Solution for a Single Particle in a Semi-Infinite Quantum Well

Claimed by Adam Barletta 4/21/22

Introduction

As you may have already learned, the Single Particle in a Box problem is a greatly intuitive way to begin to understand some of the major concepts connecting classical mechanics and quantum mechanics. It is recommended that you first begin by analyzing the "Infinite Well" solution prior to this "Semi-Infinite Well". Once you have a solid understanding of the concepts utilized in that example, you will find yourself better able to understand this specific example and explore the nuances that come with it. Like the "Infinite Well" example, we begin with a particle in a well where the potential energy where [math]\displaystyle{ x \lt 0 }[/math] is infinity. We will call the region of infinite potential "Region I". The region where [math]\displaystyle{ 0 \lt x \lt L }[/math] ([math]\displaystyle{ L }[/math] is any arbitrary distance from [math]\displaystyle{ x }[/math]) possesses a potential energy of [math]\displaystyle{ 0 }[/math] ("Region II"), and the area [math]\displaystyle{ x \gt L }[/math] has an unchanging potential equal to the positive constant [math]\displaystyle{ V_{0} }[/math] ("Region III").

Background

Of course, we will start with Schrödinger's Equation: [math]\displaystyle{ \frac{-\hbar ^{2}}{2m}\frac{\partial ^{2}\Psi}{\partial x^{2}}+V\Psi=E\Psi }[/math]. This specific form of Schrödinger's Equation is known as a "1-Dimensional Time-Independent Schrödinger Equation" due to its singular spacial dimension and its lack of dependency on time. The reason for the singular spacial dimension is for simplicity. The equation does not depend on time due to the the potential regions being only dependent on position, allowing us to ignore time for now. The variable [math]\displaystyle{ \hbar }[/math] represents Planck's constant [math]\displaystyle{ h }[/math] ([math]\displaystyle{ 6.62607015\times 10^{-34} \frac{\text{m}^{2}\text{kg}}{\text{s}} }[/math]) divided by [math]\displaystyle{ 2\pi }[/math], [math]\displaystyle{ m }[/math] represents the mass of the singular particle, [math]\displaystyle{ \Psi }[/math] represents the wave function we are trying to find, [math]\displaystyle{ V }[/math] represents the potential energy, and [math]\displaystyle{ E }[/math] represents the energy of the particle. Due to the piece-wise nature of our potential, we will solve this equation in each region separately at first.

Method to Solve

The procedure we will use to solve this problem will go as follows:

1. The general solution to the Schrödinger Equation will be found in all regions
2. Each region extending to positive or negative infinity will be made finite to comply with the requirements of a proper wave function
3. The solution will be made continuous at all boundaries and made differentiable at non-infinite potential boundaries
4. Assure the equation is properly normalized

Following each of these steps is crucial for finding an applicable solution. Although this is an impossible real world scenario, following these steps will assure that this example is intuitive and beneficial to our understanding of quantum wave functions.

Solution

Step 1: General Solutions in Each Region

Region I

First substituting the given infinite potential into the Schrödinger Equation:

[math]\displaystyle{ \frac{-\hbar ^{2}}{2m}\frac{\partial ^{2}\Psi}{\partial x^{2}}+(\infty) \Psi=E\Psi }[/math]

Thinking back on the "Infinite Well" solution, we can intuitively find that in order for this equation to make sense, the wave function:

[math]\displaystyle{ \Psi_{I}=0 }[/math]

The reason for this is due to the fact that a theoretically infinite potential would be not allow for a particle to exist in given it has a finite energy. The likely hood of the particle existing in this region is represented by the wave function, so given that it would be impossible we can infer that the wave function would equal zero in this specific region.

Region II

Again, we will first substitute the given potential, zero, into the Schrödinger Equation:

[math]\displaystyle{ \frac{-\hbar ^{2}}{2m}\frac{\partial ^{2}\Psi}{\partial x^{2}}+(0) \Psi=E\Psi }[/math]

This equation then turns into:

[math]\displaystyle{ \frac{-\hbar ^{2}}{2m}\frac{\partial ^{2}\Psi}{\partial x^{2}}=E\Psi }[/math]

Due to the constants in this equations, we can determine that the general solution of the wave function can be written using trigonometric functions or exponentials. For this example we will use the trigonometric general solution:

[math]\displaystyle{ \Psi_{II}=Asin(kx)+Bcos(kx) }[/math]

