Theory of Scattering according to Majorna

7/27/2019 Theory of Scattering according to Majorna

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6

THE THEORY OF SCATTERING

6.1. SCATTERING FROM A POTENTIALWELL

The author studied here the problem of the scattering of a plane wavefrom a one-dimensional square potential well. All the physically inter-esting cases were treated.

e= h/2=m = 1.

2 + 2(E V)= 0.V = 0:

y + 2Ey= 0.

2E=k2,

y1= eikx, y2 = e

ikx.

y + 2(E U)y= 0,U= V,

y + 2(E+ V)y= 0.

2(E+ V) =2, 2E=k2,

= k

1 +

V

E.

By imposing the matching conditions for the wavefunction and its deriva-tive, one obtains:

311


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312 E. MAJORANA: RESEARCH NOTES ON THEORETICAL PHYSICS

y=

1 + 1 + VE

2 e

ik(x+a)ia

+

1 1 + VE

2 e

ik(x+a)ia

,x < a,

eix,

g= 1 +

1

2

V

E

a < x < a,

1 +

1 + VE

2 eik(xa)+ia +

1

1 + VE

2 eik(x+a)+ia,

a < x,

E >0,

E=1

2 2 V, E=1

2 k2,

= k

1 +

V

E, k=

1 + VE

.

g gives the ratio of the wave amplitude inside and outside the well. 1

E V: 1 +

V

E =i

V

E 1,

= ik

V

E 1 =k1

V

E 1, k1=ik.

1@That is: g is given by the ratio a2+ b2/c2 wherea [b] is the coefficient of the first [second]wave term in the first or third row, whilec is the coefficient of the wave term in the secondrow (c = 1). Note that the quantity we call g, here and in what follows, is in the originalmanuscript denoted byy, the same as the symbol there used for the wave function.


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THE THEORY OF SCATTERING 313

y=

1 + i VE 1

2 e

k1(x+a)ia

+

1 i VE 1

2 e

k1(x+a)ia

,x < a,

eix, a < x < a,

1 + i

VE 1

2 ek1(xa)+ia +

1 i

VE 1

2 ek1(xa)+ia,

a < x,

V < E


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y=

1 +

V

E cos a sin k(x + a) sin a cos k(x + a),

x 0:1st approximation:

= 2ih

k0e2i(h/2m)(220)t,

= 2mh2(2 20 )k0

e2i(h/2m)(2

2

0)t 1 + ( 0) .4@Here the author denotes with0 =p0/hthe momentum (divided byh) of the free particle,while(x) signifies the Dirac delta-function.


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2nd approximation:

=

2i

h

k0e2i(h/2m)(220)t +

4im

h3 kk0

1

2

2

0

e2i(h/2m)(22

0)t e2i(h/2m)(22)t

d,

= 2mh2(2 20 )

k02 20

e2i(h/2m)(

220)t 1

+4m2

h4

kk0

e2i(h/2m)(

220)t 1

(2 20 )(2 20 )

e2i(h/2m)(22)t

1(2 2)(2 20 )

d + ( 0) .

In first approximation, for =0, we have:

||2 = 16m2

h4(2 20 )2|k0 |2 sin2

h(2 20 )t2m

.

Neglecting constant terms, for t we get:

||2 =82mh3

|k0|2 t

2 20

,

and the transition probability is:

P0=82m

h3 |k0|2

2 20

.

In second approximation:

||2 = 16m2h4(2 20 )2

|k0 |2 sin2h(2 20 )t2m

+ 32m3

h6(2 20 ) sin

h(2 20 )t2m

k0kk0+ k0kk0

sin h(2 2)t/2m(2 2)(2 20 )

sin h(2 20 )t/2m

(2 20 )(2 20 )

d.

6.3.1 Coulomb Field

For a Coulomb field:

V =C

r =

Ve

2i(xx+yy+zz) dx dy dz,


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V=C

e2iq

r dq= C

0

2

sin2r dr=

C

2;

k

=V

=

C

| |2 .In first approximation:

P0= 8mC2

h3| 0|4

2 20

= mC2

2h340sin4 /2

2 20

.

In second approximation, for =0: 5

=

2m

h2

1

2 20C

( 0)2 e2i(h/2m)(220)t

1

+4m2

h4

C2

2| |2| 0|2

e2i(h/2m)(22

0)t 1(2 20 )(2 20 )

e2i(h/2m)(22)t 1(2 2)(2 20 )

d.

