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Paul Avery PHZ4390 Sep. 25, 2013

Scattering in Quantum Mechanics 1 Rutherford Scattering ................................................................................................................ 1

1.1 Derivation of the classical Rutherford scattering formula .................................................. 1 1.2 Using the momentum transfer q ......................................................................................... 3

2 General Formula for Scattering in Nonrelativistic QM ............................................................. 3 3 QM Scattering from a Coulomb Potential ................................................................................. 5

3.1 Derivation of scattering cross section ................................................................................. 5 3.2 Matrix element and Feynman diagram for coulomb scattering .......................................... 5 3.3 Example: large angle scattering in the Rutherford experiment .......................................... 6 3.4 Coulomb scattering of two finite mass particles ................................................................ 7

4 QM Scattering from a Potential with Massive Particle Exchange ............................................ 8 4.1 Fixed potential .................................................................................................................... 8 4.2 Two interacting particles: weak interactions ...................................................................... 9

5 Appendix: Density of States .................................................................................................... 10 5.1 Derivation of density of states formula ............................................................................ 10 5.2 Example 1: Total density of states for p pmax ................................................................ 10 5.3 Example 2: Density of states and Maxwell velocity distribution ..................................... 10 5.4 Example 3: Density of states and Planck photon energy distribution .............................. 11 5.5 Example 4: Fermi momentum at low temperature ........................................................... 12

1 Rutherford Scattering Our understanding of scattering goes back to the Rutherford-Geiger-Marsden scattering experi-ments of 1909 which aimed to measure the structure of atoms. The experimenters used radium and a collimation system to produce a narrow alpha particle beam that struck a thin gold foil tar-get in an evacuated chamber. Although most alphas were gently deflected as expected (the pre-vailing atomic model assumed smooth charges embedded throughout the atom), the experiment-ers were astonished to find that a small fraction were scattered at large angles by what they later surmised was a massive, positively charged nucleus occupying a region less than 1/4000 the di-ameter of the atom. In their 1911 paper they derived an equation, known today as the Rutherford scattering formula, that explained their results in terms of the scattering of charged particles by a heavy point charge.

1.1 Derivation of the classical Rutherford scattering formula Consider a beam of particles, each of mass m and charge e, incident on a heavy nucleus of charge Ze, where the nucleus is assumed not to move ( M m ). Each beam particle will be deflected, depending on its momentum p and impact parameter b, through an angle , as shown in Figure 1.

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Figure 1: Scattering of a particle of charge e by a heavy nucleus of charge Ze.

The force acting on the beam particle is F = Ze2 / 40r

2 = Z / r2 in natural units, where r is the distance between the particles. We can determine the angle of scattering from the following argument. After the scattering the beam particle momentum p is unchanged (elastic scattering), but it is deflected by an angle . Thus py = psin . But we can also calculate py by integrating the force along the y direction. Let be the angle measured from the nucleus to the particle start-ing from the x-axis, with . Then using

py = F sin dt , we obtain

py =

Zr2

sin dt

=

Zr2

sin dtd

d

=

Zbv

cos

= Zbv

1+ cos( )

where r2 = bv from angular momentum conservation. We thus obtain a relation between the

scattering angle , momentum p and impact parameter b:

psin = Z

bv1+ cos( ) tan 12 =

Zpvb

We are interested in finding the cross sectional area d corresponding to a particle scattering into the small angular region to + d. This will happen if the particle has an impact parameter be-tween b and b + db, i.e. if it lies in an annulus of radius b centered on the nucleus. The area of the annulus is d = 2bdb and using the relation just derived for

tan 12

b = Z

pvcot 12 db =

Z2 pv

1sin2 12

we obtain

d = 2bdb = 2Z( )2

2 p2v2cos 12

sin3 12=

Z( )2p2v2

2 sind

1 cos( )2

!

bp

3

The right-most expression was obtained by multiplying numerator and denominator by sin 12

and applying the half-angle trig identities sin = 2sin 12 cos

12 and 1 cos = 2sin

2 12 . Since

the scattering process does not depend on (axially symmetric scattering),

d = 2 and we can write the numerator of the rightmost fraction as 2 sind d , the differential of solid angle. We thus obtain the well-known Rutherford differential cross section:

dd

=Zc( )2p2v2

1

1 cos( )2

where I have restored the c term. Two things should be noted: (1) c = e2 / 40 so this clas-

sical result does not depend on the QM-related quantity ; (2) integrating over d = 2d cosd gives an infinite total cross section, a consequence of the infinite range of the coulomb force. 1.2 Using the momentum transfer q

Its conventional to write the cross section in terms of the momentum transfer q = pi p f (ini-

tial minus final momentum), where q2 = 2 p2 1 cos( ) . The Rutherford formula then becomes

dd

=4 Zcm( )2

q4

In QM scattering theory we normally express the differential cross section in terms of q2 or the

Lorentz invariant quantity q2 . These are the same up to a sign for elastic scattering (

Ei E f ).

