In the literature at least two barriers are studied, for which the third-order contribution to the phase-integral asymptotic approximation of the transmission coe¢ cient do not contribute

SEMICLASSICAL TUNNELING EFFECT

ABSTRACT

This work is a contribution to the theory of the quantum tunneling e¤ect. In the literature at least two barriers are studied, for which the third-order contribution to the phase-integral asymptotic approximation of the

transmission coe¢ cient do not contribute. These are the parabolic barrier and the inverse Morse barrier. In the present work we will show that with a proper choice of the so called base function there is at least one more barrier in this category namely the Eckart-Epstein potential. The fact that the third-order contribution vanishes is a good indication that we have found an optimal choice of the base function, and the treatment to …nd an optimal base function may be possible to generalize to other classes of potential barriers.

For particles of a low energy compared to the energy near the top of the barrier we obtain a very low transmission coe¢ cient, which means that the probability for tunneling to occur is very low. There exist some cases, for exam- ple that with a double barrier which is transparent, even for certain relatively low energies but no evidence for this kind of transparency for a single barrier has been found. The present work does not give any such evidence. At the same time there are still speculations on cold fusion like e¤ects, which would demand a higher probability for tunneling through for a single barrier.

This bachelor thesis is written by Johan Öhman 19830414-1537 Student of Uppsala University

CONTENTS

1. Introduction 3

1.1 Tunneling e¤ect 3

1.2 Cold fusion 4

1.3 The phase - integral asymptotic approximation and the potential 4

2. Theory of barrier penetration, higher and third order phase-integral con-

nection formula 6

3. Application to the symmetric Eckart Epstein potential 10

4. Conclusion 18

5. References 19

6. Appendix 20

Figure 1 Show a tunneling e¤ect through a potential barrier. The peak is the maximum energy for the potential barrier needed for a classical particle to pass the barrier. For the quantum case, the dashed line is approximately energy required for the particle to tunnel through the potential barrier. You can also see that the wave function amplitude will be di¤erent before and after the tunneling.

1. INTRODUCTION 1.1 TUNNELING EFFECT

The quantum tunneling e¤ect, or quantum barrier penetration, is an e¤ect where a particle can tunnel through a potential barrier. It is impossible for a classical physical particle to tunnel through a potential barrier. The classical particle must have enough kinetic energy to overcome the potential barrier.

In …g. 1 a particle tunnels through a potential barrier and the di¤erence between the classical and the quantum physical case is illustrated. For the quantum case we have a wave function generating a ‡ux (probability current density) which di¤ers for the incident and transmitted waves.

The transmission coe¢ cient is the probability of particle transmission and the re‡ection coe¢ cient is the probability of particle re‡ection. The relation between the transmission coe¢ cient and re‡ection coe¢ cient must be

T + R = 1 there 0 T 1 and 0 R 1,

because there is no decays, interaction with other particles and the probabilies must add to one. That the wave can be partially transmitted or re‡ected is

di¤erent from the situation for a classical particle. The transmission coe¢ cient is de…ned as the transmitted ‡ux divided with the incident ‡ux and the re‡ection coe¢ cient is the ratio of the re‡ected ‡ux and the incident ‡ux and this is a probability assuming the amount of ‡ux is big enough.

There exist two kinds of barriers, the superdense barrier and the underdense barrier depending on the energy of the particle see ref. 3. The …rst one is when the particle has an initial kinetic energy (far from the barrier) smaller than the height of the potential barrier and can be seen in the …g. 2 with the z - axis. For the second one the particle initially has more kinetic energy than the potential barrier indicated in the …gure by introducing a dashed z-axis. The superdense barrier for the present work is the most interesting. An example is the alleged cold fusion when the particle has much lower energy than the potential barrier.

One can …nd the quantum tunneling e¤ect in many real cases. Some of them are the nuclear decay, the semiconductors, thin insulator and fusion.

1.2 COLD FUSION

Fusion is a nuclear reaction between two particles which form a new nucleus.

