to move in one dimension without any potential. In this case, the ground state energy is simply the energy of the particle in the lowest energy state, which can be represented by the wavefunction Yo = (2/a)1/2 sin(nπx/a).

Next, let's define our perturbation, which is the potential in the attachment. This potential introduces an extra term in the Hamiltonian, which we can represent as Vo.

To find the ground state energy of the perturbed system, we can use the first order perturbation theory formula: E = E0 + <Y|Vo|Y>, where E0 is the ground state energy of the unperturbed system and <Y|Vo|Y> is the expectation value of the potential in the ground state wavefunction.

In your attempt at a solution, you have correctly set up the equation using braket notation. However, you have not yet accounted for the potential Vo. To do so, we can express the potential in terms of the position operator x, and then evaluate the expectation value using the ground state wavefunction Yo.

I hope this helps guide you in the right direction. Please let me know if you have any further questions.