Tunneling

2
For energies closer to the top of the well, the tunneling proportion increases. Shown in Figure 3 are the
wavefunctions corresponding to the fourth and fifth energy levels for this system.
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
ψ(x)
0.0
0.0 2.0 4.0 6.0 8.0 10.0
-0.2
ψ(x)
0.0
0.0 2.0 4.0 6.0 8.0 10.0
-0.2
-0.4
-0.4
-0.6
-0.6
-0.8
-0.8
x (bohr)
x (bohr)
(a) (b)
Figure 3. (a) Third excited state (E 4 =2.75 hartrees) and (b) fourth excited state (E 5 =4.16 hartrees)
wavefunctions for the particle in a half-infinite well.
Notice that for the fourth excited state, with an energy very close to the top of the barrier (E 5 =4.16 hartrees
compared to a barrier of 4.5 hartrees), the tunneling is substantial. For this system, we can determine the tunneling
probability by integrating the probability density from x=L to x= ∞. For the 1 st energy level the tunneling probability
is 0.0025, while for the 5 th energy level the tunneling probability is 0.18.
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Example 2: Tunneling Through a Barrier of Finite Width
Another simple example of tunneling is the case of a barrier of finite width, illustrated in Figure 4.
V=V 0
I II III
x=0 x=L
Figure 4. Potential energy for a particle interacting with a finite barrier in one dimension.
The potential energy of the system may be described by the following equation,
V (x) =
⎧⎧ 0,
⎪⎪
⎨⎨ V 0 ,
⎪⎪
⎩⎩ 0,
x < 0
0 ≤ x ≤ L
x > L
⎫⎫
⎪⎪
⎬⎬ .
⎪⎪
⎭⎭
The method of solution is similar to what we did for the particle in a half-infinite well. However, there are two
degenerate solutions, one in which the particle is initially traveling to the right and one in which the particle is
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initially traveling to the left. Only one of these degenerate cases needs to be considered. Here, we will assume that
the particle is traveling to the right, so it is initially incident on the barrier from region I.