Single well

Setup

We have $V(x)=-C{1}_{|x|>\alpha }$, having $\alpha \in {\mathbb{R}}^{+}$. The goal is to find the stable energy Quantum state with this potential

Solution

We have three intervals $[-\infty ,-\alpha ]$, $[-\alpha ,\alpha ]$ and $[\alpha ,\infty ]$, each of being equivalent to the Constant Potential problem on their domain. The solution depends on which is bigger, the Energy $E$ or the Potential $V$. As such we get five cases: $E<-C$, $E=C$, $E\in [-C,0]$, $E=0$ and $E>0$. In the sections below we use:

• 🔴 for no solutions
• 🔵 for discrete solutions
• 🟢 for continuous solutions

1️⃣ $E<-C$

The solution on each interval is given by a sum of exponentials. In order for the function to be Normalizable, we need ${\lim }_{x\to \pm \infty }\psi (x)=0$. This gives us the solution

$$\psi (x)=\{\begin{array}{ll}{c}_1{e}^{\frac{\sqrt{-2mE}}{\hslash }x},& x<-\alpha \\ c_2{e}^{\frac{\sqrt{-2m(C-E)}}{\hslash }x}+c_3{e}^{-\frac{\sqrt{-2m(C-E)}}{\hslash }x},& -\alpha <x<\alpha \\ c_4{e}^{-\frac{\sqrt{-2mE}}{\hslash }x},& x>\alpha \end{array}$$

Since $V(x)$ is an even function, the functions ${\psi }_E(x):=\frac{1}{2}(\psi (x)+\psi (-x))$ and ${\psi }_O(x):=\frac{1}{2}(\psi (x)-\psi (-x))$ are also solutions of the original equation, and can be used to rebuild $\psi$ as $\psi =\frac{1}{2}({\psi }_E+{\psi }_O)$. Let us consider even and odd functions separately.

Even solutions

$${\psi }_E(x)=\{\begin{array}{ll}A{e}^{\frac{x\sqrt{2m(C-E)}}{\hslash }},& x<-\alpha \\ B{e}^{\frac{x\sqrt{-2mE}}{\hslash }}+B{e}^{-\frac{x\sqrt{-2mE}}{\hslash }},& -\alpha <x<\alpha \\ A{e}^{\frac{-x\sqrt{2m(C-E)}}{\hslash }},& x>\alpha \end{array}$$

Odd solutions

$${\psi }_O(x)=\{\begin{array}{ll}A{e}^{\frac{x\sqrt{2m(C-E)}}{\hslash }},& x<-\alpha \\ B{e}^{\frac{x\sqrt{-2mE}}{\hslash }}-B{e}^{-\frac{x\sqrt{-2mE}}{\hslash }},& -\alpha <x<\alpha \\ -A{e}^{\frac{-x\sqrt{2m(C-E)}}{\hslash }},& x>\alpha \end{array}$$

2️⃣ $E=-C$ 🔴

If the Energy equals $-C$, we get that $\psi$ is linear on $|x|>\alpha$. In order for $\psi$ to be Normalizable, it must be identically zero on $|x|>\alpha$. On $|x|<\alpha$ we have that $\psi$ is a sum of exponentials. In the interval $[-\alpha ,\alpha ]$, the generic solution for the Constant Potential gives us that $\psi (x)=A{e}^{\frac{\sqrt{2mC}}{\hslash }x}+B{e}^{-\frac{\sqrt{2mC}}{\hslash }x}$. In order for the end of the interval to match at $\alpha$, we have

