Phase shift analysis Department of Physics, SUNY at...
99
1 Phase shift analysis Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: March 13, 2017) Two methods for treating scattering problems are discussed: the Born approximation and the method of partial waves. The Born approximation is most applicable when the kinetic energy of the incoming beam is large compared with the scattering potential, whereas the method of partial waves is most readily applied when the energy of the incoming particles is low. The two methods thus tend to complement each other. The relation of virtual levels to the resonant scattering of appropriate partial waves is discussed here. Ramsauer-Townsend effect and Frank-Hertz experiment Levinston's theorem S matrix element Effective potential range Scattering length Breit-Wigner formula 1. Scattering by potential The scattered wave function has the general form; )] ( sin ) ( [cos ) ( ) ( kr n kr j a r r u r R l l l l l kl kl , except at the origin, in the case when the potential energy is zero. Since ) (kr n l diverges at the origin, the wave function has the form ) ( ) ( ) ( kr j d r r u r R l l kl kl , in the vicinity of the origin. We need to determine the phase shift from appropriate boundary conditions; the continuity of the wave function and the derivative of the wave function. ((Classical theory)) The angular momentum is conserved; kb l , where b is the impact parameter and k is the wave number of the incident particle. The scattering occurs when b is lower than the radius of the target; b<R. Then we have max l kR kb l
Transcript of Phase shift analysis Department of Physics, SUNY at...
1
Phase shift analysis Masatsugu Sei Suzuki
Department of Physics, SUNY at Binghamton (Date: March 13, 2017)
Two methods for treating scattering problems are discussed: the Born approximation and the method of partial waves. The Born approximation is most applicable when the kinetic energy of the incoming beam is large compared with the scattering potential, whereas the method of partial waves is most readily applied when the energy of the incoming particles is low. The two methods thus tend to complement each other. The relation of virtual levels to the resonant scattering of appropriate partial waves is discussed here.
Ramsauer-Townsend effect and Frank-Hertz experiment Levinston's theorem S matrix element Effective potential range Scattering length Breit-Wigner formula
1. Scattering by potential
The scattered wave function has the general form;
)](sin)([cos)(
)( krnkrjar
rurR lllll
klkl ,
except at the origin, in the case when the potential energy is zero. Since )(krnl diverges at the
origin, the wave function has the form
)()(
)( krjdr
rurR ll
klkl ,
in the vicinity of the origin. We need to determine the phase shift from appropriate boundary conditions; the continuity of the wave function and the derivative of the wave function. ((Classical theory))
The angular momentum is conserved; kbl , where b is the impact parameter and k is the wave number of the incident particle. The scattering occurs when b is lower than the radius of the target; b<R. Then we have
maxlkRkbl
2
When energy is low such that 1kR , lmax is small is equal to zer0, S wave). Partial waves for higher l are, in general, unimportant. ___________________________________________________________________________ 2. Hard sphere scattering (I)
We consider the scattering from the repulsive potential
0
)(rV Rr
Rr
Rb
p=Ñk
3
The wave function is given by
)](sin)([cos)( krnkrjerR lllli
kll .
for r>R. The wave function must vanish at r = R because the sphere is impenetrable.
)](sin)([cos0|)( kRnkRjerR lllli
Rrkll ,
or
)(
)(tan
kRn
kRj
l
ll .
Thus the phase shifts are known for any l. The values of l in the limit of kR<<1 are as follows
for each l.
((Mathematica)) Series expansion of )(
)()(
l
ll n
jf around = 0
O Rr
V
V=¶
4
Let us now consider the low energy limit ( )1kR For 1 kR
!)!12()(
lj
l
l
( )0
1
!)!12()(
ll
ln
, ( )0
where
)12)(12(97531!)!12( lll . Then we have
0!)!12(
1
!)!12()(
)(tan
12
lln
j l
l
ll
,
or
0l for any l (in the limit of 0 ).
