The objective of the potential barrier problem is to find a solution for the wave function, ψ(x), (in all three regions) which satisfies the time-independent Schroedinger wave equation.   An electron may occupy region I, II, or III, but a requirement of the wave equation states that the wave function and its derivative, dψ(x)/dx, must be finite, single-valued, and continuous.   If ψ(x) had any discontinuities, then dψ(x)/dx would diverge to infinity.   dψ/d(x) must also be continuous because the second derivative d2ψ(x)/dx2 must also be finite, given that E and V(x) are also finite.   The time-independent wave equation is


If the wave function and its derivate must be continuous everywhere, then the wave function and its derivate at the boundary of regions I and II must be equal.


Likewise, the continuity of the wave function across the other boundary between regions II and III implies that


The four equations above must be used later to solve for the complete solution to the wave equation.   First, the general solutions for the wave function must be found in each individual region.

Region I

In region I, a particle does not have any potential energy (i.e. V(x) = 0 for x ≤ 0).   Therefore, the wave equation becomes


The wave equation above is a standard second order differential equation which can be solved through normal techniques.


Solving for r gives


The general solution to the wave equation is then


It is advantageous to compact the solution by substituting a constant K for the root expression in the exponent


The expressions ejKIx and e-jKIx can be expressed through Euler’s formula as

This shows that a particle which behaves according to the wave function acts as a wave and the constant K represents the wave number.   The expressions ejKIx and e-jKIx indicate two directions of wave travel.   e-jKIx indicates a wave traveling in the -x direction while ejKIx indicates travel in the +x direction.   Therefore, e-jKIx represents the part of the wave which is reflected off of the potential barrier, and ejKIx represents the incident or incoming wave to the barrier.

Region II

In region II, a particle does have a potential energy (i.e. V(x) = V0 for 0 ≤ x ≤ a).   Therefore, the wave equation becomes


The solution to this second order differential equation is found in a similar fashion as region I.


Notice that the negative sign was absorbed; therefore, the result is not complex as was the case in region I.   The general solution for region II, then, is


Because the solution is not complex as in region I, the particle’s behavior is not wave-like.

Region III

The form of the wave function in region III mimics that of region I with one slight change.   Once again there is no potential (i.e. V(x) = 0 for x ≥ a).   The wave equation for region III is


The solution for the wave function is


Recall that the expressions ejKIx and e-jKIx represent waves traveling in the +x direction and -x direction, respectively.   However, examining region III will show that there is no other potential barrier to produce a reflection which will cause a wave in the -x direction.   Therefore, G must be zero, and the wave function is simplified to


Application of Boundary Conditions

The boundary conditions for the continuity of the wave function across the edges of the potential barrier (i.e. x = 0 and x = a) allow the coefficients in the general solutions (A, B, C, D, and G) to be found.   The boundary conditions at x = 0 are as follows:


Evaluating the wave functions in regions I and II at x = 0 gives the relation,


Another relation can be found by evaluating the derivatives at x = 0.


The boundary conditions at x = a are given by,


Evaluating the wave functions in regions II and III results in


(Note: Although G is 0, it is included here because it allows the system of equations to be solved more easily.)   The other relation due to the first derivate at x = a is given by,


The first two equations relating A, B, C, and D can be written in matrix form as follows:


Solving in terms of C and D gives


Combining the last two equations into a matrix gives


Solving in terms of F and G gives an expression for C and D which can be substituted into the matrix equation relating A and B to C and D.


Substituting this result into the matrix equation relating A and B to C and D will form another expression relating A and B to F and G (i.e. relating the wave function in region I to region III).


Since G is 0 as stated earlier, the incident wave in region I can be related to the transmitted wave in region III by


A transmission coefficient can then be found to show how well a particle transmits through the potential barrier and appears in region III.   The transmission coefficient, T, is given by the ratio of the transmitted flux to the incident flux.


Substituting for F gives


Multiplying for M11 results in the following:


M11 is then given by,


and the magnitude of M11 is


Therefore, the transmission coefficient can be found.


