WKB Method and Some Applications

Gilad Amar - Student No. 319288Supervisor - Alan S. Cornell

October 31, 2012

Abstract

In quantum mechanics the wave function gives a complete descriptionof a system through time. This function can be described by the famousSchrodinger Equation. The second order differential equation poses achallenge to be solved for various different potentials. In this project,the aim is to investigate the WKB method, an approximation methodto solve the time-independent Schrodinger Equation, and to employ it innumerous examples giving an insight into the physics involved in quantummechanics.

Contents

1 Introduction 21.1 Mathematical Introduction . . . . . . . . . . . . . . . . . . . . . 21.2 Validity of the WKB Method . . . . . . . . . . . . . . . . . . . . 5

2 Example 1: Tunneling 6

3 Matching Generalization 8

4 Example 2: Potential Well 11

5 Example 3: Alpha Decay 14

6 Example 4: Spherical Solution 18

7 Concluding Remarks 21

1

1 Introduction

The WKB Approximation is named after three physicists Gregor Wentzel, HansKramers and Lon Brillouin who discovered it in 1926. However the often over-looked mathematician Harold Jeffreys had already developed this method forsolving linear, second-order differential equations such as the Schrodinger Equa-tion(SE) in 1926. The important contribution from Wentzel, Kramers and Bril-louin was how to ‘patch’ two good solutions together at ‘turning points’ wherethe WKB solution is invalid, which will be explained in detail later.[1] Themethod provides ‘quasi classical’ means for obtaining an approximate solutionto the time-independent one-dimensional SE. The limitation of only workingwith one dimension is not as restrictive as one might expect, because in manyhigher dimensional cases there are symmetries in the potential which make theproblem effectively one dimensional, such as the radial variable in the sphericallysymmetric potential of the Hydrogen atom. The WKB approximation methodprovides a very close match to the actual wavefuntion which other approxima-tion tools such as perturbation theory, do not, but can only give an upper boundon the ground state energy.[2] This makes the WKB method useful for calcu-lating expectation values, bound state energies, tunneling probabilities and inthe case of alpha decay, the expected half-life.

I will begin this project with the mathematical introduction of the WKBmethod. How the 1st order terms are derived and a detailed description as towhere these terms are valid. In the following example, I will tackle the commonone-dimensional potential barrier with a stepwise potential. This example willbe used as a basis for the next section titled ‘Matching Generalization’. Thissection details the specifics of how to patch over a region in which the WKBterms are invalid for all the remaining examples. All that remains are a furtherthree problems: The Potential Well, Gamow’s model of Alpha Decay and myown solution for a specified Spherical Potential.

1.1 Mathematical Introduction

The way the SE is solved in the WKB method is by assuming that the answeris of an infinite converging series. The one-dimensional SE is:

− h2

2mψ′′(x) + [V (x)− E]ψ(x) = 0. (1)

The ansatz is:eiS(x)h (2)

Where

S(x) = S0(x) +h

iS1(x) +

(h

i

)2

S2(x) +

(h

i

)3

S3(x)... (3)

To justify why this is a reasonable guess consider a case where the potential isa constant V . [3] Restricting ourselves for the moment to where kinetic energy

2

is positive.−h2

2mψ′′(x) + [V − E]ψ(x) = 0 (4)

where E > V We have the simple solution where p =√

2m(E − V )/h:

ψ = A± e±ipx (5)

The general solution to the differential equation would be a linear combinationof the two independent solutions. ψ = A+ e

ipx corresponds with a wave movingto the right and ψ = A− e

−ipx with a wave moving to the left. Here the symbolp is used as this is the momentum. These solutions are oscillatory functionswith wavelength λ = 2π

p and amplitude A in both the real and imaginary plane.The crucial point that gives the hint to our choice of ansatz is this: Supposethat V (x) is not a constant but varies only slightly with λ. It makes sensethat the solution is still largely sinusoidal except that both its wavelength andamplitude are now functions of x. In fact by comparing both the ansatz withthe solution for a constant V we see that S(x) now describes ψ as an infiniteconverging series. The many resulting terms add up to give an effective varyingamplitude and wavelength as one would expect.[4] Substituting the ansatz intothe SE one obtains, after canceling the common exponential factors:

[S′(x)]2 = 2m[E − V (x)] + hiS′′(x) (6)

and:

S′′(x) = S′0(x) +

(h

i

)S′1(x) +

(h

i

)2

S′2(x) +

(h

i

)3

S′3(x) + ... (7)

Now for

(S′(x))2

=

[S′0(x) +

(h

i

)S′1(x) +

(h

i

)2

S′2(x) +

(h

i

)3

S′3(x) + ...

]2

(S′(x))2 = S′0(x)

[S′0(x) + (

h

i)S′1(x) +

(h

i

)2

S′2(x) +

(h

i

)3

S′3(x) + ...

]

+h

iS′1(x)

[S′0(x) + (

h

i)S′1(x) +

(h

i

)2

S′2(x) +

(h

i

)3

S′3(x) + ...

]

+

(h

i

)2

S′2(x)

[S′0(x) + (

h

i)S′1(x) +

(h

i

)2

S′2(x) +

(h

i

)3

S′3(x) + ...

]

+

(h

i

)3

S′3(x)

[S′0(x) + (

h

i)S′1(x) +

(h

i

)2

S′2(x) +

(h

i

)3

S′3(x) + ...