The nuclear fusion exists for example in the stars as thermonuclear fusion. But what about fusion in practise? There exist research programs for fusion e . g.

ITER.

However, what is the di¤erence between the thermonuclear fusion and the alleged cold fusion? The di¤erence is in the temperature of the systems. For a thermonuclear reaction the temperature of the plasma is extremely high (mil- lions of degrees). The particles in the plasma have enough mean kinetic energy to overcome the potential barrier, which in reality is a coulomb barrier [ref. 6.]

When the particles have overcome the coulomb barrier the nuclear strong force will take over.

Cold fusion is fusion occuring at low temperatures, room temperature or little above, and small pressure. The mean kinetic energy for the particle is then too small to overcome the potential barrier and the only possibility for fusion is the tunneling e¤ect. Already in the end of the 1980´s there was a consensus among scientists that fusion through tunneling e¤ect is not realistic [ref. 7]. The probability for a tunneling e¤ect to occur is astronomically small (as an example, nuclear decay happens because the number of nucleons is very large).

1.3 THE PHASE - INTEGRAL ASYMPTOTIC APPROXIMATION AND THE CHOICE OF THE ENERGY POTENTIAL BARRIER

To calculate the transmission coe¢ cient for a tunneling e¤ect of the Coulomb barrier the phase - integral asymptotic approximation is a well-known alternative to solve the semi - classical problem [ref. 2]. The third order contribution from the phase – integral asymptotic approximation for any single barrier will be discussed in the present work.

Figure 2 Shows the function q²(z) which is an unspeci…ed function for a superdense barrier (solid z-axis) and underdense barrier (dashed z-axis). The t´ and t´´ are called the turning points and occur on the barrier walls of the potential.

An asymptotic series is divergent. However the asymptotic series is also semiconvergent which gives a possibility to use it. In fact many important results in theoretical physics come from asymptotic approximations. The …rst terms in an asymptotic series typically improve the accuracy, but after a certain order the contributions begin to increase and the accuracy deteriorates. To use an asymptotic series we must break the series before that order. The third order contribution to the phase – integral asymptotic approximation will typically work well and e. g. transform the JWKB method from an approximation method to something close to a precision method.

Di¤erent examples of potential barrier models are the parabolic, inverse Morse and symmetric Eckart –Epstein barrier, all of them found in ref. 1. For the present work we want to use the Eckart – Epstein barrier which has many interesting properties.

The phase - integral asymptotic approximation is a method where you use the asymptotic series in your ansatz of the unspeci…ed function q(z). The function q (z) is a part of the wave function and is also a part when you do the ansatz to the asymptotic series.

2. THEORY OF BARRIER PENETRATION

2.1 HIGHER - ORDER PHASE - INTEGRAL APPROXIMATIONS AND THE PLATFORM FUNCTION

In the present chapter the third-order correction, which comes from the third term in an asymptotic series, will be found in a systematic way from the one- dimensional Schrödinger equation. As a tool we shall use a related auxiliary di¤erential equation. In this way the transmission coe¢ cient will be found for the third order general single-hump potential barrier. The one-dimensional Schrödinger equation can be written as

d²

dz² + R(z) = 0:

(2.1) In quantum problems when we consider the Schrödinger equation

R(x) = ^2m_h2[E V (x]:

(2.2) In the present work we assume that R(x) can be continued from real values of x into complex values of z. We now introduce the auxiliary di¤erential equation

d² dz² +h

Q²(z)

2 + R(z) Q²(z)i = 0:

(2.3) It is easily seen that the latter di¤erential equation becomes the original Schrödinger equation when the (auxiliary) parameter is put equal to 1. The formal solution for this di¤erential equation can be written on the form

y (z) = q ¹⁼²(z) exp i Z z

q (z) dz ;

(2.4) where q(z) is the unspeci…ed function.