Utilizing the hyperbolic trig identity, cosh2(x) = 1 + sinh2(x), the expression can be simplified further.



As the barrier width decreases, it is evident from the figure below that the probability of an electron tunneling through the barrier increases.   For larger barrier widths and for an energy less than the potential barrier height (V0), the probability of tunneling is negligible.

The potential energy of an electron in an atom is inversely proportional to the distance from the atom.   Electrons held close to the nucleus have higher energies, and vice versa for electrons further from the nucleus.   When atoms are situated close together as in a crystal lattice, each atom’s energy levels will overlap with a nearby atom’s energy levels.   It is evident in that the limit of the potential energy as r decreases to zero is -∞.


The following image shows the result of adding the overlapping energies together.   The energy nearby the nucleus still proceeds to -∞, but between the atoms, there is a finite energy.  
As more atoms are brought closer together, the “well” becomes more evident.  


Region I: x < -a/2

Schrodinger’s equation is given by,


When the electron is inside the well (i.e. -a/2 < x < a/2) it is free to move just as the electron can freely move in free space.   Therefore, we know V(x) = 0 inside the well.   Outside of the well, however, it is improbable that the electron exists because energy barrier at the edges of the well are infinite.   For the sake of calculation, let V(x) = V0 → ∞.


Since E is negligible with respect to the infinite potential, E can be ignored, and the wave equation becomes


Solving this second order, linear, homogenous differential equation gives


Recall that the wave function must remain finite for all x. Therefore, B = 0 so that ekIx will not cause ψ(x) → ∞.   Because kI → ∞, Ae-kIx → ∞ and the final solution is


The result shows that an electron does not exist in regions I and III.

Region II:-a/2 < x < a/2

In region II, V(x) = 0, so the wave equation is simplified to


The characteristic equation for the differential equation above is


Roots of the characteristic equation which have the form r = α ± jβ result in a general solution of


The roots for the characteristic equation in region II have α = 0 and


To find the coefficients, C and D, two equations are needed.   The two equations result from employing the boundary conditions at the edges of the well.   The wave function must remain continuous, finite, and single-valued for all x. For x = -a/2, the conditions give


For x = -a/2, the result is


Adding the last two equations yields


and subtraction produces


These two equations must be satisfied simultaneously for the wave function and its derivative to be continuous, finite, and single-valued.   We know that the cosine and sine cannot be zero simultaneously and that C and D cannot both be zero (otherwise the wave function would be zero and the particle would not exist anywhere).   Therefore, the only ways to satisfy both equations are to either set C to zero and choose kII so that the sine will be zero or set D to zero and choose kII so that the cosine will be zero.  
The values of kII which make the cosine zero are given by


The values of kII which make the sine zero are


Combining the results for kII from both the sine and cosine gives


We know that kII is related to a particle’s energy by


Substituting kII = nπ/a in the equation above yields
The results from the boundary conditions show that the energy of a particle is discrete within the well.


A plot of the wave function in the infinite potential well shows a standing wave.   Because of the boundary conditions (i.e. &psi(x) = 0 at x = -a/2 and x = a/2), every solution for the wave function will have at least two fixed nodes.   Particles which have larger energies (i.e. energy increases as n increases) have larger frequencies and, therefore, have more nodes.   Specifically, for a particular energy En, there are n + 1 nodes.






A very important result of the potential well problem shows the advantage of decreasing a material's size to the nanoscale.   As the dimensions of the well decrease (in one dimension, it is just the width of the well), the energy of the particle in the well increases and the energies collectively become more spread.   A quantum dot is an excellent example of how decreasing size can benefit its properties.   A quantum dot, when excited by an energy source such as infrared light, will emit radiation in the form of visible light.   The frequeny of the emitted light depends upon the width of the energy bandgap (i.e. the distance in energy between the conduction band and the valence band). A larger quantum dot will emit radiation in the red spectrum. However, a smaller quantum dot has a larger energy bandgap resulting in a smaller frequency of emitted radiation (blue light).   The animation below shows why radiation from quantum dots is described as being "blue shifted."

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