6.4. THE BORN METHOD

The scattering from a given center was studied here by means of the Bornmethod, and approximated expressions for the scattered partial waveswere obtained.

2 + k2 = F .

=0+ 1+ 2+ . . . ,

2 0+ k

20 = 0,

2 1+ k

21 =F 0,

2 2+ k

22 =F 2,

. . . ,

2 n+ k

2n=F n1,

. . . .

5@Probably, the author started to evaluate the transition probability for Coulomb scatteringin a second approximation, but succeeded only in obtaining an expression for the coefficient.


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n(q) = 14

eik|qq

|

|q q| F(q) n1(q

) dq.

0(q) = eiku0q,

1(q) = 14

eik|qq

|

|q q| eiku0q F(q) dq,

2(q) = 1

162

eik|qq

|

|q q|eik|q

q|

|q q| eiku0q F(q) F(q) dq dq,

|u0| = 1.|q| =r :

1(q) = 14r

eik|qq

| eiku0q

F(q) dq.

|q| =r, q=r u, |u| = 1;|q| =r , q =r u, |u| = 1,

r :|q q

| =r r

u u

,

1(q) = eikr

4r

eikr

(u0u)u F(q) dq.

2(q) = eikr

162r

eik|q

q|

|q q| eikru0u eikr

uu F(q) F(q) dq dq.

q =q +:

2(q) = eikr

162r

eik(u0u)q

F(q) dq

eik||

|| eiku0 F(q +) d.

F = 1

2

Fe

iq d,

F=

Feiq dq,

eiq

eikr

r dq=

4

2 k2 ,


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r :1(q) = e

ikr

4 Fk(uu0),

2(q) = eikr

162r

eik(u0u)q

F(q) dq

2

|ku0+ |2 k2Feiq d

= eikr

162r

2

|ku0+ |2 k2 Fd

ei(ku0ku+)q

F(q) dq,

2(q) = eikr

82r FFk(uu0)

|ku0+

|2

k2

d

2(q) = eikr

82r

Fku0Fku

2 k2 d.

6.5. COULOMB SCATTERING

The Schrodinger equation for the scattering of a wave from a Coulombpotential is solved and, in particular, the phase advancement is evaluated.

Ze charge of the scatterer;

Ze charge of the incident particle;

Mmass of the incident particle.

We adopt units such that M= 1, ZZe2 = 1, h/2= 1. It follows that:

the length unit ish2/42M ZZe2 = (m/M) (1/ZZ) a0; 6

the energy unit is 42M Z2Z2e4/h2 = 2(M/m) Z2Z2 Rh; 7

the velocity unit is 2Z Ze2/h = ZZ/137c, where 1/137 =e2/(1/2)hc.

The Schrodinger equation is:

2

+ 2

E1

r

= 0.

6Here m denotes the electron mass and a0 0.529109 the Bohr radius.

71 Rh = 13.54 V [Remember that the symbolV used by Majorana should more appropriatelyunderstood as eV].


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=n

=0

X(r)

r P(cos ),

X +

k2 2r ( + 1)

r2

X= 0,

k2 = 2E (the velocity of the ingoing particle in large units is v =(ZZ/137)c k).

X = X1 + X2,

X1

= x+1

eikx

F

+ 1 +

i

k , 2 + 2,2ikx ,X2 = x+1 eikx F

+ 1 i

k, 2 + 2, 2ikx

,

F( ,,x) = 1 +

x +

( + 1)

2!(+ 1)x2 +

( + 1)( + 2)

3!(+ 1)(+ 2)x3 + . . . .

Alternative solution

X +

k2 2r ( + 1)

r2

X = 0,

takes non-integer values greater than1/2,

X =r+1 u,

u + 2 + 1r

u +

k2 2r

u= 0.

[8]

u +

0+

1r

u +

0+

1r

u= 0,

0= 0, 1 = 2( + 1), 0 =k2, 1= 2:

u eiktr (t 1)+i/k (t + 1)i/k dt.8@ This equation is a particular case of the more general one reported just after it, and isalso considered by the author in another place; see Appendix6.10.


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[...]9

|Im log(1 t)| ,|Im log(1 + t)| :

u=

1

1eiktr (1 t)+i/k (1 + t)i/k dt .

