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Chapter 9
SCATTERING THEORY
9.1 General Considerations
In this chapter we consider a situation of considerable experimental and theoretical inter- est, namely, the scattering of particles o¤ of a medium containing some type of scattering centers, such as atoms, molecules, or nuclei. The basic experimental situation of interest is indicated in the …gure below.
An incident beam of particles impinges upon a target, which maybe a cell con- taining atoms or molecules in a gas, a thin metallic foil, or a beam of particles moving at right angles to the incident beam. As a result of interactions between the particles in the initial beam and those in the target, some of the particles in the beam are de‡ected and emerge from the target traveling along a direction (μ; Á) with respect to the original beam direction, while some are left unscattered and emerge out the other side having undergone no de‡ection (or undergo “forward scattering”). The number of particles de‡ected along a given direction are then counted in a detector of some sort. The kinds of interactions and the analyis of general scattering situations of this type can be quite complicated. We will focus in the following discussion on the scattering of incident particles by scattering centers in the target uner the following conditions:
- The incident beam is composed of idealized spinless, structureless, point particles.
- The interaction of the particles with the scattering centers is assumed to be elastic so that the energy of the scattered particle is …xed, the internal structure of the scatterer (if any) and, thus, the potential seen by the scattered particle does not change during the scattering event.
- There is no multiple scattering, so that each incident particle interacts with at most one scattering center, a condition that can be obtained with su¢ciently thin or dilute targets.
264 Scattering Theory
- The scattering potential V (~r 1 ; ~r 2 ) = V (j~r 1 ¡ ~r 2 j) between the incident particle and the scattering center is a central potential, so we can work in the relative coordinate and reduced mass of the system.
Under these conditions, the picture of interest reduces to that depicted below, in with an incident particle characterized by a plane wave of wavevector ~k = k^z along a direction that we take parallel to the z-axis, and scattered particles emerging through a in…nitesimal solid angle d along some direction (μ; Á). We may characterize the incident beam by its (assumed uniform) current density J^ ~i along the z (^) axis. Classically J~i = n~vi
where n = dN=dV is the particle number density characterizing the beam. The incident particle current through a speci…ed surface S is then the surface integral
I =
dNS dt
Z
S
J^ ~i ¢ d~S:
The scattered particle current dIS into a far away detector subtending solid angle d along (μ; Á) is found to be proportional to (i) the magnitude of the incident ‡ux density Ji, and (ii) the magnitude of the solid angle d subtended by the detector. We write
dIS =
d¾(μ; Á) d
Jid
where the constant of proportionality d¾(μ; Á)=d is referred to as the di¤erential cross section for elastic scattering in the direction (μ; Á). This quantity contains all information experimentally available regarding the interaction between scattered particles and the scattering center. We also de…ne the total cross section ¾ = ¾tot in terms of the total scattered current IS through a detecting sphere centered on the scattering center:
IS =
Z
dIS = Ji
Z
d¾(μ; Á) d
d = Ji¾tot
so that the total cross section is simply the integral over all solid angle
¾tot =
Z
sphere
d¾(μ; Á) d
Figure 1
266 Scattering Theory
Figure 2
Similar arguments in 3 -dimensions lead us to seek solutions to the eigenvalue equation of the form Á²(~r) = eikz^ + ÁS (~r)
in which the …rst part on the right clearly corresponds to the incident part of the beam and the second term corresponds to the scattered part. The subscript " indicates the energy of the incoming and outgoing particle, which is related to the wavevector of the incoming particle through the standard relation " = ~^2 k^2 = 2 m: We expect that at large distances from the scattering center, where the potential vanishes, the scattered part of the wave takes the form of an outward propagating wave. Hence, as r! 1; we anticipate that ÁS has the asymptotic behavior
