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The WKB Approximation

Marcus A. M. de Aguiar

November 12, 2013

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Contents

1 Introduction 3

2 General form of the WKB wave-function 7

3 The WKB wave-function for bound states 9

4 Solution near a turning point withV > 0 104.1 Statement of the problem and solution . . . . . . . . . . . . . 104.2 Solving the equation . . . . . . . . . . . . . . . . . . . . . . . 11

5 Solution near a turning point withV < 0 15

6 Semiclassical quantization 17

7 The WKB wave-function for scattering states 20

8 The other Airy function 21

8.1 The Airy equation . . . . . . . . . . . . . . . . . . . . . . . . 218.2 The function Bi . . . . . . . . . . . . . . . . . . . . . . . . . . 228.3 Asymptotic form of Bi . . . . . . . . . . . . . . . . . . . . . . 22

9 Semiclassical tunneling 24

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1 Introduction

The Schroedinger equation for a particle in a potential can be solved ex-plicitly only if the potential has a sufficiently simple form, otherwise nu-merical methods have to be used. In many cases it is preferable to haveapproximate explicit formulas, that can be further used to calculate otherquantities, than to have exact numerical solutions. The WKB approxima-tion, developed by Wentzel, Brillouin and Keller, is a semiclassical methodto solve Schroedingers equation that does not require the potential to bea perturbation of a solvable problem. Instead, it only assumes that certainclassical quantities having the dimension of action (energytime) are muchlarger than the Plancks constant. In this paper I will present the method in

a simple but complete way, paying particular attention to the famous con-nection formulas, that are the main source of difficulty in the presentation.

To make things clear from the beginning, let me be very specific aboutthe problems I have in mind. I want to find stationary solutions for a particlemoving in a one-dimensional potential. I will consider two situations: (a) thepotential has bound states, as illustrated in figure 1, or; (b) the potential hasscattering states and is different from zero only in a small region around theorigin, as illustrated in figure 2. In the first case I am particularly interested incomputing the discrete energy levels of the system, whereas in the second caseI want to compute the tunneling coefficient. In both cases the Schroedingerequation reads

22m

2x2

+V(x)(x, t) =it

. (1)

The wave-function also satisfies the continuity equation

t +

j

x= 0 (2)

where(x, t) = |(x, t)|2 (3)

is the probability density and

j(x, t) =

mIm

x

(4)

is the probability current. The continuity equation can be derived by multi-plying eq.(1) by and subtracting the resulting equation from its complexconjugate.

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V(q)

E

BA q

III III

Figure 1: A particle with energyEin a bound potential. The turning pointsA and B, where V = E, separate the classically allowed region II from theforbidden regions I and III.

V(q)

q

E

A B

II IIII

Figure 2: A particle with energyE in a scattering potential. In our calcu-lations I will assume that the particle is incident from the left side of thebarrier and that E < V0.

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In developing the WKB approximation it is important to write the com-

plex wave-function in terms of its modulus and phase as(x, t) =

(x, t)e

iS(x,t) (5)

where the explicit dependence on in the phase is introduced by convenience.In terms of and Sthe probability current becomes

j =

m

S

x. (6)

The Schroedinger equation, on the other hand, reads

2

2m 1

2

2

x2 1

4

x2

+ i

x

S

x+

i

2S

x2 1

2

S

x2

+V(x)

= i

1

2

t+

i

S

t

(7)

where I have already canceled the global factor exp (iS/). In terms of ex-plicit powers of this equation has three types of factors, involving 0, 1

and 2. I am going to show that the factors in 1 vanish exactly because ofthe continuity equation. Indeed, the two terms of this type on the left sideof the equation are

i2m

x

S

x i

2m

2S

x2 (8)

which can be re-organized as

i2

x

m

S

x

= i

2

j

x=

i

2

t (9)

and exactly cancel the term in on the right side.Dividing the remaining terms in eq.(7) by

I obtain

1

2m

S

x

2+VQ(x) = S

t (10)

where I have defined the quantum potential

VQ(x) =V(x) 24m

2x2

+ 28m2

x

2

=V(x) 2

4m

x

1

x

.

