Download APPENDIX C1: PARTIAL WAVE METHOD OF QM SCATTERING

APPENDIX C1: PARTIAL WAVE METHOD OF QM SCATTERING
Scattering cross section and scattering length
Solid angle dΩ Incident particle flux J z θ dNscat Area A Thickness t (n scattering centres per unit volume) From Appendix A1 we recall the definitions for scattering cross sections:
dNscat = I (θ, φ) A t n J dΩ
for the differential scattering cross section I (θ, φ), and
Nscat = σ A t n J
for the total scattering cross section σ.
We also had
!
=
" I(", # ) d!
and
d!
= I
d!
In a QM treatment, using Schrodinger’s equation, we had
# !2 2
&
" + V (r)(! (r) = E ! (r)
%!
$ 2m
'
After rearranging, this became
#$! 2 + k 2 " U %&! = 0
where
k2
For a typical scattering event:-
=
2mE
!2
;
U
=
2mV (r)
!2
Scattering centre z Incident wave/particle Scattered wave/particle Using spherical polar coordinates with origin at the scattering centre and the z axis along the direction of the
incident particle, we concluded that at large distances r (so V (r) ≈ 0) the total wave function will be a
superposition of eigenfunctions of
1 "#! 2 + k 2 $%! = 0
that represent the incident plane wave and spherical scattered wave. Hence for large r we expect
eikr
! ! eikz +
f (" )
r
The angular dependence of the scattering is described by the function f (θ), assuming for simplicity no φ
dependence.
The simple relation between the differential scattering cross section and the scattering length f (θ) was
2
I(! ) = f (!
Method of partial waves
This uses the above result for I (θ), together with a comparison of solutions for the wave function ψ for
Schrodinger’s equation at large distances r (where the potential U ≈ 0) and small distances r (where U is
important). This can be done using a wave-like expansion for the states with orbital angular momentum
quantum number  (just like in the nuclear Shell Model).
The wave function at large r for the incoming particle is (using z = r cosθ) is
! = eikz = eikr cos"
The wave function in spherical polar coordinates can be written in a separable form as
!
u (r)
! = " A! ! P! (cos! )
kr
! =0
where the P(cosθ) for integer values of the orbital angular momentum quantum number  are the Legendre
polynomials of degree  with cosθ as the variable. The first few are
3
P0 (cos! ) = 1 , P1 (cos! ) = cos! ,
P2 (cos! ) = (3cos2 ! !1) / 2 , P3 (cos! ) = (5cos ! ! 3cos! ) / 2
Note that we do not need to use the quantum number m here (by contrast with the Shell Model, but similar to
the multipole expansion in electromagnetism).
The factors A are amplitudes specifying how much of angular momentum state  is present. Just as in the Shell
Model (see Section B), the radial function satisfies
"
d 2 u!
! (! +1) %
+ $k 2 !
' u! = 0
2
#
dr
r2 &
The solutions for general r are complicated, but as r → ∞ the asymptotic solutions can be found and lead to
"
sin(kr ! ! "2 )
! = # (2! +1)i !
P! (cos# )
kr
! =0
Next we include effects of the scattering centre. We again have an expansion of the general form
!
v (r)
! = " B! ! P! (cos" )
kr
! =0
where there are new amplitude coefficients B and the new radial functions satisfy
"
%
d 2 v!
! (! +1)
+ $k 2 !
!U(r)' v! = 0
2
2
#
&
dr
r
2
For large enough r, such that the terms (+1)/r and U(r) are negligible, the general asymptotic solutions are
v! ~ sin(kr ! ! !2 + "! )
[You can verify this by substituting back]
2 where the constants η represent phase shifts compared with the solutions in the absence of scattering. This
phase shift will usually depend on the value of .
To summarize, we now have two expressions for ψ :
"
sin(kr ! ! "2 + #! )
! = # B!
P! (cos$ )
kr
! =0
AND
"
eikr
sin(kr ! ! !2 )
eikr
! = e
+
f (" ) = # (2! +1)i !
P! (cos! ) +
f (! )
r
kr
r
! =0
By comparing the two right-hand sides, we can eventually solve for the unknown B and f (θ) as follows:
ikz
We first have to write the sine functions in terms of complex exponentials using
1 ix !ix
sin x =
(e ! e ) for any x
2i
All the terms in both expressions are then found to be proportional to either eikr or e-ikr.