2 General Formula for Scattering in Nonrelativistic QM We can compute cross section formulas in nonrelativistic QM. The rate dJ for a process to pro-ceed from an initial quantum state i to a final state

f is given by the Fermi Golden Rule:

dJ = f M i

2 dN f 2 E f Ei( )

where f M i =

f M id

3x describes the QM transition, dNf is the differential number of states for final state f (derived in the Appendix) and

E f Ei( ) is a delta function enforcing

energy conservation. The term phase space is used for the dN f term. In relativistic scattering

phase space generalizes to multiple particles in the final state and has a delta function for 4-momentum conservation.

For scattering processes, we assume that we have a free particle in the initial state before scatter-ing and a free particle in the final state after scattering. Thus, for a scattering potential U(x),

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i x( ) = 1V eipix f x( ) = 1V e

ip f x

f M i =1V

U x( )eiqxd3x

dN f =Vd3p f

2( )3

where q = pi p f is the momentum transfer and I have used natural units. To get a cross sec-

tion, we use the standard expression dJ = jiNtd (derived in a previous note) with one target

particle ( Nt = 1) and a plane wave flux density ji = nivi = vi / V (a single beam particle in the volume moving at velocity vi). When we divide the rate by the initial flux the volumes cancel and we obtain:

d = dJjiNt

= 1vi

M fi q( )2 d3p f

2( )32 E f Ei( )

= 1vi

M fi q( )2 p f

2 d f4 2

dp f E f Ei( )

M fi q( ) = eiqxU x( )d3x , the Fourier transform of the potential, is called the matrix element.

We can integrate out the delta function using a standard variable change

dp f E f Ei( ) = dp fdE f dE f E f Ei( ) =

dp fdE f

= 1v f

using dp f / dE f = 1/ v f , which holds true even for relativistic momenta. We finally obtain a

general formula for the differential cross section:

dd

= M fi q( )2 p f

2

4 2viv f

This formula allows the scattered particle to change species with different masses, though nor-mally

vi = v f . Note that all the physics is in the matrix element M fi q( ) ; everything else is kin-

ematics and phase space.

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3 QM Scattering from a Coulomb Potential 3.1 Derivation of scattering cross section

Here the potential is U x( ) = Z / r . We calculate M fi q( ) eiqx Z / r( )d3x = 4Z / q2 (left as

an exercise1) and get the cross section (using vi = v f in the final step)

dd

= M fi q( )2 p f

2

4 2viv f=

4Z( )2q4

p f2

4 2viv f= 4Z

2 2m2

q4

which agrees exactly with the classical Rutherford formula. The graph in Figure 2 shows the Rutherford differential cross section for particles on gold for 1 cos 0.75 .

Figure 2: Rutherford differential cross section (barn/steradian) vs cos

3.2 Matrix element and Feynman diagram for coulomb scattering Scattering from a fixed coulomb potential is represented by the Feynman diagram in Figure 3. The diagram gives the essential elements of the matrix element. The upper and lower vertices have coupling constants e and Ze, respectively, while the massless photon propagator brings in a factor 1/ q

2 . Putting it all together yields M fi Ze

2 / q2 = 4Z / q2 , which happens to be the

1 This can most easily be shown using spherical coordinates, with eipx = eipr cos and d

3x = 2r2dr d cos .

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correct answer! For relativistic scattering in quantum electrodynamics (QED), there are precise rules that allow one to exactly calculate the matrix element from a Feynman diagram, including all constants and possible internal loops (which we defer to a later discussion). Once the matrix

element M fi is known, the cross section can be calculated by multiplying

Mfi2 by phase space

and dividing by the flux, as discussed in Section 2.

Figure 3: Feynman diagram of particle scattering from fixed potential

3.3 Example: large angle scattering in the Rutherford experiment When Rutherford did his classic experiments about 100 years ago, he used a collimated beam of alpha particles from radium emission to strike a gold target 400 nm thick. Lets calculate the fraction of alpha particles scattered at 90 or more from his formula. Since gold has an atomic mass of ~197, we can safely neglect its recoil. At a density of 19.3 and a thickness of 400 nm, the foil is only about 1600 atoms thick (Rutherford chose gold foil because it can be processed into extremely thin sheets which minimizes multiple scattering).

