SLIDinelscatt Skipping

8/11/2019 SLIDinelscatt Skipping

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1 Elastic scattering

We begin by considering scattering as a stationary problem, that is, we seek

the solution of the time-independent Schrdinger equation, which we write as( 2 + k2 )(r ) = U (r )(r ) (1)

Asymptotically, we expect the solution to be of the form,

(r ) = in + sc. (r ). (2)Here, in is a plane wave,

in (r ) = Cei k r , (3)

whilst the outgoing (scattered) part, sc. , is a spherical wave

sc. (r ) = C |f (, )|eikr

r . (4)

The factor f is called the scattering amplitude , and describes directional vari-ations. Note that the wave number k is the same for both incoming andscattered part, since we only consider elastic scattering.


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1.1 Currents and cross-sections

It can quite easily be shown, using probability currents , that the differential

cross-section is related to the scattering amplitude bydd

= |f (, )|2 (5)

2 The Greens function method

As a step on our way to the solution of the problem, we look at the Greensfunction method. Basically, we wish to cast equation ( 1) into an integral form,which has boundary conditions built-in. It turns out that if we use the Greens function for a free particle, G(r r ), given by

( 2 + k2 )G(r r ) = 3 (r r ) (6)

then the solution to equation ( 1) can be written as

(r ) = 0 (r ) + G(r r )U ( r )( r )d3 r (7)


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where 0 (r ) is the solution of the homogeneous equation ( 2 + k2 )(r ) = 0.Before we can get any further, we must determine G(r r ). Some calculationwill yield two independent solutions

G(r r ) = e ik |r r |

4|r r |(8)

But which solution should we choose? It so happens that the asymptoticbehavior of the solution with the +sign in the exponent is eik |r r | / 4

|r

r

| eikr /r , an outgoing spherical wave. A glance at equation ( 4) tells us that thisis the one we want. As for the homogeneous solution, we choose0 (r ) = ei k r , (9)

as this gives us the incoming plane wave solution when the potential is zero.Thus, when r is sufficiently large, equation ( 7) becomes1

(r ) ei k r

14

eikr

r e i k r U ( r )( r )d3 r . (10)1 Here we have used a proper expansion of eik |r

r | / 4|r r |


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The primed k appearing under the integral is k = krr . Upon comparison of equation ( 10) with eqs. (2) and ( 4), we nd an expression for the scatteringamplitude

f (, ) = 14 ei

k

r U ( r )( r )d

3

r (11)So far we seemingly havent got very far, because the (r ) appearing in theexpression for the scattering amplitude is as unknown as ever. However, as wenow shall see, equation ( 10) is very well suited for iterations, which will giveus solutions to put into ( 11).

2.1 The Born approximation

When the potential in equation ( 10) is sufficiently weak, we can use it foriteration. We do this by inserting the n. order approximation n (r ) into thethe equation for the (n+1). approximation. If the 0. order approximation istaken to be 0 (r ) = ei k r , we get

1 (r ) = ei k r 14

eikr

r ei k r U ( r )ei k r d3 r , (12)


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and so forth for higher orders. This neat procedure is called the Born ex-pansion. We, however, will take the easy way out and only use the 0. orderapproximation. Plugging this into equation ( 11), we get the rst order Born

approximation, which is usually just called the Born approximationf B (, ) =

m2 h2 V (r )ei qr d3 r, q = k k. (13)

We have reinstated the potential V using U = 2mV/ h2 . The Born approxima-tion can be used when the potential is weak and nite or the particle energyis large (although it must be nonrelativistic).

3 Atomic collisions

We now consider elastic collitions of charged particles with atoms.


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3.1 Charge density

A useful consept is that of the electron cloud, where the total electron chargeof the atom is thought to be smeared out around the nucleus. The charge density of a cloud made up of n electrons, is

(r ) = e i P i ( r i ) = e i |i ( r i )|2 (14)

where i (r ) is the normalized wave function of the ith electron, i.e.

|i (r )|

2 d =

1. If the cloud is comprised of only one electron, we can drop the summation.

3.2 Single-electron atom collisions

When an electron (projectile, P) is scattered off a one-electron atom like hy-

drogen, it will feel the attraction from the nucleus (target, T), as well as the re-pulsion from the bound electron (e). The total potential is thus V = V P T + V P e .If the nucleus has charge Ze, the total potential becomes:

V (R) = Ze2

| R| V e(r )

|r R|(15)


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3.3 The Yukawa potential

The potential we found in the last section is cumbersome to work with. But forlarge distances, we can approximate ( 15) using a shielded Coulomb potential,known as the Yukawa potential ,

V y (r ) = e2

4 0e r

r . (16)

