# Understanding the Energy States of a Particle in a Finite Potential Well

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In summary: However, given values for ##V_o## and ##L##, we can solve by graphing the curves given implicitly by the formulas.In summary, the conversation discusses finding energy states of a particle by defining the wave function over three discrete domains and solving the time independent Schrodinder equation. The potential term is zero inside the finite well and vanishes to zero outside the well. The wave functions are divided into even and odd parity, and boundary conditions are imposed to obtain the energy levels. However, the solutions do not have analytical forms and must be solved by graphing the given expressions.

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Homework Statement
Relevant Equations

To find the energy states of the particle, we define the wave function over three discrete domains defined by the sets ##\left\{x<-L\right\}##, ##\left\{L<x\right\}## and ##\left\{|x|<L\right\}##. The time independent Schr\"odinder equation is
\frac{\hbar^2}{2m}\frac{d^2\psi}{dx^2}=[E-V(x)]\psi

Outside of the finite well, the ##V(x)## term vanishes to zero to give the following linear second order ODE
\frac{d^2\psi}{dx^2}=\frac{2m}{\hbar^2}|E|\psi

whose characteristic polynomial is
$$r^2-\frac{2m|E|}{\hbar^2}=0$$ with the corresponding roots $$r=\pm \sqrt{\frac{2m|E|}{\hbar^2}}=\pm \alpha$$
The general solution to the ODE is

\psi=C_1e^{\alpha x}+C_2e^{-\alpha x}

To obtain the wave function inside of the finite well, we set ##V(x)=V_0## and solve the following time independent Schr\"odinder equation

\frac{d^2\psi}{dx^2}=\frac{2m}{\hbar^2}[V_0-|E|]\psi

whose characteristic polynomial is

r^2+\frac{2m}{\hbar^2}[V_0-|E|]=0

with the corresponding roots

r=\pm \sqrt{\frac{-2m}{\hbar^2}[V_0-|E|]}\Rightarrow\pm i\sqrt{q}

The general solution to the ODE is

\psi=C_1e^{iqx}+C_2e^{-iqx}\Rightarrow \psi= C_3cos(qx)+iC_4sin(qx)

Taking only the real parts of ##\psi## we have

\psi=A_1cos(qx)+A_2sin(qx)

So the even parity wave functions are

\psi=Ce^{\alpha x}|_{\left\{x<-L\right\}}

\psi=Acos(qx)|_{\left\{|x|<L\right\}}

\psi=Ce^{-\alpha x}|_{\left\{L<x\right\}}

and the odd parity wave functions are

\psi=-Ce^{\alpha x}|_{\left\{x<-L\right\}}

\psi=Asin(qx)|_{\left\{|x|<L\right\}}

\psi=Ce^{-\alpha x}|_{\left\{L<x\right\}}

For even solutions, we impose the boundary condition ##\psi \in C^1(\pm L)## such that ##\psi## is continuous and once differentiable at ##x=\pm L##

\psi_{in}(\pm L)= \psi_{out}(\pm L)\Rightarrow Acos(qL)=Ce^{-\alpha L}

\psi'_{in}(\pm L)= \psi'_{out}(\pm L)\Rightarrow Aqsin(qL)=\alpha Ce^{-\alpha L}

We divide equation (16) by (15) and then make substitutions using $$y=qL$$ and $$r=\sqrt{\frac{2mV_0L^2}{\hbar^2}}$$ The solutions to the energy states are given implicitly by the following expression.

ytan(y)=\sqrt{r^2-y^2}

We impose the same boundary condition on the odd functions

\psi_{in}(\pm L)= \psi_{out}(\pm L)\Rightarrow Asin(qL)=Ce^{-\alpha L}

\psi'_{in}(\pm L)= \psi'_{out}(\pm L)\Rightarrow Aqcos(qL)=-\alpha Ce^{-\alpha L}

We divide equation (20) by (21), make the same substitutions as above, and obtain the following expression.

-ycot(y)=\sqrt{r^2-y^2}

The expressions do not have analytical solutions, so there are no closed forms of the energy levels of the even and odd states. However, given values for ##V_o## and ##L##, we can solve by graphing the curves given implicitly by the formulas.

It looks like you got the potential backwards. It's zero inside the well and ##V_0 > 0## outside the well.

docnet
To find the energy states of the particle, we define the wave function over three discrete domains defined by the sets ##\left\{x<-L\right\}##, ##\left\{L<x\right\}## and ##\left\{|x|<L\right\}##. The time independent Schrodinder equation is
\frac{-\hbar^2}{2m}\frac{d^2\psi}{dx^2}=[E-V(x)]\psi

Outside of the finite well, ##V(x)=V_0## gives the following linear second order ODE
\frac{d^2\psi}{dx^2}=\frac{2m}{\hbar^2}[V_0-|E|]\psi

whose characteristic polynomial is
$$r^2-\frac{2m[V_0-|E|]}{\hbar^2}=0$$ with the corresponding roots $$r=\pm \sqrt{\frac{2m[V_0-|E|]}{\hbar^2}}=\pm \alpha$$
The general solution to the ODE is

\psi=C_1e^{\alpha x}+C_2e^{-\alpha x}

To obtain the wave function inside of the finite well, we set ##V(x)=0## and solve the following time independent Schrodinder equation

\frac{d^2\psi}{dx^2}=\frac{-2m}{\hbar^2}|E|\psi

whose characteristic polynomial is

r^2+\frac{2m}{\hbar^2}|E|=0

with the corresponding roots

r=\pm \sqrt{\frac{-2m}{\hbar^2}|E|}\Rightarrow\pm i\sqrt{q}

The general solution to the ODE is

\psi=C_1e^{iqx}+C_2e^{-iqx}\Rightarrow \psi= C_3cos(qx)+iC_4sin(qx)

