The WKB Approximation

Marcus A. M. de Aguiar

September 8, 2011

1

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 with V ′ > 0 104.1 Statement of the problem and solution . . . . . . . . . . . . . 104.2 Solving the equation . . . . . . . . . . . . . . . . . . . . . . . 11

5 Solution near a turning point with V ′ < 0 15

6 Semiclassical quantization 17

7 The WKB wave-function for scattering states 20

8 The other Airy function 218.1 The Airy equation . . . . . . . . . . . . . . . . . . . . . . . . 218.2 The function Bi . . . . . . . . . . . . . . . . . . . . . . . . . . 228.3 Asymptotic form of Bi . . . . . . . . . . . . . . . . . . . . . . 22

9 Semiclassical tunneling 24

2

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 Schroedinger’s 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 (energy×time) are muchlarger than the Planck’s constant. In this paper I will present the method ina 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

− ~2

2m

∂2ψ

∂x2+ V (x)ψ(x, t) = i~

∂ψ

∂t. (1)

The wave-function also satisfies the continuity equation

∂ρ

∂t+∂j

∂x= 0 (2)

whereρ(x, t) = |ψ(x, t)|2 (3)

is the probability density and

j(x, t) =~mIm

(ψ∗∂ψ

∂x

)(4)

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

3

V(q)

E

BA q

III III

Figure 1: A particle with energy E in 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 energy E 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.

4

In developing the WKB approximation it is important to write the com-plex wave-function ψ in terms of its modulus and phase as

ψ(x, t) =√ρ(x, t)e

i~S(x,t) (5)

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

j =ρ

m

∂S

∂x. (6)

The Schroedinger equation, on the other hand, reads

− ~2

2m

[1

2√ρ

∂2ρ

∂x2− 1

4ρ√ρ

(∂ρ

∂x

)2

+i

~√ρ∂ρ

∂x

∂S

∂x+i

~√ρ∂2S

∂x2− 1

~2√ρ

(∂S

∂x

)2]

+V (x)√ρ = i~

[1

2√ρ

∂ρ

∂t+i

~√ρ∂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, ~1and ~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

− i

2m√ρ

∂ρ

∂x

∂S

∂x− i

2m

√ρ∂2S

∂x2(8)

which can be re-organized as

− i

2√ρ

∂

∂x

(ρ

m

∂S

∂x

)= − i

2√ρ

∂j

∂x=

i

2√ρ

∂ρ

∂t(9)

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

√ρ I obtain

1

2m

(∂S

∂x

)2

+ VQ(x) = −∂S∂t

(10)

where I have defined the quantum potential

VQ(x) = V (x)− ~2

4mρ

∂2ρ

∂x2+

~2

8mρ2

(∂ρ

∂x

)2

= V (x)− ~2

4m√ρ

∂

∂x

(1√ρ

∂ρ

∂x

).

(11)

5

Equation (10) is exact and, together with the continuity equation, it iscompletely equivalent to Schroedinger’s equation. If it where 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, that VQ depends on the wave-function itself. Given ψ(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 ψ(x, t) = 0.

In either case equation (10) is a partial differential equation that can besolved for S(x, t) given S(x, 0). However, our interest here is to calculatestationary states, whose time dependence is of the form

ψ(x, t) = φ(x)e−iEt~ . (12)

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

ρ = ρ(x) S(x, t) =W (x)− Et (13)

so thatψ(x, t) =

√ρ(x)e

iW (x)~ e−

iEt~ .

In terms of W and E equation (10) becomes

1

2m

(∂W

∂x

)2

+ VQ(x) = E. (14)

6

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

ρ = ρ(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 Hamilton’s principal function, whereas W (x)is Hamilton’s 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 aparticle with energy E at 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’ ρ,that we compute below using the continuity equation.