For r= 0, on setting 1 t= 2x:

u(0) =

11

(1 t)+i/k (1 + t)i/k dt= 1

0(2x)i/k (2 2x)+i/k 2dx

= 22+1 1

0

(x)i/k (1

x)+i/k dx,

u(0) = 22+1 ( + 1 i/k) ( + 1 + i/k)

(2 + 2) . (1)

|r| >0:u= u1 + u2,

u

1

= e

i(/2)(+1+i/k)

e

ikx 0 e

krp

p

+i/k

(2 + ip)

ik

dp,

u2 = ei(/2)(+1i/k) eikx

0ekrp p+i/k (2 ip)+ik dp.

For real r we have u2 =u1.

u1 = (kr)(+1) ei(/2)(+1+i/k)(i/k)log kr eikr

0 ep

p+i/k

(2 + ip/kr)ik

dp.

For r :u1 = (kr)(+1) e/2k ei(/2)(+1) eikr(i/k)log kr 2i/k ( + 1 + i/k)

= 2(kr)(+1) e/2k ei(/2)(+1) eikr(i/k) log2kr ( + 1 + i/k).

Now, replace with ; the phase advancement becomes then:

k= 2 arg ( + 1 + i/k)( + 1 + i/k) .

9@ The author then evaluates u and u and verifies that the assumed form for u satisfiesthe previous differential equation.


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1/4> a >0:

( ) ( + 1 ) = ( + 1) a, +1

2 2 = +1

2

2 a,

= +1

2

+

1

2

2 a.

6.6. QUASI COULOMBIAN SCATTERINGOF PARTICLES

Let us assume a scattering potential of the form:

kr2 + a2

, (1)

a being the magnitude of the radius of the scatterer. By denoting withT the kinetic energy of the incident particles, let us define the minimumapproach distance10 b in the limit Coulomb field (a= 0) as:

k

b =T; b=

k

T. (2)

The scattering intensity under an angle will be obtained on multi-plying that appearing in the Rutherford formula by a numerical factordepending on the mutual ratios ofa,b,/2 (being the wavelength of

the free particle) and. Let us set:

i= f( ,,) iR, (3)

whereiR is the intensity calculated from the Rutherford formula (a= 0)and

= a

/2, =

b

/2. (4)

Since for a= 0 the Rutherford formula is exact, we have:

f(0, , ) = 1. (5)

10@ That is, the scattering parameter.


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Let us now consider a fixedand take the limit 0. At zeroth orderapproximation, i.e., exactly for = 0, we can use the Wentzel method.By choosing as mass unit M, wavelength unit /2 and velocity unit v

for the incident particles, from = h/Mv it follows that h= 2 in ourunits. Moreover, the kinetic energy of the incident particle is 1/2.From Eqs. (4) and(2) it follows that b = , k = /2 and a = . Bysubstituting these into Eq. (1), we get the expression for the potentialenergy, and the Schrodinger equation corresponding to the eigenvalue1/2 will be:

2 +

1

r2 + 2

= 0. (6)

Let us set:

=0+ 1+ 2+ . . . ,

where:

2 n+ n =

r2 + 2

n1. (7)

In order to avoid convergence problems, instead of /

r2 + 2 let us

consider the expression

1/r2 + 2 1/R for r < R and 0 for r >R; in the final results we will take the limit R . Eq. (7) is thenreplaced by: 11

2 n+ n =P n1; (8)

P =

1r2 + 2

1R

, for r < R;

0, for r > R.

Setting0= eiz, we have:

2 1+ 1 =Pe

iz. (9)

For an univocal solution of Eq. (9) we will choose 1 to represent a

diverging wave. In this case Eq. (9) can be integrated and, putting

r12 =

(x x)2 + (y y)2 + (z z)2,

11@ In the original manuscript, the factor is lacking.


