Tunneling

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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. € Example 2: <strong>Tunneling</strong> 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 € 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.
3 For the case E < V 0 , the general solutions may be given as € ψ I ( x) = A e ik1x + Be −ik 1x ψ II ( x) = C e k2x + De −k 2x ψ III ( x) = F e ik 1 x . The wave vectors k 1 and k 2 are defined as € € € k 1 = 2mE and k 2 = 2m ( V 0 − E) . In region I, and in general for unbounded regions where the potential energy is zero, we use exponentials with imaginary exponents as the preferred form of the wavefunction. The two parts of the wavefunction written in this € form can be related to waves traveling in the positive and negative x directions, respectively. Notice that in region II, we have used exponentials with real rather than imaginary exponents. This is the preferred form in regions where the total energy E is less than the potential energy V 0 . Also notice that in region III, there is no wave traveling to the left. This is because we started with a wave traveling to the right, and in region III, no wave traveling to the left can form because there is nothing for the wave to reflect € from. Matching the wavefunctions and their first derivatives at the boundaries x=0 and x=L yields conditions among the arbitrary constants A, B, C, D, and F. Of particular interest is a quantity called the transmission coefficient T. The transmission coefficient T is related to the ratio of the probability density current that is transmitted through the barrier to the incident probability density current. The probability density current S is defined as S = v ψ *ψ , where v is the particle velocity. Thus, the transmission coefficient T is € T = S tr S in , € where S tr is the transmitted probability density current and S in is the incident probability density current. For the specific case here, the transmitted wave in region € III has the form Fe i k 1x . Since this form has a momentum eigenvalue given by k 1 , the velocity is € v € tr = p tr m = k 1 m . € Similarly, the incident wave in region I has the form k 1 , the velocity is € Ae i k 1x . Since this form has a momentum eigenvalue given by € € v in = p in m = k 1 m . €
Page 1: Chemistry 460 Spring 2013 Dr. Jean
Page 5: 5 Example 3: Double Well Potential

Tunneling

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