(11)

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Equation (10) is exact and, together with the continuity equation, it

is completely equivalent to Schroedingers equation. If it were not for theterms in 2 in the quantum potential, eq.(10) would be the Hamilton-Jacobiequation of classical mechanics, whose solution is given by the action func-tion (see below). The WKB approximation consists exactly in discard theseterms. David Bohm proposed that we could solve these equations as theyare, and find the trajectories of the particle in the quantum potential. No-tice, however, thatVQdepends on the wave-function itself. Given(x, t0) wewould calculate VQ and propagate the wave-function to t0+dt. This wouldallows us to re-calculate VQ and proceed with the integration. The difficultyis that VQ is singular at the nodes of the wave-function, where (x, t) = 0.

In either case equation (10) is a partial differential equation that can be

solved for S(x, t) given S(x, 0). However, our interest here is to calculatestationary states, whose time dependence is of the form

(x, t) =(x)eiEt . (12)

Comparing with equation (5) we see that this amounts to impose

= (x) S(x, t) =W(x) Et (13)

so that(x, t) = (x)e

iW(x) e

iEt .

In terms ofW and Eequation (10) becomes

1

2m

W

x

2+VQ(x) =E. (14)

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2 General form of the WKB wave-function

In the limit where certain classical quantities (to be more clearly specifiedbelow) become much larger than , it is legitimate to discard the quantumcorrections to the potential VQ and replace it by the classical potential. Inthis case eq.(10) reduces to the classical Hamilton-Jacobi equation

1

2m

S

x

2+V(x) = S

t. (15)

For stationary states, where

= (x) and S(x, t) =W(x)

Et

it reduces to1

2m

W

x

2+V(x) =E. (16)

The function S(x, t) is called Hamiltons principal function, whereas W(x)is Hamiltons characteristic function. Equation (16) can be solved for W as

W(x) = x

2m(E V(x))dx (17)

where the integrand is readily recognized as the momentum p(x, E) of a

particle with energy Eat position x, since

p2

2m+ V(x) =E p(x, E) =

2m(E V(x)). (18)

Comparing the classical Hamilton-Jacobi equation (16), which is inde-pendent of, with its fully quantum version (14), we can see that quantumcorrections to W are proportional to 2. The same is true for classical ,that we compute below using the continuity equation.

Sinceis time independent, the current must be constant, since j/x=0 in this case. Using eqs.(6), (13) and (17) we see that

j= m

p(x, E) =const.

and, therefore,

(x) = const

p(x, E). (19)

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Putting all this information together we obtain the general form of the

time independent, stationary, wave function(x) as

(x) = cp(x, E)

ei

xp(x,E)dx. (20)

In the next sections I will now apply these ideas to the specific prob-lems illustrated in figures 1 and 2. In both cases, for a given total energyE the x-axis is divided into three regions by the turning points A and B.In the classically forbidden regions V(x) > E and p(x, E) becomes purelyimaginary. In these regions it is convenient to define the real function

q(x, E) = 2m(V(x) E) = |p(x, E)| (21)and re-write the general form of the wave-function as

(x) = cq(x, E)

e1

x q(x,E)dx . (22)

The lower limit of integration in equations (20) and (22) will be fixed later.

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3 The WKB wave-function for bound states

I will now consider the specific problem of bound states illustrated in figure1. In the classically forbidden regions I and III the WKB wave-functioninvolves real exponentials and we need to be careful not to let it diverge.In the classically allowed region II, on the other hand, both solutions (withplus or minus sign in the exponent) are well behaved and should be included.Taking this considerations into account we write

(x) =

D12

qe1

x q(x,E)dx x I

Cpcos

1

xp(x, E)dx+

x II

D22

qe

1

x q(x,E)dx x III

(23)

where I replaced the combination of the two exponentials in region II by acosine with a phase.

We now get to the hard part of the WKB procedure, which is to matchthe constants D1, C, D2 and . These constants cannot be arbitrary, since(x) needs to be an approximation for the true wave-function in all space

and all these constants must be related to the single undetermined constantthat we could find later by normalization.