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we get:

1(x,y,z) = 1

4

P(x

, y

, z

)

ei(r12+z)

r12 dx

dy

dz

. (10)

Assuming the point (x,y,z) to be far from the origin, we have (r is thedistance from the origin, the angle between the vector radius and thez-axis):

r : 1(r, ) = 14r

P(x, y, z) eir

eir(cos cos cos sin sin cos ) dxdydz,

cos (1 cos ) sin sin cos

= 2 sin /2

sin /2 cos cos /2 sin cos = 2 sin /2

cos(/2 /2)cos + sin (/2 /2)sin cos ,

r : 1(r, ) = eir

2 r sin /2

0rP(r) sin

2 sin /2r

dr,

(11)whence we easily deduce:

f(, 0, ) = 2

sin /2

0

rP(r) sin

2 sin /2 r

dr. (12)

In we simply replace P with /r2 + 2, the integral in Eq. (12) doesnot converge; however, we can circumvent this difficulty by keeping in-determinate the upper integration limit and assuming, for the resultingintegral, its mean value which for the upper limit tends to infinity. Wethus find:

f(, 0, ) = 2 sin /2

0

rr2 + 2

sin (2 sin /2r) dr

=

0

x sin x dxx2 + 42 sin2 /2

= ( sin /2) .(13)


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6.6.1 Method Of The Particular Solutions

u +

1

r2 + 2 ( + 1)

r2

u = 0. (14)

For the hydrogen atom we consider the values = 0.4, 0.5, 0.6, 0.7and = 0, 0.2, 0.4, 0.6, 0.8, 1. The solution of Eq. (14) is reportednumerically in the following tables for= 0 and= 0.4. 12

= 0 = 0.2 = 0.4

r

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

1.3

u u

u

0 1 0.400

1.019

0.1018 0.305

1.049

0.2067 0.207

1.070

0.3137 0.109

1.080

0.4217 0.000

1.080

0.5297 -0.106

1.069

0.6366 -0.212

u u

u

0 1 0

u u

u

0 1 0

12@ The author uses a numerical algorithm (unknown to us) in order to infer the solutionu(r) of Eq. (14) from its second (and first) derivative, and the first few results obtained aredisplayed in the tables.


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= 0.6 = 0.8 = 1.0

r

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1.0

1.1

1.2

1.3

u u

u

0 1 0

u u

u

0 1 0

u u

u

0 1 0

6.7. COULOMB SCATTERING: ANOTHERREGULARIZATION METHOD

Let us assume the potential to be as follows:

V =

V0, for r < R,

k/r, for r > R.(1)

r

V

R

V0


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Denoting with T the kinetic energy of the incident particles, the mini-mum approach distance in the Coulomb field will be:

b= k

T. (2)

The scattering intensity under an angle will be given by the productof the intensity scattering due to the Coulomb field, obtained from theRutherford formula, times a numeric function depending on,R/,b/,V0/T:

13

f

V0T

, R

/2,

b

/2,

, (3)

where is the wavelength of the free particle. Let us choose as massunitM, velocity unitv and length unit/2relative to the free particle.In such units, h= M v 14 is equal to 2, while T is 1/2. Moreover, letus set:

A=V0

T , =

R

/2, =

b

/2, (4)

so that:

i

iR =fV0

T ,

R

/2 ,

b

/2 ,

=f(A, ,,). (5)

In our units we have:

V0=A

2, R= , b= , k=

1

2, (6)

and the Schrodinger equation corresponding to the eigenvalue 1/2 takesthe form:

2 + (1 A) = 0, for r < R,

2 +

1

r

= 0, for r > R.

(7)

For the hydrogen we have:

= 0.4, 0.5, 0.6, 0.7;

= 0.4, 0.5, 0.6, 0.7, 0.8;

A= (2), (1.5), 1, 0.5, 0,

0.5,

1,

1.5, 2,

2.5,

3,

3.5, 4, 4.5, 5, 5.5, 6, 6.5, 7, 7.5, 8.

13@ In the original manuscript, the first dependent variable in Eq. (3) is V0/2T rather thanV0/T. However, in the following the author considered the latter parametrization.14@ In the original manuscript, the author wrote erroneouslyh= /Mv.


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6.8. TWO-ELECTRON SCATTERING

v, v be the velocities of the two beams;

n0, n0 be the rest numberdensities of the two beams;

n= n0/

1 v2/c2, n =n0/

1 v2/c2 be thenumberdensitiesin the laboratoryreference frame;

vr be the relative velocity according to the relativistic kinematics;

S(vr) be the cross section.

The number Nof collisions for unit volume and time can be written as:

N=a

v, v

n n =S(vr) n0 n0

vr1 v2r /c2

.

In terms ofa we thus have:

S= a

vr

1 v2r /c2

1 v2/c2

1 v2/c2(classically(that is: non relativistically), we have insteadS=a/|vv|).Without considering the resonance in the scattering cross section, let u

(0 u vr) be the velocity of the first electron after the collision in itsinitial reference frame; we have:

dS=S(vr, u) du.

