Module 10 – Infinite and Finite Wells
I.Particle In A Box With Infinite Walls
A.Why Solve An Unphysical Problem?
There is no quantum system in the real world with infinite walls (implies an infinite force). So why should we care about the solution to this problem?
All but the simplest of quantum systems are too complicated to solve by hand. Thus, we shall start by solving idealized quantum problems that can both be solved by hand and be used as models to approximate real systems in the same way that we modeled real projectiles by neglecting air friction in classical physics. In quantum physics this practice of modeling is even more important as it builds our intuition for the quantum world which differs markedly with our everyday experience. This modeling also allows us to gain familiarity with the mathematical techniques needed to solve harder quantum problems.
B.Solving the T.I.S.E.
We start by considering a particle of total energy E that is trapped in a region of space from x = 0 to x = L. Inside the box, the potential is zero while outside the box the potential is infinite.
The problem can be broke into three regions of constant potential:
Region I (- to 0), Region II (0 to L), and Region III (L to ).
Because the particle has no probability of being in either Region I or Regions III (i.e. outside the well), we know that the wave function in each of these two regions is zero!!
The problem is therefore reduced to solving the Time Independent Schrodinger Equation in Region II where the particle’s kinetic energy (E-U) is positive. Using the fact that U = 0 in this region, the Time Independent Schrodinger Equation becomes:
where
Thus, we have a traveling wave with a real linear momentum and a real wave number whose solution we considered in Lesson 9. Since we are dealing with a bounded problem, we choose to use the cosine-sine solution form from Lesson 9 as it will make the math easier when applying boundary conditions.
Next, we must consider the boundaries where we broke our problem into constant potential regions. The wave function must be continuous through out all of space. Thus, we have the following boundary conditions at x =0 and x=L:
and
Using the first condition, we get
Thus, our wave function in Region II has been reduced to only sine terms.
We now apply our second boundary condition and obtain
Since the wave function must be non-zero in some region of space for the particle to exist, the constant A and the constant k can not be zero!! Note that this condition on k implies that the particle must have some kinetic energy (i.e. a stationary particle is not allowed). This is consistent with the uncertainty principle. If we confine the particle’s location then we can’t know its momentum is exactly zero!! In order to make the equation true, k L must take on non-zero values that make the sine function zero.
Thus, we see that only certain wave numbers are allowed. This is exactly the same result that you would get by considering a bounded guitar string in classical physics (only certain notes can be played). It also means from deBroglie’s relationship that the particle has a limited number of linear momentum values (i.e. it is quantized).
By rearranging the relationship between wave number and particle energy and inserting our new quantization condition for particle’s wave number, we see that the particle’s energyis likewise quantized.
We can multiply top and bottom of the equation by c2 to obtain
Again, the product hc sets the basic size of the energy levels. If you squeeze the box (perhaps by placing atoms closer together), the energy levels get larger. This simple model explains why nuclear levels where the box has a length of femtometers is much greater than atomic energy levels where the width is on the order of tenths of nanometers. We also see how the rest mass energy of the particle affects the particle’s energy levels.
It is also important that you realize that each energy eigenvalue is associated with a specific wave function called an energy eigenfunction (remember from previous lesson that eigen is german for “proper” and means that only certain solutions are possible). From our work, we have that the wavefunction corresponding to the nth energy level is
We have one final requirement for the wavefunction that we can use to solve the remaining constant An. The wave function must be normalized in order to relate the square of the wave function to a probabilitydensity.Thus, the nth wave function must meet the requirement that
There is nothing to be gained by making the constant A complex. Thus, the absolute value sign can be dropped.
You should have previously evaluated this integral in physics 2424 for AC circuits and in Calculus II using integration by parts. Its value is the length of the region over which you are integrating divided by 2. You could also of course look it up in the Schaum’s Math Handbook. Thus, we find that the constant is the same for all the wave functions (this is a special property of the square well and not true for other potentials).
This gives us our final results.