ÁS (~r) » f (μ; Á)eikr r
r! 1:
Note that this form satis…es the eigenvalue equation for large r beyond the range of the potential, where the Hamiltonian reduces simply to the kinetic energy H! H 0 = P 2 = 2 m: Thus we seek solutions to the eigenvalue equation
(H 0 + V ) jÁ²i = "jÁ"i
which have the asymptotic form
Á"(~r) » eikz^ + f(μ; Á)
eikr r
where the quantity f (μ; Á) which determines the angular distribution of the scattered part of the wave is referred to, appropriately as the scattering amplitude in the (μ; Á) direction or, alternatively, as the scattering length^ since it is readily determined by dimensional analysis that f(μ; Á)^ has units of length. The obvious question that arises at this point is the following: what is the relation between the scattering length f(μ; Á)^ and the scattering cross section d¾(μ; Á)=d? To anwer this question we note that, by de…nition, the current into the detector can be written
dIS = ¾(μ; Á)Jid :
General Considerations 267
But we can also write this in terms of the scattered current density J~S = J~S (r; μ; Á); i.e.,
dIs = J~S ¢ d~S = r^2 JS d :
Thus, classically,
Ji
d¾(μ; Á) d
= r^2 Js(r; μ; Á):
Now for quantum systems these conditions will be obeyed by the corresponding mean values taken with respect to the stationary state of interest, so that
d¾(μ; Á) d
= r^2
hJs(r; μ; Á)i hJii
Thus, we need operators corresponding to the current density. Classically, for a single particle at ~r,
J^ ~(~r 0 ) = n(~r 0 )~v = ±(~r ¡ ~r 0 )~v = ±(~r ¡ ~r 0 ) ~p m
In going to quantum mechanics we replace ~r and ~p by operators and symmeterize to ensure Hermiticity. Thus,
J(~r 0 ) =
2 m
[±( R~ ¡ r 0 ) P~ + P ±~ ( R~ ¡ ~r 0 )]:
After a short calculation, the mean value of J~(~r 0 ) in the state jÁi is found to be
hÁj J~(~r 0 )jÁi =
m Re
Á¤(~r 0 )
i r~Á(~r 0 )
Using this, the incident ‡ux, associated with the plane wave part of the eigenstate, can be written
hJii =
m
Re
e¡ikz^
i
r~eikz
¯ =^
~k m
By contrast, the scattered ‡ux is then given by the expression
h J~S i = hj J~(r; μ; Á)i =
m
Re
Á¤ S (~r)
i
r~ÁS (~r)
which is most conveniently expressed in spherical coordinates, for which
rr =
@r
rμ =
r
@μ
rÁ =
r sin μ
@Á
Using these along with the assumed asymptotic form for ÁS we …nd that
hJS ir = ~k m
r^2
jf(μ; Á)j^2
hJS iμ =
m
r^3
Re
i
f ¤^
@f @μ
hJS iÁ =
m
r^3 sin μ
Re
i
f¤^
@f @Á
which shows that asymptotically the angular components of current density become negli- gible compared to the radial component. Thus, from the radial component of the current density we deduce that
d¾(μ; Á) d
= r^2
hJs(r; μ; Á)ir hJii
= jf(μ; Á)j^2 :
An Integral Equation for the Scattering Eigenfunctions 269
since H 0 has no non-real eigenvalues, this operator exists for all non-real values of the complex parameter z. We then de…ne the (causal) Green’s function operator G+(") as the limit, if it exists, of G(z) as z! " + i´; where ´ = 0+^ is a positive in…nitesimal:
G+(") = lim ´! 0 +
(" + i´ ¡ H 0 )¡^1 ´ ("+ ¡ H 0 )¡^1 :
where "+ = " + i´:
Thus, if the limit exists, then
lim ´! 0 +^
G(" + i´)(" ¡ H 0 ) = G+(")(" ¡ H 0 ) = 1
and we can write G+(")(" ¡ H 0 )jÁS i = G+(")jÁ"i
or, more simply, jÁS i = G+(")jÁ"i
Adding the incident part of the state jÁ 0 i to both sides, we then obtain the so-called Lipmann-Schwinger equation
jÁ"i = jÁ 0 i + G+V jÁ"i
which is itself a representation independent form of what is often referred to as the in- tegral scattering equation. The latter follows from the Lipmann-Schwinger equation by expressing it in the position representation. Multiplying on the left by the bra h~rj we obtain h~rjÁ"i = h~rjÁ 0 i + h~rjG+V jÁ"i
or, inserting a complete set of position states,
Á"(~r) = eikz^ +
Z
d^3 ~r^0 G+(~r; ~r^0 )V (~r^0 )Á"(~r^0 ):
Thus, we obtain an integral equation for the scattering eigenfunction Á"(~r) which has, we hope, the correct asymptotic behavior. To make this useful,we need to (i) evaluate the matrix elements G+(~r; ~r^0 ) = h~rjG+(")j~r^0 i; and (ii) actually solve the integral equation, at least in the asymptotic regime.
