Particle In A Box
by Kristen Adams
Discussion of Concept
Part A: Why do we care about examining a particle in a box?
A particle in a box resembles an electron in a stable orbit around a nucleus. Such an electron exhibits a standing wave pattern much like the standing wave pattern that can be produced on a string that is fixed at both ends.
Figure 1 Diagram of deBroglie matter waves of an electron in a stable orbit.
Source of figure: http://online.cclt.org/physicslab/content/PhyAPB/lessonnotes/dualnature/deBroglie.asp
The particle in a box is free (there are no forces acting upon it) but is limited spatially. We can use this model to examine and define the wave that an electron makes in orbit around a nucleus. Once we know the wave function of a particle we can then find the energy and momentum of the particle. Models, such as this one, can aid us in interpreting data gathered from actual experiments.
Part B: How does a particle in a box behave?
There are certain requirements of behavior for a particle in a box. We can visualize these requirements by again looking at the behavior of a standing wave on a string.
Figure 2 A standing wave with points of minimum amplitude (nodes) and maximum amplitude (antinodes)
Source of figure: http://www.cord.edu/dept/physics/p128/lecture99_35.html
A standing wave must meet two conditions. 1) At both ends of the string, a standing wave exhibits a node (point of zero amplitude). 2] The length of the standing wave is broken up into an integral number of half-wavelengths. The matter wave of a particle in a 3-D box has these same characteristics.
A particle starts from one side of the box at zero amplitude, hits the opposite side of the box (also at zero amplitude) and must return to its starting point, continuing the pattern. To describe the wave of the particle we must find a wave function that properly describes the motion of the particle. What kind of wave starts at zero amplitude and ends at zero amplitude? A sine wave! The matter wave of a particle inside a box is a sine wave just as the standing wave on a string is a sine wave. We have found a wave function that meets the first condition (from above) which is sin x. The wave function of form sin x should describe the wave at any point x in 1-D. If our box is 3-D, our wave function would be of the form sin(x)sin(y)sin(z) and would describe the wave at any point (x,y,z) in 3-D.
In order for the sine wave to be at a point of zero amplitude at each side of the box, the length of the sine wave in the box must be limited to ½ a wavelength, 1 wavelength, 1 ½ wavelengths, etc. which can be more succinctly written as n/2 wavelengths where n is 1, 2, 3, etc. We need to include this characteristic in our wave function. Defining the wave number, k (the number of radians of the wave cycle per unit length), to be (where is 180 degrees of the wave or one-half of a full wave and L is the length of one side of the box) we restrict the number of wavelengths in the box to an integer number of half wavelengths. Our wave function is now of the form or sin(kx). This wave function meets both conditions one and two.
Figure 3 Infinite potential well with wavelengths
Source of figure: http://hyperphysics.phyastr.gsu.edu/hbase/quantum/pbox.html
The final step in properly defining the wave function of a particle in a box is to normalize the wave function. The probability of finding a particular particle in all space is 1 (the particle exists). [Aside: Probability is a range from 0 to 1 where 0 means that a particular event will not occur and 1 indicates that a particular event is certain to occur.] A quantum mechanical property of a wave function is that the probability of finding a particle at some point in space is the absolute value of the wave function squared at the point of interest (x, y, z). The sum of the probabilities over all points in space should equal one. Thus,
where is our wave function in 3-D
In order for our wave function to meet this requirement we must tack on a coefficient (normalization coefficient). Our normalized wave function for a 1-D box is Asin(kx) where A is our normalizing coefficient and k = n/L where n=1,2,3, For a 3-D box, our normalized wave function would be Asin(kx) Bsin(ky) Csin(kz) with A,B,C acting as normalizing coefficients.
Notice that the requirements needed to produce a wave function suitable to a box has produced quantization. The wave number (k) increases by , but no values in between. Whereas an unbounded free particle has no such restrictions on k. Likewise, the energy of a particle, which depends on k, is also quantized for a particle in a box and continuous for an unbound free particle.
The particle in a box problem will be solved with more detail in the next section.
This example will illustrate a method of solving the 3-D Schrodinger equation to find the eigenfunctions for a infinite potential well, which is also referred to as a box.
A particle of mass m is captured in a box. This box can also be thought of as an area of zero potential surrounded by walls of infinitely high potential. The particle cannot penetrate infinitely high potential barriers. The box is of length a along the x axis, length b along the y axis and length c along the z axis.
This potential is described as follows:
V(x,y,z)=0 if 0<x<a , 0<y<b, 0<z<c Region I
V(x,y,z)= elsewhere Region II
Figure 4 Slice of 3-D infinite potential well
Source of figure: http://www.chembio.uoguelph.ca/educmat/chm386/rudiment/models/piab1/piab1prb.htm
The energy operator, , (the quantum mechanical operator we will use to find the energy of the particle) for a single particle of mass m, in a potential field V(x,y,z) is:
where is the momentum operator and is equal to
Total Energy = Kinetic Energy + Potential Energy
In symbolic form: E = T + V . In operator form: .
Kinetic Energy (T) =
The time-independent Schrodinger equation, which is used to find the possible energies, E, that the particle may have, is
is called the eigenfunction. It is the wave function that satisfies Eq-2.