The real parts of ##\psi## are a solution to our ODE

\psi=A_1cos(qx)+A_2sin(qx)

So the even parity wave functions are

\psi=Ce^{\alpha x}|_{\left\{x<-L\right\}}

\psi=Acos(qx)|_{\left\{|x|<L\right\}}

\psi=Ce^{-\alpha x}|_{\left\{L<x\right\}}

and the odd parity wave functions are

\psi=-Ce^{\alpha x}|_{\left\{x<-L\right\}}

\psi=Asin(qx)|_{\left\{|x|<L\right\}}

\psi=Ce^{-\alpha x}|_{\left\{L<x\right\}}

For even solutions, we impose the boundary condition ##\psi \in C^1(\pm L)## such that ##\psi## is continuous and once differentiable at ##x=\pm L##

\psi_{in}(\pm L)= \psi_{out}(\pm L)\Rightarrow Acos(qL)=Ce^{-\alpha L}

\psi'_{in}(\pm L)= \psi'_{out}(\pm L)\Rightarrow Aqsin(qL)=\alpha Ce^{-\alpha L}

We divide equation (16) by (15) and then make substitutions using $$y=qL$$ and $$r=\sqrt{\frac{2mV_0L^2}{\hbar^2}}$$ The solutions to the even functions are given implicitly by the following expression.

ytan(y)=\sqrt{r^2-y^2}

We impose the same boundary condition on the odd functions

\psi_{in}(\pm L)= \psi_{out}(\pm L)\Rightarrow Asin(qL)=Ce^{-\alpha L}

\psi'_{in}(\pm L)= \psi'_{out}(\pm L)\Rightarrow Aqcos(qL)=-\alpha Ce^{-\alpha L}

and divide equation (20) by (21). Then, we make the same substitutions as above to obtain the following expression.

-ycot(y)=\sqrt{r^2-y^2}

The expressions do not have analytical solutions, so there are no closed forms of the energy levels of the even and odd states.

When you solve for the wave function inside the well, it's not so much that you take only the real part of the solution but you choose the arbitrary constants so that the wave function is real.

docnet
vela said:
When you solve for the wave function inside the well, it's not so much that you take only the real part of the solution but you choose the arbitrary constants so that the wave function is real.
Thanks. It turns out I misquoted a theorem I read many months ago. I was supposed to write the real and imaginary parts of the complex valued solution form a fundamental set of solutions to the ODE. I went back to my ODE textbook and found the following theorem that I was trying to remember.

Question: since the Schrodinger equation has a second derivative operator, shouldn't the boundary condition for the wavefunction be that ##\psi## is twice continuously differentiable at the boundary, not just once continuously differentiable?

Last edited:
Classically, if you have a step potential, the momentum of the particle changes abruptly at the boundary. Physically, an infinite force imparts a finite momentum to the particle, and the kinetic energy of the particle is undefined at the boundary. In quantum mechanics, it's similar, so the second derivative of the wave function isn't defined at the boundary.

Note you have a similar situation with the infinite square well. In that problem, you don't even require the first derivative to be continuous at the boundaries because of the infinite potential outside the well. With the finite square well, you've gotten rid of that infinity so you can again require ##\psi'## be continuous, but there's still the remaining infinity in ##V'(x=\pm L)##.

docnet and PeroK
vela said:
Classically, if you have a step potential, the momentum of the particle changes abruptly at the boundary. Physically, an infinite force imparts a finite momentum to the particle, and the kinetic energy of the particle is undefined at the boundary. In quantum mechanics, it's similar, so the second derivative of the wave function isn't defined at the boundary.

Note you have a similar situation with the infinite square well. In that problem, you don't even require the first derivative to be continuous at the boundaries because of the infinite potential outside the well. With the finite square well, you've gotten rid of that infinity so you can again require ##\psi'## be continuous, but there's still the remaining infinity in ##V'(x=\pm L)##.
wow! thanks for the explanation.

## 1. What is a finite potential well?

A finite potential well is a concept in quantum mechanics that refers to a region in space where a particle is confined by a potential energy barrier. This barrier creates a finite well, meaning that the particle is only allowed to exist within a certain range of energy states within the well.

## 2. How does a particle behave in a finite potential well?

In a finite potential well, a particle can exist in discrete energy states, meaning that it can only have certain specific energies. These energy states are determined by the depth and width of the potential well. The particle's behavior is described by the Schrödinger equation, which predicts the probability of finding the particle at a certain energy level within the well.

## 3. What is the significance of understanding the energy states of a particle in a finite potential well?

Understanding the energy states of a particle in a finite potential well is crucial in many areas of physics, such as solid-state physics and nuclear physics. It allows us to predict and explain the behavior of particles in confined systems, which has important applications in technology and materials science.

## 4. How does the depth of the potential well affect the energy states of a particle?

The depth of the potential well determines the energy levels that a particle can occupy. A deeper well will have more energy levels available, while a shallower well will have fewer energy levels. This is because the depth of the well affects the strength of the potential barrier that the particle must overcome to exist outside of the well.

## 5. Can a particle escape from a finite potential well?

Yes, a particle in a finite potential well can escape if it has enough energy to overcome the potential barrier. This is known as tunneling, and it is a phenomenon that is unique to quantum mechanics. The probability of tunneling depends on the width and height of the potential barrier, as well as the energy of the particle.