9.2.1 Evaluation Of The Green’s Function
To evaluate the matrix elements of the Green’s function it is most convenient to begin the calculation in the wavevector representation in which the operator (" ¡ H 0 ) is diagonal. Note that in k-space h~kj("+ ¡ H 0 )j~k^0 i = ("+ ¡ "k) ±(~k ¡ ~k^0 )
where "k = ~^2 k^2 = 2 m
so that
("+ ¡ H 0 ) =
Z
d^3 q j~qi ("+ ¡ "q) h~qj
and hence, as is easily veri…ed by direct multiplication,
("+ ¡ H 0 )¡^1 =
Z
d^3 q
j~qih~qj ("+ ¡ "q )
270 Scattering Theory
If we set
k+ =
r 2 m"+ ~^2
= k + i´ (´ = 0+)
where the positivity of ´ in this last equation follows from the positive imaginary part of "+; then we can write
G+(") =
2 m ~^2
Z
d^3 q
j~qih~qj k^2 + ¡ q^2
so that
G+(~r; ~r^0 ) = 2 m ~^2
Z
d^3 q h~rj~qih~qj~r^0 i k+^2 ¡ q^2
=
2 m ~^2
Z
d^3 q (2¼)^3
ei~q¢(~r¡~r
(^0) )
k+^2 ¡ q^2
= 2 m ~^2
Z 2 ¼
0
dÁ
Z ¼
0
dμ sin μ
Z 1
0
dq q^2 (2¼)^3
ei~qR^ cos^ μ k^2 + ¡ q^2
where R~ = ~r ¡ ~r^0 and R =
¯ R~
¯ :^ The angular integrations are readily evaluated and give
G+(~r; ~r^0 ) =
m ~^2 ¼^2 R
Z 1
0
dq
q sin(qR) k^2 + ¡ q^2
m 2 ~^2 ¼^2 R
Z 1
¡
dq
q sin(qR) k^2 + ¡ q^2
where we have used the fact that the integrand is an even function of q. Splitting sin(qR) into exponentials and setting q^0 = ¡q in the second we …nd that
G+(~r; ~r^0 ) =
m ¼~^2 R
2 ¼i
Z 1
¡
dq
qeiqR k+2 ¡ q^2
This integral can be evaluated by contour integration in the complex q-plane using Cauchy’s theorem,which states that for a function f(z)^ that is analytic in and on a closed contour ¡ (^) in the complex z-plane enclosing the point z = a,
1 2 ¼i
I
¡
f (z)dz z ¡ a
= f(a):
In our case we choose a closed path in which q runs from ¡Qto +Q and then circles back around on a semicircle in the upper half plane, which is ultimately taken to occur at jQj = 1 : Since the contribution from the integrand vanishes along this latter part, the integral over the closed contour coincides with the one of interest. To proceed, we note that
k+^2 ¡ q^2 = (k+ ¡ q) (k+ + q)
which generates simple poles at q = k+ = k + i´
and q = ¡k+ = ¡k ¡ i´
only the …rst of which is enclosed by our contour. Setting
f (q) = qeiqR k+ + q
272 Scattering Theory
and so our integral equation provides a solution of the form
Á"(~r) ' eikz^ ¡ m 2 ¼~^2
eikr r
Z
d^3 r^0 e¡ik^r¢~r
0 V (~r^0 )Á"(~r^0 )
' eikz^ + f (μ; Á)
eikr r
where
f (μ; Á) = ¡
m 2 ¼~^2
Z
d^3 r^0 e¡ikr^¢~r
0 V (~r^0 )Á"(~r^0 )
is independent of r; but depends only upon r^ = ^r(μ; Á); as it should. Thus, a solution to this equation should indeed have the correct asymptotic properties associated with the stationary scattering states of interest.