]+ ... (8)

3

Substituting ε = hi for convenience and rearranging in powers of ε

(S′(x))2 = [S′0(x)]2

+ ε[2S′0(x)S′1(x)]

+ ε2[S′1(x)2 + 2S′0(x)S′2(x)]

+ ε3[2S′0(x)S′3(x) + 2S′1(x)S′2(x)]

+ ... (9)

We see that in order for the left hand side to be equal to the right hand sideof SE, each power of h, term must add to zero. This gives us the followingequations:

[S′0(x)]2 = −p(x)

2S′0(x)S′1(x) + S′′0 (x) = 0

[S′1(x)]2 + 2S′0(x)S′2(x) + S′′1 (x) = 0

2S′0(x)S′3(x) + 2S′1(x)S′2(x) + S′′2 (x) = 0

... (10)

That means each term of S(x) can be found by going down the equations andsolving individually for S0(x), then S1(x) ad infinitum. Of course only the firstfew terms are needed for good accuracy and the more terms one chooses towork with the more difficult it becomes to apply to solving the SE for differentpotentials and this approximation method is no longer practical. Solving forthe first four terms with respect to the potential we obtain: Using Q(x) =2m[E − V (x)] = p(x)2 for convenience

S0(x) = ±∫ x

x0

√Qdx′

S1(x) =−1

4lnQ

S2(x) = ∓ i8

∫ x

x0

[Q′′

Q32

−( 5

4 )Q′2

Q52

]dx′

S3(x) =1

16

[Q′′

Q2− 5

4

Q′2

Q3

](11)

Clearly the expressions for higher order terms become more and more complex.The 1st order solutions to the SE by substituting for S0(x) = ±

∫ xx0dx′ p(x′)

are:

ψ+(x) =a+√p(x)

exp

[i

h

∫ x

x0

dx′ p(x′)

](12)

ψ−(x) =a−√p(x)

exp

[− ih

∫ x

x0

dx′ p(x′)

](13)

One thing which can be noticed right away is what happens when p(x) becomesequal to zero, at the points where E = V . This makes all the terms diverge and

4

we see the WKB method is invalid at those points. The divergence at classicalturning points (where a classical particle would stop and turn around becauseof the potential hill) where E = V is the reason why ‘good’ solutions fromeither side must be ‘patched’ over the WKB breakdown region. For some added

insight note that ψ2 ≈ |C|2

p(x) which shows that in regions where the momentum

is high the probability of the particle being found there is smaller than otherregions where the momentum is low. This has a simple physical interpretationas the odds of finding a particle is proportional to the time it spends there andtherefore inversely proportional to its velocity.

1.2 Validity of the WKB Method

Concerning the validity of WKB method’s solution we need to make sure thatthe magnitude higher order terms decreases rapidly, where the the 1st orderterm is most influential. [5] Expressed mathematically we need to know where:∣∣∣∣dS0

dx

∣∣∣∣� ∣∣∣∣hdS1

dx

∣∣∣∣ (14)

Some work needs to be done to check what conditions must be met for theabove equation to hold true. So referring to Eq. 11 from the mathematicalintroduction for h of order 1:

d

dx

dS0

dx= 2

dS0

dx

dS1

dx(15)

Using the previously determined result dS0

dx = ±√Q = ±p(x) and substituting

into the previous equation and rearranging for dS1 = 12dp(x)p(x) and putting this

in out inequality

|p(x)| �∣∣∣∣ 1

2p

dp

dx

∣∣∣∣ (16)

The de Broglie wavelength is λ = 2πp Therefore | λ4π

dλdp | � λ This means that

the change in λ is small over a distance of λ4π . This is true where the potential

varies slowly; or, stated differently, where the momentum of the particle is nearlyconstant over numerous wavelengths. As mentioned before, the WKB solutionsare invalid at points where the momentum is equal to zero. Approaching thesepoints as p(x) becomes small, the wavelength becomes large and the aboveinequality is no longer true. This validity concern is the motivating feature thatcompels us to ‘patch’ over the point in which V = E and the very close regionnearby.

5

2 Example 1: Tunneling

Here we shall consider the simple case of a particle with momentum p0 =√

2mEmoving from left to right in the following potential:

V (x) =

0 −∞ < x < x1 region IV (x) x1 ≤ x ≤ x2 region II

0 x2 < x < +∞ region III(17)

E is greater than V (x) everywhere except x1 < x < x2 where E < V . Fora finite potential step which is also a turning point, the WKB solutions canbe used right up to the discontinuity, provided that the potential varies slowlyas always. This means that the magnitudes and slopes of the WKB solutionscan be matched at the turning point. This will not be the case in the latterexamples as the WKB solutions break down near the turning point and so areinadmissible for use. In the next section we shall use this example to generalizewhat to do for more difficult potentials where there is a point at which E = V .

Classically the transmission coefficient giving a measure of the likelihood oftraversing the potential barrier would be zero. Lets investigate this with theWKB method. In both regions I and III we use the solution we used earlierfor where V = 0.

ψI(x) = Aeip0x/h +B e−ip0x/h (18)

ψIII(x) = E eip0x/h . (19)

In region III the same solution exists as in Region I, however there is only atransmitted, right traveling wave. Now in Region II, using the combination ofsolutions obtained earlier:

ψII(x) =C√p(x)

exp

(− 1

h

∫ x

x1

dx′ p(x′)

)+

D√p(x)

exp

(1

h

∫ x

x1

dx′ p(x′)

),

(20)where p(x) =

√2m(E − V (x)). Here we make D = 0 for physical reasons.

