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 experiments 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 target in an evacuated chamber. Although most alphas were gently deflected as expected (the prevailing atomic model assumed smooth charges embedded throughout the atom), the experimenters 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 diameter 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 θ, as shown in Figure 1.
1
!
p
b
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 / 4πε 0r 2 = Zα / r 2 in natural units, where r is
the distance between the particles. We can determine the angle of scattering θ from the following
argument. After the scattering the beam particle momentum p is unchanged (elastic scattering),
but it is deflected by an angle θ. Thus p y = psin θ . But we can also calculate p y by integrating
the force along the y direction. Let ψ be the angle measured from the nucleus to the particle starting from the x-axis, with −π ≤ ψ ≤ θ . Then using p y = ∫ F sinψ dt , we obtain
∞
Zα
−∞
r2
py = ∫
sinψ dt = ∫
θ
Zα
θ
−π
⎤
dt
Zα
Zα
sinψ
dψ =
cosψ ⎥ =
(1+ cosθ )
2
dψ
bv
r
⎦ −π bv
where r 2ψ = bv from angular momentum conservation. We thus obtain a relation between the
scattering angle θ, momentum p and impact parameter b:
psin θ =
Zα
Zα
1+ cosθ ) ⇒ tan 12 θ =
(
bv
pvb
We are interested in finding the cross sectional area dσ corresponding to a particle scattering into
the small angular region θ to θ + dθ. This will happen if the particle has an impact parameter between 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 dσ = 2π bdb and using the relation just derived for tan 12 θ
b=
Zα
cot 12 θ
pv
db =
Zα
1
2
2 pv sin 1 θ
2
we obtain
2
2
Zα ) cos 12 θ ( Zα ) 2π sin θ dθ
(
dσ = 2π bdb = 2π 2 2 3
= 2 2
2 p v sin 12 θ
p v (1− cosθ )2
2
The right-most expression was obtained by multiplying numerator and denominator by sin 12 θ
and applying the half-angle trig identities sin θ = 2sin 12 θ cos 12 θ and 1− cosθ = 2sin 2 12 θ . Since
the scattering process does not depend on φ (axially symmetric scattering),
∫ dφ = 2π
and we
can write the numerator of the rightmost fraction as 2π sin θ dθ → dΩ , the differential of solid
angle. We thus obtain the well-known Rutherford differential cross section:
dσ ( Zα c )
1
=
dΩ
p 2v 2 (1− cosθ )2
2
where I have restored the c term. Two things should be noted: (1) α c = e2 / 4πε 0 so this classical result does not depend on the QM-related quantity  ; (2) integrating over
dΩ = 2π d cosθ dθ gives an infinite total cross section, a consequence of the infinite range of the
coulomb force.
1.2
Using the momentum transfer q
It’s conventional to write the cross section in terms of the momentum transfer q = pi − p f (initial minus final momentum), where q2 = 2 p 2 (1− cosθ ) . The Rutherford formula then becomes
dσ 4 ( Zα cm)
=
dΩ
q4
2
In QM scattering theory we normally express the differential cross section in terms of q2 or the
Lorentz invariant quantity q 2 . 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 proceed from an initial quantum state ψ i to a final state ψ f is given by the Fermi Golden Rule:
dJ = ψ
where ψ
f
f
M ψi
2
(
× dN f × 2πδ E f − Ei
)
M ψ i = ∫ ψ ∗f Mψ i d 3x describes the QM transition, dNf is the differential number
(
)
of states for final state f (derived in the Appendix) and δ 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 4momentum conservation.
For scattering processes, we assume that we have a free particle in the initial state before scattering and a free particle in the final state after scattering. Thus, for a scattering potential U(x),
3
ψ i (x ) =
ψ
f
1
V
e
ipi ⋅x
M ψi =
dN f = V
ψ
f
1
(x ) =
V
e
ip f ⋅x
1
U ( x ) eiq⋅x d 3x
∫
V
d3p f
( 2π )3
where q = pi − p f is the momentum transfer and I have used natural units. To get a cross section, we use the standard expression dJ = ji N t dσ (derived in a previous note) with one target
particle ( N t = 1 ) and a plane wave flux density ji = ni vi = 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:
3
2 d pf
dJ
1
dσ =
= M fi ( q )
2πδ E f − Ei
3
ji N t vi
2
π
( )
(
)
2
2 p f dΩ f
1
= M fi ( q )
dp f δ E f − Ei
vi
4π 2
(
)
M fi ( q ) = ∫ eiq⋅xU ( x ) d 3x , 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 δ E f − Ei = ∫
dp f
dE f
(
)
dE f δ E f − Ei =
dp f
dE f
=
1
vf
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:
2
pf
2
dσ
= M fi ( q )
dΩ
4π 2v v
i f
This formula allows the scattered particle to change species with different masses, though normally vi = v f . Note that all the physics is in the matrix element M fi ( q ) ; everything else is kinematics and phase space.
