PHYS 590  Study Guide


The Potential Barrier  -  TUNNELING


By Davide Donato  (Spring 2004)




The tunnel effect is a purely quantum effect since there is not a classical explanation: particles can escape from regions surrounded by potential barriers even if their kinetic energy is less than the potential energy. This effect can be studied in detail when you consider the problems of a rectangular barrier (in our case the one-dimensional barrier). This kind of problems is a useful simplification to describe observed phenomena as a-decay, cold emission of electrons, fission, and fusion.

1.     Conceptual Discussion


One-D barrier problems require the definition of current densities for a beam of particles: incident, reflected, and transmitted. The current density is related to the particle density through the continuity equation:

This means that the current is defined as the net number of particles that flows through a region (usually a surface in the case of a 3-D discussion).

If we consider a wavefunction y that represents a beam of particles, then  represents the number of particles in a given interval [a,b]. In this way |y|2 can be considered as a density of particles (identified by r). But not only the density, but also the current Jx is related to |y|2. Jx is the current along the x direction.

In fact the particles in the beam obey to the Schrödinger equation and so:

The time derivative of the particle density y*y is

and considering the typical one-dimensional Hamiltonian

we can obtain 

This equation can be compared with the continuity equation and we can define the current along the x direction as follow:


The wavefunctions (that describe the incident, reflected, and transmitted beams) change in function of the sign and of the intensity of the momentum: while the incident beam has a momentum of p=ħk1, the reflected momentum is p=-ħk1 (it moves on the opposite direction), while the transmitted momentum is p=ħk2 (same direction of the incident current but with different momentum since some kinetic energy is converted in potential). In this way we can write the wavefunctions as

The energy is conserved across the potential barrier so that the frequency remains constant (w1=w2), while the wavenumbers change due to changes in momentum and kinetic energy.

Once defined the current Jx, we can obtain the expression for the incident, transmitted and reflected currents:


It is immediate to introduce the transmission and reflection coefficients:

These two coefficients follow the rule that their sum must be equal to 1 (conservation of the number of particles) and represent the probability that a beam of particles is transmitted or reflected when it finds a potential barrier. Note that we always speak of the behavior of a beam of particles and not of a single particle.

The general formulas for the transmission and reflection coefficients (substituting the expression of the currents) are:


Now we apply these concept to some more "practical" problems in which we have a beam of particles that interact with two different kind of potential: a step potential and a finite rectangular potential. In the first case the potential is 0 for negative x and is constant to a finite value for the entire positive x. In the second case the potential is always 0 except for a finite region (between -a and a). For each problem, two sub-problems can be considered: the first with energy of the particles bigger than the potential energy, while the second with energy less than the potential energy.

The solutions are different for each problem since we have different shapes of the potential (finite and infinite) and since we can have different values for the quantity E-V.

If this value is positive then the solutions can be expressed as combination of sin(kx) and cos(kx) (or alternative as a combination of eikx and e-ikx) and this means that the solutions represent an oscillating unperturbed signal. If the quantity E-V is negative then the solutions can be expressed as combination of ekx and e-kx and this means that the solutions represent a signal that decays or increases exponentially in the positive or negative direction along the x-axis.

These 4 kinds of problems are very interesting since they show aspect that only quantum mechanic is able to explain and they don't have any comparison with the classical mechanic. The last kind of problem (rectangular barrier potential and particles with energy less than the potential) will give the explanation of the tunnel effect.





For this case we have to consider the case with negative x, for which the energy E is completely kinetic (V=0) and with positive x, for which the energy is E-V. The quantity E and E-V are defined as follow:

The time independent Schrödinger equation will be for the first and the second regions:

Or equivalently


For this kind of problem a possible representation can be

while the solutions for the two regions are:

The term with amplitude A represents the incident beam, the term with B the reflected beam and the term with C the transmitted beam. Two things must be noted: the first is that the D term has to be 0 since there is not any beam coming from the positive x. The second, and more relevant thing, is that there is a reflected wavefunction. This means that not all the particle can cross x=0 (T<1) even if they have an energy greater than the potential. To solve the problem (finding the values for the transmission and reflection coefficients) the first step is consider that there must be continuity between the solution in region I and the solution in region II and that there must be continuity also for the first derivate. It is not possible to impose the continuity of the second derivate due to the presence of a step in the potential.

Imposing these conditions we have a system of two equations ( and ) with 3 unknown (A, B, and C). This brings to have a not homogeneous system and the solutions will form a continuum spectrum of values.