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

j =ρ

mp(x,E) = const.

and, therefore,

ρ(x) =const

p(x,E). (19)

7

Putting all this information together we obtain the general form of thetime independent, stationary, wave function φ(x) as

φ(x) =c√

p(x,E)e±

i~∫ x p(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

φ(x) =c√

q(x,E)e±

1~∫ x q(x′,E)dx′

. (22)

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

8

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

φ(x) =

D1

2√qe

1~∫ x q(x′,E)dx′

x ∈ I

C√pcos

(1

~

∫ x

p(x′, E)dx′ + ξ

)x ∈ II

D2

2√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 ξ. These constants cannot be arbitrary, sinceφ(x) needs to be an approximation for the true wave-function in all spaceand 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 φ(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-Sommerfeldquantization rule that discretizes the energy levels.

9

4 Solution near a turning point with V ′ > 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 of x = B for bound states (seefig.1) or x = A for scattering states (see fig.2). In the vicinity of x0 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,

p2

2m|φ〉+ [E + V ′(x0)(x− x0)]|φ〉 = E|φ〉 (25)

orp2

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

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

φ(x) = c

∫ +∞

0

cos (t3/3− ty)dt (27)

where c is a normalization constant and

y =

(2mV ′(x0)

~2

)1/3

(x0 − x). (28)

This is exact, but not very helpful. The variable y 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

φ(x) ≈

c√

p(x,E)cos

(1~

∫ x

x0p(x′, E)dx′ + π/4

)x < x0

c

2√

q(x,E)e− 1

~∫ xx0

q(x′,E)dx′x > x0.

(29)

10

4.2 Solving the equation

In the momentum representation 〈p|φ〉 = φ(p) equation (26) reads

p2

2mφ(p) + i~V ′(x0)

∂φ(p)

∂p= V ′(x0)x0φ(p). (30)

This equation has a simple solution

φ(p) = A exp (iαp3 − ipx0/~) (31)

with

α =1

6m~V ′(x0)(32)

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

φ(x) =A√2π~

∫ +∞

−∞e(iαp

3+ip(x−x0)/~)dp

which can be easily re-arranged as

φ(x) = A′∫ +∞

0

cos (αp3 + p(x− x0)/~)dp.

Finally, changing variables to t = p(3α)1/3 and defining

y =

(2mV ′(x0)

~2

)1/3

(x0 − x) (33)

we obtain

φ(x) = c

∫ +∞

0

cos (t3/3− ty)dt ≡ cAi(−y) (34)

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

the previous section. However, as it is, it is of not much use. In order to seethat equation (34) indeed resembles the WKB forms we need to do a littlemore work.

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

11

larger than 1. In this case the cosine in the Airy function oscillates very fastbetween +1 and −1 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) = 1

2

∫ +∞

−∞eiγ(t,y)dt (35)

whereγ(t, y) = t3/3− ty. (36)

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

t± = ±√y. (37)

Since the important contributions to the integral come from the neighbor-hood of these are the points, we may expand γ to second order around them:

γ(t, y) ≈ γ(t±, y) +12γ

′′(t±, y)(t− t±)

2

= ∓23y3/2 ±√

y(t− t±)2.

(38)

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

Ai(−y) ≈ 1

2e∓i 2

3y3/2

∫ +∞

−∞e±i

√y(t−t±)2dt (39)

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

√y is real, there is no risk of divergence in the exponents

and both stationary points contribute to the integral and we must sum theircontributions. We get

Ai(−y) ≈ 12

√iπ√ye−i 2

3y3/2 + 1

2

√−iπ√

yei

23y3/2

=√

π√ycos

(23y3/2 − π/4

).

(40)

12

For x > x0 we find that y < 0 and√y = i

√−y. In this case only the

stationary point t+ contributes, since t− would lead to a diverging integral.We get

Ai(−y) ≈ 12e−

23(−y)3/2

∫ +∞−∞ e−

√−y(t−t±)2dt

= 12

√π√−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 relate y with p(x,E). We startfrom

p(x,E) =√

2m(E − V (x)) ≈√

−2mV ′(x0)(x− x0) =√2mV ′(x0)(x0 − x)

and∫ x

x0

p(x′, E)dx′ ≈√

2mV ′(x0)

∫ x

x0

√(x0 − x′)dx′ = −2

3

√2mV ′(x0)(x0−x)3/2.

