<<

LectureLecture 44 ScatteringScattering theorytheory

SS2011: ‚Introduction to Nuclear and Particle , Part 2‘ SS2011: ‚Introduction to Nuclear and , Part 2‘ 1 I.I. ScatteringScattering experimentsexperiments

Scattering experiment: A beam of incident scatterers with a given or (number of particles per unit dA per unit dt ) impinges on the target (described by a potential); the flux can be written as

The number of particles per unit time which are detected in a small region of the solid , dΩ, located at a given angular deflection specified by (θ, φ), can be counted as

2 ScatteringScattering crosscross sectionsection

The differential cross-section for scattering is definedσ as the number of particles scattered into an element of dΩ in the direction (θ,φ)ϑperφ unit time : Ω d ( , ) 1 dN (1.1) = sc d J inc dΩ [dimensions of an area] Jinc - incident flux The total cross-section corresponds to through any scattering angle:

(1.2)

Most scattering experiments are carried out in the laboratory (Lab) frame in which the target is initially at rest while the projectiles are moving. Calculations of the cross sections are generally easier to perform within the center-of- (CM) frame in which the center of mass of the projectiles–target system is at rest (before and after collision) Î one has to know how to transform the cross sections from one frame into the other. Note: the total σ is the same in both frames, since the total number of collisions that take place does not depend on the frame in which the observation is carried out. However, the differential cross sections dσ/dΩ are not the same in both frames, since the scattering (θ,φ) are frame dependent. 3 ConnectingConnecting thethe aanglesngles inin thethe LabLab andand CMCM framesframes

Elastic scattering of two structureless particles in the Lab and CM frames:

Note: consider for simplicity NON-relativistic kinematics!

r r V CM || V 1 L

To find the connection between the Lab and CM cross sections, we need first to find how the scattering angles in one frame are related to their counterparts in the other.

If denote the of m1 in the Lab and CM frames, respectively, and if denotes the position of the center of mass with respect to the Lab frame, we have A time derivative of this relation leads to (1.3) where are the of m1 in the Lab and CM frames before collision and is the of the CM with respect to the Lab frame. Similarly, the velocity of m after collision is 1 (1.4) 4 ConnectingConnecting thethe aanglesngles inin thethe LabLab andand CMCM framesframes

Since r r (1.5) V CM || V 1 L (1.6)

Dividing (1.6) by (1.5), we end up with (1.7)

Use that CM momenta r r r (1.8) pCM = p1 L + p 2 L r r r ( m 1 + m 2 )V CM = m 1V 1 L + m 2V 2 L

Since from (1.8) Î

(1.9) or from (1.9) Î (1.10)

(1.11)

5 ConnectingConnecting thethe aanglesngles inin thethe LabLab andand CMCM framesframes

Since the center of mass is at rest in the CM frame, the total momenta before and after collisions are separately zero: r r v r r r pC = p1C + p 2 C = pC′ = p1′ C + p′2 C = 0

(1.12)

(1.13)

Since the kinetic is conserved: (1.14)

(1.15)

In the case of elastic collisions, the of the particles in the CM frame are the same before and after the collision;

From (1.11) Î (1.16)

Dividing (1.9) by (1.16) Î (1.17)

6 ConnectingConnecting thethe aanglesngles inin thethe LabLab andand CMCM framesframes

Finally, a substitution of (1.17) into (1.7) yields

(1.17) and using we obtain

(1.18)

Note: In a similar way we can establish a connection between θ2 and θ. From (1.4) we have + using Î the x and y components of this relation are

(1.19)

(1.20)

7 ConnectingConnecting thethe LabLab andand CMCM CrossCross SectionsSections

The connection between the differential cross sections in the Lab and CM frames can be obtained from the fact that the number of scattered particles passing through an infinitesimal cross section is the same in both frames: (1.21) What differs is the solid angle dΩ : in the Lab frame: in the CM frame:

(1.22)

Since there is a cylindrical symmetry around the direction of the incident beam (1.23)

From (1.18) Î (1.24)

8 ConnectingConnecting thethe LabLab andand CMCM CrossCross SectionsSections

Thus : (1.25)

From (1.23) and (1.20) Î

(1.26)

Limiting cases: 1) the Lab and CM results are the same,

since (1.17) leads to θ1=θ Î (1.27)

2) then from (1.20) Î

from (1.25) Æ (1.28)

9 FromFrom classicalclassical toto quantumquantum scatteringscattering theorytheory

1. Classical : - scattering of hard ‚‘ - individual well-defined trajectories

2. theory: - scattering of pakeges Å wave–particle duality - probabilistic origin of scattering process

10 II.II. ClassicalClassical ttrajectoriesrajectories andand ccrossross--ssectionsections

Scattering trajectories, corresponding to different impact parameters, b give different scattering angles θ. All of the particles in the beam in the hatched region of area dσ = 2π bdb are scattered into the angular region (θ, θ + dθ) The equations of for incident particle: (2.1)

+ initial conditions: Î defines the trajectory (2.2)

(2.3) since initial kinetic energy

For a particle obeying classical :  the trajectory for the unbound motion, corresponding to a scattering , is deterministically predictable, given by the interaction potential and the initial conditions; the path of any scatterer in the incident beam can be followed, and its angular deflection is determined as precisely as required 11 ClassicalClassical ttrajectoriesrajectories andand ccrossross--ssectionsections

The number of particles scattered per unit time into the angular region (θ, θ+dθ) with any value of φ can be written as

(2.4)

(2.5)

The knowledge of b(θ), obtained directly from ’s laws or other methods, is then sufficient to calculate the scattering cross-section.

