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ψ II ( x ) = C e β x + D e + β x .

The two types of solutions in the three regions are illustrated in [link] .

A solution to the barrier potential U of x is plotted as a function of x. U is zero for x less than 0 and for x greater than L. It is equal to U sub 0 between x =0 and x=L. The wave function oscillates in the region x less than zero. The wave function is labeled psi sub I in this region. It decays exponentially in the region between x=0 and x=L, and is labeled psi sub I I in this region. It oscillates again in the x greater than L region, where it is labeled psi sub I I I. The amplitude of the oscillations is smaller in region I I I than in region I but the wavelength is the same. The wave function and its derivative are continuous at x=0 and x=L.
Three types of solutions to the stationary Schrӧdinger equation for the quantum-tunneling problem: Oscillatory behavior in regions I and III where a quantum particle moves freely, and exponential-decay behavior in region II (the barrier region) where the particle moves in the potential U 0 .

Now we use the boundary conditions to find equations for the unknown constants. [link] and [link] are substituted into [link] to give

A + B = C + D .

[link] and [link] are substituted into [link] to give

C e β L + D e + β L = F e + i k L .

Similarly, we substitute [link] and [link] into [link] , differentiate, and obtain

i k ( A B ) = β ( D C ) .

Similarly, the boundary condition [link] reads explicitly

β ( D e + β L C e β L ) = i k F e + i k L .

We now have four equations for five unknown constants. However, because the quantity we are after is the transmission coefficient, defined in [link] by the fraction F / A , the number of equations is exactly right because when we divide each of the above equations by A , we end up having only four unknown fractions: B / A , C / A , D / A , and F / A , three of which can be eliminated to find F / A . The actual algebra that leads to expression for F / A is pretty lengthy, but it can be done either by hand or with a help of computer software. The end result is

F A = e i k L cosh ( β L ) + i ( γ / 2 ) sinh ( β L ) .

In deriving [link] , to avoid the clutter, we use the substitutions γ β / k k / β ,

cosh y = e y + e y 2 , and sinh y = e y e y 2 .

We substitute [link] into [link] and obtain the exact expression for the transmission coefficient for the barrier,

T ( L , E ) = ( F A ) * F A = e + i k L cosh ( β L ) i ( γ / 2 ) sinh ( β L ) · e i k L cosh ( β L ) + i ( γ / 2 ) sinh ( β L )

or

T ( L , E ) = 1 cosh 2 ( β L ) + ( γ / 2 ) 2 sinh 2 ( β L )

where

( γ 2 ) 2 = 1 4 ( 1 E / U 0 E / U 0 + E / U 0 1 E / U 0 2 ) .

For a wide and high barrier that transmits poorly, [link] can be approximated by

T ( L , E ) = 16 E U 0 ( 1 E U 0 ) e 2 β L .

Whether it is the exact expression [link] or the approximate expression [link] , we see that the tunneling effect very strongly depends on the width L of the potential barrier. In the laboratory, we can adjust both the potential height U 0 and the width L to design nano-devices with desirable transmission coefficients.

Transmission coefficient

Two copper nanowires are insulated by a copper oxide nano-layer that provides a 10.0-eV potential barrier. Estimate the tunneling probability between the nanowires by 7.00-eV electrons through a 5.00-nm thick oxide layer. What if the thickness of the layer were reduced to just 1.00 nm? What if the energy of electrons were increased to 9.00 eV?

Strategy

Treating the insulating oxide layer as a finite-height potential barrier, we use [link] . We identify U 0 = 10.0 eV , E 1 = 7.00 eV , E 2 = 9.00 eV , L 1 = 5.00 nm , and L 2 = 1.00 nm . We use [link] to compute the exponent. Also, we need the rest mass of the electron m = 511 keV / c 2 and Planck’s constant = 0.1973 keV · nm / c . It is typical for this type of estimate to deal with very small quantities that are often not suitable for handheld calculators. To make correct estimates of orders, we make the conversion e y = 10 y / ln 10 .

Solution

Constants:

2 m 2 = 2 ( 511 keV / c 2 ) ( 0.1973 keV · nm / c ) 2 = 26 , 254 1 keV · ( nm) 2 ,
β = 2 m 2 ( U 0 E ) = 26 , 254 ( 10.0 eV E ) keV · ( nm) 2 = 26.254 ( 10.0 eV E ) / eV 1 nm .

For a lower-energy electron with E 1 = 7.00 eV :

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Source:  OpenStax, University physics volume 3. OpenStax CNX. Nov 04, 2016 Download for free at http://cnx.org/content/col12067/1.4
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