Should D 6= 0 the second term which is an increasing exponential will meet atx2 with the solution in region III so ψIII will have an amplitude greater than

in region I. When substituted into the Transmission coefficient T ≡ |ψtrans||ψinc| =

|E|2|A|2 . it would be greater than one. This would not make sense as the particles

motion is hindered not enhanced by the potential barrier. To solve for thecoefficients A, B, C and D, in order to determine the Transmission coefficientwe need to apply the boundary conditions at x1 and x2.There the wave functionmust be continuous, and so must its derivative. At x1 we have

∫ x1

x1dx′ p(x′) = 0

:

ψII(x1) =C√p(x)

(21)

ψI(x1) = Aeip0x1/h +B e−ip0x1/h (22)

6

ThereforeC√p(x)

= Aeip0x1/h +B e−ip0x1/h (23)

We require that the first derivatives must be continuous at x1:

ψ′I(x1) =ip0

h(Aeip0x1/h −B e−ip0x1/h) = ψ′II(x1) (24)

ψ′II(x1) =−C p(x)

h√p(x)

(25)

where p′(x1) ≈ 0. It follows:

ip0

h

(Aeip0x1/h −B e−ip0x1/h

)= − p(x)C

h√p(x)

(26)

Similarly for point x2 we obtain:

C√p(x2)

exp

(− 1

h

∫ x2

x1

dx′ p(x′)

)= E eip0x2/h (27)

p(x2)C

h√p(x2)

exp

(− 1

h

∫ x2

x1

dx′ p(x′)

)=ip0

hE eip0x2/h (28)

With the objective being to find EA . Using Eq. 23 times ip0

h plus Eq. 26 to getC in terms of A:

C =2A

1− a/(ip0)

√a eip0x1/h (29)

Now substituting this result into the third equation, and keeping things tidy wesay a = p(x1) and b = p(x2)

E

A=

2

1− a/(ip0)

√a

beip0(x1−x2)/h exp

(− 1

h

∫ x2

x1

dx′ p(x′)

)(30)

which allows to compute the transition coefficient

T =|E|2

|A|2=

4

b/a+ ab/p20

exp

(− 2

h

∫ x2

x1

dx′ p(x′)

). (31)

And so we can see the Transmission is hindered exponentially. As expected bythe analytic solution for a constant potential.

The wave functions are not completely solved. For this example, and all theremaining examples, the normalization constant has not been determined. Thiswould be done by using the wave function requirement that:∫ +∞

−∞dx |ψ(x)|2 = 1 (32)

7

3 Matching Generalization

It was seen that the 1st order WKB solution is

ψ±(x) =a±√p(x)

exp

[±ih

∫ x

x0

dx′ p(x′)

](33)

where p(x) is real. In a region where a particle would not be allowed classically,i.e. where p(x) is imaginary, the solutions will be exponential functions

ψ±(x) =a±√|q(x)|

exp

[±1

h

∫ x

x0

dx′ q(x′)

](34)

Where q(x) =√

2m[V (x)− E]. The i in the solutions ‘disappears’ since it is

multiplied from the i coming from the root of√p(x). Here we begin to see

a peep of the ‘quantum physics’, as the WKB method allows for non-classicalsolutions.

In the Tunneling example done previously there was no issue of the momen-tum going to zero as V → E and so the solution on either side of potential wallworks fine without the WKB solution breaking down. Now we shall look atwhat happens when the classical region (E > V ) joins the non-classical region(V > E). At E = V the WKB method breaks down. In this section we shallprovide a general method of how to patch over that discontinuity in the wavefunction at x = 0 for all further examples. Like for the tunneling example, thesolutions for either side of the wall are:

1√p(x)

[Be

ih

∫ 0

xdx′ p(x′)

+ Ce−ih

∫ 0

xdx′ p(x′)

]if x < 0

1√q(x)

De−ih

∫ x0dx′ q(x′)

if x > 0(35)

Except the potential V (x) is not a constant on either side of the turning pointand is not yet defined so the integral as the exponent of e remains. The potentialremains greater than E for all x > 0 and the constants B, C and D have yet tobe determined.

Patching regions x < 0 and x > 0 is done by directly solving the SE at thispoint. However we will not solve the actual SE, instead the SE is solved byexpanding around x = 0 where the turning point is, and assuming the changein the potential around that point to be very small. After solving for the wavefunction at these turning point locations the solutions’ asymptotic expressionscan be matched to the WKB solutions further away and thereby determine thecomplete wave function. Taking the term V (x) this can be expanded as a Taylorseries[6] at the point of the discontinuity x = 0 losing no generality should thepoint be at x1.

V (x) = V (0) + (x)V ′(0) +1

2!(x)2V ′′(0)... (36)

Locally, in the vicinity of turning points, we reasonably assume the variationin V (x) will be approximately linear up the point where the short wavelength

8

description is acceptable. So we ignore all second order and higher order terms.Therefore we have E − V (x) = −(x)V ′(0). Taking the appropriate change

of variables in order to solve this equation z = 2mh2 V ′(x)

13 (x) And taking the

moment to define α =[

2mh2 V ′(x)

] 13 for convenience. We obtain:

d2ψpdz2

− zψp = 0 (37)

The underscore ‘p’ refers to the patching region. The Airy Functions “Ai(z)”and “Bi(z)” represented in integral form are

Ai(z) =1

π

∫ ∞0

[cos

(s3

3+ sz

)]ds (38)