Equating the coefficients of e-ikr gives
B! = (2! +1) i ! ei!!
Equating the coefficients of eikr gives
1 "
f (! ) =
(2! +1) (e 2i!! !1) P! (cos! )
#
2ik ! =0
1 !
f (! ) =
(2! +1) ei!! sin !! P! (cos! )
"
k ! =0
Finally,
Results for total cross section
The total scattering cross section is
!
= 2"
=
2!
k2
!
"
0
!
I(# ) sin # d#
!
"
0
2
f (# ) sin # d#
2
!
# "(2! +1) e
0
= 2"
i"!
sin "! P! (cos# ) sin # d#
! =0
2! ! # #
(2! +1)(2! ! +1) ei"! e"i"! ! sin "! sin "! ! P! (cos# ) P! ! (cos# ) sin # d#
2 %0 $$
k
! =0 ! !=0
Next we can do the integral over θ by using the orthogonality property that
$& 2 / (2! +1) if ! ! = !
"
" 0 P! (cos! )P! ! (cos! ) sin! d! = % 0
if ! ! # !
'&
=
This gives
!
=
4" !
(2! +1)sin 2 !!
2 "
k ! =0
The method of partial waves is useful if the phase shifts can be deduced from solutions of Schrodinger’s
equation. Typically, if the interaction has an approximate range r0, then the above sum over  converges very
rapidly and only the first term or the first few terms are needed. The condition is roughly that
 < r0 k
where we recall that k is related to the energy E by k =
2mE !
In simple cases, only  = 0 will be appreciable
⇒
S wave scattering.
If  = 1
⇒
P scattering ; and if  = 2
⇒
D scattering ; etc.
3 Another useful result can be found by putting θ = 0 in the previous result for f (θ) giving
1 !
f (0) =
(2! +1) ei!! sin !! P! (1)
"
k ! =0
1 !
"(2! +1) (cos!! + isin !! )sin !!
k ! =0
where we used a general property that P(1) = 1 in the last line. Then, taking the imaginary parts:
1 !
k
Im f (0) =
(2! +1)sin 2 !! =
#
"
k ! =0
4"
where we used the partial wave expression for σ in the last step above. Hence we get the relationship that
4"
which is sometimes known as the optical theorem.
! =
Im f (0)
k
=
Application to neutron-proton (n-p) scattering
We consider the 2-particle system consisting of n and p. If their separation r is small enough, the particles will
interact through short-range nuclear forces (as in the Shell Model).
V(r)
r
0
The 3D potential V(r) will be approximated by a square
well as shown.
We use centre-of-mass coordinates for the n-p system, so
the reduced mass is
V0
since the individual particles masses are roughly equal.
R
Schrodinger’s equation is
! 2! +
2M
[E "V (r)]! = 0
!2
There are 2 possibilities for the energy eigenvalue E :
If E < 0 , we have a bound state of n and p (if the potential well is deep enough), and ψ falls off exponentially
outside the well.
If E > 0 , we have the case corresponding to scattering of the particles (ψ is wave-like for both r < R and r > R).
It is the scattering case that we will consider here.
In fact, we also need to take spin into account. There are 2 situations depending on whether the spins of n and p
are parallel or antiparallel:1/2
n
1/2
p
Spin S = 1
1/2
p
1/2
n
Since the spin degeneracy factor is 2S+1,
the S=1 and S=0 states are often referred to
as the triplet and singlet states respectively.
Spin S = 0
The values of V0 and R for the potential well are found to be different in the triplet and singlet states:
V0 = 39 MeV and R = 1.4 fm for the triplet state (S = 1)
V0 = 14 MeV and R = 2.5 fm for the singlet state (S = 0)
The larger V0 in the triplet case can lead to the formation of a bound state (2H1 known as the deuteron), but this
is not so for the singlet state. However, both states can contribute to the scattering.
4 In the scattering experiment we have schematically:
p in target material (e.g., gas
of hydrogen molecules Incident beam of n
Studies have been made mainly over the energy range for E ~ 10-3 eV to 103 MeV.