Let us now denote with u1 the relative velocity between the frame ofthe first electron before (after) the collision and that of the second elec-tron after (before) the collision. By taking into account the resonancebetween the two electrons, u andu1 are indistinguishable. Putting, con-

ventionally, u u1, the maximum value ofu is given by 15

:

umax=umin=c

1 4 (1 v2r /c2)1 +

1 v2r /c2

2 = y2

2 y2

c2

y=c

1

1 v2r /c2

1 +

1 + v2r /c2

.

The relation between u andu1 is the following:

11 u2/c2 +

11 u21/c2

= 1 + 11 v2r /c2

.

15In this case we have u= u1


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6.9. COMPTON EFFECT

n= n0/1 v2/c2, number of electrons per cm3;

n0, rest numberdensities of the electron beams;

N, number of photons 16 per cm3;

N0 =N 0/, number of photons per cm3 in the electron frame;

h, energy of one photon;

h0, energy of one photon in the electron frame (before or after

the collision);u1, relative velocity between the ingoing electron frame and theoutgoing one (according to relativistic kinematics);

S(0), cross section.

The number of collisions for unit volume and time is thus:

S(0) n0 N0 c= a n N,

so that, in terms ofa,

S(0) =a

c

11 v2/c2

0.

The differential cross section can be written as:

dS=F(0, u) du,so that

S=

0

F(0, u) du.

Classically, the cross section is given by:

Sclass =8

3

e4

m2c4.

16@ For the sake of clarity, here and in the following we have translated with photonswhat was termed quanta in the original manuscript.


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6.10. QUASI-STATIONARY STATES

The author considered the transition from a discrete (unperturbed) state0 with energy E0 to a continuum (perturbed) state , assuming thatthe unperturbed system has two continuum spectra W and W withenergyE0+W. What here reported are the scratch calculations whichprepared the Sect. 28 of Volumetto IV, to which we refer the reader

for notations and further explanations of the arguments treated by theauthor. However, a further generalization is present here with respect towhat considered after Eq. (4.499) of Volumetto IV.

= 12/Q2 + 2Q2

IQ2

0+ 2|I|2Q4

W |I|2Q2

W

W WdW

+2IL

Q4 W IL

Q4

W

W WdW

+I

e2i(

)t/hdW

(2/Q2 + 2Q2)(W W) 0

+ |I

|2

Q2 e2i()t/hW

(2/Q2 + 2Q2)(W W) dW

|I|2

e2i()t/hdW

(2/Q2 + 2Q2)(W W)

WdW

W W

+IL

Q2

e2i(

)t/hW

(2/Q2 + 2Q2)(W W) dW

IL e2i(

)t/hdW

(2/Q2 + 2Q2)(W

W)

WdW

W

W

+|L|2Q2

W ILQ2

W.

=

A 0+ B W+ CW+

b WdW

+

c WdW

e2iEt/h,

QuantityA:

1

(2/Q2 + 2Q2)( ) = Q2

( + iQ2)( iQ2)( ) ,

R1 = e2it/h e2

2Q2t/h 1

2i( + iQ2),


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R2 = 1

2/Q2 + 2Q2,

2i R1+1

2 R2

=

1

2/Q2 + 2Q2

e

2it

e

22Q2t/h

Q2

+ i

i

,

A= 1

2/Q2 + 2Q2

I

Q2

1 e2it/h et/2T

Ii

1 e(2i/h)t et/2T

,

A= I + iQ2

1 e2it/h et/2T

.

QuantityB:

1

2/Q2 + 2Q22|I|2

Q4 +

2|I|22/Q2 + 2Q2

+|L|2

Q2 =

1

Q2 |I|2 + |L|2

= 1,

B = 1.

QuantityC:

1

2/Q2 + 2Q22IL

Q4 +

2IL

2/Q2 + 2Q2 IL

Q2 = 0,

C= 0.


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Quantityb: 17

e2i(

)t/hd

(2/Q2 + 2Q2)(

)(

)

=

Q2e2i(

)t/hd

( + iQ2)( iQ2)( )( ) ,

2i

R0+1

2R1+

1

2R2

,

R0= e2it/h e2

2Q2t/h 1

2i( + iQ2)( + ipiQ2) ,

R1 = 12/Q2 + 2Q2

1 ,

R2= e2i(t/h 1

2/Q2 + 2Q21

.