Bi(z) =1

π

∫ ∞0

[e−

s3

3 +sz + sin

(s3

3+ sz

)]ds (39)

These have the asymptotic forms of:

Ai(z) ∼ 1

2√πz

14

e−23 z

32 (40)

Bi(z) ∼ 1√πz

14

e23 z

32 (41)

for z � 0 and;

Ai(z) ∼ 1√π(−z) 1

4

sin

(2

3(−z) 3

2 +π

4

)(42)

Bi(z) ∼ 1√π(−z) 1

4

cos

(2

3(−z) 3

2 +π

4

)(43)

for z � 0.The analytic solution in the region where the potential is approximately

linear is ψp(x) = aAi(x) + bBi(x) where a and b are constants. The asymp-totic expressions are of little use unless we can connect the oscillating WKBsolution on the one side to the exponential WKB solution on the other sideof the classical turning point where the potential can be modeled as linear.[7]Finding the integral of

∫ xx1

[p′(x)]dx which is part of the WKB solution onthe right hand side can be done by substituting in for the potential with α inp(x) =

√2m[E − V (x)] this gives us:

p(x) = hα32

√−x (44)∫ x

0

[p(x′)]dx′ = hα32

∫ x

0

√x′dx′ =

2

3h(αx)

32 (45)

Giving us the WKB solution as:

ψ(x) =D√

hα34x

14

e−23 (αx)

32 (46)

9

This needs to be compared to the asymptotic forms of Airy’s functions at largez which is

ψ(x) =a

2√π(αx)

14

e−23 (αx)

32 +

b√π(αx)

14

e23 (αx)

32 (47)

Comparing the two solutions we see that they will match if b = 0 and a =√

4παhD

But now the comparison must be made on the left hand side of x = 0. Here the

integral∫ 0

xp′(x)dx′ = 2

3 h(−αx)32 . The resulting WKB wave function at the

patching region is:

ψ(x) =1√

hα34 (−x)

14

[Bei

23 (−αx)

32 + Ce−i

23 (−αx)

32

](48)

Comparing this again to the patching Airy’s solution for negative large Z

ψp(x) =a

√π(−αx)

14

sin

(2

3(−αx)

32 +

π

4

)(49)

Which can be rewritten as

ψp(x) =a

√π(−αx)

14

1

2i

[eiπ4 ei

23 (−αx)

32 − e−iπ4 e−i 2

3 (−αx)32

](50)

Comparing these with our WKB solution on the left hand side we find

a

2i√πeiπ4 =

B√hα

(51)

and−a

2i√πe−i

π4 =

C√hα

(52)

that for the constants B and C:

B = −ieiπ4 D (53)

C = ie−iπ4 D. (54)

Substituting for both B,C and A into the WKB solution of the left hand sidewe find

ψI(x) =2D√p(x)

sin

(1

h

∫ x2

x

p′(x)dx′ +π

4

)(55)

⇒ ψI(x) =2D√p(x)

cos

(1

h

∫ x2

x

[p′(x)]dx′ − π

4

)(56)

In general if we have a turning point x1 for a barrier on the right with

E > V (x) for x < x1

E < V (x) for x > x1 . (57)

10

The matching relations at the turning point is given by:

2√p(x)

cos

[1

h

∫ x1

x

dx′ p(x′)− π

4

]↔ 1√

q(x)exp

[− 1

h

∫ x

x1

dx′ q(x′)

](58)

1√p(x)

sin

[1

h

∫ x1

x

dx′ p(x′)− π

4

]↔ − 1√

q(x)exp

[1

h

∫ x

x1

dx′ q(x′)

].

In the same fashion we can find the general connection formula for a barrier onthe left:

1√q(x)

exp

[− 1

h

∫ x1

x

dx′ q(x′)

]↔ 2√

p(x)cos

[1

h

∫ x

x1

dx′ p(x′)− π

4

](59)

1√q(x)

exp

[1

h

∫ x1

x

dx′ q(x′)

]↔ − 1√

p(x)sin

[1

h

∫ x

x1

dx′ p(x′)− π

4

].

In practice, it is of little importance to know the particular form of the wavefunction in the small region of the turning point, provided the WKB solutionsfor either side are joined properly. In fact it is no longer necessary to refer tothe Airys functions anymore now that matching has been done.

4 Example 2: Potential Well

Using the results of the WKB patch generalization we shall tackle the potentialwell which has two classical turning points at x1 and x2. We may now considerthree regions defined by the intervals (−∞, x1)(x1, x2)(x2,∞) as regions I, IIand III respectively. In Region I we have the simple, exponential solution of

ψI(x) =C√q(x)

e− 1h

∫ x1

xq(x′)dx′

(60)

Picking the solution that goes to zero as x→∞ so that ψ(−∞) = 0 This is di-rectly matched to region II by the use of the generalization relations. Thereforethe wave function in region II is:

ψII(x) =2C√p(x)

cos

(1

h

∫ x

x1

p(x′)dx′ − π

4

)(61)

With no hard work as this has been done already. However, matching regionII to region III is slightly more difficult. Looking at the equation we havethe integral in the argument of cosine of

∫ xx1p(x′)dx′ this is the wrong way

round in comparison to out matching relations. If one were to try correctingthis by using cos(θ) = cos(−θ) then the constant in the argument of cosinewould be π

2 ; again not being the correct sign to use in out matching relations.This difficulty is overcome using the cosine addition formula cos(A + B) =

11

cos(A) cos(B)−sin(A) sin(B) in such a way that the correct integral is preservedwith the correct accompanying constant. We make the substitution:

ξ =1

h

∫ x2

x1

p(x′)dx′ − π

2. (62)

Then it follows by splitting the integral:

1

h

∫ x

x1

p(x′)dx′ = ξ +π

2−∫ x2

x

p(x′)dx′ (63)

Substituting this result into the wave function

ψII(x) =2C√p(x)

cos

(1

h

∫ x2

x

p(x′)dx′ − π

4− ξ)

(64)

And using the addition formula mentioned

ψII(x) =2C√p(x)

[cos(ξ) cos

(1

h

∫ x2

x

p(x′)dx′ − π

4

)+ sin(ξ) sin

(1

h

∫ x2

x

p(x′)dx′ − π

4

)](65)

Using the matching relations we can ‘convert’ this into the equation for regionIII

ψIII(x) =C√p(x)

[cos(ξ)e

− 1h

∫ xx2q(x′)dx′

− 2 sin(ξ)e1h

∫ xx2q(x′)dx′

](66)

At this point it might seem acceptable to have both terms in the solution for anyvalue of ξ. However an important subtlety is that we require the wave funtion togo to zero as x approaches infinity. This means that the second term, a growingexponential function, is not allowed. This imposes the limit that sin(ξ) = 0 orξ = nπ, where n is an integer. Rearranging for the integral and using h = h

2π :∫ x2

x1

p(x′)dx′ =(n+ 1

2 )h

2(67)

This result corresponds with the Bohr-Sommerfeld quantization condition wheren can be interpreted as the number of nodes of the wave function betweenthe classical turning points. This gives an insight into the physics, boundaryconditions naturally impose quantized energies. To be more specific, we applythe WKB method to a commonly known problem in undergraduate quantummechanics courses we consider; the harmonic oscillator where the potential isthe parabolic equation of V (x) = 1

2ωx2 To evaluate the quantization condition

we must know the turning points x1 and x2 where the momentum is zero.

p(x) =√

2mE −mωx2 = 0 (68)

x2 =2mE

mω(69)

12

x1 = −√

2mE

ω(70)

x2 =

√2mE

ω(71)

However noticing we are integrating an even function∫ x2

x1

p(x)dx = 2

∫ x2

0

p(x)dx =

∫ x2

0

√2mE −mωx2dx (72)

Using the solution for integrals[8]:∫ √a2 − u2du =

u

2

√a2 − u2 +

a2

2arcsin

(ua

)+ C (73)

In this case

a =

√2E

ω(74)

u = x (75)

Therefore∫ x2

x1

p(x′)dx′ = 2√mω

[x

2

√2E

ω− x2 +

E

ωarcsin

(x√ω√

2E

)+ C

]x2

0

(76)

Using the result for x2 we conclude with:

En =

(n+

1

2

)hυ (77)

where

υ =1

π√

mω

(78)

This semiclassical approximation gives us the same result for the possible en-ergies as the exact solution. In general we expect the WKB method to giveaccurate solutions several wavelengths from the turning point, or for very highquantum numbers ie. ‘near classical’. So the exact expression for the energylevels of the harmonic oscillator can, in a way, be viewed as something of an acci-dent considering that it even works exactly for the ground state energy (n = 0).

13

5 Example 3: Alpha Decay

In 1928 George Gamow used the idea of quantum tunneling to explain alphadecay by certain radioactive nuclei. The alpha particle, consisting of two pro-tons and two neutrons can be considered to be repelled by the remainder ofthe nucleus charge electrically once it gets far enough from the nucleus that itexperiences the Coulomb Potential[9] and not the Strong and Weak forces of thenucleus. The process can now be viewed as an alpha particle tunneling throughthe potential barrier. This is incidentally the first time that quantum mechanicshad been used to explain results in nuclear physics. The potential at far fromthe nucleus will be:

V (r) =2(Z − 2)e2

r. (79)

using basic electrostatics. Take note that we are working with a particle in threedimensions, however due to spherical symmetry the radial differential equationcan be separated from the angular components and so r is the effective onedimensional variable[10]. It must be said that we are also considering only thecase of zero orbital angular momentum (l = 0), otherwise there would be anangular momentum barrier in addition to the coulomb barrier. This will allbe shown explicitly in the next example. The potential can be divided intothree qualitatively different regions. Region I where V is a negative constant,representing the approximate finite square well of the attractive nuclear force.Region II, where E < V and tunneling occurs followed by region III wherethe potential is of the same Coulomb potential as region II but E > V whereclassically the alpha particle would be repelled from the nucleus.

V (x) =

−V0 E > V region I 0 < r < b

2(Z−2)e2

r E < V region II b < r < c2(Z−2)e2

r E > V region III c < r <∞(80)

‘b’ is the radius of the nucleus and shall represent the width of the potentialwell. Point ‘c’ is the classical turning point on the outside of the nucleus. Inregion I we have as the general WKB solution

ψI(r) =A√p0exp(

i

h

∫ r

0

p(r′)dr′ ) +B√p0exp(− i

h

∫ r

0

p(r′)dr′ ) (81)

Defining p0 =√

2mα(E − V0) and using Euler’s formula to express this in termsof sin and cos:

ψI(r) =A√p0

[cos(p0r

h

)+ i sin

(p0r

h

)]+

B√p0

[cos(p0r

h

)− i sin

(p0r

h

)](82)

As∫ r

0p(r′)dr′ =

∫ r0

√2mα(E − V0)dr′ = p0r We know that, due to the spher-

ical symmetry ψI(r) = ψI(−r).