Know these results and be able to (1) apply them to solve problems both analytically and graphically.
C.Superposition
The Time Independent Schrodinger Equation also possesses the property that any superposition of individual solutions is also a solution of the equation. This again is a property that all waves share and therefore all wave equations must posses. This means that the most general solution is an infinite series of the individual energy eigenstates:. The square of the individual constants, an, give the probability that a measurement will find the particle with an energy En. This probability requirement also means that the constants must obey the requirements that
Let us consider a specific example in order to make this clearer. A particle in an infinite square well might have a wavefunction of
Such a wavefunction is called an ensemble of states and can be interpreted as the particle being in multiple energy states at the same time!!
If we wanted to graph the wavefunction, we would need to graph the function:
This would be somewhat painful to graph by hand, but easy using a graphing calculator, MathCad, Excel, etc.
What value would we get if we tried to measure the particle’s energy? The answer is that we can’t know for certain what energy value we would get!! In fact the general interpretation of quantum mechanics (Copenhagen Interpretation) is that the particle has no energy (i.e. energy has no reality) till we measure it. The coefficients tell us that we have a 36% chance of measuring E2 and a 64% chance of measuring E3. No other energy would be measured as the particle has no probability of being in any state besides states 2 and 3.Even more amazingis that the very act of measuring the particle’s energy changes the particle’s wave function. This is called “collapsing the wavefunction.” If we measure the particle’s energy to be E2 and then make a second measurement of the particle’s energy we are guaranteed to get E2 again. This means that the particle must be 100% in the 2nd state and the wave function for the particle must be
This is a very profound fact. It means that you can never observe the quantum world without disturbing the system that you are trying to measure no matter how good you make your instruments!! To see what strange consequences this fact has upon our philosophy of physics, read In Search of Schrodinger’s Cat by Gibbons or Quantum Reality by Nick Herbert. They are excellent non-technical books on Quantum Physics and its impact on our way of thinking about science.
II.Particle In A Finite Well
We next modify our infinite well to a more realistic potential that can be used to model real physical systems like the nucleus.
We consider a particle of energy E trapped in a finite well of width L and finite height Uo as shown below.
A.Solution Method
- Since the potential is not a function of time, we already know the time dependent part of the solution and can solve the Time Independent Schrodinger Equation to find the spatial part of the solution.
where
- We can simplify the problem by breaking the problem into regions where the wave number is constant (i.e. regions of constant U).
In this case, we can break the problem into following three regions:
Region I:
Region II:
Region III:
The advantage to this approach is that we know the solution to the problem in each region by inspection. The disadvantage is that we have to use properties of the wave function and its derivatives at the boundaries between the various regions to merge our individual solutions into the total wave function solution.
B.Solution for Region I
In this region, we have the differential equation
where
As we have discussed previously in this course, the general solution to this differential equation is
C.Solution for Region III
In this region, we have the same differential equation as region I
where
Thus, the general solution for region III is of the same form as the solution for region I.
D.Solution for Region II
In this region, we have the differential equation
where
As we have discussed previously in this course, the general solution to this differential equation is
E.Applying Boundary Conditions To and
There are four boundaries in this problem. The boundaries are located at , , , and .
To represent a quantum system, the wave function must have the following required properties:
- The wave function must be continuous and finite.
- The wave function must go to zero at .
- The wave function must be normalizable.
- The spatial derivative of the wave function must be continuous at all points in space where the potential function is finite.
F.Applying Boundary Condition At
Using our wave function conditions and our solution in Region I, we have
The only way that this condition can be obtained is if term involving A in the general solution is not retained (i.e. A =0).
Thus, our wave function solution for region I is reduced to
G.Applying Boundary Condition At
Using our wave function conditions and our solution in Region III, we have
The only way that this condition can be obtained is if term involving F in the general solution is not retained (i.e. F =0).
