8.3: Particle in a One-Dimensional Box

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.

Introduction

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.

recipe for Quantum Mechanics

To solve the problem for a particle in a 1-dimensional box, we must follow our Big, Big recipe for Quantum Mechanics:

1. Define the Potential Energy, \(V\)
2. Solve the Schrödinger Equation
3. Solve for the wavefunctions
4. Solve for the allowed energies

Step 1: Define the Potential Energy V

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.

Figure \(\PageIndex{1}\): Cartoon of a particle trapped in a 1-D infinite potential of length L.

Step 2: Solve the Schrödinger Equation

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

\[-\dfrac{\hbar^2}{2m} \dfrac{d^2 \psi(x)}{dx^2} + V(x)\psi(x) = E\psi(x) \label{5.5.1}\]

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:

\[ -\dfrac{\hbar^2}{2m} \dfrac{d^2\psi(x)}{dx^2} = E\psi(x) \label{5.5.2}\]

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

\[\psi(x) = A\sin(kx) + B\cos(kx) \label{5.5.3}\]

where A, B, and k are constants.

Step 3: Define the Wavefunction

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 \(x=0\), then \(\sin(0)=0\) and \(\cos(0)=1\); therefore, \(B\) must equal 0 to fulfill this boundary condition giving:

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

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

Solving for \(k\)

Differentiate the wavefunction with respect to \(x\):

\[\dfrac{d\psi}{dx} = kA\cos(kx) \label{5.5.5}\]

Differentiate the wavefunction again with respect to \(x\):

\[\dfrac{d^{2}\psi}{dx^{2}} = -k^{2}A\sin(kx) \label{5.5.6}\]

Since \(\psi(x) = A\sin(kx)\), then

\[\dfrac{d^{2}\psi}{dx^{2}} = -k^{2}\psi \label{5.5.7}\]

Substituting this for the second derivative in the Schrödinger equation (\(\ref{5.5.2}\)) yields:

\[-\frac{\hbar^2}{2m}k^2\psi(x) = E\psi(x)\label{5.5.8a}\]

If we then solve for k, we find:

\[k = \left( \dfrac{8\pi^2mE}{h^2} \right)^{1/2} \label{5.5.8}\]

Now we plug \(k\) into our wavefunction (Equation \ref{5.5.4}):

\[\psi = A\sin\left(\dfrac{8\pi^{2}mE}{h^{2}}\right)^{1/2}x \label{5.5.9}\]

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:

\[0 = A\sin\left(\dfrac{8\pi^{2}mE}{h^{2}}\right)^{1/2}L \label{5.5.10}\]

This is only true when

\[\left(\dfrac{8\pi^{2}mE}{h^{2}}\right)^{1/2}L = n\pi \label{5.5.11}\]

where \(n = 1,2,3, …\)

Plugging this back in gives us:

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

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.

\[\int^{L}_{0}\psi^{2}dx = 1 \label{5.5.13}\]

For our system, the normalization looks like:

\[A^2 \int^{L}_{0}\sin^2\left(\dfrac{n\pi}{L}\right)x\,dx = 1 \label{5.5.14}\]

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

\[A = \sqrt{\dfrac{2}{L}} \label{5.5.15}\]

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

\[\psi_n = \sqrt{\dfrac{2}{L}}\sin{\dfrac{n\pi}{L}}x \label{5.5.16}\]

where \(n = 1,2,3, …\)

Step 4: Determine the Allowed Energies

Solving for the energy of each \(\psi\) requires substituting Equation \(\ref{5.5.16}\) into Equation \(\ref{5.5.2}\) to get the allowed energies for a particle in a box:

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

Equation \(\ref{5.5.17}\) is a very important result and tells us that:

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

Zero point energy is one manifestation of the Heisenberg Uncertainty Principle, which states that you cannot know a particle's exact momentum and exact position simultaneously.

Heisenberg uncertainty principle²

The Heisenberg uncertainty principle is a general quantum phenomenon, where certain pairs of physical quantities cannot be measured exactly simultaneously. In other words, the quantities are coupled in such a way that the uncertainty in one is related to the uncertainty in the other quantity. These pairs are called complementary observables. Momentum and position are one set of complementary observables. In the case of momentum and position of a particle the Heisenberg uncertainty principle quantifies this relationship as:

\[\Delta x \Delta p \ge \frac{\hbar}{2}\label{HeisUnc}\]

where \(\Delta x\) is the uncertainty in position and \(\Delta p\) is the uncertainty in momentum. Thus, for a particle confined to a 1-D box of length L (\(\Delta x ≈ L\)):

\[\Delta p \gtrsim \frac{\hbar}{2L}\label{partinboxHeisUnc}\]

Since kinetic energy  is expressed in terms of momentum by the equation \(KE=\frac{p^2}{2m}\) this means a confined particle must have some non-zero amount of kinetic energy to satisfy the Heisenberg uncertainty principle.

What does all this mean?

The wavefunction for a particle trapped in an infinite 1-D well are sine waves with nodes at the walls as shown in figure \(\PageIndex{2}a\).

a)

b)

Figure \(\PageIndex{2}\): Calculated energies, wavefunctions and probability densities for an electron in a 1.0 nm long infinite well. a) Energy levels and wavefunctions. b) Energy levels and probability densities.