The total cross section for scattering 90 or more is easily calculated, with Z = 79 for gold and z = 2 for alpha particles, to be

90( ) = Zzc( )2

p2v22d cos

1 cos( )210 =

Zzc( )2p2v2

Alpha particles from radium emission have kinetic energies of 4.87 MeV. Using p2v2 = 4K 2 we

use energy units with c = 0.197 GeV fm to obtain

90( ) = Zzc( )

2

4K 2=

3.14 79 2 0.197 / 137( )24 0.004872

= 1700 fm2 = 17b

The probability of scattering through this angle is P = nAux 4105 , so approximately 1 in

25,000 alpha particles will scatter through this angle. Rutherfords team measured 1 in 20,000 (the scatterings had to be counted by hand!). Figure 4 shows the integrated Rutherford cross sec-tion for 1 x cos( ) .

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Figure 4: Integrated Rutherford cross section for 1 x cos vs cos

3.4 Coulomb scattering of two finite mass particles We can generalize the previous scattering result to two particles with finite mass. The process is shown in the Feynman diagram in Figure 5.

Figure 5: Coulomb scattering of two particles

Here it is more natural to work in the CM frame. Let primed quantities refer to quantities in the final state. The initial flux is

j = v1 v2 / V , where v = p / m for each initial particle. In the CM

frame the momenta have equal magnitudes pi and opposite directions, which gives a flux of

j = v1 + v2( ) 1V = pi

1m1

+ 1m2

1V

=pi

1V

1

2

!1

!2

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where is the reduced mass. Likewise, the integration over final states has to be modified be-cause the energy delta function is

E 1 + E 2 E1 E2( ) and both final state energies depend on

p 1 = p 2 p f . Integration over

dp f yields:

d3p f

2( )32 E 1 + E 2 E1 E2( ) =

p f2 d

4 2 v1 + v2( )=

p f2 d

4 2p fm1

+p fm2

=p f

4 2d

So the differential scattering cross section for two particles in the CM frame becomes:

dd

= 4Z2 2

q4p f

2

v1 + v2( )2= 4Z

2 22

q4= Z

2 22

p4 1 cos( )2

This reduces to the standard Rutherford scattering formula when m2 is infinite.

4 QM Scattering from a Potential with Massive Particle Exchange 4.1 Fixed potential

Consider the potential U x( ) = gemgr / r , which is similar to the Coulomb potential for

r 1/ mg but falls exponentially at large distances. It corresponds to the exchange of a particle

of mass mg with coupling constant g = 4 g , in contrast to the Coulomb potential which in-

volves the exchange of a massless photon with coupling constant e = 4 . The matrix element is

M fi q( ) = eiqx gr

emgrd3x =4 g

q2 + mg2( )

which you can verify. This leads to the differential cross section for a fixed potential

dd

= M fi q( )2 p f

2

4 2viv f=

4 g2

q2 + mg2( )2

p f2

viv f

Unlike the Coulomb potential, the total cross section for massive particle exchange is finite. Note that I express the cross sections in terms of

p f

2 / viv f so that we can approximately extend the formulas to relativistic energies later.

When mg

2 q2 we have approximately

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dd

4 g

2

mg4

p f2

viv f

16 g2

mg4

p f2

viv fFor mg

2 q2

Thus when a heavy particle is exchanged, the angular distribution is approximately uniform, un-like Coulomb scattering which strongly peaks in the forward direction because of the

1 cos( )2 term. But the

mg4 term also leads to very small cross sections when

mg is large, which is exactly the case for weak interactions which have a coupling constant comparable to but an exchanged mass mW ~ 80 GeV. 4.2 Two interacting particles: weak interactions Extending the previous result to two finite mass particles interacting in their CM frame is easy and follows the method we used for coulomb scattering. The CM cross section is

dd

=4 g

2

q2 + mg2( )2

p f2

v1 + v2( )2

For massive exchanged particles like W bosons, we can extend this (approximately) to highly relativistic particles:

dd

4W

2

mW4

E2

4=

4W2

mW4

s16

=W

2

mW4 s

To approximate the weak interactions, we set W = and mW = 80 GeV. This yields

4.11012s . The correct answer for neutrino scattering from another fermion using the ac-

cepted Standard Model theory is = GF2 s / 4.31011s , which involves a fully relativistic

formulation, including spin effects. For antineutrinos the total cross section is

= GF2 s / 3 1.41011s . So our approximate treatment is good to an order of magnitude.

Note that in reality W 0.74 .

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5 Appendix: Density of States Cross sections in nonrelativistic and relativistic scattering are proportional to a quantity known as the density of states, which is essentially the number of quantum states possible for each particle at a given energy and direction. But the density of states plays an essential role in many areas of physics, where it is used to calculate the Maxwell distribution of gas molecular energies, the Planck photon energy distribution, degeneracy pressure in white dwarfs and neutron stars, electrical and thermal conductivity in materials, etc. We derive in the next subsection the density of states for a particle using elementary QM.