For the scattering from nontrivial central , the trajectory can be obtained from the equations of motion by using energy- and angular - conservation methods E.g., one can rewrite L -

in the form (2.6)

12 ClassicalClassical ttrajectoriesrajectories andand ccrossross--ssectionsections

The angular momentum can be written via the

Îfrom (2.6) (2.7)

Î the angle through which the particle moves between two radial r1 and r2 :

(2.8)

Scattering trajectories in a central potential :

rmin - the of closest approach, Θ – deflection angle (2.9)

Use that the initial angular momentum and

(2.10) Î from (2.8), (2.9)

13 ClassicalClassical CoulombCoulomb scatteringscattering

The potential corresponding to the scattering of charged particles via Coulomb’s law can be written in the general form (2.11)

where the constant corresponds to the electrostatic interaction of point-like charges Z1e and Z2e

Î from (2.10) (2.12)

Note that for large (E→∞) or vanishing charges (A → 0), one has rmin → b

Use that (2.13)

Î from (2.12)

(2.14)

14 Classical Coulomb scattering

Use that (2.15)

Î Rutherford formula - characteristic for Coulomb scattering from a point-like charge:

(2.16)

Note: If one attempts to evaluate the total cross-section using Eq. (2.16), one obtains an infinite result. This divergence is due to the infinite range of the 1/r potential, and the “infinity” in the integral comes from scatterings as θ → 0 which corresponds to arbitrarily large impact parameters where the Coulomb potential causes a scatter through an arbitrarily small angle, which still contributes to the total cross-section.

15 III.III. QuantumQuantum ScatteringScattering

 The of spinless, nonrelativistic particles: Consider the between two spinless, nonrelativistic particles of m1 and m2. During the scattering process, the particles interact with each other. If the interaction is time independent, we can describe the two-particle system with stationary states (3.1)

where ET is the total energy and is a solution of the time-independent Schrödinger equation: (3.2)

is the potential representing the interaction between the two particles.

In the case where the interaction between m1 and m2 depends only on their relative distance one can reduce the eigenvalue problem (3.2) to two decoupled eigenvalue problems:

one for the center of mass (CM), which moves like a free particle of mass M= m1+m2 (which is of no concern to us here) and another for a fictitious particle with a reduced mass which moves in the potential (3.3) 16 ScatteringScattering ofof spinlessspinless particlesparticles

In the incident particle is described by means of a wave packet that interacts with the target. After scattering, the consists of an unscattered part propagating in the forward direction and a scattered part that propagates along some direction (θ,ϕ)

17 ScatteringScattering ofof spinlessspinless particlesparticles

One can consider (3.3) as representing the scattering of a particle of mass μ from a fixed scattering center described by V(r), where r is the distance from the particle μ to the center of V(r): (3.3)

⎧ ≠ 0 ,if r ≤ a assume that V(r) has a finite range a: Vˆ ( r ) = ⎨ ⎩ = 0 ,if r > a

 Outside the range, r > a, the potential vanishes, V(r)= 0; the eigenvalue problem (3.3) then becomes (3.4)

In this case μ behaves as a free particle before the collision and can be described by a plane wave (3.5)

where is the wave vector associated with the incident particle;A is a normalization factor. Thus, prior to the interaction with the target, the particles of the incident beam are independent of each other; they move like free particles, each with a momentum 18 ScatteringScattering ofof spinlessspinless particlesparticles

When the incident wave (3.5) collides or interacts with the target, an outgoing wave is scattered out. In the case of an isotropic scattering, the scattered wave is spherically symmetric, having the form In general, however, the scattered wave is not spherically symmetric; its amplitude depends on the direction (θ,ϕ) along which it is detected and hence

(3.6) where f (θ,ϕ) is called the scattering amplitude, is the wave vector associated with the scattered particle, and θ is the angle between After the scattering has taken place, the total wave consists of a superposition of the incident plane wave (3.5) and the scattered wave (3.6):

(3.7)

where A is a normalization factor; since A has no effect on the cross section, as will 19 be shown later, we will take it equal to one. ScatteringScattering amplitudeamplitude andand differentialdifferential crosscross sectionsection

Introduce the incident and scattered flux densities:

(3.8)

(3.9)

Inserting (3.5) into (3.8) and (3.6) into (3.9) and taking the magnitudes of the expressions thus obtained, we end up with

(3.10)

We recall that the number dN(θ,ϕ) of particles scattered into an element of solid angle dΩ in the direction (θ,ϕ) and passing through a surface element dA= r2dΩ per unit time is given as follows: (3.11)

Using (3.10) Î (3.12)

20 ScatteringScattering amplitudeamplitude andand differentialdifferential crosscross sectionsection

By inserting (3.12) and in (1.1), we obtain:

(3.13)

Since the normalization factor A does not contribute to the differential cross section, it will be taken to be equal to one.

For elastic scattering k0 is equal to k; so (3.13) reduces to

(3.14)

Î The problem of determining the differential cross section dσ/dΩ therefore reduces to that of obtaining the scattering amplitude f (θ,ϕ).

21