The problem is that we cannot simply match the different parts of thewave-function (x) because it diverges right at the matching points x = Aand x = B! At these points p(x, E) = q(x, E) = 0. The trick to overcomethis difficulty is to go back to the original Schroedinger equation and tosolve it in the neighborhood of these points by approximating the potentialby a straight line. The next two sections are dedicated to that. Once wehave these solutions we can compare them with our WKB forms and relatethe coefficients to each other. We will find that the matching cannot bealways performed and the condition for it to work is the Bohr-Sommerfeld

quantization rule that discretizes the energy levels.

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4 Solution near a turning point withV>0

4.1 Statement of the problem and solution

Consider a turning point x0 with V(x0)> 0, where the prime means differ-entiation with respect to x. This is the case ofx= B for bound states (seefig.1) or x= A for scattering states (see fig.2). In the vicinity ofx0 we canexpand the potential as

V(x) V(x0) +V(x0)(x x0) =E+V(x0)(x x0). (24)

The time-independent Schroedinger equation becomes, in operator form,

p22m

| + [E+V(x0)(x x0)]| =E| (25)

orp2

2m| +V(x0)x| =V(x0)x0|. (26)

I am going to solve this equation step by step in this section. The finalresult will be written as

(x) =c

+0

cos(t3/3 ty)dt (27)

wherecis a normalization constant and

y=

2mV(x0)

2

1/3(x0 x). (28)

This is exact, but not very helpful. The variabley is zero at the turning pointx= x0. If we assume that the ratio

2/mV(x0) is so small that y becomesvery large as we move away from x0, without leaving the region where thelinear approximation of the potential is valid, we can derive approximateformulas for this integral that directly relate to our WKB wave-functions. Inthis case I will show that

(x)

cp(x,E)

cos1

xx0

p(x, E)dx+/4

x < x0

c

2

q(x,E)e1

xx0

q(x,E)dxx > x0.

(29)

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4.2 Solving the equation

In the momentum representationp| =(p) equation (26) readsp2

2m(p) +iV(x0)

(p)

p =V(x0)x0(p). (30)

This equation has a simple solution

(p) =A exp(ip3 ipx0/) (31)

with

= 1

6m

V(x0)

(32)

as the reader can check by substitution. The corresponding wave-function inposition representation is obtained by Fourier transforming:

(x) = A

2

+

e(ip3+ip(xx0)/)dp

which can be easily re-arranged as

(x) =A +0

cos(p3 +p(x x0)/)dp.

Finally, changing variables tot = p(3)1/3 and defining

y=

2mV(x0)

2

1/3(x0 x) (33)

we obtain

(x) =c

+0

cos(t3/3 ty)dt cAi(y) (34)

whereAi(z) is the Airy function.The solution (34) should be compared with the WKB solution derived in

the previous section. However, as it is, the comparison is not immediate. Inorder to see that equation (34) indeed resembles the WKB forms we need todo a little more work.

I am going to assume that the ratio 2/mV(x0) is so small that, forvalues ofx where the linear approximation is still good,y can become much

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larger than 1. In this case the cosine in the Airy function oscillates very fast

between +1 and1 as t changes, and the integral seems to tend to zero.However, there are points along the t axis where the phase of the cosinebecomes nearly constant, and in the vicinity of these points the cosine stopsoscillating and contributes some finite value to the integral. All we need todo is to find these points and expand the cosine around them.