5.1 Derivation of density of states formula Consider a particle normalized to lie within a box of length L. Then boundary conditions force the wavefunction to be zero on the boundary, leading to quantized momentum states satisfying

Nx / L = k = px / , where n is a positive integer and px is positive. Thus the number of states between px and px + dpx is dNx = Ldpx / . Repeating the argument for y and z gives

dN =Vd3p / 33 for the number of states in a small momentum slice. However, if we consider

that momentum can range equally over positive and negative values, then the differential number

of states in a volume V is dN =Vd3p / 2( )3 . Thus the differential density of states is

dn = d3p / 2( )3 .

5.2 Example 1: Total density of states for p pmax

Lets calculate the total density of states in a range of momentum p pmax . Integrating using

spherical variables ( d3p = p2dpd ) gives

n = d3p / 2 3( )3 = p2 dp / 2 230pmax = pmax3 / 6 23

5.3 Example 2: Density of states and Maxwell velocity distribution At thermal equilibrium in a gas at temperature T, the relative probability of a single gas molecule to have energy E is e

E /kBT . The distribution of velocities at thermal equilibrium is thus

dN v( ) d3p

2( )33eE /kBT p2eE /kBT dp v2e

12 mv

2 /kBT dv

The normalization constant can be obtained by integration over all velocities. The Maxwell ve-locity distribution is shown in Figure 6.

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Figure 6: Maxwell velocity distribution showing the most probable velocity and rms velocity

5.4 Example 3: Density of states and Planck photon energy distribution For a region at temperature T, we can use density of states reasoning to calculate the distribution

of photon energies (and frequencies). Using the photon probability function 1/ eE /kBT 1( ) and

the density of states gives the Planck photon number density vs energy:

dn =2d3p

2( )331

eE /kBT 1

dndE

=E

2

2 c( )31

eE /kBT 1

with E = p c . The factor of two in the density of states accounts for both photon polarization

states which must be counted. Integrating this over energy gives the photon number density

n = 2 3( ) kBT / c( )3 / 2 , where 3( ) 1.202 is the Riemann zeta function for x = 3. The total energy density is

u = E n dE0

= 2 kBT( )4 / 15 c( )3 and the average photon energy

E = u / n =

4kBT / 30 3( ) 2.701kBT . The most probable photon energy, obtained by dif-ferentiation, is

E MP = 1.594kBT . The Planck energy distribution is shown in Figure 7.

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Figure 7: Planck energy distribution for photons, showing the most probable and mean energies

5.5 Example 4: Fermi momentum at low temperature Half-integer particles (fermions) follow Fermi-Dirac statistics in which the wave function for two identical fermions is the antisymmetric product of their wavefunctions,2 i.e

x1,x2( ) = 12 1 x1( ) 2 x2( ) 2 x1( )1 x2( )

An important consequence is that fermions obey the Pauli exclusion principle, which says that no more than one fermion of a given type can occupy the same quantum state. The exclusion principle explains why atomic orbitals never have more than two electrons (same spatial wave function but opposite spin states) and additional electrons are forced to occupy higher energy levels as the lower orbitals fill up. Without the exclusion principle atoms would have all their electrons in the ground state and chemistry would be impossible.

The exclusion principle also affects the behavior of free electrons in close proximity with one another, a situation that occurs in conditions as diverse as metals and white dwarf stars. As they are packed together, additional electrons have to be placed in higher momentum states as the lower ones are filled.

At low temperature (satisfied for most interesting situations) the differential number density for electrons is given by the density of states

2 The antisymmetrization applies to the entire wavefunction, including nonspatial components such as spin. For sim-licity, only the spatial antisymmetrization is shown here.

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dne =2d3p

2( )33

where the factor of 2 accounts for the two spin states of each electron that must be counted. So the total electron number density is obtained by integrating this to the maximum momentum achieved, pF , known as the Fermi momentum. This yields a total electron density of

ne =2d3p

2( )33= p

2dp 230

pF =pF

3

3 23

The Fermi momentum can therefore be determined directly from the electron number density. Additional electrons can be added to the volume only if they have momenta p > pF .

We can apply this analysis to metals which always have one or more free electrons per atom, forming an electron gas within the material. Copper, for example, has 1 free electron per atom. With a mass density of = 8.94 g / cm

3 and an atomic mass of A = 63.54, copper has a free elec-

tron density of ne = 8.51028 / m3 . This corresponds to an electron Fermi momentum of 2690

eV/c or a velocity of vF 1.6106 m/s. This velocity is an order of magnitude faster than the

average thermal electron velocity at room temperature of

vthermal = 3kBT / me 1.2105 m/s.

Thus the fastest electrons in the metal are moving at extremely high speed, which explains why the electrical and thermal conductivity of metals is so high compared to other materials.