Substituting the latter expression into the auxiliary di¤erential equation gives

d²

dz²q ¹²(z) exp 2 4 i

Zz

q (z) dz 3 5

+h

Q²(z)

2 + R(z) Q²(z)i

q ¹²(z) exp 2 4 i

Zz

q (z) dz 3 5 = 0,

() ³4q ⁵²(z) ^dq_dz ^{2 1}₂q ³²(z)^d_dz^2q2 q³²(z)

+ hQ²(z)

2 + R(z) Q²(z) i

q ¹²(z) = 0

() q²(z) q¹²(z)_dz^d22q ¹²(z) + Q²(z) 1 ¹2 R(z) = 0;

(2.5) or

q(z) Q(z)

2 ₂

Q²(z)(q(z) )¹² _dz^d22(q(z) ) ¹² + ^{2 Q2(z) R(z)}_Q2(z) = 1:

(2.6) Introducing a new complex variable which can be written

= Z z

Q (z) dz, from which ^d_dz = Q(z),

(2.7) one can …nd out that the relation can be expressed as

q(z) Q(z)

2 2 q(z)

Q(z)

1 2 d²

d ² q(z) Q(z)

1

2 + ^{2 Q2(z) R(z)}_Q2(z)

2Q ³²(z)_dz^d22Q ¹²(z) = 1:

(2.8) The calculation can be found in the appendix A. The last term on the left hand side can be rewritten

Q ³²(z)_dz^d22Q ¹²(z) = _Q(z)¹ _dz^d h1

2f (z)_dz^d _{f (z)Q(z)}¹ i

h1

2f (z)_dz^d _{f (z)Q(z)}¹ i2

+

1 2 d dz[^{f (z)}dz^d

1

f (z)]⁺[¹2f (z)_dz^d_{f (z)}¹ ]²

Q²(z) :

Here a new function f (z) appears, which can be called the form function.

From this last relation one can also see that one can write Q(z) = ¹₂f (z)_dz^d _{f (z)Q(z)}¹ ;

(2.9) and

Q

1(z) = ¹₂f (z)_dz^d _{f (z)}¹ :

(2.10) The function (z) is called the platform function, 1(z) being the platform function when we formally put Q²(z) = 1. Hence

Q ³⁼²(z)_dz^d22Q ¹⁼²(z) = _Q(z)¹ _dz^d Q

(z) Q2

(z) + ^12dzd

Q

1(z)+Q2 1(z) Q²(z)

= _d^d Q

(z) Q2

(z) + ^12dzd

Q

1(z)+Q₂

1(z) Q²(z) :

(2.11) The equation (2.8) can hence be written

q(z) Q(z)

2 2 q

Q(z)

1 2 d²

dz² q(z)

Q(z)

1 2 +^Q

2(z) R(z) ¹_2dz^d Q

1(z)+Q2 1(z) Q²(z)

2n

d d

Q(z) Q2

(z) o

= 1:

(2.12) Let us now make an Ansatz

q (z) = Q(z) X1 n=0

Y2n(z) ²ⁿ;

(2.13) Set the to unity and use the …rst two terms in equation (2.13). Introducing this Ansatz into the q relation (2.12), the …rst contributions become

Y₀(z) = 1 and Y2(z) = ^{R(z) Q}

2(z) ¹_2dz^d Q

1(z)+Q2 1(z)

2Q²(z) +_d^{d 1}₂Q

(z) ¹₂Q2

(z) X₂(z) +_d^{d 1}₂Q

(z) ¹₂Q2

(z):

(2.14)

The solution of the Auxilliary equation can be written on the form

(z) =h

Q(z) 1 + Y₂(z) ² i 1=2

exp i Z zh

Q(z) 1 + Y₂(z) ² i dz:

(2.15) We now put the auxiliary parameter equal to one. Thereafter the third- order formal solution can be written

(z) = [Q (z) (1 + Y₂(z))] ¹⁼²exp i Z z

[Q(z) (1 + Y₂(z))] dz :

(2.16) The transmission coe¢ cient for Eckart - Epstein potential is

T = _1+exp¹_f2Kg:

(2.17) For the derivation of the transmission coe¢ cient see ref. 8. In chapter 3 we want to calculate if the third order approximation gives a contribution to tunneling e¤ect with the symmetric Eckart-Epstein potential or not. There K is

K = K0+ K2= Z z

[Q(z) (1 + Y2(z))] dz;

(2.18) there K2is the third order contribution and Y2 is the third order correction.