In order to do that I will re-write the Airy function as an integral from to + as

Ai(y) =12

+

ei(t,y)dt (35)

where(t, y) =t3/3

ty. (36)

We need to find the points where /t = 0. This condition defines thepoints where becomes stationary. We find

t= y. (37)Since the important contributions to the integral come from the neighbor-hood of these points, we may expand to second order around them:

(t, y) (t, y) + 12

(t, y)(t t)2

= 23

y3/2 y(t t)2.(38)

(the primes now denote differentiation with respect to t and remember that(t, y) = 0). This approximation takes us to Gaussian integrals aroundeach stationary point that we know how to do:

Ai(y) 12

ei23y3/2

+

eiy(tt)2dt (39)

However we must be careful: y >0 only ifx < x0. This corresponds tothe classically allowed region (both in the case of bound states and scatteringstates). In this case

yis real, there is no risk of divergence in the exponents

and both stationary points contribute to the integral and we must sum their

contributions. We getAi(y) 1

2

iy

ei23y3/2 + 1

2

iyei

23y3/2

=

ycos

23

y3/2 /4. (40)

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For x > x0 we find that y < 0 and

y = iy. In this case only the

stationary point t+ contributes, since t would lead to a diverging integral.We getAi(y) 1

2e

23(y)3/2+

ey(tt+)2dt

= 12

ye

23(y)3/2.

(41)

The asymptotic analysis presented above is not rigorous, since we have notshown that the contour of integration can be deformed so as to pass throughthe stationary points. In any case, it gives an idea of the general procedure.

To complete the analysis we only need to relatey withp(x, E). We startfrom

p(x, E) =

2m(E V(x)) 2mV(x0)(x x0) =

2mV(x0)(x0 x)

and xx0

p(x, E)dx

2mV(x0) xx0

(x0 x)dx= 2

3

2mV(x0)(x0x)3/2.

Therefore,

2

3y3/2 =

2

32mV(x0)(x0 x)3/2 = 1

x

x0

p(x, E)dx

and also

y1/2 = p(x, E)

[2mV(x0)]1/3.

Therefore, equation (40) can be re-writen as

Ai(y) cp(x, E)

cos

1

xx0

p(x, E)dx+/4

. (42)

forx < x0, wherec= [2mV(x0)]1/3 and we used that cos () = cos ().

Similarly, for x > x0,

q(x, E)

2mV(x0)(x x0), xx0

q(x, E)dx 23

2mV(x0)(x x0)3/2,

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2

3

(

y)3/2

1

x

x0

q(x, E)dx

and

(y)1/2 = q(x, E)[2mV(x0)]1/3

.

This gives

Ai(y) cq(x, E)

e 1

xx0

q(x,E)dx. (43)

These results are summarized in equations (27) and (29) at the beginning

of this section.

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5 Solution near a turning point withV x0, in the classically allowed region, we get

Ai(y)

y

cos

2

3y3/2 /4

. (49)

For x < x0, in the classically forbidden region, we find

Ai(y) 12

y e

23(y)3/2. (50)

Finally we need to relate y with p(x, E). We find

p(x, E)

2m|V(x0)|(x x0), xx0

p(x, E)dx 23

2m|V(x0)|(x x0)3/2,

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2

3

y3/2

1

x

x0

p(x, E)dx

and

y1/2 = p(x, E)

[2m|V(x0)|]1/3 .

Therefore, the asymptotic solutions can be re-writen as

(x)

cp(x,E)

cos1

xx0

p(x, E)dx /4

x > x0

c

2

q(x,E)e1

xx0

q(x,E)dxx < x0.

(51)

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6 Semiclassical quantization

We are now in a position to match the different parts of the wave-functionfor bound states, as given by equation (23) that we repeat here:

(x) =

D12

qe1

x q(x,E)dx x I

Cp

cos

1

xp(x, E)dx+

x II

D2

2qe

1

x q(x,E)dx x

III

We start by connecting regions II and III, where the turning point B satisfiesV(B) > 0. Comparing with the approximate solutions near the turningpoint, equation (29), also repeated below,

(x)

cp(x,E)

cos1

xx0

p(x, E)dx+/4

x < x0

c

2

q(x,E)e1

xx0

q(x,E)dxx > x0.

we find that D2 = C, the lower limit of the integrals should be fixed as Band =/4.On the other hand, regions I and II, connected by the turning point A

whereV(A) x0

c

2

q(x,E)e1

xx0

q(x,E)dxx < x0.

In this case we find that D1 =C, the lower limit of the integrals should be

fixed as A and = /4.

