finite square wear Suppose instead we have a well with zo = 8; find the allowed energies En and plot the energy eigenfunctions on (x). (Hint: You might have to change the root-finding code - you'll have more than four states, and you should adapt your "guesses" accordingly.)5.5 0 FINITE SQUARE WELL Now let's make the problem a little more realistic by having the potential energy outside the well be finite instead of infinite. We still assume that the well is square, which still results in an infinite force at the walls. However, this new problem illustrates several important features of bound energy states that were not evident in the infinite well. A finite well can be used to model many real systems, such as an electron in a thin semiconductor. In Section 5.8, we use this model to discuss quantum well semiconductor lasers. The finite square well potential energy is shown in Fig. 5.15 and is written as Vo: V(x) = -a a, where we have deliberately chosen a different position origin from the infinite well case in order to give you practice and also for convenience. For now, we look for bound state solutions, that is, for energies below the potential Vo. Energies above Vo correspond to unbound states that we will discuss in the next chapter. With this new potential energy function, the energy eigenvalue equation is 2 m dxz to ) PE(x) = EVE(x ). inside box (5.74) ( # d + Vo QE ( x ) = ExE(x). outside box. V(x) Vo X -a FIGURE 5.15 Finite square potential energy well.In the infinite well problem, we found it useful to use the wave vector k 2mE k = (5.75) In this case, it is also useful to define a similar constant outside the well 2m ( Vo - E). (5.76) For bound states, 0 a The odd solutions are He'll, I -=: a gum-[r] = Csinlk'r], a E x E a {5.81) Ae'q'. I 3- 1:. Let's rst do the even solutions. The boundary conditions at the right side ofthe well (I = a} give pma}: D cos[ka] = ale'3" d {5.32) M :kD sinUl'a] = q.4e"l'\". (ix I-a The boundary conditions at the left side ofthe well [I = a) yield the same equations. which must be true because of the symmetry. The two equations above have three unknowns: the amplitudes A and D and the energy E. which is contained in the parameters It and q. The normalization condition provides the third equation required to solve for all three unknowns. We nd the energy condition rather simply by dividing the two equations. which eliminates the amplitudes and yields ktan[ka) = q. {5.83) Because both it and q are functions of the energy, this equation ves us a formula to nd the allowed energies. It is independent of the constants A and D. which are found by applying the normalization condition and using Eq. (3.32) again. As usual with these types of problems, the eigenvalue condi- tion is obtained rst. and then the eigenfunctions are obtained later. To make the energy dependence explicit. we use Eqs. {5.75) and [5.76] to write Eq. (5.33} as 2 2 film VigEtan(Vi?TEa) = \\l 51 (PE, E]. {5.84) The next step is to solve this transcendental equation for the energy E. For the odd solutions. a similar argument leads to the transcendental equation (Problem 5.15} k cotfkn) = q. {5.85) A graphical solution for the allowed energies using these two transcendental equations is most useful here. There are many ways of doing this. One way involves defining some new dimensionless parameters: 2mEa z = ka = 2m Via2 (5.86) Zo 2m(Vo - E)a 90= where the variable z parameterizes the energy of the state and the constant zo characterizes the strength of the potential energy well. These definitions lead to the convenient expressions (ka) + (qa)' = 27 (5.87) (qa)? = =3 - (ka)? = =3 - z. This allows us to write the transcendental equations in this form: ka tan(ka) = qa - z tan(2) = V z 2 2 (5.88) -ka cot(ka) = qa - -zcot(z) = V z - 23. In each of these new transcendental equations, the left side is a modified trig function, while the right side is a circle with radius zo. These functions are plotted in Fig. 5.16 as a function of the parameter z. The intersection points of these curves determine the allowed values of z and hence the allowed energies E, through Eq. (5.86). Because the constant zo is the radius of the circle, there are a limited number of allowed energies, and that number grows as Zo gets larger. Wells that are deeper and wider have more allowed bound energy states. That's it for the energies. There is no simple formula-the transcendental equations must be solved graphically or numerically for each different well. For example, the curves in Fig. 5.16 corre- spond to a well with zo = 6, which results in four intersection points and hence four bound states. The intersection points and four allowed energies are 21 = 1.34 - E = 1.81 2ma2 Z2 = 2.68 E, = 7.18 2ma (5.89) 23 = 3.99 - Ex = 15.89- 2ma2 ZA = 5.23 Ex = 27.31 2ma2 The energy eigenstate wave functions are characterized by the allowed values of the parameters k and q from Eq. (5.86). All that remains to do is normalize the wave function, which is straightfor- ward but tedious (Problem 5.16). Once again, we use a unified diagram to show the potential energy well, the allowed energies, and the allowed eigenstate wave functions superimposed in Fig. 5.17.f(z) z tanz -Z cotz 2T Zo 2 26 - Z2 TT 0 TT ZOOT 2 JRE 5.16 Graphical solution of the transcendental equations for the allowed energies of a finite e well (Zo = 6). - n = 3 - n = 2 - n = 1 * X -a URE 5.17 Unified schematic diagram of the finite potential energy well and the bound state itions, showing the well, the allowed energies, and the energy eigenstate wave functions.Note that the finite well eigenstates share many features with the infinite well states, with one major exception-they extend into the classically forbidden region. Quantum mechanical particles have a finite probability of being found where classical particles may not exist! This is a purely quantum mechanical effect and is commonly referred to as barrier penetration. The ability of the particle to penetrate the potential energy barrier leads to the phenomenon of tunneling, an example of which is radioactive decay. We'll say more about these wave functions in a bit, but let's first check that our solution is consistent with the solution we derived earlier for the infinite energy well case. The limit of an infinitely deep well corresponds to the radius zo in Fig. 5.16 going to infinity, in which case the allowed values of z become the asymptotes of the modified trig functions, shown by the dashed lines in Fig. 5.16. These limits are the same as for the simple trig functions and yield In = n- -= kna = 17 (5.90) KA = 2a from which we recover the infinite well energy eigenvalues: En = (5.91) 2m(2a)- Note that the width of the well is 2a here, whereas we called the width L in the infinite well case. The infinite well eigenstate wave functions for this symmetric well position are Pa(x) = 2 -COS n = 1, 3, 5. ... 2a 2a (5.92) On( x) = 2 -sir 2a n = 2, 4, 6. ... . 2a There are two sets of solutions because we chose a different coordinate system to solve the problem. In the limit zo - co, the decay length q becomes zero and the energy eigenstates are zero outside the well, as expected. The infinite well eigenstates are shown in Fig. 5.18(a) for this new choice of coor- dinates. Comparing the wave functions in Fig. 5.18(a) with those from Fig. 5.14, though, we see that these are the same eigenstate wave functions that we found before. In Fig. 5.18(b) we show the finite well states for comparison