Fig 3 An illustration of the Eckart - Epstein potential

3. APPLICATION TO THE SYMMETRIC ECKART EPSTEIN POTEN- TIAL

Let us …nd out what the third order contribution K2 from (2.17) is with the turning points t´ and t´´ if one has a symmetric Eckart –Epstein potential barrier, see Fig 3. The integral which we can see below means one half of a loop integral

K₂=

Z (t´ ´ ) (t´ )

Q (z) Y₂(z) dz = ¹₂ I

Q (z) Y₂(z) dz

When the numerator in the integrand is the same as zero, the turningpoints t´ and t´´ is de…ned after the value of z.

From ref. 5 the Eckart - Epstein potential can be found as V (z) = 4V0 exp(z=z0)

(exp(z=z0)+1)²;

(3.1) where z0 and V0are positive real constants.

The corresponding function R (z) is according to (2.8) R (z) = ^2m_h2 [E V (z)] = ^2m_h2

h

E 4V_0(exp(z=z^exp(z=z0)

0)+1)²

i :

(3.2)

The contour integrals will be done for the third order, Z (t´ ´ )

(t´ )

Q (z) Y₂dz

= ¹₂ Z (t´ ´ )

(t´ )

Q (z)Q2

(z) dz+

Z (t´ ´ ) (t´ )

Q (z) X2(z) dz+¹₂ Z (t´ ´ )

(t´ )

Q (z)_d^d Q (z) dz

(3.3) The last term can be rewritten using (2.14)

() Z (t´ ´ )

(t´ )

Q (z) Y2dz

= ¹₂ Z (t´ ´ )

(t´ )

Q (z)Q2

(z) dz + Z (t´ ´ )

(t´ )

Q (z) X₂(z) dz +¹₂ Z (t´ ´ )

(t´ ) d dz

Q(z) dz

= ¹₂ Z (t´ ´ )

(t´ )

Q (z)Q2

(z) dz + Z (t´ ´ )

(t´ )

Q (z) X₂(z) dz:

(3.4) In (3.4) we have made use of the fact that the integral of a total derivative of a single-valued function over a closed loop is zero.

Let us now …nd out what the (Q²= 1) platform functionQ

1(z) is when the form function guess is f (z) = exp (z=z0) + 1.

Q

1(z) = ¹₂f (z)_dz^df ¹(z) = ¹₂f (z) f ²(z)^{df (z)}_dz = ¹₂f ¹(z)^{df (z)}_dz

= ¹₂[exp (z=z0) + 1] ^{1 d}_dz[exp (z=z0) + 1] = _2z¹

0

exp(z=z0) exp(z=z0)+1:

(3.5) And for the term third and forth in (2.14) we have

d dz

Q

1(z) = _2z¹

0

d dz

exp(z=z0) exp(z=z0)+1 =_2z¹

0

n1 z0

exp(z=z0) exp(z=z0)+1

1 z0

exp(2z=z0) [exp(z=z0)+1]²

o

= _2z¹2 0

nexp(z=z0)[exp(z=z0)+1] exp(2z=z0) [exp(z=z0)+1]²

o

=_2z¹2 0

exp(z=z0) [exp(z=z0)+1]²;

(3.6) Q2

1(z) =n

1 2z0

exp(z=z0) exp(z=z0)+1

o2

= _4z¹2 0

exp(2z=z0) [exp(z=z0)+1]²:

(3.7) Thereafter with the explicit expression for R(z) (3.2) we see that X2(z) becomes

