Link to return to Modern Physics front page
Links to specific sections in the text:
6.a. Potential barrier
6.b. Penetration for large barriers for
particles
6.c. Applications
We started the discussion of quantum mechanics with free particles. A monochromatic free particle (particle with definite momentum but completely undetermined coordinate is given by a plane wave.
y(x,t) = A exp{ i 2p ( px - E(p) t )/h}.
Remember, p is just a parameter of the wavefunction. If p>0, then this describes a wave propagating towards positive values of x. For a nonrelativistic particle E(p) = p2 / 2m. This form satisfies the time dependent Schrodinger equation with vanishing potential energy. In fact, if you remember we constructed the Schrodinger equation so that this wavefunction would satisfy it. Though this wavefunction cannot be normalized, because its absolute squared
|y(x,t)|2 = A2
and the integral over over the x axis is infinite, we may imagine that the wavefunction describes a monochromatic beam (an infinite stream) of particles coming in each after from left to right. We will use now similar wavefunctions to investigate the phenomenon of reflection and transmission through potential barriers. We will do this by solving the Schrodinger equation for a state of definite energy, E. If the energy is definite, then the time dependence of the wavefunction is very simple, it has the form
y(x,t) = exp{- i 2p E t / h}f(x),
where f(x) satisfies the time independent Schrodinger equation
E f(x) = - (h2 / 8p2 m)(d2/dx2)f(x) + V(x) f(x)
Let us assume that the potential energy forms a simple barrier of thickness L and hight V0. In other words the potential is defined by
V= 0 if x<0, or x>L, and
V= V0 if 0< L < x.
Many physical systems can be approximated by this potential. Imagine particles impinging on a thin foil, or two conductors connected with a thin insulator.
We need to solve the Schrodinger equation in three regions independently. First for negative x there are two solutions of the equation, both having the same energy but one moving left, the other moving right
f(x) = exp{ ikx} + A exp{- ikx }, if x<0.
where k = 2p p / h. Remember, when multiplied by the common time dependent factor, the second term of the complete wavefunction descibes a left moving wave. We set the amplitude of the wave coming at the barrier from the left to unity. In other words we set the intensity of the incoming particle beam such that there is one particle per unit distance. Then A2 will provide the intensity of the reflected beam. One expects that R = A2 < 1, i.e. only a fraction of the particles is reflected another fraction will be transmitted. The quantity R is called the reflection coefficient.
We still need to solve the equation inside the barrier and on the right side of the barrier. We know the solution inside the barrier, it is similar to the solution for the finite potential well:
f(x) = B exp{ -k x} + C exp{k x }, if 0 < x < L,
where k = [ 2m ( V0 -E) ]1/2
Since the interval where the wavefunction has this form is finite, we do not have the right to set C=0.
In the third interval the general solution of the Schrodinger equation for energy E also gives a general combination
f(x) =D exp{ ikx} + F exp{- ikx }, if x>L.
Now physically we can set F=0, because the problem we wish to solve is that of a particle beam coming in from the left. If is not equal to zero, then there would be another particle beam that would come in from the right. Thus the wavefunction in the third region can be chosen as
f(x) =D exp{ ikx }, if x>L.
This wavefunction represents a particle beam passed the barrier, that is a transmitted beam. T = D is the transmission coefficient that provides the intensity of the transmitted beam. The probability requires that 1= R + T. This must be so because the probability is conserved and a particle cannot vanish but is either reflected or transmitted. The correct solution of the Schrodinger equation insures the conservation of probability, but we have not solved the Schrodinger equation yet in a global sense, because a true solution must be continuous, with continuous derivatives everywhere, including at points where the potential jumps.
Let us investigate first the point x=0. the continuity of the wavefunction and its derivative impies that
1 + A = B + C, (1)
ik ( 1 - A) = k ( C - B ). (2)
The continuity of the wavefunciton and its derivative at x=L requires that
B exp{ -k L} + C exp{k L } = D exp{ ikL}, (3)
k [ - B exp{ -k L} + C exp{k L } ] = ik D exp{ ikL }. (4)
Since we do not need the coefficients B and C for calculating either the transmission of the reflection coefficient, we can eliminate them. Indeed, by dividing (2) by k and adding and subtracting (1) and (2) we obtain expressions for B and C
C = (1 + ik/k )/2 + A(1 - ik/k )/2,
B = (1 - ik/k )/2 + A(1 + ik/k )/2.