ψI(−r) =A√p0

[cos(p0r

h

)− i sin

(p0r

h

)]+

B√p0

[cos(p0r

h

)+ i sin

(p0r

h

)](83)

14

This will be true should A = −B Therefore the wave function in region I reads:

ψI(r) =2iA√p0

sin(p0r

h

)(84)

Absorbing the 2i into the normalization constant

ψI(r) =A√p0

sin(p0r

h

)(85)

Again we encounter the same difficulty as in the potential well. Our wavefunction is not in a form conducive to using the matching relations. So weemploy the same ‘trick’ and use the addition formula. Rewriting the wavefunction as:

ψI(r) =A√p0

sin

{(p0(r − b)

h+π

4

)+

(p0b

h− π

4

)}(86)

Applying the trigonometric addition formula:

ψI(r) =A√p0

[− sin

(p0(b− r)

h− π

4

)+ cos

(p0b

h− π

4

)+ cos

(p0(b− r)

h− π

4

)+ sin

(p0b

h− π

4

)](87)

using sin(−θ) = − sin(θ). Applying the matching relations as mere routine atthis point, returning the wave function in region II:

ψII(r) =A√p

[cos

(p0b

h− π

4

)exp

(1

h

∫ r

b

dr′ p(r′)

)+

1

2sin

(p0b

h− π

4

)exp

(− 1

h

∫ r

b

dr′ p(r′)

)], (88)

The only item to be careful with is that nowp0 is a function of ‘r’.

p =√

2mα (V (r)− E) . (89)

This in turn is matched to the wave function in region III in the usual way:

ψIII(r) =A√p

[2 cos

(p0 b

h− π

4

)exp

(1

h

∫ c

b

dr p(r)

)cos

(1

h

∫ r

c

dr′ p(r′)− π

4

)− 1

2sin

(p0 b

h− π

4

)exp

(− 1

h

∫ c

b

dr p(r)

)sin

(1

h

∫ r

c

dr′ p(r′)− π

4

)],

(90)

Wherep =

√2mα (E − V (r)) , (91)

The minor subtlety to take care of; in region II the exponential parts exp(± 1h

∫ rbdr′ p(r′))

‘would’ extend into region III if not for the matching relations. In region IIIthese exponentials could be written as

exp(± 1

h

∫ c

b

dr′ p(r′)± 1

h

∫ r

c

dr′ p(r′)) , (92)

15

It is only the piece of the exponential extending from ‘c’ to ‘r’ which is trans-formed according to the matching relations. The piece extending from ‘b’ to ‘c’is a constant. ψII is the sum of two exponential functions; one increasing andthe other decreasing. While this may make consistent ‘mathematical’ sense itis lacking in physical sense. This is because the increasing exponential functionwill quickly come to dominate over the decreasing one for any non-zero coeffi-cient. This would mean that the sinusoidal wave in region III would have agreater amplitude than in region I and is more likely to be there.[11] Mean-ing we must impose a constraint of the system such that the coefficient of theincreasing term must be zero:

cos(p0b

h− π

4) , (93)

And so the metastable state condition is:

p0 =(4n+ 3)πh

4b, (94)

Giving the lowest (n = 0) state requiring that:

p0 =3πh

4b, (95)

As we established in the Tunneling example, the transmission is related to thebarrier potential in the following way:

T ∼= e−2γ , (96)

where

γ =1

h

∫ c

b

dx′ p(x′) , (97)

The variable ‘c’ can be determined because we know that at the turning pointV (c) = E

⇒ c =2e2(z − 2)

E(98)

Substituting for c into the integral we see:

γ =1

h

∫ c

b

dr√

2mE

√c

r− 1 , (99)

This can be easily solved using the trigonometric substitution r = c sin2(u) andthe approximation c� b and so c− b ≈ c; skipping to the answer:

γ =

√2mE

h[cπ

2− 2√bc] , (100)

At this point Gamow considered the alpha particle to be rattling inside theinside the nucleus with a velocity ν.[12] This would the the average time between

16

‘collisions’ with the potential barrier will be 2bν so the frequency of collisions is

ν2b . Hence the probability of the particle escaping per second is:

ν

2be−2γ , (101)

Using the definition of the lifetime of the particle as the inverse of the previousfunction:

τ =2b

νe2γ , (102)

Taking the natural logarithm of this obtain after some manipulation:

ln(τ) = ln

(2b

ν

)+

2e2π(z − 2)

h

√2mα

E− 8

h

√(z − 2)e2bmα , (103)

Finally we can see how well this theory correlates with the experimentally deter-mined half-lives. First we shall need to make a few approximations. The velocityν can be estimated by using the classical kinetic energy E = 1

2mαν2. The value

for b can be approximated by using the liquid drop model of the nucleus.[13]

This means b ' 1.2A13 .10−13cm In fact the entire logarithmic expression can

be further approximated by noticing that the ln(b√

2mαE

)is effectively con-

stant for the very small range of E (about 4 to 10MeV). The result is a linearrelationship, when plotting log(τ) against 1√

Ewhere:

log(τsec) ≈148√E− 54 (104)

The above graph is the graph of log(τ) against 1√E

for experimentally measured

values. We clearly see that the Gamow’s theory of alpha decay compares ex-tremely well for the isotopes of Uranium and Thorium. A very useful result ofemploying the WKB method.