4
3 QM Scattering from a Coulomb Potential
3.1
Derivation of scattering cross section
Here the potential is U ( x ) = Zα / r . We calculate M fi ( q ) ∫ eiq⋅x ( Zα / r ) d 3x = 4π Zα / q2 (left as
an exercise1) and get the cross section (using vi = v f in the final step)
pf
pf
2
4π Zα )
(
dσ
4Z 2α 2 m2
= M fi ( q )
=
=
dΩ
4π 2vi v f
q4
4π 2vi v f
q4
2
2
2
which agrees exactly with the classical Rutherford formula. The graph in Figure 2 shows the
Rutherford differential cross section for α particles on gold for −1 ≤ cosθ ≤ 0.75 .
Figure 2: Rutherford differential cross section (barn/steradian) vs cosθ
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/ q2 . Putting it all together yields M fi  Ze2 / q2 = 4π Zα / q2 , which happens to be the
1
This can most easily be shown using spherical coordinates, with eip⋅x = eipr cosθ and d 3x = 2π r 2 dr d cosθ .
5
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 M
2
fi
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. Let’s 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
0
( Zzα c )2
−1
p 2v 2
σ (θ ≥ 90° ) = ∫
2π d cosθ
(1− cosθ )
2
=
π ( Zzα c )
2
p 2v 2
Alpha particles from radium emission have kinetic energies of 4.87 MeV. Using p 2v 2 = 4K 2 we
use energy units with c = 0.197 GeV ⋅ fm to obtain
σ (θ ≥ 90° ) =
π ( Zzα c )
4K
2
2
=
3.14 × ( 79 × 2 × 0.197 / 137 )
4 × 0.00487
2
2
= 1700 fm 2 = 17b
The probability of scattering through this angle is P = σ n Au Δx  4 × 10−5 , so approximately 1 in
25,000 alpha particles will scatter through this angle. Rutherford’s team measured 1 in 20,000
(the scatterings had to be counted by hand!). Figure 4 shows the integrated Rutherford cross section for σ ( −1 ≤ x ≤ cosθ ) .
6
Figure 4: Integrated Rutherford cross section for −1 ≤ x ≤ cosθ vs cosθ
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.
1!
1
2
2!
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 − v 2 / 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
⎛
) V1 = pi ⎜⎝ m1
1
7
+
1 ⎞ 1 pi 1
=
m2 ⎟⎠ V µ V
where µ is the reduced mass. Likewise, the integration over final states has to be modified because the energy delta function is δ E1′ + E2′ − E1 − E2 and both final state energies depend on
(
)
p1′ = p2′ ≡ p f . Integration over dp f yields:
d3p f
(
p 2f dΩ
)
2πδ E1′ + E2′ − E1 − E2 =
3
( 2π )
(
4π 2 v1 + v2
)
p 2f dΩ
=
⎛ pf pf ⎞
4π 2 ⎜
+
⎟
⎝ m1 m2 ⎠
=
pf µ
4π 2
dΩ
So the differential scattering cross section for two particles in the CM frame becomes:
2
pf
dσ 4Z 2α 2
=
dΩ
q4
v1 + v2
(
)2
=
4Z 2α 2 µ 2
q4
=
Z 2α 2 µ 2
p 4 (1− cosθ )
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 ) = α g e
−mg r
/ 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 = 4πα g , in contrast to the Coulomb potential which involves the exchange of a massless photon with coupling constant e = 4πα . The matrix element
is
M fi ( q ) = ∫ eiq⋅x
αg
r
e
−mg r 3
d x=
4πα g
(q
2
+ mg2
)
which you can verify. This leads to the differential cross section for a fixed potential
2
pf
2
dσ
= M fi ( q )
=
dΩ
4π 2v v
i f
(q
4α g2
2
+ mg2
)
p 2f
2
vi v 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 2f / vi v f so that we can approximately extend the
formulas to relativistic energies later.