Clear"Global`";
f1L1_, z1_ :
SeriesSphericalBesselJL1, z1SphericalBesselYL1, z1 , z1, 0, 12 Normal;
PrependTableL, f1L, z, L, 0, 4,
"L", " fL,z" TableForm
L fL,z0 z z3
3 2 z5
15 17 z7
315 62 z9
2835 1382 z11
155 925
1 z3
3 z5
5 z7
7 8 z9
81 34 z11
495
2 z5
45 z7
189 z11
2673
3 z7
1575 z9
10 125 z11
185625
4 z9
99225 z11
848 925
5
((Note))
00tan ,
3tan
3
11
,
45tan
5
22
x=kR
y=d0
-1.0 -0.5 0.0 0.5 1.0
-1.0
-0.5
0.0
0.5
1.0
6
Fig. ContourPlot of the phase shift l vs x = kR for l = 0, 1, and 2. ________________________________________________________________________ 3. Hard sphere scattering (II): Low energy case (kR<<1)
x=kR
y=d1
-1.0 -0.5 0.0 0.5 1.0
-0.3
-0.2
-0.1
0.0
0.1
0.2
0.3
x=kR
y=d2
-1.0 -0.5 0.0 0.5 1.0
-0.02
-0.01
0.00
0.01
0.02
7
We consider the l = 0 case (S-wave scattering). For l = 0,
)tan()(
)(tan
0
00 kR
kRn
kRj
or
kR0 (0 < 0).
Then we have
)sin(
)]cos(sin)sin([cos
)](sin)([cos)(
0
00
00000,
0
0
0
krkr
e
krkrkr
e
krnkrjerR
i
i
ilk
where
x
xxj
sin)(0 ,
x
xxn
cos)(0 .
Since kR0 , we have
)](sin[)(0
0, Rrkkr
erR
i
lk
Then we have
22220
22
44
sin4
RRkkktot ,
which is four times the geometric cross section R2. In this case tot is the total surface area of
the sphere with a radius R. The waves feel their way around the whole sphere. ((Note-1))
0
22 sin12
4
lltot l
k .
8
((Note-2)) Direct calculation of differential equation
Here we derive the above solution directly from solving the differential equation.
0)(])1(
)([)(" 22
ru
r
llrUkru , (1)
where
)()( rrRru . When l = 0 and U(r) = 0, we get the differential equation
0)()(" 2 rukru ,
)sin()( 0 krCrrRu
For r = R,
u(r) = 0, or
0kR .
________________________________________________________________________ 4. Hard sphere scattering (III): High energy case ( 1kR )
We consider the semi-classical situation where 1kR
kRl
lltot l
k 0
22 sin12
4 (1)
with
22
2
2
22
)}({)}({
)}({
tan1
tansin
kRnkRj
kRj
ll
l
l
ll
with
)(
)(tan
kRn
kRj
l
ll
The asymptotic form is given by
9
)2
sin(1
)(l
kRkR
kRjl ,
)2
cos(1
)(l
kRkR
kRnl ,
for 1kR . Then we have
)2
(sinsin 22 lkRl , (
2
lkRl )
We also have
)2
(cos)2
)1((sinsin 22
12 l
kRl
kRl
.
For an adjacent pair of partial waves, we have
1)2
(cos)2
(sinsinsin 221
22 l
kRl
kRll .
With so many l-values contributing to Eq.(1),
2
1sin2 l ,
and
222
02
0
22
2)1(2
122
sin124
RkRk
lk
lk
kRl
l
kRl
lltot
,
which is twice larger than the geometric cross section 2R . 5. Origin of 22 Rtot for kR>>1 (Sakurai and Napolitano)
The scattering amplitude is given by
10
kR
ll
kR
ll
i
kR
ll
i
lll
i
shadowreflection
Plk
iPel
ik
Pelik
Pelk
fff
l
l
l
00
2
0
2
0
)(cos122
)(cos122
1
)(cos)1(122
1
)(cossin121
)(
where
kR
ll
ireflection Pel
ikf l
0
2 )(cos122
1 ,
kR
llshadow Pl
k
if
0
)(cos122
.