9.3 The Born Expansion
Now that we have an explicit representation for the Green’s function G+(") we can attempt a solution to the Lipmann-Schwinger equation
jÁ"i = jÁ 0 i + G+V jÁ"i:
The traditional method of solving this kind of equation, or its integral equation equivalent is by iteration. Any approximation to jÁ"i can be substituted into the right-hand side of the equation to generate a new approximation. Moreover, we can formally write
jÁ"i = jÁ 0 i + G+V [jÁ 0 i + G+V jÁ"i] = jÁ 0 i + G+V jÁ 0 i + G+V G+V jÁ"i
Proceeding in this way generates the so-called Born expansion
jÁ"i = jÁ 0 i + G+V jÁ 0 i + G+V G+V jÁ 0 i + ¢ ¢ ¢
=
X^1
k=
(G+V )n^ jÁ 0 i:
The Born expansion gives a an expansion in powers of the potential V; and obviously requires for its convergence that the e¤ect of the perturbation V on the incident wave be small, hence jjÁS jj << jjÁ 0 jj; jjÁ"jj: The solution obtained by truncating the series at order n is referred to as the nth order Born approximation to the scattered state. We defer till later an exploration of the approximate solutions obtained in this fashion, and instead introduce additional ways of looking at the problem.
9.4 Scattering Amplitudes and T-Matrices
The form that we have developed for the scattering amplitude
f (μ; Á) = f(^r) = ¡
m 2 ¼~^2
Z
d^3 r^0 e¡ik^r¢~r
0 V (~r^0 )Á"(~r^0 )
describes the amplitude for measuring a de‡ected particle along the direction r^(μ; Á)
with wavevector k:^ Thus, it measures a state of wavevector ~kf =^ k^r;^ so we can write f(μ; Á) = f (^r) = f(~kf ; ~k 0 ) in the form
f (~k; ~k 0 ) = ¡
m 2 ¼~^2
Z
d^3 r^0 e¡ikf^ ¢~r
0 V (~r^0 )Á"(~r^0 )
Scattering Amplitudes and T-Matrices 273
which has the form of a projection of V jÁ"i onto the plane wave state j~kf i associated with
the wave function h~rj~kf i = ei~kf^ ¢~r^ (Note that the normalization of this plane wave state is a little di¤erent than usual, but is consistent with our choice of jÁ 0 i:) Thus, e.g.,
h~kf jV jÁ"i =
Z
d^3 r^0 e¡ikf^ ¢~r
0 V (~r^0 )Á"(~r^0 )
f (~kf ; ~k 0 ) = ¡ m 2 ¼~^2
h~kf jV jÁ"i
which appears to be a matrix element of V; except that it is taken between states of di¤erent type. The state on the left is part of an ONB of free particle states, that on the right is an eigenstate of the same energy in the presence of the potential, which is “smoothly connected” to the plane wave state jÁ 0 i = jk 0 i as V! 0 : It is useful to introduce an operator T; referred to as the T matrix, or transition operator which is de…ned so that V jÁ"i = T jÁ 0 i = T j~k 0 i:
This allows us to express the scattering amplitude
f (μ; Á) = f(~k; ~k 0 ) = ¡
m 2 ¼~^2
h~kf jT j~k 0 i
as the matrix element of the transition operator between initial and …nal free particle states (which are the ones that we deal with in the laboratory, outside of the target region). Hence the T -matrix contains all information regarding the scattering transitions induced by the potential. How do we evaluate T? We generate an integral equation for it from the Lipmann-Schwinger equation, which gives
jÁ"i = jÁ 0 i + G+V jÁ"i: = jÁ 0 i + G+T jÁ 0 i = ( 1 + G+T )jÁ 0 i
We can compare this with the Born series to obtain