17

6 Example 4: Spherical Solution

To find a solution for SE in a spherical potential it makes sense to adopt sphericalco-ordinates to exploit the symmetry. The SE for multiple dimensions is:

h2

2m∇2ψ + [E − V ]ψ = 0 , (105)

The Laplacian in spherical co-ordinates is:

∇2 =1

r2

∂

∂r

(r2 ∂

∂r

)+

1

r2 sin(θ)

∂

∂θ

(sin(θ)

∂

∂θ

)+

1

sin2(θ)

(∂2

∂φ2

), (106)

Where θ is the polar angle and φ, the azimuthal angle. In spherical coordinatesthe SE reads as:

− h2

2m

[1

r2

∂

∂r

(r2 ∂ψ

∂r

)+

1

r2 sin(θ)

∂ψ

∂θ

(sin(θ)

∂

∂θ

)+

1

sin2(θ)

(∂2ψ

∂φ2

)]+V ψ = Eψ ,

(107)In fact this can be solved by using the procedure known as Separation of Vari-ables. [electrodynamics textbook]Following this procedure we assume the solu-tion for ψ(r, θ, φ) is separable into products of functions such that ψ(r, θ, φ) =R(r)Y (θ, φ) substituting this the SE equation we obtain:

− h2

2m

[Y

r2

∂R

∂r

(r2 ∂R

∂r

)+

R

r2 sin(θ)

∂Y

∂θ

(sin(θ)

∂Y

∂θ

)+

R

sin2(θ)

(∂2Y

∂φ2

)]+V RY = ERY ,

(108)

Multiplying by −2mr2

RY h2 :{1

R

∂

∂r

(r2 ∂R

∂r

)− 2mr2

h2 [V (r)− E]

}+

1

Y

{1

sin(θ)

∂Y

∂θ

(sin(θ)

∂Y

∂θ

)+

1

sin2(θ)

∂2Y

∂φ2

}= 0 ,

(109)The first term is dependant on only the variable r. The second dependanton only θ and φ. Therefore each term must be constant for the equation toalways be true. Using the ”separation constant” as l(l+ 1). The reason for theconstant to be exactly that is only due to foresight of the solutions and the mostconvenient way of expressing them.{

1

R

∂

∂r

(r2 ∂R

∂r

)− 2mr2

h2 [V (r)− E]

}= l(l + 1)

1

Y

{1

sin(θ)

∂

∂θ

(sin(θ)

∂Y

∂θ

)+

1

sin2(θ)

∂2Y

∂φ2

}= −l(l + 1) , (110)

The interest of the report is to investigate the WKB method. To do this werequire that the wave function is effectively one dimensional, in particular forspherical symmetry that the wave function be dependant solely on r. Since r is

18

the variable of interest I will not consider the whole argument to solve for thegeneral solution of Y (θ, φ), but will instead quote the general solution as:

Y ml (θ, φ) = (−1)m

√(2l + 1)(l − |m|)!

4π(l + |m|)!eimφPml (cos θ) , (111)

In the derivation to the solution it is discovered that l, the azimuthal quantumnumber, turns out to be only integer values (l = 0, 1, 2...) and the magneticquantum number m = −l,−l + 1...l − 1, l. In addition Pml (cos θ) is the Leg-endre polynomial as one would expect in spherical symmetry, with cos θ as itsargument. To suit our requirement that ψ be entirely independent on of bothθ and φ we must work with the quantum numbers l and m, equal to zero. Thismakes out P 0

0 = 1 and so:

Y 00 (θ, φ) =

√1

4π, (112)

The entire wave function, as a result, is:

ψ(r, θ, φ) = R(r) , (113)

After having absorbed the the constant√

14π into R(r). Now to solve for the ra-

dial part of the equation using the WKB method. The radial equation simplifies

if we use the change of variable χ(r) = rR(r). [14] Then R = χr , dR

dr =r dχdr−χr2

and ddr = r2(dRdr ) = r2 d

2χdr2 , with some manipulation gives us:

d2χ

dr2+

2m

h2

[E − V (r)− h2l(l + 1)

2mr2

]χ = 0 , (114)

As we said, l must equal zero. Of course if we were satisfied with finding onlythe radial part of the wave function, by using the WKB method, we could relaxthe condition thatY ml be equal to a constant. But the result would not bethe wave function, only its one product R(r) from the separation of variables.In that case, the equation looks just like the one dimensional case, except the

potential, V (x) is replaced by the effective potential. Veff (r) = V (r) + h2l(l+1)2mr2 ,

where the additional term is known as the ‘centrifugal barrier term’, alluded toin the Alpha Decay example. In fact we can see that Gamow’s explanation ofAlpha Decay is merely the solution to this SE for the specified Coulomb andNuclear Force potential.

To make things interesting, the solution that we shall consider is for a po-tential:

V (r) = K|r| , (115)

an ‘absolute’ potential wall, where the constant K is positive and of the appro-priate dimensions. Determining the wave function in Region I, where E > Vfollows exactly the same as for Gamow’s Alpha Decay except the integral,

19

∫ r0p(r′)dr′ 6= p0r where p0 =

√2m(E −K|r|). Therefore in region I, we

have:

ψI(r) =A√p(r)

sin

(1

h

∫ r

0

p(r′)dr′), (116)

Requiring the same ‘trick’ as in Alpha Decay to make this function amenableto matching to the wave function in region II. Rewriting ψI :

ψI(r) =A√p(r)