Substituting these into (3) and (4) we get
(1+A) cosh k L + (1-A) ik/k sinh k L = D exp{ ikL},
(1+A) sinh k L + (1-A) ik/k cosh k L = ik/k D exp{ ikL},
or equivalently,m by rearranging we get
cosh k L + ik/k sinh k L + A [ cosh k L - ik/k sinh k L ] = D exp{ ikL} (5)
sinh k L + ik/k cosh k L + A [ sinh k L - ik/k cosh k L ] = ik/k D exp{ ikL}. (6)
Multiplying (5) by [ sinh k L - ik/k cosh k L ] and (6) by [ cosh k L - ik/k sinh k L ] and subtracting the two equations we can eliminate A to obtain
[ cosh k L + ik/k sinh k L ][sinh k L - ik/k cosh k L ] - [ sinh k L + ik/k cosh k L ][ cosh k L - ik/k sinh k L ] =
-2ik / k = D exp{ ikL} [ sinh k L - ik/k cosh k L - ik/k ( cosh k L - ik/k sinh k L ) ]
= D exp{ ikL}[ (1 - k2/k2 ) sinh k L - 2 i k /k cosh k L ].
where we used the relation cosh2 k L - sinh2 k L = 1. Now we can write
D exp{ ikL}= -2ik / k [ (1 - k2/k2 ) sinh k L - 2 i k /k cosh k L ]-1
and take tha absolute value of both sides to get
T = |D|2 = 4k2/k2 [ (1 - k2/k2 )2 sinh2 k L + 4k2/k2 cosh2 k L ]-1
Then dividing both the numerator and the denominator by 4k2/k2 and using cosh2 k L = 1 + sinh2 k L we obtain
T = [ (k2 + k2 )2 / (4k2k2 ) sinh2 k L + 1]-1
or substituting the expression of k2 and k2 we obtain
T = { V02 / [4E( V0 - E )] sinh2 k L + 1}-1
It is worth while to investigatre some properties of the transmission coefficient. Notice first that if V0 =0 then T=1. This should be so because in that case the barrier vanishes and the waves proceeds unimpeded. Since, as one check by explicit calculation of R = |A|2 that R + T =1, then of course, if T=1, the A =0, there is no reflected wave. Another limiting case: if the barrier gets wider and wider i.e.k L is large then we can approximately write 4 sinh2 k L = e2kL and we obtain
T µ e-2kL, vanishes exponentially with the thickness of the barrier. This is also understandable on physical grounds: even quantum mechanically it is harder and harder to pass a larger and larger barrier. As the thickness of the barrier increases the transmission coefficient decreases and the reflection coefficient approaches one. At an infinitely large L one has a potential step rather than a barrier. Then the reflection coefficient is one, but there is a nontrivial phase difference between the incoming and reflected wave.
The classical limit of quantum mechanics is attained in a formal way if we take the limit h -> 0. Think of the harmonic oscilator (or Planck). When h -> 0 the quanta of the harmonic oscillator, hf -> 0 as well. that means the spectrum of the harmonic oscillator becomes continuous. Also the limit of h -> 0 implies that the right hand side of the Heisenberg uncertainty relation vanishes, thus as in classical physics the momentum and coordinate become exactly measurable. One can study the limit of h ->0 in the framework of the Schrodinger equation if one uses the so-called semiclassical approximation. This approximation is a systematic Taylor expansion of the wave function in a series of h around h=0. It is easy to see that to achieve this one should write the wavefunction in the trial form of
f(x) = A exp{ i 2 p S(x)/h } + B exp{ - i 2 p S(x)/h }
in the classically allowed region, where E>V(x) and
f(x) = A exp{ 2 p S(x)/h } + B exp{ - 2 p S(x)/h }
in the classically forbidden region, where E<V(x). An example for this is of course the particle inside a barrier. Take for example the latter one and only the second term substitute into the Schrodinger equation. Note that
f"(x) = { [ 2 p S'(x)/h ]2 - 2 p S"(x)/h } f(x)
If S(x) is independent of h then the second term in the curly bracket can be neglected and the Schrodinger equation becomes
E f(x) = - (h2 / 8p2 m) [ 2 p S'(x)/h ]2 f(x) + V(x) f(x).