Therefore,

2

3y3/2 =

2

3~√

2mV ′(x0)(x0 − x)3/2 = −1

~

∫ x

x0

p(x′, E)dx′

and also

y1/2 =p(x,E)

[2m~V ′(x0)]1/3.

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

Ai(−y) ≈ c√p(x,E)

cos

(1

~

∫ x

x0

p(x′, E)dx′ + π/4

). (42)

for x < x0, where c =√π[2mV ′(x0)~]1/3 and we used that cos (−θ) = cos (θ).

Similarly, for x > x0,

q(x,E) ≈√2mV ′(x0)(x− x0),∫ x

x0

q(x′, E)dx′ ≈ 2

3

√2mV ′(x0)(x− x0)

3/2,

13

2

3(−y)3/2 ≈ 1

~

∫ x

x0

q(x′, E)dx′

and

(−y)1/2 = q(x,E)

[2m~V ′(x0)]1/3.

This gives

Ai(−y) ≈ c√q(x,E)

e− 1

~∫ xx0

q(x′,E)dx′. (43)

These results are summarized in equations (27) and (29) at the beginningof this section.

14

5 Solution near a turning point with V ′ < 0

This is very similar to what we did in the previous section, but with someimportant sign changes. The time-independent Schroedinger’s equation isthe same

p2

2m|φ〉+ V ′(x0)x|φ〉 = V ′(x0)x0|φ〉 (44)

and the solution in the p representation is now written as

φ(p) = A exp (−iαp3 − ipx0/~) (45)

with

α = − 1

6m~V ′(x0)=

1

6m~|V ′(x0)|. (46)

Following the same steps of the last section we can write φ(x) as

φ(x) = c

∫ +∞

0

cos (t3/3− ty)dt ≡ cAi(−y) (47)

where we defined t = p(3α)1/3 and

y =

(2m|V ′(x0)|

~2

)1/3

(x− x0). (48)

The asymptotic form of this expression for large y is identical to the ones wederived in section 4: for x > x0, in the classically allowed region, we get

Ai(−y) ≈√

π√ycos

(2

3y3/2 − π/4

). (49)

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

Ai(−y) ≈ 1

2

√π√−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),∫ x

x0

p(x′, E)dx′ ≈ 2

3

√2m|V ′(x0)|(x− x0)

3/2,

15

2

3y3/2 ≈ 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

φ(x) ≈

c√

p(x,E)cos

(1~

∫ x

x0p(x′, E)dx′ − π/4

)x > x0

c

2√

q(x,E)e

1~∫ xx0

q(x′,E)dx′x < x0.

(51)

16

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:

φ(x) =

D1

2√qe

1~∫ x q(x′,E)dx′

x ∈ I

C√pcos

(1

~

∫ x

p(x′, E)dx′ + ξ

)x ∈ II

D2

2√qe−

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,

φ(x) ≈

c√

p(x,E)cos

(1~

∫ x

x0p(x′, E)dx′ + π/4

)x < x0

c

2√

q(x,E)e− 1

~∫ xx0

q(x′,E)dx′x > x0.

we find that D2 = C, the lower limit of the integrals should be fixed as Band ξ = π/4.

On the other hand, regions I and II, connected by the turning point Awhere V ′(A) < 0, must match equations (51):

φ(x) ≈

c√

p(x,E)cos

(1~

∫ x

x0p(x′, E)dx′ − π/4

)x > x0

c

2√

q(x,E)e

1~∫ xx0

q(x′,E)dx′x < x0.

In this case we find that D1 = C, the lower limit of the integrals should befixed as A and ξ = −π/4.

