The Tunneling Phenomenon - Treatment

Let's consider a particle with mass m which is flying with energy E (from left side) up to potential barrier having width a:

Zone 1 | Zone 2 | Zone 3
h²/2md²/dx²y1 + Ey1 = 0 y1 = A1eik1x + B1e−ik1x k12 = 2m E/h² | h²/2md²/dx² y2+(E -V)y2 = 0 y2 = A2 eik2x + B2 e−ik2x k22 = 2m (E−V)/h² | h²/2md²/dx²y3 + Ey3 = 0 y3 = A3 eik3x + B3 e−ik3x k32 = k12 = 2m E/h²

If the particle energy E is smaller than V in zone 2: k22< 0, i.e. k2 is imaginary number. If we write down k2 = ik where k is real number now k2= 2m (V−E)/h² then we will obtain the following wavefunction in zone 2:

y2 = A2 e−kx + B2 ekx

This function doesn't vibrate but shows the exponential dependence on x.

Now we will calculate coefficients Ai and Bi (i = 1, 2, 3). Since particles come from left side the coefficient B3 should be 0 because the wavefunction B3 e−ik1x corresponds to particles moving from right to the left. However there are no such particles ! The coefficient B1 should not be 0 since particles can be reflected by potential barrier and then move back to the left. Another coefficients can be obtained from boundary conditions and normalization: y and y' should be continuous

| x = 0 | x = a
y continuous: | A1 + B1 = A2 + B2 | A2 e−ka + B2 eka = A3 eik1a
y' continuous: | ik1A1 − ik1B1 = −kA2 + kB2 | -kA2 e−ka + kB2 eka = ik1A3 eik1a

The coefficient B1 determines the probability (|B1|2) for the particle to be reflected and the coefficient A3 corresponds to the probability (|A3|2) for the particle to pass through potential barrier V and then move along the right side with hk. The transmission probability T = |A3|2/|A1|2 corresponds to the chance for particle to cross a barrier. Simple but detailed transformations of the above-mentioned boundary conditions give finally:

T = {1 + [eka− e−ka]2/[16 E/V (1-E/V)]}-1 where k = (2m(V-E)/h²)½ or T = {1 + sinh2(ka)/[4 E/V (1-E/V)]}-1

Although E < V the transmission probability T isn't equal to 0, i.e. a particle can cross the potential barrier that isn't allowed in classical physics.
The wavefunction isn't equal to 0 at point x=0 since V is finite in height and width. One can write down the following approximation for ka > 1:

T ≈ 16 E/V (1 − E/V) e−2ka

For E > V T can be < 1 , which means that particles are reflected back even if the particle energy is greater than the barrier height. Here one can write down (k' = ik):

T = [1 - sin² k' a/4 E/V (1 − E/V)]-1 where k' = (2m(E-V)/h²)½

P.S. One can also write down for E = V (after Taylor expansion of numerator and denominator):

T = [1 −ma²V/2h²]-1