jÁ"i = jÁ 0 i + G+V jÁ 0 i + G+V G+V jÁ 0 i + ¢ ¢ ¢
to obtain the operator relation
G+T = jÁ"i = G+V + G+V G+V + ¢ ¢ ¢
from which we deduce that
T = V + V G+V + V G+V G+V + ¢ ¢ ¢
which gives a Born expansion for T in powers of the potential V. Formally we can write
T = V [ 1 + G+V + G+V G+V + ¢ ¢ ¢ ] = V + V G+T
which is the integral equation obeyed by the T -matrix that generates the Born series. The nth order Born approximation to the T -matrix is then obtained by truncating the series ath the nth order term. The …rst Born approximation to the T -matrix is just the scattering potential T = T (1)^ = V
and so we obtain to this order
fB (~kf ; ~ki) = ¡
m 2 ¼~^2 h~kf jV j~kii:
Scattering Amplitudes and T-Matrices 275
then
V^ ~ (q) = 4 ¼e
2 q
Z 1
0
dr sin(qr) e¡®r
4 ¼e^2 ®^2 + q^2
Thus, in this case, f (μ; Á) = f(μ) = f (q); where q = 2k sin μ= 2 ; and
f (μ) = ¡
2 me^2 ~^2
®^2 + q^2
2 me^2 ~^2
®^2 + 4k^2 sin^2 μ= 2
and the cross section becomes
d¾ d
4 m^2 e^4 ~^4
®^2 + 4k^2 sin^2 μ= 2
By taking the limit that ®! 0 we obtain the corresponding cross section, in the Born approximation, for the Coulomb potential
d¾ d
m^2 e^4 4 ~^4 k^4 sin^4 μ= 2
e^4 16 "^2 sin^4 μ= 2
As a second example, considering the elastic scattering of electrons by a neutral atom in its ground state with initial electron energies that are too small to excite the atom to any of its excited states. For an atom of atomic number Z; the charge density ½(~r) (^) can be written ½(~r) = e [Z±(~r) ¡ n(~r)]
where n(~r) is the number density of electrons in the atom at ~r and can be written
n(~r) = hÃj
X
i
±(~r ¡ ~ri)jÃi '
X
i
ä ni (~r)Ãni (~r)
where the second form holds in an independent electron approximation. The electric potential '(~r) at a point ~r due to the charge density of the bound electrons and the nucleus satis…es Poisson’s equation
r^2 ' = ¡ 4 ¼½(~r) = ¡ 4 ¼e [Z±(~r) ¡ n(~r)]
where charge neutrality implies that
R
d^3 r n(~r) = Z. The corresponding potential energy seen by an incoming electron is given by
V (~r) = ¡e'(~r)
so that r^2 V (~r) = 4¼e^2 [Z ¡ n(~r)] :
We now note that if
V (~r) =
Z
d^3 q (2¼)^3
ei~q¢~r^ V~ (~q) V~ (~q) =
Z
d^3 q e¡i~q¢~r^ V~ (~r)
then
r^2 V (~r) =
Z
d^3 q (2¼)^3
q^2 e¡i~q¢~r^ V~ (~q)
276 Scattering Theory
has as its Fourier transform the function ¡q^2 V~ (q): We write, therefore, as the Fourier transform of Laplaces equation
¡q^2 V~ (q) = 4¼e^2 [Z ¡ F (q)]
where the atomic form factor
F (~q) =
Z
d^3 r e¡i~q¢~rn(~r)
is the Fourier transform of the electronic charge density. Thus, we can solve for V~ (~q) to obtain
V^ ~ (~q) =^4 ¼e
(^2) [F (q) ¡ Z] q^2
which allows us to evaluate the scattering amplitude in the Born approximation
f(q) = ¡
2 e^2 m ~^2
[F (q) ¡ Z] q^2
and the corresponding cross section
d¾ d
4 e^4 m^2 ~^4
jF (q) ¡ Zj^2 q^4
Thus, measurement of the cross section for all momentum transfer allows information to be inferred about the distribution of charge in the atom [as contained in the form factor F (q)]. For a spherically symmetric charge density it is possible, in principle, to invert this relation to determine the charge density ½(~r) directly.
