If you like us, please share us on social media.

The latest UCD Hyperlibrary newsletter is now complete, check it out.

ChemWiki: The Dynamic Chemistry E-textbook > Physical Chemistry > Quantum Mechanics > 5.5: Particle in Boxes > Particle in a 1-dimensional box

MindTouch

Copyright (c) 2006-2014 MindTouch Inc.

http://mindtouch.com

This file and accompanying files are licensed under the MindTouch Master Subscription Agreement (MSA).

At any time, you shall not, directly or indirectly: (i) sublicense,
resell, rent, lease, distribute, market, commercialize or otherwise
transfer rights or usage to: (a) the Software, (b) any modified version
or derivative work of the Software created by you or for you, or (c)
MindTouch Open Source (which includes all non-supported versions of
MindTouch-developed software), for any purpose including timesharing or
service bureau purposes; (ii) remove or alter any copyright, trademark
or proprietary notice in the Software; (iii) transfer, use or export the
Software in violation of any applicable laws or regulations of any
government or governmental agency; (iv) use or run on any of your
hardware, or have deployed for use, any production version of MindTouch
Open Source; (v) use any of the Support Services, Error corrections,
Updates or Upgrades, for the MindTouch Open Source software or for any
Server for which Support Services are not then purchased as provided
hereunder; or (vi) reverse engineer, decompile or modify any encrypted
or encoded portion of the Software.

A complete copy of the MSA is available at http://www.mindtouch.com/msa

A particle in a 1-dimensional box is a fundamental quantum mechanical approximation describing the translational motion of a single particle confined inside an infinitely deep well from which it *cannot *escape.

The particle in a box problem is a common application of a quantum mechanical model to a simplified system consisting of a particle moving horizontally within an infinitely deep well from which it cannot escape. The solutions to the problem give possible values of E and \(\psi\) that the particle can possess. E represents allowed energy values and \(\psi(x)\) 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 1-dimensional box, we must follow our **Big, Big recipe for Quantum Mechanics:**

- Define the Potential Energy,
*V* - Solve the Schrödinger Equation
- Define the wavefunction
- Define the allowed energies

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

The time-independent Schrödinger equation for a particle of mass *m* moving in one direction with energy *E* is

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

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
- V(x) is the potential energy as a function of position
- E is the energy, a real number

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

\[ -\dfrac{\hbar^2}{2m} \dfrac{d^2\psi(x)}{dx^2} = E\psi(x) \]

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

\(\psi(x) = Asin(kx) + Bcos(kx)\)

where A, B, and k are constants.

The solution to the Schrödinger equation we found above is the general solution for a 1-dimensional system. We now need to apply our **boundary conditions** to find the solution to our particular system.

According to our boundary conditions, the probability of finding the particle at x=0 or x=L is zero. When x=0 sin(0)=0 and cos(0)=1; therefore, *B must equal 0* to fulfill this boundary condition giving:

\(\psi(x) = Asin(kx)\)

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

**Solving for k**

Differentiate the wavefunction with respect to x:

\(\dfrac{d\psi}{dx} = kAcos(kx)\)

\(\dfrac{d^{2}\psi}{dx^{2}} = -k^{2}Asin(kx)\)

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

\(\dfrac{d^{2}\psi}{dx^{2}} = -k^{2}\psi\)

If we then solve for k by comparing with the Schrödinger equation above, we find:

\[k = \left( \dfrac{8\pi^2mE}{h^2} \right)^{1/2} \]

Now we plug k into our wavefunction:

\(\psi = Asin\left(\dfrac{8\pi^{2}mE}{h^{2}}\right)^{1/2}x\)

**Solving for A**

To determine A, we have to apply the boundary conditions again. Recall that the *probability of finding a particle at x = 0 or x = L is zero.*

When x = L:

\(0 = Asin\left(\dfrac{8\pi^{2}mE}{h^{2}}\right)^{1/2}L\)

This is only true when

\(\left(\dfrac{8\pi^{2}mE}{h^{2}}\right)^{1/2}L = n\pi\), where n = 1,2,3…

Plugging this back in gives us:

\(\psi = Asin{\dfrac{n\pi}{L}}x\)

To determine A, recall that the total probability of finding the particle inside the box is 1, meaning there is no probability of it being outside the box. When we find the probability and set it equal to 1, we are *normalizing* the wavefunction.

\(\int^{L}_{0}\psi^{2}dx = 1\)

For our system, the normalization looks like:

\(A^2 \int^{L}_{0}sin^2\left(\dfrac{n\pi}{L}\right)xdx = 1\)

Using the solution for this integral from an integral table, we find our normalization constant, A:

\(A = \sqrt{\dfrac{2}{L}}\)

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

\(\psi = \sqrt{\dfrac{2}{L}}sin{\dfrac{n\pi}{L}}x\)

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

\(E_n = \dfrac{n^{2}h^{2}}{8mL^{2}}\)

This is a very important result; it tells us that:

- The energy of a particle is quantized.
- The lowest possible energy of a particle is
zero. This is called the**NOT****zero-point energy**and means the particle can never be at rest because it always has some kinetic energy.

This is also consistent with the Heisenberg Uncertainty Principle: if the particle had zero energy, we would know where it was in both space and time.

The wavefunction for a particle in a box at the n=1 and n=2 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 n=1 and n=2 energy levels looks like this:

Notice that the number of **nodes**(places where the particle has zero probabily of being 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 energy levels get closer together and overlap, forming a continuum. This continuum means the particle is free and can have any energy value. At such high energies, the classical mechanical model is applied as the particle behaves more like a continuous wave. Therefore, the particle in a box problem is an example of Wave-Particle Duality.

- The energy of a particle is quantized. This means it can only take on discreet energy values.
- The lowest possible energy for a particle is
zero (even at 0 K). This means the particle**NOT***always*has some kinetic energy. - The square of the wavefunction is related to the probability of finding the particle in a specific position for a given energy level.
- The probability changes with increasing energy of the particle and depends on the position in the box you 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

- Draw the wave function 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 m between \(x = L/4\) and \(x = L/2\) in a 1-D box of length \(L\)? Assume the particle is in the \(n=1\) energy state.
- 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.
- Suggest where along the box the \(n=1\) to \(n=2\) electronic transition would most likely take place.

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

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

Last Modified

12:25, 10 Jul 2014

**Analytical Chemistry**

**Biological Chemistry**

**Inorganic Chemistry**

**Organic Chemistry**

**Physical Chemistry**

**Theoretical Chemistry**

**Cal Poly Pomona**

**Diablo Valley College**

**Florida State U**

**Hope College**

**Howard University**

**Purdue**

**Sacramento City College**

**UC Davis**

**UC Irvine**

**Zumdahl 9 ^{ed}**

An NSF funded Project

- © Copyright 2014 Chemwiki

Unless otherwise noted, content in the UC Davis ChemWiki is licensed under a Creative Commons Attribution-Noncommercial-Share Alike 3.0 United States License. Permissions beyond the scope of this license may be available at copyright@ucdavis.edu. Questions and concerns can be directed toward Prof. Delmar Larsen (dlarsen@ucdavis.edu), Founder and Director. Terms of Use