We express B and C in function of A and we find the value of T and R that are:

If we substitute the value of k1 and k2 using the definitions of the quantities E and E-V, we find that

therefore we can express T and R as a function of E and V.


We can immediately note that for energy much larger than the potential, the particles are almost all transmitted, while for energy close to the potential the particle are almost all reflected. This last limits brings directly to the second case.





This problem has the same solution of the Schrödinger equation for the negative x that we found in the previous problem. For the positive x the solution is

where the new term k can be express as

Since the value (E-V) is less than zero, the solution of the Schrödinger equation in the 2 regions is:

As in the previous problem, we have not any signal from the right of the graphic and so the solution for the region II does not contain the term with the D amplitude and we need continuity for the solutions and for their first derivates at x=0.

The representation is

Also for this problem we find something completely different from the classical mechanic, for which the region II is considered "forbidden" and there is no possibility to find any particle. In quantum mechanic we have the possibility to find particles also in region II, though the possibility is very low. This function decrease rapidly to 0 but also with big value in the exponent, the function never reaches 0.

Considering that the absolute value of a complex number is defined as

and that in this case

we can calculate the T and R coefficients. We find that the reflection coefficient is always equal to 1 and the transmission coefficient is always 0. This means that there is total reflection: the particles that enter in region II at the end will be reflected to region I.





For the case of rectangular barrier we have to consider 3 regions along the x axis: regions I and III for negative and positive x respectively where the potential is 0, and region II for the x range (-a;a) for which the potential V is constant value not zero.

In regions I and II we will have the incident/transmitted beam but also the reflected, while for region III we have only the transmitted. The domain can be represented as

while the solution of the time-independent Schrödinger equation are:

The last equation relates the width and the height of the barrier to the values of the wavenumber k1 and k2 (and so the energies) of the region I-III and II respectively:


The solutions are oscillating wavefunctions for all the 3 regions and as for the step potential, there is the quantum effect of a reflected wavefunction for both the regions I and II.

The transmission coefficient will be the ratio of the output wavefunction  (region III) and the input wavefunction (region I), that means the ratio of the amplitude F and A. The reflection coefficient will be the ratio of the wavefunction reflected from the first boundary condition x=-a (B amplitude) and the input wavefunction (A amplitude). 

Also in this case we have to impose the continuity of the solution and their derivatives. This brings to have 5 unknown (A, B, C, D, E, and F) but with a system of 4 equations:



We can express all the unknown in function of A. Without considering all the steps of the calculation, the expression for the transmission coefficient (in terms of wavenumbers and energies) is:


Also in this case if we consider the limits we obtain expected results: for E much greater than V we have that the coefficient is close to 1, while for E close to V we can substitute the value E-V in terms of k2 and obtain that the coefficient is oscillating (there is a sin function) with values in the range (0,1]. In fact there is the possibility that the transmission factor can be equal to 1 and that happens when the value of the sin function is 0 (so when 2*k2*a is a multiple of p). When the barrier width (2a) is an integral number of half wavelengths (n*l/2), the barrier becomes transparent to the incident beam.






The last case is similar to the previous one, but the energy now is less than the potential. In the region II the solution is not a combination of oscillating functions, but a combination of exponential functions, since in this region the beam is absorbed.  If we consider an incoming beam with some energy we can note that the signal is able to enter in the region where we have the potential and the signal decreases rapidly (exponentially) to zero without reaching this value. So the incredible results is that when the wavefunction reaches the boundary condition x=a, the probability to observed a signal is not zero and a beam, de-amplified, can cross this boundary and travel for x>a. In this region, the wavefunction will travel once again with an oscillating behavior. But another thing happens at x=a. At this point there is a change in potential and so a reflected wavefunction appears. This wavefunction has the D amplitude and, as the incoming wavefunction with amplitude C, it can be described with an exponential law, since it is absorbed (but not completely because of the finiteness of the barrier). Once reached the next boundary condition (x=-a), the wave split in two: one transmitted to the negative x and the other one reflected inside the potential until it reaches x=a. This view is not correct since it gives a temporal vision while the solution of the problem is time-independent. We have only spatial information.

The domains are described in the next figure.


Also in this case there is a relation between the dimensions of the rectangular potential and the energy of the beam through the wavenumbers:

As in the cases before, we can express the transmission coefficient using the values of the wavenumbers or the energies:


We can evaluate the limit of T for E close to V. It is possible to find that the coefficient T is a function of the dimensions of the rectangular barrier simply substituting the value of k given above. This means that the probability to observe tunneling depends on the intensity of the potential energy and the length of the rectangular barrier.