17

Putting all this information together we may write

φ(x) =

C ′

2√qe

1~∫ xA q(x′,E)dx′

x ∈ I

C ′√pcos

(1

~

∫ x

A

p(x′, E)dx′ − π

4

)=

C√pcos

(1

~

∫ x

B

p(x′, E)dx′ +π

4

)x ∈ II

C

2√qe−

1~∫ xB q(x′,E)dx′

x ∈ III

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

~

∫ x

A

p(x′, E)dx′ − π

4=

1

~

∫ B

A

p(x′, E)dx′ − π

2+

1

~

∫ x

B

p(x′, E)dx′ +π

4.

We now impose that

1

~

∫ B

A

p(x′, E)dx′ − π

2= nπ (52)

so that the arguments of the two cosine become related by

cos

(1

~

∫ x

A

p(x′, E)dx′ − π

4

)= (−1)n cos

(1

~

∫ x

B

p(x′, E)dx′ +π

4

).

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

φ(x) = C

(−1)n

2√qe

1~∫ xA q(x′,E)dx′

x ∈ I

1√pcos

(1

~

∫ x

B

p(x′, E)dx′ +π

4

)x ∈ II

1

2√qe−

1~∫ xB q(x′,E)dx′

x ∈ III

(53)

18

where C is related to the normalization of φ(x). This global matching ispossible only if the quantization condition (52) is satisfied. This can bere-writen as ∫ B

A

p(x,E)dx =

(n+

1

2

)π~ (54)

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

19

7 TheWKB 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 with E smallerthan the height of the barrier. The general form of the WKB wave-functionfor this setting is

φ(x) =

A√pe

i~∫ x p(x′,E)dx′

+R√pe−

i~∫ x p(x′,E)dx′

x ∈ I

C√qe

1~∫ x q(x′,E)dx′

+D√qe−

1~∫ x q(x′,E)dx′

x ∈ II

T√pe

i~∫ x p(x′,E)dx′

x ∈ III

. (55)

Notice that the forbidden region is now region II, which is finite. There-fore the exponentials with both + and − signs 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.

20

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|φ〉+ V ′(x0)x|φ〉 = V ′(x0)x0|φ〉. (56)

we find

− ~2

2m

∂2φ

∂x2+ V ′(x0)(x− x0)φ(x) = 0. (57)

In terms of

y =

(2mV ′(x0)

~2

)1/3

(x0 − x) (58)

we obtain∂2φ

∂y2+ yφ = 0 (59)

which is the Airy equation. Let us check that

φ(x) =

∫ +∞

0

cos (t3/3− ty)dt = Ai(−y) (60)

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

φ′′

= −∫ +∞0

t2 cos (t3/3− ty)dt

= −∫ +∞0

{∂∂t[sin (t3/3− ty)]− y cos (t3/3− ty)

}dt

= sin (t3/3− ty)|∞0 − yφ = −yφ.

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

21

8.2 The function Bi

The Airy equation is a second order equation and must have two independentsolutions. Looking at Ai(−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

φ(x) =

∫ +∞

0

[e−t3/3−yt + sin (t3/3− ty)

]dt ≡ Bi(−y). (62)

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

φ′′

=∫ +∞0

[t2e−t3/3−yt − t2 sin (t3/3− ty)

]dt

=∫ +∞0

{− ∂

∂t[e−t3/3−yt] + ∂

∂t[cos (t3/3− ty)]− ye−t3/3−yt − y sin (t3/3− ty)

}dt

= −e−t3/3−yt|∞0 + cos (t3/3− ty)|∞0 − yφ = −yφ.(63)

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

8.3 Asymptotic form of Bi

Far 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,

φ(x) ≈

c√

p(x,E)cos

(1~

∫ x

x0p(x′, E)dx′ − π/4

)x < x0

c√q(x,E)

e1~∫ xx0

q(x′,E)dx′x > x0

(64)

and, for V ′(x0) < 0,

φ(x) ≈

c√

p(x,E)cos

(1~

∫ x

x0p(x′, E)dx′ + π/4

)x > x0

c√q(x,E)

e− 1

~∫ xx0

q(x′,E)dx′x < x0.