X2(z) = _2Q¹2(z)

h

R (z) Q²(z) _dz^d Q

1(z) +Q2 1(z)

i

= _2Q¹2(z)

h2m h²

h

E 4V0 exp(z=z0) (exp(z=z0)+1)²

i

Q²(z) +_2z¹2 0

exp(z=z0) [exp(z=z0)+1]²

i

+_2Q¹2(z)

h 1 4z²₀

exp(2z=z0) [exp(z=z0)+1]²

i

() X²(z) = _2Q¹2(z)

h2m h²

h

E 4V0 exp(z=z0) (exp(z=z0)+1)²

ii

+_2Q¹2(z)

h

Q²(z) +_4z¹2 0

exp(z=z0)

[exp(z=z0)+1]² +_4z¹2 0

exp(z=z0) [exp(z=z0)+1]

i :

(3.8) Now, we shall choose the simplest possible X₂(z). The …rst choice which comes to mind is X₂(z) = 0 . However for the present potential as we shall see is that choice a¤ects the boundary properties of the wave function in such a way that the boundary conditions at 1 will not be in strong form. This can be showed for Q²(z)

X₂(z) = 0 () Q²(z) = ^2m_h2

h

E 4V0 exp(z=z0) (exp(z=z0)+1)²

i +_4z¹2

0

exp(z=z0)

[exp(z=z0)+1]² +_4z¹2 0

exp(z=z0) [exp(z=z0)+1]

Let the z goes to in…nity and we …nd out that,

zlim!1Q²(z) = ^2mE_h2 +_4z¹2 0

The results show that this will not be in strong form because of the constant z₀which can be shosen arbitrarily. This means that we must choose di¤erently.

Choosing the Fröman - Thylwe - Yngve expression for Q²(z) in ref 1 which is

Q²(z) = ^2m_h2

h

E 4V₀ ^exp(z=z0)

(exp(z=z0)+1)²

i +_4z¹2

0

exp(z=z0) [exp(z=z0)+1]²;

(3.9) which gives

zlim!1Q²(z) = ^2mE_h2

Then X2(z) becomes X₂(z) = _8z2¹

0Q²(z)

exp(z=z0) [exp(z=z0)+1]:

(3.10) We see that the Fröman - Thylwe - Yngve expression for Q²(z) has the same properties at in…nity as the usual JWKB expression for Q²(z) whereas the choice corresponding to X₂(z) = 0 would constitute a "displacement of the potential"

at in…nity, and hence would not correspond to the boundary properties on strong form.

The expression for Q²(z) can be written Q²(z) = _z¹2

0

h 2 2 exp(z=z0) [exp(z=z0)+1]²

i

;

(3.11) where

2=^2m_h2Ez₀² and ²=^8mV_h2⁰z²₀ ¹₄. With these parameters

X2(z) = _2Q¹2(z)

h 1 4z₀²

exp(z=z0) [exp(z=z0)+1]

i

= _8z¹2 0

exp(z=z0) [exp(z=z0)+1]

n1 z₀²

h 2 2 exp(z=z0) [exp(z=z0)+1]²

io 1

= ¹_8[exp(z=z^exp(z=z0)

0)+1]

nh 2 2 exp(z=z0) [exp(z=z0)+1]²

io 1

:

(3.12) Let us now …nd out what theQ2

(z) is. The platform relation (2.13) was Q(z) = ¹₂f (z)_dz^df ¹(z) Q ¹(z)

For an easier calculation, use the variable which can be written as

w (z) = exp (z=z0) and ^dw_dz = ^w(z)_z

0 .

(3.13) The form factor f (z) and Q²(z) is here

f (z) = exp (z=z0) + 1 = w(z) + 1;

(3.14)

and Q²(z) = _z¹2

0

h 2 2 exp(z=z0) [exp(z=z0)+1]²

i

=_z¹2 0

h 2 2 w(z) [w(z)+1]²

i :

(3.15) Using the relation (3.12) one can write

d

dz = _dw^{d dw}_dz =^w(z)_z

0

d dw.

