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We divide the real axis into three regions with the boundaries defined by the potential function in [link] (illustrated in [link] ) and transcribe [link] for each region. Denoting by ψ I ( x ) the solution in region I for x < 0 , by ψ II ( x ) the solution in region II for 0 x L , and by ψ III ( x ) the solution in region III for x > L , the stationary Schrӧdinger equation has the following forms in these three regions:

2 2 m d 2 ψ I ( x ) d x 2 = E ψ I ( x ) , in region I: < x < 0 ,
2 2 m d 2 ψ II ( x ) d x 2 + U 0 ψ II ( x ) = E ψ II ( x ) , in region II: 0 x L ,
2 2 m d 2 ψ III ( x ) d x 2 = E ψ III ( x ) , in region III: L < x < + .

The continuity condition at region boundaries requires that:

ψ I ( 0 ) = ψ II ( 0 ) , at the boundary between regions I and II and

and

ψ II ( L ) = ψ III ( L ) , at the boundary between regions II and III .

The “smoothness” condition requires the first derivative of the solution be continuous at region boundaries:

d ψ I ( x ) d x | x = 0 = d ψ II ( x ) d x | x = 0 , at the boundary between regions I and II ;

and

d ψ II ( x ) d x | x = L = d ψ III ( x ) d x | x = L , at the boundary between regions II and III .

In what follows, we find the functions ψ I ( x ) , ψ II ( x ) , and ψ III ( x ) .

We can easily verify (by substituting into the original equation and differentiating) that in regions I and III, the solutions must be in the following general forms:

ψ I ( x ) = A e + i k x + B e i k x
ψ III ( x ) = F e + i k x + G e i k x

where k = 2 m E / is a wave number and the complex exponent denotes oscillations,

e ± i k x = cos k x ± i sin k x .

The constants A , B , F , and G in [link] and [link] may be complex. These solutions are illustrated in [link] . In region I, there are two waves—one is incident (moving to the right) and one is reflected (moving to the left)—so none of the constants A and B in [link] may vanish. In region III, there is only one wave (moving to the right), which is the transmitted wave, so the constant G must be zero in [link] , G = 0 . We can write explicitly that the incident wave is ψ in ( x ) = A e + i k x and that the reflected wave is ψ ref ( x ) = B e i k x , and that the transmitted wave is ψ tra ( x ) = F e + i k x . The amplitude of the incident wave is

| ψ in ( x ) | 2 = ψ in * ( x ) ψ in ( x ) = ( A e + i k x ) * A e + i k x = A * e i k x A e + i k x = A * A = | A | 2 .

Similarly, the amplitude of the reflected wave is | ψ ref ( x ) | 2 = | B | 2 and the amplitude of the transmitted wave is | ψ tra ( x ) | 2 = | F | 2 . We know from the theory of waves that the square of the wave amplitude is directly proportional to the wave intensity. If we want to know how much of the incident wave tunnels through the barrier, we need to compute the square of the amplitude of the transmitted wave. The transmission probability    or tunneling probability    is the ratio of the transmitted intensity ( | F | 2 ) to the incident intensity ( | A | 2 ) , written as

T ( L , E ) = | ψ tra ( x ) | 2 | ψ in ( x ) | 2 = | F | 2 | A | 2 = | F A | 2

where L is the width of the barrier and E is the total energy of the particle. This is the probability an individual particle in the incident beam will tunnel through the potential barrier. Intuitively, we understand that this probability must depend on the barrier height U 0 .

In region II, the terms in equation [link] can be rearranged to

d 2 ψ II ( x ) d x 2 = β 2 ψ II ( x )

where β 2 is positive because U 0 > E and the parameter β is a real number,

β 2 = 2 m 2 ( U 0 E ) .

The general solution to [link] is not oscillatory (unlike in the other regions) and is in the form of exponentials that describe a gradual attenuation of ψ II ( x ) ,

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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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