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

Created by Adam Barletta - Spring 2022, needs a computational model

## 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:

- The general solution to the Schrödinger Equation will be found in all regions
- Each region extending to positive or negative infinity will be made finite to comply with the requirements of a proper wave function
- The solution will be made continuous at all boundaries and made differentiable at non-infinite potential boundaries
- Analyze and find intuitive concepts

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].

#### Boundary: [math]\displaystyle{ x = 0 }[/math]

As stated earlier, any boundary point that involves an infinite potential must only be continuous. When [math]\displaystyle{ x \lt 0 }[/math], potential energy equals infinity. This allows us to only need to fulfill the condition that:

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

Substituting in the given x value for each wave equation:

[math]\displaystyle{ 0=Asin(k(0))+Bcos(k(0)) }[/math]

[math]\displaystyle{ 0 = A(0) + B(1) }[/math]

[math]\displaystyle{ B = 0 }[/math]

Now knowing that our constant [math]\displaystyle{ B = 0 }[/math], we can rewrite our wave equations so they fit the boundary conditions.

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

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

#### Boundary: [math]\displaystyle{ x = L }[/math]

This boundary condition lies on the border of two finite potentials, requiring us to have continuity and differentiability between our two wave functions. We will start by applying these conditions:

[math]\displaystyle{ \Psi_{II}(L)=\Psi_{III}(L) }[/math]

[math]\displaystyle{ \Psi_{II}'(L)=\Psi_{III}'(L) }[/math]

For mathematical simplicity we can express these equations as ratios:

[math]\displaystyle{ \frac{\Psi_{II}'(L)}{\Psi_{II}(L)}=\frac{\Psi_{III}'(L)}{\Psi_{III}(L)} }[/math]

Differentiating each wave equation:

[math]\displaystyle{ \Psi_{II}'=Akcos(kx) }[/math]

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

Substituting into the ratio we created previously and plugging in our given value of x:

[math]\displaystyle{ \frac{Akcos(kL)}{Asin(kL)}=\frac{-Dk'e^{k'L}}{De^{k'L}} }[/math]

Which then conveniently simplifies down to:

[math]\displaystyle{ kcot(kL)=-k' }[/math]

Now that we have a clear representation of the relationship between [math]\displaystyle{ k }[/math] and [math]\displaystyle{ k' }[/math], we can begin to better understand the implications our calculations have in the domain of quantum mechanics.

## Final Analysis

After many rigorous calculations, we have finally created a solution for the "Semi-Infinite Well" problem. Given the regions of [math]\displaystyle{ x \lt 0 }[/math] ("Region I" with infinite potential), [math]\displaystyle{ 0 \lt x \lt L }[/math] ("Region II" with zero potential), and [math]\displaystyle{ x \gt L }[/math] ("Region III" with finite, postive potential [math]\displaystyle{ V_{0} }[/math]), Our final wave functions are as follows:

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

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

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

Which consist of the following relationships:

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

Now what does all this tell us? To start, If we were to plug in our known values of [math]\displaystyle{ k }[/math] and [math]\displaystyle{ k' }[/math] in for our third listed relationship, we would notice something interesting. The equality only is true for discrete values of [math]\displaystyle{ E }[/math]. This relationship displays the idea of the quantization of energy! Our wave function changes at these different energy levels, and the particle can only possess the quantized energy predicted by our relationships. Secondly, you may have noticed that unlike the infinite potential region, the non-zero potential region has a decaying wave function. We know that there is no such thing as an infinite potential in the real world, so this decaying function has startling implications in the world of quantum physics. In this non-zero potential region, the particle's likelihood of existing is extremely low but never zero. If we were to reduce the width of this region and have it act more as a wall between two zero potential regions, we could possibly see the particle pop out on the other side! This is of course was not thought to be possible in classical physics. Classically, no matter how long you stare at a ball in a valley, for example, it would never miraculously roll all the way up one of the sides and escape the valley. This is not true for quantum physics! As we enter the domain of particles we observe this exact behavior and it is known as Quantum Tunneling.

References:

Dr. John M. Standard's 2013 "The Particle in a Half-Infinite Well", url: https://www.yumpu.com/en/document/read/34504139/the-particle-in-a-half-infinite-well

https://www.chegg.com/homework-help/questions-and-answers/energy-values-first-part-proble-e1-1187-ev-e2-4667-ev-q31361434

Paul D'Alessandris's 2021 "Solving the 1D Semi-Infinite Square Well" url: https://phys.libretexts.org/Bookshelves/Modern_Physics/Book%3A_Spiral_Modern_Physics_(D'Alessandris)/6%3A_The_Schrodinger_Equation/6.6%3A_Solving_the_1D_Semi-Infinite_Square_Well