If we stuff this into the Born approximation, we get

f B (, ) = me2

8 2 0 h2 ei qr e r

r d3 r, q = kf kb (17)

where kb and kf are the wave numbers of the incoming and outgoing wave,respectively (and they are equal since we have elastic scattering). We can do

this integral. First note that q r = qr cos( ). Then put in d3

r = r2

drd dand use the trick sin ( )d = d(cos( )). Skipping the boring details of thecalculation, we have the result

f B (, ) = me2

2 0 h21

2 + q 2 (18)


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The scattering cross section becomes

dBd

= |f B (, )|2 =

me2

2 0 h21

2 + q 2

2

(19)

We can get rid of the q in the equation by using k sin( 2 ) = q/ 2 and k = 2mE/ h. These substitutions givedBd

= me2

2 0 h21

2 + 8 mE h 2 sin2 ( 2 )

2

(20)

Note that as we let 0, so that the Yukawa potential tends to the Coulombpotential, we getdBd

= e2

16 0 E

21

sin4 ( 2 ). (21)

We obtained this result using the very crude Born approximation. The incred-ible thing is that our result coincides with both the exact quantum mechanicaland the classical one, and yet the Coulomb potential does not fulll any of the conditions we stated as necessary for the Born approximation to be valid.The equation is called the Rutherford cross section , named after he who rstderived it (classically).


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4 Inelastic scattering on objects with struc-ture

Until now we have been discussing elastic collisions between a particle and anatom (typically a hydrogen atom). In the case of only one incomming elemen-tary particle elastic scattering is fully possible. The particles may well interactwithout the incoming particle having the suitable energy quanta to exite anyof the atoms inner degrees of freedom. But in the case of for example an

incomming electron beam of millions of particles with an energy distributionexitations are unavoidable. We need therefore consider the possibility of theincoming particle and the atom changing states(not swapping).Again we treat the inelastic case with only one incoming charged elementaryparticle being the projectile and a single atom being the target. For simplicitylets say we have a hydrogenlike atom (we leave to someone else to worry aboutthe many body case).The relation between the scattering cross-section and the scattering amplitudeis easily derived using the probability density currents just as above only theks dont cancel because kf = kb. So we get:


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dd

= kf kb |f inel (, )|

2

But how does the f inel

look like?Lets assume that the projectile doesnt interact inelastically with the nucleus.Then we need to consider the system consisting of the projectile and the singleelectron of the atom. For this system the energy is conserved but the compo-nents (the energy of the projectile and the energy of the electron) may change.The hamiltonian for this system is:

H = T p( R) + T e(r ) + V T p ( R) + V T e (r ) + V pe ( R,r )

Where p stands for the projectile, e for the electron and T for the target (thenucleus). And we have as usual assumed an innite heavy nucleus resting qui-etly at the origin.

We now expand the solutions for the system using the solutions (r )for thehydrogenlike atom as a basis. Sum over all orbitals:

( R,r ) =n

n ( R)n (r )


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Before the collision when the particles are far far from each other we have

b( R,r ) = ei k i R

b(r )

The projectile being a free particle and the atom being in any of its eigenstates.After the collision we have:

f ( R,r ) = f ( R)f (r )

Far away the wavefunction for the projectile should look like:

( R) f inel (, )eik f R

R (22)

This we plug this expansion into the eigenvalue eguation H = E and useFouriers trick.(leftmultiply by j | ) The details are tiring to write in Latex


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and are left as an exercise. Muhahaha.We get

{T p( R) + V T p ( R) + j E } j ( R) = n j |V pe |n n ( R) j = 1 , 2, 3, . . .

E is the energy of the system, n is the energy of the bounded electron and E nis the energy of the projectile

E = n + E n.

Lets now forget about the potential from the nucleus V T p ( R) to simplify mat-ters. The potential could of course be included in the potential on the rightside, which would be a better way to go about it.Now assume only one transition is possible, that is


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b( R,r ) = b( R)b(r ) f ( R,r ) = f ( R)f (r )Then the nasty system of coupled equations reduce to this one little equation:

{T p( R) + f E }f ( R) = f |V pe |b b( R)

{T p( R) + E f }f ( R) = g( R)b( R)g( R) = f |V pe |b

This can be written in form of an inhomogenous Helmholtz equation and solvedby means of a Greens function.

f ( R) = 2mh2 G( R R )g( R )b( R )d3 R


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where

( 2 + k2f )G( R R ) = ( R R )G( R R ) = e

ik | R R |

4| R R |and

kf = 2mE f / hThat this is a solution is readily checked. With the asymptotic form of theGreens function (derived earlier) we get:

f ( R) = 2m

4 h2

eik f R

R

e i k f R g( R )ei kb

R d3 R

By comparision to the desired form ( 22) of f ( R) we nally get:

f inel (, ) = 2m4 h2 e i k f R g( R )ei kb R d3 R


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Writeg( K ) = ei K R g( R )d3 R

so that

f inel (, ) = 2m4 h2

g( kb kf )

and the scattering cross-section becomes:

inel = m2

(2)2 h4kf kb d|g( kb kf )|2

The end

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