Thus, our wave function solution for region I is reduced to
H.Boundary Conditions at x = 0
At the boundary between region I and region II, we require that both the wave function and the spatial derivative of the wave function is continuous.
and
Using our wave solutions for regions I and II, continuity of the wave function requires that
Using our wave solutions for regions I and II, the continuity of the spatial derivative of the wave function requires that
Thus, our wave function in region II is
H.Boundary Conditions at x = L
At the boundary between region II and region III, we require that both the wave function and the spatial derivative of the wave function is continuous.
and
Using our wave solutions for regions II and III, continuity of the wave function requires that
Solving this equation for E, we obtain that
Using our wave solutions for regions II and III, the continuity of the spatial derivative of the wave function requires that
Using our equation for E, we have that
Substituting our definitions of and k, we obtain the following constraint on the allowed energies of the system.
We can simplify this condition by writing the energy as where is a number greater than 0 and less than 1. Our energy constraint condition now becomes
The only possible and therefore energies are those where the plot of the left-hand side are equal -1.
For an electron trapped in a well with a width of 0.1 m and a height of 4 eV, I have plotted the constraint as a function of energy using Excel. We see that only one energy state is allowed (E = 3.32 eV, eta = 0.83).
If we increase the height of the well to 40 eV and the width to 0.5 nm, we get the following graph
We see that both the numerator and denominator are oscillatory leading to several places where the left-hand side passes through -1. Thus, we have several allowed energy states for this system. These energies are approximately 5.84eV, 12.6eV, 19.5eV, 26.4eV, 33.2eV, and 39.4eV.
The graph is somewhat misleading because of the denominator going to zero. The vertical lines are an artifact of using Excel to graph points for discrete values of eta. No line should be drawn between points on two sides of the asymptote where the denominator went to zero.
I have placed this spreadsheet on the website for you to download and investigate the effect of the height and width of the well upon the energy eigenstates.
How are the results of your investigation similar to what we found for the particle in an infinitely deep well?
How are the results of your investigation different from what we found for the particle in an infinitely deep well?
J.Wave Function and Normalization
We can now write our wave function solution for all three regions in terms of a single constant B.
Once the width and height of the potential well is specified, this constant can be found by normalizing the wave function for an given energy state!
For a given well, the normalization constant, B, is different for each energy state since the particle’s energy effects both alpha and k. Although this is somewhat painful to do b y hand, it is relatively easy to do numerically with a computer.
K.Graphing the Wave Function
Although computational work is difficult without a computer, it is still easy to graph the energy eigenfunctions for a particle trapped in a finite well by hand if we take into account that our results must converge with the infinite well as we increase the depth of the well.
- In region I & III, we have a wave function that is decreasing exponentially.
- In region II, we have a wave function exhibiting oscillatory behavior.
1st Allowed Energy States
2nd Allowed Energy States
The existence of a particle in regions I & III is prohibited in classical physics as the kinetic energy in this region would be negative according to conservation of energy. However in quantum mechanics, the particle does have a non-zero probability of being located in such regions according to our graphs above. This leads to a strange and important phenomenon called tunneling which is common to all waves.
In regions I & III according to quantum mechanics it is possible for a particle to violate conservation of energy without experimental detection due to the uncertainty principle.
One can picture a particle borrowing a small amount of energy for a very short amount of time that is less than h-bar over 2. If a particle must borrow more energy, then the particle can travel less distance into the forbidden region.
L.Final Thoughts
Although our work may seem mathematical and tedious, the problem is one of the simplest that we could have attempted. For instance, the well was of constant height in regions I & II instead of being a function of position as in the case of the Coulomb potential. Also, we made the well symmetrical in height and isolated from any other well. Real systems are much more complicated. Thus, you should see the difficulty in performing detailed quantum mechanics calculations on real systems, the importance of numerical techniques in physics and engineering, and the importance of knowing the results of simple models to guide you to quick qualitative results.
The finite well is an essential starting block to understanding band theory for electrical engineers and radioactivity for nuclear physicists.