The condition  < R k from the partial waves method (using the given values of R) leads to the conclusion that if
E < 10 MeV it is a good approximation to take only  = 0 (S wave scattering). Therefore we restrict attention to
this simple case.
As in the Shell Model (but putting  = m = 0) the solutions of Schrodinger’s equation can be expressed as
u(r) 0
u(r)
! (r, " , # ) =
Y0 (" , # ) =
r
r
where the radial term satisfies
d 2u
for r < R [where V(r) = -V0]
+ (k02 + k 2 )u = 0
2
dr
and
d 2u
for r > R [where V(r) = 0]
+ k 2u = 0
dr 2
2MV0
2ME
with
.
k02 =
, k2 =
2
!
!2
The solutions for u have the form
!#
a sin[(k02 + k 2 )1/2 r]
for r < R
u(r) = "
b sin(kr + ! )
for r > R
#$
where a, b and φ are constants. Now, according to the general partial wave theory for scattering
u(r) ~ sin(kr + !0 )
for very large r
[S wave scattering]
Thus we have the identity that:
φ = phase shift η 0
The other constants a and b are found from the QM boundary conditions at r = R (i.e., u and du/dr must be
continuous at that value of r). These conditions give
and
a sin[(k02 + k 2 )1/2 R] = b sin(kR + !0 )
a (k02 + k 2 )1/2 cos[(k02 + k 2 )1/2 R] = b k cos(kR + !0 )
Eliminating a and b (by dividing) gives
k
tan(kR + !0 ) =
tan[(k02 + k 2 )1/2 R]
2
2 1/2
(k0 + k )
This can be thought of as an equation for the phase shift η 0 in terms of k, k0 and R. It can be solved numerically
to find η 0 if we know the well depth and width and the kinetic energy E of the particle being scattered.
Recall that in a n-p scattering experiment the spins can either be parallel (triplet, total spin S = 1) or antiparallel
(singlet, S = 0). Since each allowed quantum state must be equally probable, it follows that the probability of a
triplet scattering process is 3 times that of a singlet scattering process. Hence the probabilities must be ¾ and ¼
respectively.
The previous expression for the scattering cross section in the case of S wave scattering was
4"
! =
sin 2 #0
2
k
5 Therefore in the case of n-p scattering we have
" 4"
" 4"
3%
1%
!
! total = $ 2 sin 2 #0t ! ' + $ 2 sin 2 #0s ! '
=
( 3sin 2 !0t + sin 2 !0s )
#k
#k
4&
4&
k2
where η 0t and η 0s are the phase shifts corresponding to the triplet and singlet states respectively.
For example, taking scattering at E ~ 100 eV and the well parameters quoted earlier, the above expression gives
σtotal ~ 20 b
[1 barn (b) = 10-28 m2]
Typically, experiments show an overall behavior like:σtotal (b)
40
20
1
10
100
1000
10000
E (eV)
Therefore the theory provides satisfactory agreement with experiment for E above about 1 eV.
[Note that the increase in cross section for E below 1 eV is because the protons in the target can no longer be
considered as “free” particles at these low energies, modifying the previous theory].
Applications to p-p and n-n scattering
Both of these cases are either more complicated and/or less interesting than in the n-p case. In part, this is
because we now have identical particles, so the Pauli Exclusion Principle has to be taken into account when
including the effects of spin.
Also, in the case of p-p scattering there is the (repulsive) Coulomb repulsion in addition to the (attractive)
potential well at small separations. At lower energies only the Coulomb barrier will have any effect, and this is
the regime of Rutherford scattering mentioned earlier. At particle energies of ~ few MeV or more (sufficient to
overcome the Coulomb barrier) the nuclear potential well has to be taken into account, but the method of partial
waves has only limited success.
In the case of n-n scattering, the cross section σtotal cannot be measured directly, because there is a lack of a
dense target of “free” neutrons. Instead it is necessary to use indirect methods, such as reactions that produce a
pair of neutrons in close proximity. Some simple reactions are:
n + 2 H1
→
p + 2n
or
3 1
3 1
4
2
H
+
H
→
He
+ 2n
From the energy distribution and angular distribution of the outgoing n, the n-n scattering properties can be
deduced.
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