Do Perturbations in a Quantum Well's Bottom Shape Affect Particle Eigenvalues?

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The discussion focuses on the application of perturbation theory to demonstrate that the eigenvalues of a particle in a quantum well are unaffected by the shape of the well's bottom. Specifically, it examines a scenario where the well's bottom is deeper in the middle than at the sides. Participants suggest using a scaled and shifted sine function to model the potential, emphasizing the need to adjust the sine function to achieve a "bowl-shaped" bottom. The correct approach involves shifting the sine function by adding a constant to the phase to ensure the trough aligns with the well's bottom.

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d) Show that the eigenvalues of a particle in a well aren’t affected by the structure of the bottom of the well by applying perturbation theory. As an example, investigate the case where the bottom of the well is deeper in the middle (x = a2) than at the sides (x = 0 and x= a). To do this:-
i. Find a mathematical description of the potential at the bottom of the well by scaling and shifting a
sine function to achieve the required shape. Illustrate this with a plot of your chosen function.
ii. Write down the perturbed Hamiltonian

With part (i) is this asking for a function that relates the sine wave to the potential. I know that the bottom of the well is in a shape such that it has the negative part of the sine curve as its base. I can't get an equation for the potential at the bottom of the well.

I have tried using sin((gamma + 1)*pi) where 0 < gamma < 1 but it doesn't seem to work through.

What should i do in order to find the function for the potential at the bottom of the well.

thanks in advance.
 
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First you need to scale the sine function so that it's period is such that half a cycle fits just inside the well (corresponding to a "bowl-shaped" bottom surface of the well, with the bowl being a trough of the sine wave). If the argument is (pi/a)*x, then when x = a, we'll have the argument being pi (half a cycle, as desired).

Unfortunately, the half cycle currently in the well is a crest, not a trough (due to the nature of the sine function). So, you have to shift everything a half cycle to the left or right (by pi). Remember how you shift the function...by adding a constant to the phase.
 

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