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Particle in a 1-Dimensional box

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A particle in a ane-dimensional box is a fundamental quantum mechanical approximation describing the translational motion of a single particle bars within an infinitely deep well from which it cannot escape.

Introduction

The particle in a box trouble is a common awarding of a quantum mechanical model to a simplified organisation consisting of a particle moving horizontally inside an infinitely deep well from which it cannot escape. The solutions to the problem give possible values of Due east and \(\psi\) that the particle can possess. E represents immune energy values and \(\psi(ten)\) is a wavefunction, which when squared gives us the probability of locating the particle at a certain position within the box at a given energy level.

To solve the problem for a particle in a ane-dimensional box, we must follow our Big, Big recipe for Quantum Mechanics:

Define the Potential Free energy, 5
Solve the Schrödinger Equation
Define the wavefunction
Define the allowed energies

Step one: Define the Potential Energy V

A particle in a 1D infinite potential well of dimension \(50\).

The potential free energy is 0 within the box (V=0 for 0<x<L) and goes to infinity at the walls of the box (5=∞ for x<0 or x>L). We assume the walls take infinite potential energy to ensure that the particle has zero probability of existence at the walls or outside the box. Doing so significantly simplifies our later mathematical calculations as we utilize these purlieus conditions when solving the Schrödinger Equation.

Step 2: Solve the Schrödinger Equation

The fourth dimension-independent Schrödinger equation for a particle of mass yard moving in one direction with energy E is

\[-\dfrac{\hbar^2}{2m} \dfrac{d^2 \psi(x)}{dx^2} + V(x)\psi(x) = E\psi(x) \label{5.five.one}\]

with

\(\hbar\) is the reduced Planck Constant where \( \hbar = \frac{h}{2\pi}\)
m is the mass of the particle
\(\psi(x)\) is the stationary time-independent wavefunction
5(x) is the potential energy as a office of position
\(E\) is the energy, a real number

This equation can be modified for a particle of mass thou free to move parallel to the x-axis with zero potential energy (V = 0 everywhere) resulting in the breakthrough mechanical description of free motion in one dimension:

\[ -\dfrac{\hbar^2}{2m} \dfrac{d^2\psi(ten)}{dx^2} = East\psi(x) \characterization{v.5.2}\]

This equation has been well studied and gives a general solution of:

\[\psi(x) = A\sin(kx) + B\cos(kx) \label{5.five.3}\]

where A, B, and yard are constants.

Step three: Define the wavefunction

The solution to the Schrödinger equation we found higher up is the general solution for a 1-dimensional system. Nosotros now need to utilize our boundary conditions to observe the solution to our particular arrangement. According to our purlieus conditions, the probability of finding the particle at x=0 or x=50 is null. When \(x=0\), \(\sin(0)=0\), and \(\cos(0)=ane\); therefore, B must equal 0 to fulfill this purlieus condition giving:

\[\psi(x) = A\sin(kx) \label{5.5.4}\]

We can now solve for our constants (A and k) systematically to ascertain the wavefunction.

Solving for k
Differentiate the wavefunction with respect to 10:

\[\dfrac{d\psi}{dx} = kA\cos(kx) \characterization{five.5.5}\]

\[\dfrac{d^{two}\psi}{dx^{2}} = -g^{2}A\sin(kx) \characterization{five.5.6}\]

Since \(\psi(10) = Asin(kx)\), then

\[\dfrac{d^{two}\psi}{dx^{two}} = -k^{2}\psi \characterization{five.five.7}\]

If we then solve for m by comparing with the Schrödinger equation in a higher place, nosotros find:

\[k = \left( \dfrac{8\pi^2mE}{h^2} \right)^{one/two} \label{5.v.8}\]

Now we plug k into our wavefunction:

\[\psi = A\sin\left(\dfrac{8\pi^{2}mE}{h^{2}}\right)^{1/2}x \label{5.5.ix}\]

Solving for A
To decide A, we take to apply the boundary conditions over again. Recollect that the probability of finding a particle at ten = 0 or ten = 50 is zero.

