One dimensional scattering

In chapter 5 we discussed one dimensional scattering from simple barriers (rectangular).

Analytical solutions were possible in these cases. In a general case the form of the wave-function is still the same very far to the left and right of the barrier as all physical forces

CHAPTER 6. NUMERICAL TECHNIQUES IN ONE SPACE DIMENSION 67

Figure 6.2: General one dimensional scattering potential.

are expected to die out at large distances. Also, in physical problems measurements of scattered and re°ected waves are made at large distances where the wave is virtually free of forces. Hence, in the following, one dimensional space will be broken into three regions as follows (¯g. 6.2).

I. Region of incident (and re°ected) beam (V = 0).

II. Region of scattering (V 6= 0).

III. Region of transmitted beam (V =V_t).

Region I is force free and hence, has a constant potential. With a suitable choice of reference, this potential can be taken to be zero. Region III is also force free. However, as the choice of reference has already been made in keeping the potential in region I as zero, the potential in region III cannot, in general, be zero. But it will still be a constant V_t. The region of nonzero force (region II) has been chosen to be betweenx=¡aand x= 0. This choice of coordinate origin is made for future convenience. For numerical solutions of problems the position representation is usually preferred as it leads to di®erential equations and standard numerical techniques for di®erential equations are well known. Hence, we shall use the same method as given in chapter 5 with appropriate changes made for the more general potential.

The equation to be solved in the three regions is still the time independent SchrÄodinger equation as given in equation 5.61.

¡¹h² 2m

d²u

dx² +V u=Eu: (6.10)

From equation 5.63, we already know the solution in region I to be

u=Aexp(ikx) +Bexp(¡ikx); forx <¡a: (6.11)

CHAPTER 6. NUMERICAL TECHNIQUES IN ONE SPACE DIMENSION 68 In region III V (=V_t) is a constant and hence equation 6.10 can be written as

d²u

dx² =¡k_t²u; (6.12)

wherek_t is a constant given by

kt= [2m(E¡Vt)=¹h²]¹⁼²: (6.13) Hence, the solution of equation 6.12 would be

u=Cexp(iktx) +Dexp(¡iktx); for x >0: (6.14) If E < V_t, then k_t is imaginary. This would make the second term in equation 6.14 go to in¯nity at x = +1. As this is not allowed, D must be zero. Such a solution can be seen to lead to a zero transmission probability irrespective of the potential in region II (see problem 6 of chapter 5). Hence, for a numerical solution we need to consider only the case E > Vt. So kt must be real and can be chosen to be positive. In chapter 5 it was shown that the second term in equation 6.14 would be due to a particle moving backwards in region III. Such a particle is not possible as there is no source or re°ecting force in region III. Hence, D= 0 and

u=Cexp(ik_tx); for x >0: (6.15)

The solution in region II is to be found numerically for arbitrary potentials. Numerical solu-tions cannot be written in a general form as an arbitrary linear combination of two linearly independent solutions. The freedom of the two arbitrary constants needs to be removed through boundary or initial conditions before solving the equation. These conditions must come from the known solutions in regions I and III and the continuity conditions at the boundaries of the three regions (theorem 5.2). For the computation of the only measurable quantities, (the re°ection and transmission coe±cients) we have seen that only the ratios of the constantsA,Band C are needed. Hence, without loss of generality, one can set any one of these constants to be equal to 1. Depending on which of the constants is chosen to be unity, we need di®erent numerical methods. At present we are going to discuss the method in whichC = 1. This will be later seen to be the simplest.

As a computing machine, in general, cannot handle complex numbers, one must write equation 6.10 as two separate equations for the real and imaginary parts. Separating u in its real and imaginary parts one writes

u=g+ih; (6.16)

where g and h are real. From equation 6.10 one now obtains the two real equations forg and h.

d²g

dx² +2m(E¡V)g

¹

h² = 0; (6.17)

d²h

dx² +2m(E¡V)h

¹

h² = 0: (6.18)

CHAPTER 6. NUMERICAL TECHNIQUES IN ONE SPACE DIMENSION 69 For a numerical solution of these equations it is convenient to choose a dimensionless vari-able for the independent parameter. Let y be such a variable such that

x=cy: (6.19)

A choice ofc that will simplify equations 6.17 and 6.18 would be

c= ¹h=(2mE)¹⁼² = 1=k: (6.20)

Then equations 6.17 and 6.18 become d²g

dy² + (1¡V =E)g = 0; (6.21)

d²h

dy² + (1¡V =E)h = 0: (6.22)

To obtain numerical solutions of the above equations, we use the ¯nite di®erence approxi-mation for the second derivative as given in equation 6.8 with thexi replaced by theyi so thatw=y_i¡y_i¡1. It leads to the following di®erence equations.

g_i+1¡2g_i+g_i¡1+w²(1¡V_i=E)g_i = 0; (6.23) h_i+1¡2h_i+h_i¡1+w²(1¡V_i=E)h_i = 0; (6.24) where V_i is the value of the potential V aty_i. These equations are recursion relations for g and h. Ifgi, hi, gi¡1 and hi¡1 are given, these equations can be solved to ¯ndgi+1 and h_i+1 that is

g_i+1 = [w²(V_i=E¡1) + 2]g_i¡g_i¡1; (6.25) h_i+1 = [w²(V_i=E¡1) + 2]h_i¡h_i¡1: (6.26) Hence, if g₀, h₀, g₁ and h₁ are known, g and h can be found at all points. These initial values can be found from initial conditions of the di®erential equations. g₀ and h₀ are the initial values of the functions. To determine g₁ and h₁ we notice that approximate forms for the derivatives at the initial point are (see problem 1)