(65)

22

Notice that both forms diverge as x→ ±∞ in the forbidden regions. This iswhy the function Bi 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 Bi has to be included.

23

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:

φ(x) =

A√pe

i~∫ x p(x′,E)dx′

+R√pe−

i~∫ x p(x′,E)dx′

x ∈ I

C√qe

1~∫ x q(x′,E)dx′

+D√qe−

1~∫ x q(x′,E)dx′

x ∈ II

T√pe

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 point B, where V ′(B) < 0 we can writethe solution as a combination of Ai and Bi as

φ(x) ≈ FAi(−y) +GBi(−y).

In the allowed region III these functions have asymptotic forms given byequations (51) and (65):

φ(x) ≈ F√p(x,E)

cos(1~

∫ x

Bp(x′, E)dx′ − π/4

)+ G√

p(x,E)cos

(1~

∫ x

Bp(x′, E)dx′ + π/4

)= 1√

p(x,E)

[F cos

(1~

∫ x

Bp(x′, E)dx′ − π/4

)+G sin

(1~

∫ x

Bp(x′, E)dx′ − π/4

)]The desired form for region III can be recovered with the choice G = iF . Weget

φ(x) ≈ F√p(x,E)

exp

(i

~

∫ x

x0

p(x′, E)dx′ − π/4

)which gives our first coefficient T = Fe−iπ/4.

This same solution, with G = iF , must work in region II, where

φ(x) ≈ F [Ai(−y) + iBi(−y)]

≈ F

[1

2√

q(x,E)e

1~∫ xB q(x′,E)dx′

+ i√q(x,E)

e−1~∫ xB q(x′,E)dx′

].

24

Therefore, fixing the lower limit of integration in region II as x0 = B, thecoefficients must be C = F/2 = T/2eiπ/4 and D = iTeiπ/4.

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

φ(x) ≈ fAi(−y) + gBi(−y).

In region II this becomes

φ(x) ≈ f

2√

q(x,E)e−

1~∫ xA q(x′,E)dx′

+ g√q(x,E)

e1~∫ xA q(x′,E)dx′

= 1√q(x,E)

[fe−∆/~

2e−

1~∫ xB q(x′,E)dx′

+ ge∆/~e1~∫ xB q(x′,E)dx′

]where we have defined

∆ =

∫ B

A

q(x,E)dx.

Comparing with the form in region II we see that

C = fe−∆/~

2≡ T

2eiπ/4

D = ge∆/~ ≡ iTeiπ/4

orf = Te∆/~+iπ

4

g = iTe−∆/~+iπ4

Finally, we look at region I to determine A and R as a function of T . Inthis region we have

φ(x) ≈ fAi(−y) + gBi(−y)

≈ f√p(x,E)

cos(1~

∫ x

Ap(x′, E)dx′ + π/4

)+ g√

p(x,E)cos

(1~

∫ x

Ap(x′, E)dx′ − π/4

).

25

Writing the cosines as exponentials and comparing with the general form inregion I we see that

A = feiπ/4+ge−iπ/4

2

= Teiπ/2+∆/~+iTe−∆/~

2

= iT cosh (∆/~)

andR = fe−iπ/4+geiπ/4

2

= Te∆/~+iTe−∆/~+iπ/2

2

= T sinh (∆/~).

The final result is

φ(x) = T

i cosh (∆/~)√p

ei~∫ xA p(x′,E)dx′

+sinh (∆/~)

√p

e−i~∫ xA p(x′,E)dx′

x ∈ I

i1/2

2√qe

1~∫ xB q(x′,E)dx′

+i3/2√qe−

1~∫ xB q(x′,E)dx′

x ∈ II

1√pe

i~∫ xB p(x′,E)dx′

x ∈ III

.

(66)The transmission and reflection coefficients are

t =1

cosh2 (∆/~)≈ 4e−2∆/~ (67)

andr = tanh2 (∆/~) ≈ 1− 4e−2∆/~. (68)

26