Dividing by f(x) and simplifying leads to
[ S'(x) ]2 = 2m [ V(x) - E ],
or S(x) = (2m)1/2 Ú [ V(x) - E ] dx
Then the wavefunction in the leading order semiclassical approximation is
f(x) µ exp{ - 2 p S(x)/h } = exp{ - 2 p (2m)1/2 Ú [ V(x) - E ] dx / h }.
It is easy to show that when the particle penetrates the barrier then the limits of integration are the classical turning points, i.e. the points at which V(x) = E. Then this formula provides a general means to calculate how the amplitude of a wave decreases when it passes through a barrier. The square of this expression will give an approximate form for the transmission coefficient. The corresponding physical problem can be represented as follows:
In the above problem the limits of the integral in the exponent are x1 and x2. The transmssion coefficient is the sqare of the ratio of the wavefunctions before the barrier and after the barrier. Since the wavefunction is continuous this is exactly the square of the ratio by which the wavefunction decreases throug the barrier, approximately
T µ exp{ - 4 p (2m)1/2 Ú [ V(x) - E ] dx / h }.
The advantage of the semiclassical method is that one can calculate the transmisson coefficient for potentials, which do not allow for an exact solution of the Schrodinger equation. We only need to integrate to get the answer.
Problem: Apply the semiclassical form to the case we discussed earlier in an exact fashion, the square barrier.
Solution: The potential is constant, V(x) = V0. Then we can easily calculate the exponent and we obtain
T µ exp{ - 4 p (2m)1/2 Ú [ V(x) - E ] dx / h } = exp{ - 4 p[ V0 - E ] (2m)1/2 Ú dx / h }
= exp{ - 4 p[ V0 - E ] (2m)1/2 L / h } = exp{ - 2 k L },
which is exactly the behavior we obtained earlier by exact method.
Problem: Calculate the probability that a person runs towards the wall , tunnels through and appears unchanged on the other side of the wall. Negelect the interaction among the atoms of the person. The wall presents a 0.1m thick and 10 eV high barrier for every atom. Since the person is mostly water assume that the average atomic weight is (1+1+16)/3 = 6. Assume that the person is 60kg.
Solution: The number of atoms in the person is N = 60 kg / ( 6 ¥ 1.67 ¥ 10-27kg) = 6¥ 1027. The probability that a single atom tunnels through is approximately P µ e-2kL Since there is no interaction among atoms, simultaneous tunelling of atoms are independent events and their probabilitie should be multiplied to get the total probability. Think of throwing two dice. Then the probability of a double six is (1/6)¥ (1/6) = 1/36. Thus, the probability of all of the atoms tunnelling through simultaneously is P µ e-2kNL Since L=0.1m we only need to calculate
k =2 p [ 2m V0 ]1/2 / h = 2 p [ 2mc2 V0 ]1/2 / hc = (2 ¥ 6¥ 938 MeV ¥ 10 eV )1/2 / 197 eV nm = 1.7¥ 1012m-1
Then the exponent is
2kNL = 2¥ 1.7¥ 1012m-1¥ 6¥ 1027 ¥0.1m = 2.06¥ 1039
Then P µ e-2kNL µ exp{- 2.06¥ 1039}, which is a very, very small number. Even trying for every nanosecond during the age of the universe once (3 ¥ 1026 times) would only leave the probability practically the same:
P¥ 3 ¥ 1026 = exp{- 2.06¥ 1039 + log( 3 ¥ 1026) } = exp{- 2.06¥ 1039 + 61 } µ exp{- 2.06¥ 1039}
The tunnel effect has a large number of importnat physical applications. We will discuss only a few of these. The book discusses field emission, alpha decay of nuclei, the states of the ammonia molecule, the decay of black holes, and the scanning tunneling microscope. Let us discuss these briefly.
