Finite Potential Well

Transcription

1 Finite Potential Well These notes are provided as a supplement to the text and a replacement for the lecture on 11/16/17. Make sure you fill in the steps outlined in red. The finite potential well problem is a generalization of the infinite potential well that allows for particles to exist outside of the well region. The standard form of the potential is given graphically as in the text in Figure 8.1 nalytically this can be represented as follows: V = 0 x < a Region I V x a Region II 0 x > a Region III Note that the well width for the standard form finite well is twice that of the infinite well and the well itself has a finite, negative but constant potential and outside the well the potential is zero. In order to solve this problem we must first solve the eigenvalue problem. This is no different than for the infinite well case except for the constant potential term. Thus in the three regions we have Region I x < a Ĥ I = ˆp 2 x 2m =!2 d 2 Ĥ I = E!2 2m = e κ x! 2 κ 2 2m = E > 0 d 2 dx 2 = E Note that we have followed the convention of explicitly making the energy and potential to have negative values so that when we use the symbols E and V these can be positive numbers. We have also excluded the solution with a negative exponent in the exponential function because such a solution, while satisfying the differential equation can be seen to be non-physical since it will become infinite at infinite distance from the origin. Repeating for region III we have the same situation but for positive x. 1

2 Region III x > a Ĥ III = ˆp 2 x 2m =!2 d 2 Ĥ III I = E I!2 2m I = De κ x! 2 κ 2 2m = E > 0 d 2 I dx 2 = E I We can see that this is just the mirror image of Region I. For Region II, the equation is a bit more complicated because the potential is non-zero. Region II x a Ĥ II II = ˆp 2 x 2m V =!2 d 2 2m dx V 2 Ĥ II = E!2 d 2!2 d 2 = ( V E )φ II = Be ikx + e ikx! 2 k 2 V = E 2m = V E > 0 k 2 = 2m V E! 2 We can also combine the results for the energy eigenvalues inside and outside the well to see the following relationship (show): Κ 2 + k 2 = 2mV! 2 t this point we have established the form of the solutions inside and outside the well. Notice that outside the well the wave function takes on the form of a damped exponential, tending to zero at infinity. Inside the well, the solution is oscillatory since it involves an exponential with an imaginary exponent. Next we have to deal with the boundary conditions. These require the wave function and its first derivative to be continuous at the two well edges. In analytic form this requires: 2

3 ( a) = ( a) (a) = I (a) ( a) = ( a) (a) = I (a) Using the forms for the wave functions that we found earlier we may obtain a set of equations governing the constants, B, and D. Show that this reduces to: e κ a = Be ika + e ika Be ika + e ika = De κ a κ e κ a = ik Be ika e ika ik( Be ika e ika ) = κ De κ a In matrix form this set of four linear equations in the four constants, B, and D is DV = e -κa -e -ika e ika 0 0 e ika e -ika -e -κa κe -κa ike -ika ike ika 0 0 ike ika ike -ika κe -κa B D Thus we may obtain a nontrivial set of solutions for the coefficients provided the determinant of the coefficient matrix is zero: det D = 0 Before proceeding it is convenient to rewrite the matrix equation using the following definition: G = ( κ + ik)e ika = G e i( ka+φ) Show that with this definition the following relationships hold. tan( φ) = k κ G = GG * = ( κ 2 + k 2 ) 3

4 To see why this definition is helpful first return to the set of equations and write them in standard order: e κ a Be ika e ika + D 0 = Be ika + e ika De κ a = 0 + D 0 = 0 +κ De κ a = 0 κ e κ a ik Be ika e ika 0 + ik Be ika e ika Next we use the rules of linear algebra that allow us to multiply any line by a constant and add and subtract any line to any other line (since we are just adding and subtracting zero). We can eliminate the and D terms in the first two lines by this method. Show how to do this manipulation to obtain the result: (hint couple the first and third lines and the second and fourth lines): ( κ ik)be ika + ( κ + ik)e ika D 0 = 0 ( κ + ik)be ika + ( κ ik)e ika D 0 = 0 Be ika e ika κ ik e κ a + D 0 = 0 Be ika + e ika + 0 κ ik De κ a = 0 or in matrix form using the definition of G given above: DV = G * G 0 0 G G * 0 0 κ e -ika e ika ik e-κa 0 e ika e ika 0 κ ik e-κa B D Next we need to evaluate the determinant. Use the rules of determinants to show that 4

5 D = κ k 2 e ( -2κa G *2 G ) 2. Recall that we need to find the conditions under which this determinant is zero. learly this occurs when G *2 = G 2 Recalling the definition of G, we obtain G * = ±G G e i( ka+φ) = ± G e cos ka + φ i ka+φ + isin( ka + φ) = ± cos( ka + φ) isin( ka + φ) Show that this implies two conditions on the argument of the trig functions: ka + φ = 0 (positive root) and ka + φ = π 2 (negative root) Returning to the definition of the phase angle ϕ we obtain the first condition from the positive root: tan( φ) = tan( ka) k κ κ k = tan( ka) = cot( ka) κ = k cot( ka) Show that the other (negative) root leads to the condition: κ = k tan( ka) lso show that these two roots may be written alternatively: 5

6 G G * = ±1. Following the usual methods of linear algebra, show that these two roots also imply the following relationships between the coefficients of the finite well solutions and the consequent forms of these solutions given below: For the positive root: B = 1, I B = D B = 2isin( ka)e κ a ( x) = 2iBsin( ka)e κ ( x+a) ( x) = 2iBsin( kx) κ x a ( x) = 2iBsin( ka)e Likewise for the negative root: B = +1, I B = D B = 2cos( ka)e κ a ( x) = 2Bcos( ka)e κ ( x+a) ( x) = 2iBcos( kx) κ x a ( x) = 2iBcos( ka)e lso show the following: k cot( ka) = κ k tan( ka) = κ The solutions above satisfy the boundary conditions stated at the beginning of the analysis. The positive root solutions are odd under parity. The negative root solutions are even under parity. 6