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Putting all this information together we may write

(x) =

C

2

qe1

xAq(x,E)dx x I

Cp

cos

1

xA

p(x, E)dx 4

=

Cp

cos

1

xB

p(x, E)dx+

4

x II

C

2

qe

1

xBq(x

,E)dx x III

The two forms in region II are written in order to connect with regions I

and III respectively. But of course they must be representations of the samefunction and, therefore, identical. In order to assure their identity, we re-write the argument of the first cosine in terms of the argument of the secondcosine:

1

xA

p(x, E)dx 4

= 1

BA

p(x, E)dx 2

+1

xB

p(x, E)dx+

4.

We now impose that

1

B

A

p(x, E)dx 2

=n (52)

so that the arguments of the two cosine become related by

cos

1

xA

p(x, E)dx 4

= (1)n cos

1

xB

p(x, E)dx+

4

.

This last sign, (1)n is finally absorbed by imposing C= (1)nC.The final result is

(x) =C

(1)n2

q e

1

xAq(x

,E)dx x I

1p

cos

1

x

B

p(x, E)dx+ 4

x II

1

2

qe

1

xBq(x,E)dx x III

(53)

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where C is related to the normalization of (x). This global matching is

possible only if the quantization condition (52) is satisfied. This can bere-writen as BA

p(x, E)dx=

n+

1

2

(54)

which is the well known Bohr-Sommerfeld quantization condition of the oldquantum theory.

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7 The WKB wave-function for scattering states

Let us now consider the problem of scattering states illustrated in figure 1,where a particle (or beam of particles) is incident from the left withEsmallerthan the height of the barrier. The general form of the WKB wave-functionfor this setting is

(x) =

Ap

ei

xp(x,E)dx + R

pe

i

xp(x,E)dx x I

Cq

e1

x q(x,E)dx + D

qe

1

x q(x,E)dx x II

Tp

ei

xp(x,E)dx x III

. (55)

Notice that the forbidden region is now region II, which is finite. There-fore the exponentials with both + andsigns are allowed. As we shall see,this introduces a complication in the matching of the different parts of thewave-function. We will deal with that in the next section.

Once the matching of the coefficients has been completed, we will calcu-late the transmission and reflection coefficients, given by

t=|T|2

|A|2

and

r=|R|2|A|2 .

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8 The other Airy function

8.1 The Airy equation

Let us return to the Schroedinger equation in the vicinity of a turning pointx0, but this time in the x representation. Starting from

p2

2m| +V(x0)x| =V(x0)x0|. (56)

we find

2

2m

2

x2+V(x0)(x x0)(x) = 0. (57)

In terms of

y=

2mV(x0)

2

1/3(x0 x) (58)

we obtain2

y2 +y = 0 (59)

which is the Airy equation. Let us check that

(x) =

+0

cos(t3/3 ty)dt= Ai(y) (60)

is indeed a solution. We differentiate twice with respect toy and manipulatethe result as follows:

= +0 t2 cos(t3/3 ty)dt= +

0

t

[sin(t3/3 ty)] y cos(t3/3 ty) dt= sin (t3/3 ty)|0 y = y.

(61)

The boundary term is a bit delicate: at t = 0 it is clearly zero, but at t = it is not well defined. The argument is that it oscillates so fast for large t

that any average of this term will be zero, and therefore we set it to zero.

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8.2 The function Bi

The Airy equation is a second order equation and must have two independentsolutions. Looking atAi(y) we may guess that the other solution has thesame form with the cosine replaced by a sine. That, however, does not work.The other solution is given by

(x) =

+0

et

3/3yt + sin (t3/3 ty)

dt Bi(y). (62)

The need of the extra exponential becomes evident when we calculate thesecond derivative ofB iwith respect to y :

= +0

t2et3/3yt t2 sin(t3/3

ty) dt

=+0

t[et

3/3yt] + t

[cos (t3/3 ty)] yet3/3yt y sin(t3/3 ty)

dt

= et3/3yt|0 + cos(t3/3 ty)|0 y = y.(63)

If it were not for the extra exponential term, the boundary term resultingfrom the cosine att = 0 would leave a factor 1, which is canceled by a similarfactor coming from that exponential.