9.5 Partial Wave Expansions
In the last section we have not really used the fact that V^ (r)^ is a spherically symmetric potential. In this section we explore some of the simpli…cations that occur as a result of this fact. Speci…cally, if V = V (r); then H commutes with L^2 and Lz and we know that there exists a basis of eigenstates common to
H; L^2 ; Lz
: Let j"; ; mi = jk;; mi denote such a basis for the positive energy subspace of the system of interest, where, as usual, k =
p 2 m"=~^2 : We note, in particular, that the potential V (r) = 0 is spherically symmetric, so there must exist a basis of this sort for free particles. Let us denote by j"; ; mi(0)^ = jk;; mi(0)^ the corresponding free particle eigenstates common to
H 0 ; L^2 ; Lz
: Both sets of states satisfy orthonormality relations
hk; ; mjk^0 ;^0 ; m^0 i = ±(k ¡ k^0 )±;^0 ±m;m^0 = (0)hk; ; mjk^0 ;^0 ; m^0 i(0)
and have functions of the following form
Ãk;;m(r; μ; Á) = Fk;(r)Y (^) `m (μ; Á) =
Ák;`(r) r
Y (^) `m (μ; Á)
Ã(0) k;;m(r; μ; Á) = F (^) k;(0) (r)Y (^) `m (μ; Á) =
Á
(0) k;`(r) r
Y (^) `m (μ; Á)
where the functions Ák;(r) = rFk;(r) obey the radial equation
Á^00 k;` ¡
μ ( + 1 ) r^2
Ák;` = 0
278 Scattering Theory
As a preliminary simpli…cation, we note that, because of the spherical symmetry of the potenital, there is azimuthal symmetry along the z-axis associated with the incident beam, thus, only m = 0 components exist in the expansion:
f(μ; Á) = f (μ) =
X
`
fY (^)^0 (μ) (9.1)
To proceed we express the stationary scattering states of interest as an expansion
Á"(~r) » eikz^ + f (μ)
eikr r
X
`
AÃk;; 0 (~r)
in states of well-de…ned angular momentum. The th term in this expansion is referred to as theth partial wave. In terms of the spherical harmonics, this expansion takes the form
eikz^ + f (μ)
eikr r
X
`
A`
Ák;` r
Y ` 0 (μ):
To make this useful, we now use the known expansion of the function eikz^ in free particle spherical waves: eikz^ =
X
`
cÃ(0) k;; 0 (~r) =
X
`
Bj(kr) Y (^) `^0 (μ):
The Bcan be calculated exactly. The result is B = i`
p 4 ¼(2` + 1 ) so that
eikz^ =
X
`
i`
p 4 ¼(2+ 1 ) j(kr) Y (^) `^0 (μ):
Thus we can write
X
`
Bjl(kr) + feikr r
Y (^) `^0 (μ) =
X
A`
Ák;`(r) r
Y (^) `^0 (μ):
Linear independence of the Y (^) `m ’s implies that
Br j(kr) + feikr^ = AÁk;`(r):
Now asymptotically,
Br j(kr) »
B`
k
sin(kr ¡ `¼=2) =
B`
2 ik
h eikre¡i¼=^2 ¡ e¡ikrei¼=^2
i
and
AÁk;(r) » asin(kr ¡¼=2 + ±`)
=
a` 2 i
h eikre¡i¼=^2 ei±^ ¡ e¡ikrei¼=^2 e¡i±
i :
Substituting these last two equations into our previous expansion and equating coe¢cients of eikr^ and e¡ikr^ we …nd that
a` 2 i
e¡i¼=^2 ei±^ =
B`
2 ik
e¡i¼=^2 + f
and a2 i ei¼=^2 e¡i±`^ =
B`
2 ik ei`¼=^2
Partial Wave Expansions 279
which gives two equations in the two unknown quantities aand f: Solving for f` we …nd that
f` =
k
p 4 ¼(2+ 1 )ei±^ sin ±`
so that
f (μ) =
X
`
fY (^)^0 (μ) =
k
X
`
p 4 ¼(2+ 1 )ei±^ sin ±Y (^)^0 (μ)
from which follows the expansion for the di¤erential scattering cross section
d¾ d
(μ) =
k^2
X^1
`=
p 4 ¼(2+ 1 )ei±^ sin ±Y (^)^0 (μ)
2
The total cross section can then be written
¾tot =
k^2
Z
d
X^1
`=
p 4 ¼(2+ 1 )ei±^ sin ±Y (^)^0 (μ)
2
k^2
X^1
`=
X^1
`^0 =
p 4 ¼(2` + 1 )
p 4 ¼(2^0 + 1 )ei(±¡±^0 )^ sin ± sin ±`^0
Z
d Y (^) ^0 (μ)Y 00 (μ)
which reduces to
¾tot =
k^2
X^1
`=
(2+ 1 ) sin^2 ±
X^1
`=
¾`
where
¾` =
k^2
(2+ 1 ) sin^2 ± ·
k^2
(2` + 1 )
is the scattering cross section to states with angular momentum . Note that for free particles ±! 0 and ¾tot! 0 ; as we would expect.