(¢g)₀ = (g₁¡g₀)=w; (6.27)

(¢h)₀ = (h₁¡h₀)=w: (6.28)

Then

g1 = w(¢g)0+g0; (6.29)

h1 = w(¢h)₀+h0: (6.30)

If the initial values of the functions and their derivatives are known, equations 6.29 and 6.30 would give the approximations for g₁ and h₁. Hence, in this formulation of the problem,

CHAPTER 6. NUMERICAL TECHNIQUES IN ONE SPACE DIMENSION 70 initial conditions are easier to handle than boundary conditions. A set of boundary condi-tions would give values of the funccondi-tions at both ends of region II, but not their derivatives.

So values for g1 and h1 cannot be found directly.

Initial conditions in region II can be determined from the results of the analytical solutions in either region I or region III if we use the conditions of theorem 5.2. However, the solution in region I has two unknowns (AandB). One of them can be set to unity, but the other still remains unknown. The solution in region III, on the other hand, has only one unknown (viz. C). This can be set equal to 1 as mentioned earlier.

C = 1: (6.31)

Now equation 6.15 can be written as

u= exp(iktx); for x >0: (6.32) As the region of computation (region II) is to the left of this initial point (y = 0), we number the indices ofy,g andhin increasing order from right to left such that

y₀ = 0; y_i =y_i¡1¡w: (6.37)

This makes (¢g)₀ and (¢h)₀, the approximate forms of the derivatives (equations 6.27 and 6.28), change in sign. Hence, replacing approximate derivatives by the negatives of actual derivatives (equation 6.36), equations 6.29 and 6.30 will give

g₁ = 1; h₁ =¡w(1¡V_t=E)¹⁼²: (6.38) Now, equations 6.35 and 6.38 will provide all the initial values needed to solve the recursion relations in equations 6.25 and 6.26. If, in region II, there aren intervals of widthw each, then the last computed value ofg_i andh_i will beg_nandh_nrespectively. This will complete the numerical solution of the di®erential equation.

To determine the re°ection and transmission coe±cients we need to compute the constants A and B. Once the numerical solution for u is known in region II, A and B

CHAPTER 6. NUMERICAL TECHNIQUES IN ONE SPACE DIMENSION 71 can be found by matching the values of u and its derivative (with respect to y) u_y at the boundary of region I and region II. In doing this we use equations 6.11, 6.19 and 6.20 to get

u = Aexp(iy) +Bexp(¡iy) (6.39)

uy = iAexp(iy)¡iBexp(¡iy): (6.40)

This leads to

jAj² = ju_y+iuj²

4 (6.41)

jBj² = juy¡iuj²

4 : (6.42)

Then using equation 6.16 we get

jAj² = (g_y¡h)²+ (h_y+g)²

4 (6.43)

jBj² = (gy+h)²+ (hy¡g)²

4 ; (6.44)

where gy and hy are the derivatives of gand h with respect toy. Due to theorem 5.2, the values ofg,g_y,handh_y must be the same atx=¡a(y=¡ka) using either the solution in region I or the numerical solution in region II. Hence, the approximate values from region II can be introduced in equations 6.43 and 6.44 to abtain

jAj² ' (gyn¡hn)²+ (hyn+gn)²

4 (6.45)

jBj² ' (gyn+hn)²+ (hyn¡gn)²

4 ; (6.46)

whereg_nandh_nare the last computed values from the recursion relations of equations 6.25 and 6.26 i.e. they are the values ofgandhatx=¡a(y=¡ka) which also meansnw=ka.

g_yn andh_yn are the approximations of the derivatives ofg and haty=¡ka.

g_yn = (g_n¡1¡g_n)=w; h_yn = (h_n¡1¡h_n)=w: (6.47) Now, the transmission coe±cientT and the re°ection coe±cientRcan be found from the de¯nitions in equation 5.22. The currents N and n_r are the same as in equations 5.69 and 5.70. However, as k_t is no longer the same ask,n_t is given by

n_t= ¹hk_t

m jCj²: (6.48)

Hence, we obtain

T = ^k_k^tjCj2

jAj² = 4(1¡V_t=E)¹⁼²

(gyn¡hn)²+ (hyn+gn)² (6.49) R= ^jBj2

jAj² = (g_yn+h_n)²+ (h_yn¡g_n)²

(g_yn¡h_n)²+ (h_yn+g_n)²: (6.50)

CHAPTER 6. NUMERICAL TECHNIQUES IN ONE SPACE DIMENSION 72 Thus an algorithm for the computation of transmission and re°ection coe±cients will con-tain the following parts.

² Determine desired values of E,V_tand w. For better results, the function should not change much in any interval w. Hence, the wavelength of the wavefunction must be large compared to wi.e. w¿1=k.

² Set the initial valuesg0,h0,g1 and h1 using equations 6.35 and 6.38.

² Use the recursion relations in equations 6.25 and 6.26 in a loop to compute all values of gi and hi uptogn and hn. In each step of the loop, y is to be decreased by wand the value of the potential is to be computed at that value of y. The loop is to be terminated wheny∙¡ka.

² Re°ection and transmission coe±cients are to be computed using equations 6.47, 6.49 and 6.50.