Info

The contents of Exact solutions of the quantum double square well potential provide a published solution to this, from 2006

Setup

We have a particle moving in dimension, with a Potential given by

We want to find the Eigenstates in this Potential

Solution

Each interval acts as a particle in a Constant Potential, which has well-known solutions which depend on whether the Energy is higher or lower than the Potential in that interval. We have to consider:

In all of the cases below, equals zero if is not in one of the defined intervals.

WIP

2️⃣

For , the Constant Potential solution on and is a linear function. In order for to be continuous at and , this function will be identically on these intervals. As such we end up with a setup equal to the Single well setup with Energy lower than the Potential () ), which has the general solution . When we attempt continuity at both extremes , we get , hence , which is not a Quantum state. As such does not provide a solution to this setup.

3️⃣

The general solution, with , is given by

We will simplify the various constants by requiring the continuity of at the intervals’ endpoints. For continuity at we we need . Since is a linear combination of trigonometric functions, we can do some trigonometric magic to rewrite as . Doing the same thing on and re indexing the constants we get

We can deconstruct any solution as . This allow us to look for even and odd solutions of this Time-independent Schrodinger equation separately, and write the general solution as linear combinations of the 2 parts. Matching constants in the expression above ( and again changing constant indexes), we get:

Even solutions

By requiring continuity at we get

Odd solutions

4️⃣

WIP

5️⃣

Since the Energy is higher than the Potential on all intervals, we get that the solution to the Constant Potential problem on each interval is a trigonometric solution. The matching condition between intervals requires that both and match. The only way that happens at is by having , but then is identically zero.As such we get that , which is not a valid Quantum state.

  • Finish the calculations above