[− sin

(1

h

∫ a

r

p(r′)dr′ − π

4

)cos

(1

h

∫ a

0

p(r′)dr′ − π

4

)+ cos

(1

h

∫ a

r

p(r′)dr′ − π

4

)sin

(1

h

∫ a

0

p(r′)dr′ − π

4

)] , (117)

Which carries to region II like so:

ψII(r) =A√q(r)

[cos

(1

h

∫ a

0

p(r′)dr′ − π

4

)exp

(1

h

∫ r

a

q(r′)dr′)

+1

2sin

(1

h

∫ a

0

p(r′)dr′ − π

4

)exp

(− 1

h

∫ r

a

q(r′)dr′)

] , (118)

We require the first, growing exponential, term to have a coefficient of zero inorder for the wave function to make sense.

cos

(1

h

∫ a

0

p(r′)dr′ − π

4

)= 0 , (119)

1

h

∫ a

0

p(r′)dr′ − π

4=π

2+ nπ , (120)

Where n is an integer. This leads to the requirement:∫ a

−ap(r′)dr′ = h

(n+

3

4

), (121)

This differs from the Bohr-Sommerfeld quantization conditions we achieved ear-lier: ∫ a

−ap(r′)dr′ =

h

2

(n+

1

2

), (122)

The relation between the n0, (subscript 0 to indicate it comes from the require-ment that the coefficient be equal to zero) and nB , of the Bohr-Sommerfeldquantization condition , must then be n0 = 2nB + 1. I am not sure of thedifference and why there is one.

As we expect, due to boundary conditions there will be quantized energies.Utilizing the quantization condition:∫ a

−ap(r′)dr′ =

h

2

(n+

1

2

)=

∫ a

−a

√2m

h2

√E −K|r|dr′ , (123)

20

Using the fact that the integral is of an even function, and using the substitutionof u = E −K|x|, so du = Kdx, we get:

4√

2mE3

3K=n+ 1

2h

2, (124)

Therefore:

En = m−13

[3K(n+ 1

2 )h

8√

2

] 23

, (125)

As an interesting aside, should we instead have a potential:

V (r) =

K|r| |r| ≤ z region I2Kz −K|r|z z ≤ r ≤ 2z region II

0 2z < r < +∞ region III(126)

Where ’z’ is a positive constant. Essentially a triangular potential barrier. Wesee that if the particles energy is less than the maximum potential of Kz theenergy must be quantized. Higher energies however, have no such limitation.

7 Concluding Remarks

The WKB method, has been seen to provide neat and accurate solutions for arange of problems, particularly the potential barrier and the potential well whencompared to the analytically determined solutions. In fact, should one desire,the accuracy of the WKB method can be enhanced by using second or eventhird order terms. The WKB method works best for slowly varying potentials,but has the strict requirement that the problem be effectively one dimensional.This is quite a limitation to its usefulness, however by exploiting symmetries ofa problem the WKB method can actually be applied to more situations thanthis limitation might suggest. In my opinion, though the method can be appliedto many circumstances, they are all ‘qualitatively’ the same problem, requiringthem to all be reduced to a one-dimensional problem. The WKB method is notvery useful at all should the potential at hand not be convenient. Of course,this does not mean it is useless, only that it can, at best, only be one of thetools in the toolbox.

In conclusion, the WKB method has met the aims of this project, giving aninsight into the physics of quantum mechanics. It has been explicitly seen thatboundary conditions have to be imposed for the wave function to be solved ina physically meaningful way. As a consequence of these boundary conditions;quantized energies appear. In addition the WKB method was used to explainthe perplexing half-life of heavy unstable elements, as in Gamow’s alpha decay,giving a convincing argument that Quantum Mechanics theory greatly correlateswith the real world.

21

References

[1] http://www.tcm.phy.cam.ac.uk/ bds10/aqp/handout approx.pdf Date Ac-cessed 05-09-2011;

[2] Applications of Quantum Mechanics, by B J Cole, published by Universityof the Witwatersrand, 2002.;

[3] Griffiths, David J. Introduction to Quantum Mechanics (2nd Edition),Pearson, Reed College(2005);

[4] Dynamics at the Horsetooth Volume 2A, Focused Issue: ”Asymptotics andPerturbations”, 2010. Perturbation Theory and the WKB Method JaimeShinn Department of Mathematics Colorado State University Fall 2010;

[5] http://electron6.phys.utk.edu/QM1/modules/m7/wkb.htm Last Date Ac-cessed 05-09-2011;

[6] Abramowitz, Milton; Stegun, Irene A. (1970), Handbook of MathematicalFunctions with Formulas, Graphs, and Mathematical Tables, New York:Dover Publications, Ninth printing;

[7] Quantum Mechanics, Leonard I. Schiff, Mcgraw Hill, 1949;

[8] Calculus (6th Edition), James Stewart, BrooksCole Publishers, 2008 ;

[9] Introduction to Electrodynamics(3rd Edition), David J. Griffiths, PearsonPrentice Hall, 2005;

[10] Schaum’s Outline of theory and problems of Quantum Mechanics, Pelegand Pinni and Zaarur, Mcgraw Hill, 1998;

[11] Quantum Mechanics (Vol. 1), Albert Messiah, North-Holland PublishingCompany, 1961;

[12] Quantum Mechanics, Professor John W. Norbury, University of Wisconsin-Milwaukee, 2000;

[13] Concepts of Modern Physics(6th Edition), Arthur Beiser, Mcgraw Hill,2003;

[14] A Textbook of quantum mechanics, Piravonu Mathews Mathews, K.Venkatesan, Mcgraw Hill, 1979;

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