$$\begin{array}{rcll}{\psi (x)|}_{x=\alpha }=0& \wedge & {\dot{\psi }(x)|}_{x=\alpha }=0& \text{Plugging in the definition of ψ}\\ {A{e}^{\frac{\sqrt{2mC}}{\hslash }x}+B{e}^{-\frac{\sqrt{2mC}}{\hslash }x}|}_{x=\alpha }=0& \wedge & {A\frac{\sqrt{2mC}}{\hslash }{e}^{\frac{\sqrt{2mC}}{\hslash }x}-\frac{\sqrt{2mC}}{\hslash }B{e}^{-\frac{\sqrt{2mC}}{\hslash }x}|}_{x=\alpha }=0& \\ A{e}^{\frac{\sqrt{2mC}}{\hslash }\alpha }+B{e}^{-\frac{\sqrt{2mC}}{\hslash }\alpha }=0& \wedge & A\frac{\sqrt{2mC}}{\hslash }{e}^{\frac{\sqrt{2mC}}{\hslash }\alpha }-\frac{\sqrt{2mC}}{\hslash }B{e}^{-\frac{\sqrt{2mC}}{\hslash }\alpha }=0& \\ A{e}^{\frac{\sqrt{2mC}}{\hslash }\alpha }=-B{e}^{-\frac{\sqrt{2mC}}{\hslash }\alpha }& \wedge & [\frac{\sqrt{2mC}}{\hslash }=0\vee A{e}^{\frac{\sqrt{2mC}}{\hslash }\alpha }=B{e}^{-\frac{\sqrt{2mC}}{\hslash }\alpha }]& \text{Since ℏ2mC=0}\\ A{e}^{\frac{\sqrt{2mC}}{\hslash }\alpha }=-B{e}^{-\frac{\sqrt{2mC}}{\hslash }\alpha }& \wedge & A{e}^{\frac{\sqrt{2mC}}{\hslash }\alpha }=B{e}^{-\frac{\sqrt{2mC}}{\hslash }\alpha }& \\ A=0& \wedge & B=0& \end{array}$$

Hence $\psi (x)=0$, which is not a valid Quantum state.

3️⃣ $E\in [0,C]$

Applying the Constant Potential solution to the individual intervals, we have that

$$\psi (x)=\{\begin{array}{ll}A{e}^{\frac{x\sqrt{2m(C-E)}}{\hslash }},& x<-\alpha \\ B{e}^{\frac{x\sqrt{-2mE}}{\hslash }}+C{e}^{-\frac{x\sqrt{-2mE}}{\hslash }},& -\alpha <x<\alpha \\ D{e}^{\frac{-x\sqrt{2m(C-E)}}{\hslash }},& x>\alpha \end{array}$$

Were we used ${\lim }_{x\to \pm \infty }=0$ to remove terms of $\psi (x)$ for $|x|>\alpha$. Since the Potential is an even function, we can consider odd and even functions ${\psi }_E,{\psi }_O$ separately. Their general form is given by

Even solutions

$$\psi (x)=\{\begin{array}{ll}A{e}^{\frac{x\sqrt{2m(C-E)}}{\hslash }},& x<-\alpha \\ B{e}^{\frac{x\sqrt{-2mE}}{\hslash }}+B{e}^{-\frac{x\sqrt{-2mE}}{\hslash }},& -\alpha <x<\alpha \\ A{e}^{\frac{-x\sqrt{2m(C-E)}}{\hslash }},& x>\alpha \end{array}$$

Odd solutions

$$\psi (x)=\{\begin{array}{ll}A{e}^{\frac{x\sqrt{2m(C-E)}}{\hslash }},& x<-\alpha \\ B{e}^{\frac{x\sqrt{-2mE}}{\hslash }}-B{e}^{-\frac{x\sqrt{-2mE}}{\hslash }},& -\alpha <x<\alpha \\ -A{e}^{\frac{-x\sqrt{2m(C-E)}}{\hslash }},& x>\alpha \end{array}$$

4️⃣ $E=C$ 🔴

If the Energy equals $C$, we have that the solution is described by

$$\psi (x)=\{\begin{array}{rl}A{e}^{\frac{x\sqrt{2m(C-E)}}{\hslash }},& x<-\alpha \\ Bx+C,& -\alpha <x<\alpha \\ D{e}^{\frac{-x\sqrt{2m(C-E)}}{\hslash }},& x>\alpha \end{array}$$

Since no intervals have sign changes, $\psi (x)>0\forall x\in \mathbb{R}$, or $\psi (x)<0\forall x\in \mathbb{R}$. By considering $-\psi$ if needed, we can assume that $\psi (x)>0\forall x\in \mathbb{R}$.

Since $\psi (x)$ is Monotonic on $|x|>\alpha$, we have $\dot{\psi }(-\alpha )>0$ and $\dot{\psi }(\alpha )<0$. However no linear functions on $[-\alpha ,\alpha ]$ satisfy these constraints, so we have no solutions with $E=C$.

5️⃣ $E>C$ 🔴

If $E>C$, the solution in each interval is given by a sum of trigonometric functions. In order for the function to be Normalizable, it will be identically zero on $|x|>\alpha$. By continuity, this will require $x=0$ on $|x|<\alpha$ too, so we have no solutions here.