2i

R0+1

2R1+

1

2R2

=

1

2/Q2 + 2Q21

2/Q2 + 2Q2

e2it/h et/2T Q2 i

Q2 i

2

Q2+ 2Q2

i

+

2

Q2+ 2Q2

ie2i (

)t/h

,

b = |I|2

Q21

1

2/Q2 + 2Q2+

|I|2Q2

1

e2i(

)t/h

2/Q2 + 2Q2

+ |I

|2

2/Q2 + 2Q21

2/Q2 + 2Q2

e2it/h et/2T

Q2 i

Q2 i

2

Q2+ 2Q2

i

+

2

Q2+ 2Q2

ie2i (

)t/h

= |I|2 + iQ2

1

|I|

2

+ iQ21

e2i (

)t/h

+ |I|2 e2i t/ht/2T

( + iQ2)( + iQ2),

17@ Cf. the figure above.


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b = |I|2

( + iQ2)( + iQ2) 1 + e2i t/ht/2T

+ + iQ2

1 e2i ()t/h

,

b = |I|2

( + iQ2)( + iQ2) e2i ()t/h + e2i t/ht/2T

+ + iQ2

1 e2i ()t/h

.

Quantityc:

c = ILQ2

1 12/Q2 + 2Q2 +

ILQ2

1 e

2i()t/h

2/Q2 + 2Q2

+ IL

2/Q2 + 2Q21

2/Q2 + 2Q2

e2it/h et/2T

Q2 i

Q2 i

2

Q2+ 2Q2 i

+

2

Q2+ 2Q2 ie

2i ()t/h

=

IL

+ iQ21

IL

+ iQ21

e2i ()t/h

+ IL e2i t/ht/2T

( + iQ2)( + iQ2).

c = IL( + iQ2)( + iQ2)

1 + e2i t/ht/2T

+ + iQ2

1 e2i ()t/h

.


26/27


= e2iEt/h W+ I

+ iQ2

e2iEt/h e2i(E0k)t/h et/2T

0

I

+ iQ2 IW+ LW

+ iQ2 e2iEt/h

1 e2i t/ht/2T

d

+I

IW+ LW

+ iQ2 e2iEt/h

1

1 e2i()t/h

d,

= e2i Et/h W+ a e2i E0t/h 0

+

bW

IW+ LW

dW e2i Et/h,

H = Ee2iEt/h

W+ IWe2i E0t/h

0+ a E0e2i E0t/h

0

+

a e2i E0t/hIWWdW

+

a e2i E0t/hLWWdW

+

EbW

IW+ LW

e2i E

t/hdW

+Q2

bWdW 0e2i Et/h,

IW =I:

a = 2ih

e2iWt/hI+ Q2

bWe

2i Wt/hdW

,

bW = 2ih

e2i Wt/h a.

= e2i Et/hW

+ I

+ iQ2

e2iEt/h e2i(E0k)t/h et/2T

0

I + iQ2

IW+ LW

+ iQ2

e2i Et/h

1 e2i t/ht/2T

d

+ I

+ iQ2

IW+ LW

e

2i Et/h

1 e2i()t/h

d.


27/27


= + ,

= e2iEt/hW+ I

+ iQ2e2iEt/h 0

I + iQ2

IW+ LW

e

2iEt/h

1 e2i ()t/h

d,

= I + iQ2

e2i(E0k)t/h et/2T 0

I + iQ2

IW+ LW

+ iQ2

e2i Et/h

1 e2i t/ht/2T

d.

Appendix:Transforming a differential equation

u +

0+

1r

u +

0+

1r

u= 0,

= r

k

u.

u = rk u,

u = u

k

r

,

u = u

k

r

2+ u

2

2 +

k

r2

= u

2k

r

+

k(k+ 1)

r2

.

2 k

r

+

k(k+ 1)

r2

+ 0

0 k

r+

1r

k1

r2 + 0+

1r

= 0,

+

0+

1r 2 k

r

+

0+

1 k0r

+k(k+ 1) k1

r2

= 0.

k=1

2; 1 = 2k,

+ 0 +

0+

1 k0r

k(k 1)r2

= 0.

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Theory of Scattering according to Majorna