The probability of finding a particle at a certain spot in the box is determined by squaring \(\psi\). The probability distributions for a trapped particle are shown in figure \(\PageIndex{2}b\).

Notice:

• The lowest energy (n=1) quantum state has energy > 0, the zero-point energy. This is a characteristic of localized particles.
• The number of nodes (places where the particle has zero probability of being located) increases with increasing energy level, n.
• The spacing between energy levels increases as n increases. This can be made quantitative by deriving the expression for the separation between neighboring levels.

\[\Delta E = E_{n+1} - E_n = \dfrac{(n+1)^2 h^2}{8 m L^2} - \frac{n^2 h^2}{8 m L^2} = (2n+1)\frac{h^2}{8mL^2}\label{PinBdeltaE}\]

Other potentials have different energy spacing patterns.

 

Example \(\PageIndex{1}\)

What is the \(\Delta E\) between the \(n = 4\) and \(n = 5\) states for an \(F_2\) molecule trapped within in a one-dimension well of length 3.0 cm? At what value of \(n\) does the energy of the molecule reach \(¼k_BT\) at 450 K, and what is the separation between this energy level and the one immediately above it?

Solution

Since this is a one-dimensional particle in a box problem, the particle has only kinetic energy (V = 0), so the permitted energies are:

\[E_n = \dfrac{n^2 h^2}{8 m L^2} \nonumber\]

with \(n=1,2,...\)

The energy difference between \(n = 4\) and \(n=5\) is then

\[\Delta E = E_5 - E_4=\dfrac{5^2 h^2}{8 m L^2} - \dfrac{4^2 h^2}{8 m L^2}\nonumber\]

Using Equation \(\ref{5.5.17}\) with the mass of \(F_2\) (37.93 amu = \(6.3 \times 10^{-26}\;kg \)) and the length of the box (\(L= 3 \times 3.0\times10^{-2}\;\text{m}^2\):

\[\Delta E=\dfrac{9 h^2}{8 m L^2} = \dfrac{9 (6.626\times10^{-34}\;\text{kg} \cdot \text{m}^2 \cdot \text{s}^{-1})^2 }{8(6.30938414\times10^{-26}\;\text{kg})(3.0\times10^{-2}\;\text{m}^2)^2 }\nonumber\]

\[\Delta E=8.70\times10^{-39}\; J \nonumber\]

The \(n\) value for which the energy reaches \(\frac{1}{4} k_B T\):

\[\dfrac{n^2 h^2}{8 m L^2} = \dfrac{1}{4} k_B T \nonumber\]

\[n=1.79 \times 10^9 \nonumber\]

The separation between \(n+1\) and \(n\):

\[\Delta E = E_{n+1} - E_{n} = \dfrac{(n+1)^2h^2 - (n)^2h^2}{8mL^2} \nonumber\]

\[\Delta E = \dfrac{(2n+1)h^2}{8mL^2} \nonumber\]

\[\Delta E = 3.47\times10^{-30}\;J \nonumber\]

Example \(\PageIndex{2}\): Modeling a linear conjugated system 

The particle-on-a-line model provides a reasonable approximation for the electronic states of the \(\pi\) electrons in conjugated chains (nearly linear). Consider hexatriene (=–=–=). We can estimate the transition energy between the highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO) using this model. The lowest energy photon absorbed by the electronic system (longest wavelength) would be this transition.

Solution

There are 6 \(\pi\) electrons. Each orbital (quantum state) can hold two electrons. So the HOMO is the n = (# \(\pi\) electrons)/2 = 3.

Thus we are looking for the transition energy between the n = 3 and n = 4 states:

\[\Delta E = E_{4} - E_{3} = \dfrac{4^2 h^2}{8 m L^2} - \dfrac{3^2 h^2}{8 m L^2} = \dfrac{7 h^2}{8mL^2}\nonumber\]

Using an average C–C bond length of roughly 140 pm give L ≈ 700 pm. The mass of an electron is 9.109 X 10^-31 kg which yields:

\[\Delta E =  \dfrac{7 (6.626 \times 10^{-34} J s)^2}{8(9.109 \times 10^{-31} kg)(700 \times 10^{-12} m)^2} = 8.61 \times 10^{-19} J \nonumber\]

Which corresponds to a wavelength of about 218 nm, well into the UV.

Important Facts to Learn from the Particle in the Box

• The energy of a particle is quantized. This means it can only take on discreet energy values.
• The lowest possible energy for a confined particle is NOT zero (even at 0 K). This means the particle 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.
• In classical physics, the probability of finding the particle is independent of the energy and the same at all points in the box.

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: www.falstad.com/qm1d/

References

1. Chang, Raymond. Physical Chemistry for the Biosciences. Sansalito, CA: University Science, 2005.
2. Levine, Ira. Quantum Chemistry, 4th Ed. (Prentice-Hall, Inc. Englewood Cliffs, New Jersey, 1991) Pp. 80 - 83.