Briefly speaking a metal contains so-called conduction electrons which are more or less free to roam around inside the metal. They cannot escape the metal because they are in a potential well of depth V0 and their kinetic energy is not sufficiently large. Suppose now that an electrivc field is created outside of the metal. This can be done by positively charging a conductor close to the surface of the metal. For the sake of simplicity assume that the charged conductor creates a homogeneous electric field. Then the combination of the homogeneous electric field and the potential well creates a finite potential barrier for the electron of energy E, as shown below:
The potenial barrier is wedge shaped. It is drawn with a thick line. The thickest line symbolizes the metal creating the homogeneous electric field. The classical turning point is at where E=-eex, where e denotes the electric field and E denotes (the negative) energy level. The transmission coefficient is
T µ exp{ - 4p / h Ú [ 2m ( -eex - E )]1/2 }
where the limits of integration are 0 and x0=-E/ ee. Then we obtain
T µ exp{ - (8p / 3eeh )(2m)1/2 |E|3/2 } = exp{ - (8p / 3eehc )(2mc2)1/2 |E|3/2 }
To get an estimate for the current we need to estimate the number of electrons colliding with the wall. We need the velocity of electrons, v, and the density of the electrons, r. Then the number of electrons colliding with unit area of the wall per unit time is N = vr. The current density, the number of electrons leaking out per unit area and unit time is eNT = evrT. Let u estimate these quantities: The kinetic energy is of the order of 2 eV. Taking 2eV we obtain 2eV = m v2 / 2 . Then 4eV/ 511keV =(v/c)2 This gives v=.0028c=8.4¥105m/s. Now r = d A /mole, where A is Avogadro's number and d is the mass density. E.g. for aluminum mole/d = 10cm3 so the number density is r = 6 ¥1022cm-3 = 6 ¥1028m-3 Then F0 = evr, the current flux impinging on the potential wall is
F0 = 8.4¥105m/s ¥6 ¥1028m-3¥1.6 ¥10-19C = 8 ¥1015A/m2
which is a tremendously large current. On the other hand T is very small, so in effect the field emission will be very small. Take a strong field of e=3¥109V/m, which is thirty million volt per centimeter. Then assuming that |E|=2eVwe obtain
T = exp{ - [8p / (3 ¥3¥109eV/m 197 eV nm)](1MeV)1/2(2eV)3/2}= exp{ -36} = 2.3¥10-16
Then the current density is F = F0 T = 8 ¥1015A/m2¥ 2.3¥10-16 = 1.84 A/m2= 0.184 mA/cm2
A very similar problem, albeit on a very different scale is the radioactive a-decay of heavy nuclei. Many heavy nuclei are unstable. They have various channels for decay, among others a-decay in which a He4 nucleus is ejected from the decaying heavy (Z,A) nucleus to transform it into another nucleus with (Z-2,A-4). For example the most abundant isotope of Uranium (Z=92, A= 238) decays as follows
238U92 Æ 234Th90 + 4He2
The radioactive decay of Uranium and the lifetime of the Uranium
nucleus can be approximately explained by the tunneling effect.
The fact that the decay happens at all indicates that it is
energentically possible for the a-particle
to separate fromwhat is left, a Thorium nucleus. Then the question
arises how does the Uranium nucleus hold together, since two nuclei,
being positively charged, always have a strong Coulomb repulsion
between them. The answer is that there is a strong, but very short
range attractive nuclear force between these nuclear fragments. The
potential energy corresponding to this force can be approximated by a
square potential well of depth of 40MeV in which the a-particle has kinetic energy larger then 40Mev.
Still, the a-particle cannot escape
becouse interestingly enough the Coulomb force, which is repulsive,
creates a barrier that keeps the a-particle in for some time. Let us use the
approximation of neglecting the Coulomb potential that vanishes in
the origin inside the nucleus. Then the schematic model for the
potential is:
The transmission coefficient, as before, using the semiclassical approximation, and a one dimensional model, can be aproximated by
T µ exp{ - 4 p (2m)1/2 Ú [ V(x) - E ] dx / h }= exp{ - 4 p (2m)1/2 Ú [ 2Zke2 /x - E ] dx / h },
where the limits of integration are (R, x0), where x0 is the classical turning point, x0 = 2Zke2 / E. One can easily calculate the integral but it is not very illuminating. Once T is calculated one can obtain the number of tunneling per unit time as the number of collisions of the a-particle with the wall, N, multiplied by tunnelling probability the half-lifetime of parent nucleus is then t = log 2 / NT. The number of collisions can be obtained by calculating the speed of the a-particle, from
K = E + V0 µ 45 MeV and the mass of the a-particle, about 4GeV/c2. We obtain v= c (2K/mc2)1/2 = 0.15c. The number of collisions/second is approximately N = v /2R = 4.5¥107 m/s / 10-14m = 4.5¥1021 1/s.