Eq-2 is often called an eigenvalue equation. The eigenfunction for Eq-2 is to be found. This eigenfunction is used to determine the energies possible for the situation (generically referred to as the eigenvalues).
In Region II, V is infinite and the Hamiltonian, , is infinite.
The wave function and the energy are finite. Thus is zero in this
region. Since , there is zero probability that the particle will be found in this region.
In Region I, V is zero and the Hamiltonian is purely kinetic.
Plugging this Hamiltonian into the eigenvalue equation [Eq-2], we find
The subscript n is used in anticipation of discrete values that depend upon some integer n.
Now we must find a wave function that solves the eigenvalue equation, where is a number.
The wave function must be continuous across the regions. Therefore, at the walls, the wave function inside the box must equal the wave function outside the box. Since we have already determined that the wave function outside the box must be zero, the wave function inside the box must go to zero at the walls.
Figure 5 Slice of 3-D infinite potential well
Source of figure: http://scienceworld.wolfram.com/physics/InfiniteSquarePotentialWell.html
The walls of the box are located at x=0 and x=a; y=0 and y=b; z=0 and z=c. Thus,
Since the wave function does not exist beyond the walls of the box (created by the infinite potential barrier) we are not concerned with the behavior of the wave function across the barrier. However, in other types of problems where the wave function does exist on the other side of the barrier, we need to make sure that the first derivative of the wave function is continuous across the barrier. [See worked example B for an example of this procedure.]
Back to our time-independent Schrodinger equation [Eq-3],
The general solution to this homogeneous differential equation in 1-D is
+ where [Eq-5]
Our boundary conditions [Eq-4] tell us that at x=0, 0. Therefore, B must equal 0 and . The same applies for the 3-D case.
A is the normalization coefficient and the superscripts (1), (2), (3) signify that each dimension is independent.
From our boundary conditions [Eq-4] we must also make the wave function equal zero at x=a, y=b and z=c. The sine function is zero at integral multiples of . Therefore,
an; b n; c n where n = 0,1,2,3
The n subscript on k has been dropped to improve clarity, but is technically still there.
Rearranging gives; ; [Eq-7]
We can now plug k into Eq-5 to find the possible values of energy for a particle in this box. Rearranging Eq-5 to solve for E gives
[Aside: ] Substituting in values for k
from Eq-7 one finds .
As a final step, we must normalize our wave function.
Our wave function [Eq-6] is .
To normalize, 1
This becomes 1
The eigenenergies, , and the normalized eigenfunctions, , for the 3-D box problem are:
This example will illustrate a method of solving the 1-D Schrodinger equation to find the eigenfunctions for a finite potential well. The potential is defined as follows:
V(x)= 0 if x<-a Region I
V(x)=-Vo if Region II [Eq-8]
V(x)= 0 if x>a Region III
If E is greater than 0, the wave function is unbounded as there are no boundary conditions that would place any limitations on the wave function.
If E is less than zero, boundary conditions are imposed upon the wave function. This is the case that will be examined below.
Figure 6 Finite potential well
Source of figure: http://scienceworld.wolfram.com/physics/FiniteSquarePotentialWell.html
Starting with the time-independent Schrodinger equation:
Applying the conditions of Eq-8, inside the well, the Schrodinger equation is:
for Region II for Regions I and III
Let and .
Inside the well, the wave function has the general solution:
Outside the well, the general solution to the Schrodinger equation is: .
Now we must apply boundary conditions to ensure that the composite wave function behaves properly as it crosses boundaries.
There are several boundary conditions must be imposed upon our wave function.
1. The wave function for the particle outside the well must show that the likelihood of finding the particle in these regions (I and III), decreases as the distance into the barrier region increases.
2. The wave functions must be continuous across the boundaries.
3. The first derivatives of the wave function must be continuous across the boundaries.
Applying the first boundary condition, the wave function must decrease as the distance into the barrier increases. For Region I (x is negative) . Likewise, in Region III (x is positive) .
Applying the second and third boundary conditions to the boundary between Regions I and II, where x = -a,
Applying the second and third boundary conditions to the boundary between Regions II and III, where x = a,
There now exists a set of four equations ([Eq-9] - [Eq-12]) for our four unknown coefficients A, B, C, D. We can solve for the coefficients through addition and subtraction of these equations and normalization.
Adding [Eq-9] and [Eq-11] produces
Adding [Eq-10] and [Eq-12] results in
Subtracting [Eq-11] from [Eq-9] gives
Subtracting [Eq-12] from [Eq-10] yields
Dividing [Eq-16] by [Eq-13] results in .
Inserting this into [Eq-15] we obtain
In order for this equality to be true, A=0 and C=D.
Substituting these values into [Eq-13] and [Eq-15] (equations matching the wave functions at the boundaries) we get
Thus we get an even function inside the well.
Dividing [Eq-14] by [Eq-15] results in .
Inserting this into [Eq-15] we obtain .
In order for this equality to be true, B=0 and C=-D.
Substituting these values into [Eq-13] and [Eq-15](equations matching the wave functions at the boundaries) we get
Thus we get an odd function inside the well.
The precise values for the coefficients are found through normalization:
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