(3.16) Hence

Q(z) = ^z₂⁰[w (z) + 1]_dz^d [w (z) + 1] ¹

h 2 2 w(z) [w(z)+1]²

i 1=2

= ¹₂w (z) (w (z)+1)_dw^d [ ²(w (z) + 1)^{2 2}w (z)] ¹⁼²= w(z)(w(z)+1)[^{2 2(w(z)+1) 2}]

4[ ^{2(w(z)+1)2 2w(z)}]³⁼² : (3.17)

The platform function in square is Q2

(z) = ₁₆¹ [w (z) + 1]²w²(z) (^{2 2w(z)+2 2 2})² [ ^{2(w(z)+1)2 2w(z)}]³ :

(3.18) Return to the integral (3.4) with the third order integrand and do the cal- culation for respective terms. For the …rst third-order term with w(t´) = w´, w(t´´) = w´´

1 2

Z (t´ ´ ) (t´ )

Q (z)Q2

(z) dz

= ₃₂¹ Z (w´ ´ )

(w´ )

h 2 2 w(z) [w(z)+1]²

i1=2

(w + 1)²w (^{2 2w(z)+2 2 2})²

[ ^{2(w(z)+1)2 2w(z)}]³ dw;

(3.19) or

1 2

Z (t´ ´ ) (t´ )

Q (z)Q2

(z) dz

= ₃₂¹ Z (w´ ´ )

(w´ )

h 2(w (z) + 1)^{2 2}w (z) i1=2

[w (z) + 1] w (z) (2 ²w(z)+2 ^{2 2})²

[ ^{2(w(z)+1)2 2w(z)}]³ dw

= ₃₂¹ Z (w´ ´ )

(w´ )

[w (z) + 1] w (z) (^{2 2w(z)+2 2 2})²

[ ^{2w(z)2+(2 2 2)w(z)+ 2}]⁵⁼² dw

= ₃₂¹ Z (w´ ´ )

(w´ )

(1+1=w(z))(^{2 2+(2 2 2})^=w(z)]2 [ ^{2+(2 2 2)=w(z)+ 2=w(z)2}]⁵⁼²

dw w(z)

= ₈¹

Z (w´ ´ ) (w´ )

(1+1=w(z))[1+(1 ¹₂ ²)=w(z)]²

[^{1+(2 2})=w(z)+1=w(z)²]⁵⁼²

dw w(z);

(3.20) where

= ;

(3.21) Using the Cauchy residue theorem from complex analysis [ref. 4]

I

f (w) dw = 2 Z (w´ ´ )

(w´ )

f (w) dw = 2 i Xn j=1

res (wj) ;

(3.22) Z (w´ ´ )

(w´ )

f (w) dw = i Xn j=1

res (w_j) ;

(3.23) and the theorem for poles of order n.

res (w0) = lim

w !w⁰ 1 (n 1)!

d^{n 1}

dw^{n 1}[(w w0)ⁿf (w)] :

(3.24) The contour is given in …gure 4. It is not analytic in the turning points and they will be called for the branch points. The black area between the two branch points is called the branch cut.

We can not use the Cauchy Residue Theorem directly and to calculate this we de…ne a new function

u = _w(z)¹ and _dw^du = _w(z)¹ 2 = u²() dw = u¹²du:

(3.25)

Fig 4 Illustrate the contour in the complex w - plane.

The equation (3.20) is then

1 2

Z (t´ ´ ) (t´ )

Q (z)Q2

(z) dz = ₈¹ Z

C

(1+u)[1+(1 ¹₂ ²)u]² [1+(2 ²)u+u²]⁵⁼²

dw u :

(3.26) Here, we get a new contour see …gure 5. The branch points is outside of the contour and is called detours. Use the equations (3.23), (3.24) and the contour integral will thereafter be

res (0) = lim

u !0

h

u^{(1+u)[1+(1 12}

2)u]² [1+(2 ²)u+u²]⁵⁼²

1 u

i

= 1;

1 8

Z

C

(1+u)[1+(1 ¹₂ ²)u]² [1+(2 ²)u+u²]⁵⁼²

dw

u = ₈ⁱres(0) = ₈ⁱ :

The second contribution to (3.3) is Z (t´ ´ )

(t´ )

Q (z) X2(z) dz = ¹₈ Z (w´ ´ )