When \(10 = L\):

\[0 = A\sin\left(\dfrac{8\pi^{ii}mE}{h^{2}}\right)^{1/2}50 \label{five.v.x}\]

This is only true when

\[\left(\dfrac{viii\pi^{2}mE}{h^{2}}\correct)^{one/2}L = due north\pi \label{5.v.eleven}\]

where northward = 1,two,three…

Plugging this dorsum in gives us:

\[\psi = A\sin{\dfrac{n\pi}{L}}x \characterization{5.5.12}\]

To determine \(A\), recall that the total probability of finding the particle inside the box is i, meaning in that location is no probability of it being outside the box. When we detect the probability and gear up it equal to 1, nosotros are normalizing the wavefunction.

\[\int^{L}_{0}\psi^{ii}dx = 1 \label{five.5.13}\]

For our system, the normalization looks like:

\[A^2 \int^{50}_{0}\sin^2\left(\dfrac{n\pi x}{Fifty}\right) dx = ane \label{5.5.14}\]

Using the solution for this integral from an integral tabular array, we find our normalization constant, \(A\):

\[A = \sqrt{\dfrac{2}{50}} \label{5.v.15}\]

Which results in the normalized wavefunction for a particle in a 1-dimensional box:

\[\psi = \sqrt{\dfrac{2}{L}}\sin{\dfrac{n\pi}{L}}ten \label{5.v.16}\]

Step 4: Make up one's mind the Allowed Energies

Solving for E results in the allowed energies for a particle in a box:

\[E_n = \dfrac{n^{2}h^{two}}{8mL^{2}} \label{5.5.17}\]

This is an important event that tells us:

The energy of a particle is quantized and
The everyman possible energy of a particle is Not zippo. This is called the cypher-point energy and means the particle can never exist at rest because it always has some kinetic energy.

This is as well consistent with the Heisenberg Uncertainty Principle: if the particle had zero free energy, we would know where information technology was in both space and time.

What does all this mean?

The wavefunction for a particle in a box at the \(n=ane\) and \(n=ii\) energy levels look like this:

The probability of finding a particle a certain spot in the box is determined by squaring \(\psi\). The probability distribution for a particle in a box at the \(due north=1\) and \(northward=2\) energy levels looks like this:

Observe that the number of nodes (places where the particle has zero probability of existence located) increases with increasing energy n. Also note that as the energy of the particle becomes greater, the quantum mechanical model breaks down as the free energy levels get closer together and overlap, forming a continuum. This continuum means the particle is gratuitous and can take any free energy value. At such high energies, the classical mechanical model is applied equally the particle behaves more like a continuous moving ridge. Therefore, the particle in a box problem is an example of Wave-Particle Duality.

Important Facts to Learn from the Particle in the Box

The energy of a particle is quantized. This ways it tin can merely have on detached energy values.
The lowest possible free energy for a particle is Not zero (fifty-fifty at 0 Chiliad). This means the particle e'er has some kinetic energy.
The foursquare of the wavefunction is related to the probability of finding the particle in a specific position for a given free energy level.
The probability changes with increasing free energy of the particle and depends on the position in the box yous are attempting to define the energy for
In classical physics, the probability of finding the particle is independent of the energy and the same at all points in the box

Questions

Depict the wave role for a particle in a box at the \(n = 4\) energy level.
Draw the probability distribution for a particle in a box at the \(n = 3\) energy level.
What is the probability of locating a particle of mass chiliad between \(x = Fifty/4\) and \(x = 50/2\) in a 1-D box of length \(L\)? Assume the particle is in the \(n=1\) energy country.
Calculate the electronic transition energy of acetylaldehyde (the stuff that gives you a hangover) using the particle in a box model. Assume that aspirin is a box of length \(300 pm\) that contains 4 electrons.
Propose where forth the box the \(n=one\) to \(due north=2\) electronic transition would about likely take identify.

Helpful Links

Provides a live quantum mechanical simulation of the particle in a box model and allows you to visualize the solutions to the Schrödinger Equation: world wide web.falstad.com/qm1d/

References

Chang, Raymond. Physical Chemistry for the Biosciences. Sansalito, CA: University Science, 2005.