Infinite double well

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 $1$ dimension, with a Potential given by

$$V(x)=\{\begin{array}{ll}0,& x\in [-L,-\alpha ]\\ v_0,& x\in [-\alpha ,\alpha ]\\ 0,& x\in [\alpha ,L]\\ \infty ,& \text{otherwise}\end{array}$$

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 $E$ is higher or lower than the Potential in that interval. We have to consider:

1. $E<0$
2. $E=0$
3. $E\in (0,v_0)$
4. $E=v_0$
5. $E>v_0$

In all of the cases below, $\psi (x)$ equals zero if $x$ is not in one of the defined intervals.

$1\stackrel{◯}{R}⃣E<0$

WIP

2️⃣ $E=0$

For $E=0$, the Constant Potential solution on $[-L,-\alpha ]$ and $[\alpha ,L]$ is a linear function. In order for $\psi (x)$ to be continuous at $L$ and $-L$, this function will be identically $0$ on these intervals. As such we end up with a setup equal to the Single well setup with Energy lower than the Potential () $E=0<v_0$ ), which has the general solution $\psi (x)=A\exp \left(\frac{\sqrt{2m{v}_0}}{\hslash }x\right)+B\exp \left(-\frac{\sqrt{2m{v}_0}}{\hslash }x\right)$. When we attempt continuity at both extremes $\{-\alpha ,\alpha \}$, we get $A=B=0$, hence $\psi (x)=0$, which is not a Quantum state. As such $E=0$ does not provide a solution to this setup.

3️⃣ $E\in (0,v_0)$

The general solution, with $c_i\in \mathbb{C}$, is given by

$$\psi (x)=\{\begin{array}{ll}{c}_1\sin \left(\frac{\sqrt{2mE}}{\hslash }x\right)+c_2\cos \left(\frac{\sqrt{2mE}}{\hslash }x\right),& x\in [-L,-\alpha ]\\ c_3\exp \left(\frac{\sqrt{2m(E-v_0)}}{\hslash }x\right)+c_4\exp \left(-\frac{\sqrt{2m(E-v_0)}}{\hslash }x\right),& x\in [-\alpha ,\alpha ]\\ c_5\sin \left(\frac{\sqrt{2m(E-k)}}{\hslash }x\right)+c_6\cos \left(\frac{\sqrt{2m(E-k)}}{\hslash }x\right),& x\in [\alpha ,L]\end{array}$$

We will simplify the various constants $c_i$ by requiring the continuity of $\psi$ at the intervals’ endpoints. For continuity at $L$ we we need $\psi (L)=0$. Since ${\psi (x)|}_{[\alpha ,L]}$ is a linear combination of trigonometric functions, we can do some trigonometric magic to rewrite $c_5\sin \left(\frac{\sqrt{2m(E-k)}}{\hslash }x\right)+c_6\cos \left(\frac{\sqrt{2m(E-k)}}{\hslash }x\right)$ as $c_7\sin \left(\frac{\sqrt{2m(E-k)}}{\hslash }(x-(-L))\right)=c_7\sin \left(\frac{\sqrt{2m(E-k)}}{\hslash }(x+L)\right)$. Doing the same thing on $[-L,-\alpha ]$ and re indexing the constants we get

$$\psi (x)=\{\begin{array}{ll}{c}_1\sin \left(\frac{\sqrt{2m(E-k)}}{\hslash }(x+L)\right),& x\in [-L,-\alpha ]\\ c_2\exp \left(\frac{\sqrt{2m(E-v_0)}}{\hslash }x\right)+c_3\exp \left(-\frac{\sqrt{2m(E-v_0)}}{\hslash }x\right),& x\in [-\alpha ,\alpha ]\\ c_4\sin \left(\frac{\sqrt{2m(E-k)}}{\hslash }(x-L)\right),& x\in [\alpha ,L]\end{array}$$

We can deconstruct any solution $\psi (x)$ as $\frac{1}{2}[\psi (x)+\psi (-x)]+\frac{1}{2}[\psi (x)-\psi (x)]$. 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

$$\psi (x)=\{\begin{array}{ll}{c}_1\sin \left(\frac{\sqrt{2m(E-k)}}{\hslash }(x+L)\right),& x\in [-L,-\alpha ]\\ c_2\left[\exp \left(\frac{\sqrt{2m(E-v_0)}}{\hslash }x\right)+\exp \left(-\frac{\sqrt{2m(E-v_0)}}{\hslash }x\right)\right],& x\in [-\alpha ,\alpha ]\\ c_1\sin \left(\frac{\sqrt{2m(E-k)}}{\hslash }(x-L)\right),& x\in [\alpha ,L]\end{array}$$

By requiring continuity at $-\alpha$ we get

$$\begin{array}{rl}{c}_1\sin \left(\frac{\sqrt{2m(E-k)}}{\hslash }(-\alpha +L)\right)& =c_2\left[\exp \left(\frac{\sqrt{2m(E-v_0)}}{\hslash }(-\alpha )\right)+\exp \left(-\frac{\sqrt{2m(E-v_0)}}{\hslash }(-\alpha )\right)\right]\\ c_1\sin \left(\frac{\sqrt{2m(E-k)}(L-\alpha )}{\hslash }\right)& =c_2\left[\exp \left(-\frac{\alpha \sqrt{2m(E-v_0)}}{\hslash }\right)+\exp \left(\frac{\alpha \sqrt{2m(E-v_0)}}{\hslash }\right)\right]\end{array}$$

Odd solutions

$$\psi (x)=\{\begin{array}{ll}{c}_1\sin \left(\frac{\sqrt{2m(E-k)}}{\hslash }(x+L)\right),& x\in [-L,-\alpha ]\\ c_2\left[\exp \left(\frac{\sqrt{2m(E-v_0)}}{\hslash }x\right)-\exp \left(-\frac{\sqrt{2m(E-v_0)}}{\hslash }x\right)\right],& x\in [-\alpha ,\alpha ]\\ -c_1\sin \left(\frac{\sqrt{2m(E-k)}}{\hslash }(x-L)\right),& x\in [\alpha ,L]\end{array}$$

4️⃣ $E=v_0$

WIP

5️⃣ $E>v_0$

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 $\psi (x)$ and $\dot{\psi }(x)$ match. The only way that happens at ${|}_{\{-L,L\}}$ is by having $\psi (x)=\dot{\psi }(x)=0$, but then $\psi (x)$ is identically zero.As such we get that $\psi (x)=0\forall x\in \mathbb{R}$, which is not a valid Quantum state.

• Finish the calculations above