The expression O(Î) represents the sum of terms whose value goes to 0 with Î, where Î is the quantity E-V. In the evaluation of T, not considering the term O(Î) gives however a good approximation of this coefficient. For thick barrier the probability goes to 0 and for big values of thickness it is possible to reach the limit of a step potential.


The fact that a beam can penetrate the barrier also when the energy is less than the potential is called TUNNELING effect and is very important in nuclear physics. The 2 major examples are the emission of a particles (nucleus of the helium atom) from a nucleus during the a-decay and the emission of electrons from a metal surface in the presence of an electric field. Other examples are the emission of neutrons during the fission and the absorption of proton during the fusion.




2.     Real Examples of Tunneling


In general the barriers that occur in physical phenomena are not square and so we can obtain an approximate expression for the transmission coefficient through an irregular barrier. The only solution is to treat a smooth, curved barrier (the potential is a slowly function of x) as a series of square barriers (see picture below).

If we consider a series of infinitely narrow barriers we can approximate the transmission potential as:

with the integration over the region in which the square root is real (so for the tunneling conditions).




If we consider the electrons in a metal at 0°K temperature, we know that a certain amount of energy (work function) is necessary to bring the electron out of the metal. In this condition the electrons are not escaping and they are trapped by an approximate square barrier. Electrons can be removed or heating the metal, or transferring energy through photons (photoelectric effect) or applying an external electric field. In this latter case the potential seen by the electrons is not V but approximately (V-eEx) where e is the charge of the electron and E is the intensity of the electric field. The real potential should include imperfection of the metal (with local potential) and the fact that when an electron leaves the surface an image positive charge is created and this attracts the escaping electron.

Substituting in the formula given above the new potential, we can calculate the new transition coefficient:

This is called Fowler-Norheim formula and describes the emission only qualitatively.




The a-decay is the emission of a nucleus of a helium atom (a particle) from a heavy nucleus. For this process we can write the transition coefficient as T = e-G where G is defined as

where R is the nuclear radius and b is called “turning point” (where the potential V is close to the energy E of the particles). Z1 is the charge of the nucleus after the emission and Z1 is the charge of the a particle (=2). The integral can be evaluated exactly and for low energies we have that



where b=.




3.     Examples


1.      Electrons in a beam of density r=1015 electrons/m are accelerated through a potential of 100V. The resulting current then impinges on a potential step of height 50V. What are the incident, reflected, and transmitted currents?


This is clearly an example of a potential step problem in which the electrons are accelerated until they reach a potential of 100V and then they collide to a potential of only 50V. The energy is greater than the potential.


We know that the density r correspond to |y|2, therefore the square root of the value of the density correspond to the normalization of the incident wavefunction and so r=|A|2.

Using the expression of the incident current and the definition of the energy of the incident beam we find that , while using the definition of the reflected current we have . The term B/A is a function of the term k2/k1 that is related to the ratio of potential and energy of the particles.

Finally the transmitted current is .

Now we have all the elements and only a substitution of values is required:

Potential of particles = 100V => Energy=100*electric charge=100*1.6E-19

Potential step = 50V implies that k2/k1= and that B/A=0.17 .

Since for the electron  m=9.11*10-31 kg, we have that the currents are:



2.      Show that the reflection coefficients for the 2 cases depicted below are equal.

Also this example is for particles with energy greater than the potential (that has a step shape).

For the first case we have to use exactly the formulas and the procedure described in the Conceptual Discussion. The reflection coefficient is:

For the second case we have to go through the calculation, since the new formula will be , where k3 and k4 are the wavenumbers for the region on the left (V>0) and on the right (V=0) respectively.  In this case k3 and k4 are such that

   and    .

(Both the quantities E and E-V are positive and so the solutions will be oscillating, that is    and    for the two regions respectively.) Substituting the value of the two wavenumber in the formula of the reflection coefficient we can find that

The two expression of the reflection coefficient are the same.



3.      An electron beam is incident on a barrier of height 10eV. At E=10eV, T=3.37*10-3. What is the width of the barrier?


Since the value of the potential and the incident energy are the same, we have to use the formulas obtained for the tunneling effect, that is the expression of the transmission coefficient in function of the parameter g that depends on the characteristics of the barrier. The expression for the width will be the following:

Considering that h=6.58eVs, mc2=0.511*106eV, V=10eV, T=3.37*10-3, we obtain that a=0.106Å




4.     Resources


5.     Keywords


Tunneling – potential barrier – step barrier – rectangular barrier – transmission, reflection coefficients – incident, reflection, transmission currents - a-decay – cold emission -wavefunction