(w´ )

h 2 2 w(z) [w(z)+1]²

i1=2h

2 2 w(z)

(w(z)+1)²

i 1 dw w(z)+1; (3.27)

or Z (t´ ´ )

(t´ )

Q (z) X2(z) dz = ₈¹

Z (w´ ´ ) (w´ )

(1 + 1=w (z))^{2 2}=w (z) ¹⁼²_w(z)^dw :

(3.28)

Fig 5 Illustrate the contour in the complex u - plane Use the equation (3.25) and we get

Z (t´ ´ ) (t´ )

Q (z) X2(z) dz = ₈¹ Z

C

(1 + u)^{2 2}u ^{1=2 du}_u;

(3.29) and thereafter use the equations (3.24) and (3.25),

res(0) = lim

u !0

h

u (1 + u)^{2 2}u ^{1=2 1}_ui

= 1

1 8

Z

C

(1 + u)^{2 2}u ^{1=2 du}_u = ₈ⁱ res(0) = ₈ⁱ :

Hence

K = i Z (t´ ´ )

(t´ )

Q(z)[1 + Y₂(z)]dz = i Z (t´ ´ )

(t´ )

Q(z)dz +₈ⁱ ₈ⁱ = K₀.

Above K0 is the …rst-order K integral. Hence for the present potential T = _1+exp¹_f2K0g.

(3.30) Hence for the Eckart - Epstein potential with the choice of the base function as in the present paper there is no third-order contribution to the transmission coe¢ cient.

4. CONCLUSION

The third order correction does not give any contribution to the transmis- sion coe¢ cent for the present potential. However the third-order correction contributes to the wave function. We conclude that it is important to choose the base function in a proper way. Instead, if we have a freedom to choose the base function. Using such freedom, however, can give as a result that we have to bring along rather large higher-order corrections e. g. in order to have a physically reasonable transmission coe¢ cient.

We have obtained a contribution to the solution for the third order asymp- totic approximation which was zero for a particular potential. We …nd that the

…rst order is often good enough to describe the transmission coe¢ cent provided we use a proper base function not necessarily equal to the usual JWKB base function.

5. REFERENCES

1. Study of the validity of the phase-integral connection formula ... AIP Journal of Mathematical Physics, of Karl - Erik Thylwe, Sta¤an Yngve and Per Olof Fröman 2006.

2. New Phase - Integral Method Platform Function, of Sta¤an Yngve and B. Thide. Department of Physics and Astronomy, Theoretical Physics division, Uppsala University.2009

3. JWKB approximation contributions to the Theory of Nanny Fröman and Per Olof Fröman. 1965

4. Fundamentals of Complex Analyses with applications to engineering and science (third edition) of E.B. Sa¤ A.D Snider.

5. Eckart, Phys. Rev. 35 1303 (1930) 6. http://www.iter.org/

7.Introduction to Nuclear Science of Je¤ C. Bryan 8.Fröman and P.O, Nucl. Phys. A147, 606 (1970).

6. APPENDIX Appendix A

Let us test change complex variable from to z for the term ^q_Q ¹⁼²_d^d22 q Q

1=2

from (2.26) and use that ^d_dz = Q

q Q

1=2 d² d ²

q Q

1=2

= ^q_Q

1=2 d dz

d dz

q Q

1=2

Q ¹ Q ¹

= ^q_Q ¹⁼² _dz^d22

q Q

1=2

Q ² _dz^{d q}_Q ¹⁼²Q ^{3 dQ}_dz = ^{(q )}_Q2² d²

dz²(q ) ¹⁼² Q ^{3=2 d}_dz²2Q ¹⁼²

The …rst term is the same for the second term in (2.25). Now we can …nd the second and fourth term in (2.26)

(q )² Q²

d²

dz²(q ) ¹⁼² Q ^{3=2 d}_dz²2Q ¹⁼² +Q ^{3=2 d}_dz²2Q ¹⁼²= ^q_Q ¹⁼²_d^d22 q Q

1=2

+ Q ^{3=2 d}_dz²2Q ¹⁼²