Where [math]\displaystyle{ A }[/math], [math]\displaystyle{ B }[/math], and [math]\displaystyle{ k }[/math] are arbitrary constants. [math]\displaystyle{ k }[/math] can be solved for by substituting our general solution back into the original Schrödinger Equation, taking the second derivative, and preforming some algebraic manipulations. The second derivative of the Schrödinger Equation with the wave function we found, [math]\displaystyle{ \Psi_{II} }[/math], gives:

[math]\displaystyle{ \frac{\hbar ^{2}k^{2}}{2m}[Asin(kx)+Bsin(kx)]=E[Asin(kx)+Bsin(kx)] }[/math]

Manipulating to solve for [math]\displaystyle{ k }[/math]:

[math]\displaystyle{ k=\frac{\sqrt{2mE}}{\hbar } }[/math]

Region III

When solving Schrödinger Equation with a finite constant potential, [math]\displaystyle{ V_{0} }[/math], the wave function will differ given the scenario that the particles energy, [math]\displaystyle{ E \gt V_{0} }[/math] or [math]\displaystyle{ E \lt V_{0} }[/math]. Given that this is a "Semi-Infinite Well" problem, we will maintain the "well" scenario by assuming [math]\displaystyle{ E \lt V_{0} }[/math]. I personally implore you attempt to solve this problem given [math]\displaystyle{ E \gt V_{0} }[/math] and discover how these answers differ.

Once more, substituting our given potential [math]\displaystyle{ V_{0} }[/math] into the Schrödinger Equation:

[math]\displaystyle{ \frac{-\hbar ^{2}}{2m}\frac{\partial ^{2}\Psi}{\partial x^{2}}+(V_{0}) \Psi=E\Psi }[/math]

Keeping in mind that [math]\displaystyle{ E \lt V_{0} }[/math], the equation can be rewritten in the following form:

[math]\displaystyle{ \frac{\hbar ^{2}}{2m}\frac{\partial ^{2}\Psi}{\partial x^{2}}=(V_{0}-E)\Psi }[/math]

Similar to "Region II", we now find ourselves with a simple differential equation with constants. Unlike "Region II", we will instead chose to utilize the exponential form of the wave equation given that this region is classically forbidden to a particle with energy [math]\displaystyle{ E \lt V_{0} }[/math]. Our wave function must then take the form:

[math]\displaystyle{ \Psi_{III}=Ce^{k'x}+De^{-k'x} }[/math]

Where [math]\displaystyle{ C }[/math], [math]\displaystyle{ D }[/math], and [math]\displaystyle{ k' }[/math] are arbitrary constants. [math]\displaystyle{ k' }[/math] can be solved for by substituting our general solution back into the original Schrödinger Equation, taking the second derivative, and preforming some algebraic manipulations. The second derivative of the Schrödinger Equation with the wave function we found, [math]\displaystyle{ \Psi_{III} }[/math], gives:

[math]\displaystyle{ -\frac{\hbar ^{2}k'^{2}}{2m}(Ce^{k'x}+De^{-k'x})=(E-V_{0})(Ce^{k'x}+De^{-k'x}) }[/math]

Manipulating to solve for [math]\displaystyle{ k' }[/math]:

[math]\displaystyle{ k'=\frac{\sqrt{2m(V_{0}-E)}}{\hbar } }[/math]

Step 2: General Solutions at Positive and Negative Infinity

Intuition tells us that the value of the wave equation cannot "blow up" at either positive or negative infinity. This is due to the fact that the particle must have a total probability of existing at any point in space of 100% (it must exist somewhere). If our wave function "blows up" then we would have an non-applicable answer where the probability of the particle existing at a given position would not fit the requirements of quantum mechanics we've confirmed with numerous experimental trials. Applying this condition only applies to "Region I" and "Region "III".

Region I

Given that our wave function in this region currently has a value:

[math]\displaystyle{ \Psi_{I} = 0 }[/math]

It will not blow up as x approaches negative infinity. We are free to leave this region's wave function as it is.

Region III

Our wave function in this region is currently being described in the form:

[math]\displaystyle{ \Psi_{III}=Ce^{k'x}+De^{-k'x} }[/math]

In order for this equation to not blow up as x approaches infinity, we must set our constant [math]\displaystyle{ C = 0 }[/math]. This leaves us with a decaying function:

[math]\displaystyle{ \Psi_{III}=De^{-k'x} }[/math]

Where the wave function approaches 0 as x approaches infinity. Our wave function now fits this quantum wave function requirement.

Step 3: Applying Boundary Conditions

Recall that as stated earlier, this piece-wise wave function must be continuous at every point. At points where an infinite potential is not involved, it must also be differentiable. The boundaries we must adjust for are [math]\displaystyle{ x = 0 }[/math] and [math]\displaystyle{ x = L }[/math].