When mg2  q2 we have approximately
8
2
2
dσ 4α g p f

dΩ m4 vi v f
g
⇒ σ
16πα g2 p 2f
mg4
vi v f
For mg2  q2
Thus when a heavy particle is exchanged, the angular distribution is approximately uniform, unlike Coulomb scattering which strongly peaks in the forward direction because of the
(1− cosθ )−2 term. But the mg−4 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 α
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
dσ
=
dΩ
(q
4α g2
2
+ mg2
)
p 2f
2
( v1 + v2 )2
For massive exchanged particles like W bosons, we can extend this (approximately) to highly
relativistic particles:
2
2
πα 2
dσ 4αW E 2 4αW s
≅ 4
= 4
⇒ σ = 4W s
dΩ m
4
mW 16
mW
W
To approximate the weak interactions, we set αW = α and mW = 80 GeV. This yields
σ  4.1× 10−12 s . The correct answer for neutrino scattering from another fermion using the accepted Standard Model theory is σ = G F2 s / π  4.3× 10−11 s , which involves a fully relativistic
formulation, including spin effects. For antineutrinos the total cross section is
σ = GF2 s / 3π  1.4 × 10−11 s . So our approximate treatment is good to an order of magnitude.
Note that in reality αW  0.74α .
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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
N xπ / L = k = px /  , where n is a positive integer and px is positive. Thus the number of states
between px and px + dpx is dN x = Ldpx / π  . Repeating the argument for y and z gives
dN = Vd 3 p / π 33 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 = Vd 3 p / ( 2π  ) . Thus the differential density of states is
3
dn = d 3 p / ( 2π  ) .
3
5.2
Example 1: Total density of states for p ≤ pmax
Let’s calculate the total density of states in a range of momentum p ≤ pmax . Integrating using
spherical variables ( d 3 p = p 2 dpdΩ ) gives
( )
n = ∫ d 3 p / 2π 3 3 = ∫
5.3
pmax
0
3
p 2 dp / 2π 23 = pmax
/ 6π 2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/k BT
dN ( v ) ∝
. The distribution of velocities at thermal equilibrium is thus
d3p
( 2π )
3 3

e
− E/k BT
∝ p 2e
− E/k BT
dp ∝ v 2e
− 12 mv 2 /k BT
dv
The normalization constant can be obtained by integration over all velocities. The Maxwell velocity distribution is shown in Figure 6.
10
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/ e
the density of states gives the Planck photon number density vs energy:
dnγ =
2d 3 pγ
( 2π )
3 3
 e
1
Eγ /k BT
−1
⇒
dnγ
dEγ
=
Eγ2
E/k BT
)
− 1 and
1
π 2 ( c ) e
3
Eγ /k BT
−1
with Eγ = pγ 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
(
)3
nγ = 2ζ ( 3) k BT / c / π 2 , where ζ ( 3)  1.202 is the Riemann zeta function for x = 3. The total
∞
(
energy density is uγ = ∫ Eγ nγ dEγ = π 2 k BT
0
)4 / 15( c )3 and the average photon energy
Eγ = uγ / nγ = π k BT / 30ζ ( 3)  2.701k BT . The most probable photon energy, obtained by dif4
ferentiation, is Eγ
MP
= 1.594k BT . The Planck energy distribution is shown in Figure 7.
11
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
ψ ( x1, x2 ) =
1
2
( ) ( )
( ) ( )
⎡ψ 1 x1 ψ 2 x2 − ψ 2 x1 ψ 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 simlicity, only the spatial antisymmetrization is shown here.
12
dne =
2d 3 p
( 2π )3 3
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, p F , known as the “Fermi momentum”. This yields a total electron density of
ne = ∫
2d 3 p
( 2π )3 3
=∫
pF
0
p 2 dp
π 23
=
p3F
3π 23
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 > p F .
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 ρ = 8.94 g / cm 3 and an atomic mass of A = 63.54, copper has a free electron density of ne = 8.5 × 1028 / m 3 . This corresponds to an electron Fermi momentum of 2690
eV/c or a velocity of v F  1.6 × 106 m/s. This velocity is an order of magnitude faster than the
average thermal electron velocity at room temperature of vthermal = 3k BT / me  1.2 × 105 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.
13