8.3 Asymptotic form of BiFar from the turning point x0 we can derive simplified expressions for theBi functions just as we did with Ai. I will not do that here and will onlypresent the results. We find that, for V (x0)>0,

(x)

cp(x,E)

cos1

xx0

p(x, E)dx /4

x < x0

cq(x,E)

e1

xx0

q(x,E)dxx > x0

(64)

and, for V (x0)< 0,

(x)

cp(x,E)

cos1

xx0

p(x, E)dx+/4

x > x0

cq(x,E)

e 1

xx0

q(x,E)dxx < x0.

(65)

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Notice that both forms diverge as x in the forbidden regions. This iswhy the functionBi does not appear in the calculation of bound states, wherethe forbidden regions extend to infinity. However, in scattering problems, theforbidden region is finite, no divergence occurs and B ihas to be included.

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9 Semiclassical tunneling

Using the results from sections 5 and 8 we can match the different parts ofthe wave-function for scattering states, as given by equation (55) that werepeat here:

(x) =

Ap

ei

xp(x,E)dx + R

pe

i

xp(x,E)dx x I

Cq

e1

x q(x,E)dx + D

qe

1

x q(x,E)dx x II

Tpe

i

x

p(x

,E)dx

x III

We start with region III, where the solution represents a particle movingto the right only. Near the turning pointB, where V(B) < 0 we can writethe solution as a combination ofAiandB ias

(x) F Ai(y) +GBi(y).In the allowed region III these functions have asymptotic forms given byequations (51) and (65):

(x) F

p(x,E) cos 1 xBp(x, E)dx /4 + Gp(x,E) cos 1 xBp(x, E)dx+/4= 1

p(x,E)

Fcos

1

xBp(x

, E)dx /4 +G sin 1

xBp(x

, E)dx /4The desired form for region III can be recovered with the choice G = iF. Weget

(x) Fp(x, E)

exp

i

xx0

p(x, E)dx /4

which gives our first coefficient T =F ei/4.This same solution, with G = iF, must work in region II, where

(x) F[Ai(y) +iBi(y)]

F

1

2

q(x,E)e1

xBq(x,E)dx + i

q(x,E)e

1

xBq(x,E)dx

.

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Therefore, fixing the lower limit of integration in region II as x0 = B, the

coefficients must be C=F /2 =T /2ei/4

and D = iT ei/4

.

We now repeat the matching between regions I and II, where x0= A andV(A)>0. In the vicinity ofx = A we write

(x) f Ai(y) +gBi(y).

In region II this becomes

(x) f2

q(x,E)e

1

xAq(x,E)dx + g

q(x,E)e1

xAq(x,E)dx

= 1q(x,E)

fe/

2 e

1

xBq(x

,E)dx +ge/e1

xBq(x

,E)dx

where we have defined

=

BA

q(x, E)dx.

Comparing with the form in region II we see that

C = fe/

2 T

2 ei/4

D =ge/ iT ei/4

orf =T e/+i

4

g =iT e/+i4

Finally, we look at region I to determine Aand R as a function ofT. Inthis region we have

(x) f Ai(y) +gBi(y)

fp(x,E) cos 1 xAp(x, E)dx+/4 + gp(x,E) cos 1 xAp(x, E)dx /4.

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Writing the cosines as exponentials and comparing with the general form in

region I we see thatA = fe

i/4+gei/4

2

= T ei/2+/+iTe/

2

=iTcosh(/)

andR = fe

i/4+gei/4

2

= T e/+iT e/+i/2

2

=Tsinh(/).

The final result is

(x) =T

i cosh (/)p

ei

xAp(x

,E)dx +sinh(/)

p e

i

xAp(x

,E)dx x I

i1/2

2qe1

xBq(x

,E)dx +i3/2

qe

1

xBq(x

,E)dx x

II

1p

ei

xBp(x

,E)dx x III

.

(66)The transmission and reflection coefficients are

t= 1

cosh2 (/) 4e2/ (67)

andr= tanh2 (/)

1

4e2/. (68)

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