Fig. Shadow scattering (Schwabl). Now we calculate the contribution from the reflection,
11
2222
02
',0 0'
222
0 0''
1
1
222
2
)12(
12
2)1'2)(12(
4
2
)(cos)(cos)(cos)1'2)(12(4
2
'
'
RRkk
lk
lell
k
PPdellk
fd
kR
l
ll
kR
l
kR
l
ii
kR
l
kR
lll
iireflection
ll
ll
where
',' 12
2)(cos)(cos)(cos llll l
PPd
It is particularly strong in the forward direction because 1)(cos lP for = 0, and the
contribution from various l-values and add up coherently. The contribution from the shadow is obtained as
2
02
0
21
1
22
2
)12(
)](cos)[(cos)12(4
2
Rlk
Pdlk
fd
kR
l
kR
llshadow
Note that
])Re[2( *22
reflectionshadowshadowreflectionshadowreflection ffffdffd
The interference between shadowf and reflectionf vanishes:
0]Re[ * reflectionshadow ffd
The reason for this is as follows. We note that
2
lkRl ,
and
kR0 .
12
for kR>>1. Then we have
kR
ll
li
kR
ll
ireflection
Pleik
Pelik
f l
0
2
0
2
)(cos)1(122
1
)(cos122
1
0
since lile )1( . Then we get
])(cos)(cos)1'2)(12()1(4
1Re[
)(cos)1(1'22
1)(cos12
2Re[]Re[
0 0''
22
0''
2
0
*
0
0
kR
l
kR
lll
li
kR
ll
likR
llreflectionshadow
PPllek
Pleik
Plk
iff
So we have the interference between shadowf and reflectionf as
)2cos()1(
)]1()1(Re[
)]1()1(Re[
])12()1(Re[
]12
4)1'2)(12()1(
4
1Re[
])(cos)(cos)1'2)(12()1(4
1Re[
]Re[
2
22
22
0
22
0 0'
',22
0 0''
22
*
0
0
0
0
kRkR
R
kRek
kRek
lek
llle
k
PPdllek
ffdI
kR
kRikR
kRi
kR
l
li
kR
l
kR
l
llli
kR
l
kR
lll
li
reflectionshadowsr
or
kR
kR
R
I kRsr )2cos()1( 1
2
,
where we assume that kR is integer. We also use the formula
)1()1()12()1(0
kRl kRkR
l
l
.
13
We make a plot of Isr as a function of kR (=integer). We find that Isr oscillates with kR and reduces to zero for sufficiently large kR.
Fig. Plot of )/( 2RIsr as a function of kR (= integer).
6. Optical theorem and shadow part
The shadow is due to the destructive interference between the incident wave and the newly scattered wave. We now calculate
2
22
02
02
2
)1(2
122
)1(122
)]0(Im[4
)](Im[4
R
kRk
lk
Plk
fk
fk
kR
l
kR
ll
shadowshadow
where
1)1( lP
Thus we have the optical theorem;
)]0(Im[4
)]0(Im[4
shadowtot fk
fk
kR
Isr R2
10 20 30 40 50
0.15
0.10
0.05
0.05
0.10
0.15
14
since 22 Rtot .
7. Physical meaning (D. Bohm, Quantum Theory)
The total cross section is given by (i) Quantum limit
The quantum scattering occurs when 1kR ( R 2 )
24 RT . (long-wavelength) As the wavelength goes below the size of the sphere, the first effect will be to introduce waves of higher angular momentum. So that the cross section becomes angular dependent. As the wavelength made still shorter, however, and the classical region is approached, the cross section once again becomes spherically symmetrical, with a value reduced to R2, except for a region near = 0 with an angular width of the order of
R
2 .
The large projection in the forward direction is essentially a diffraction effect, containing a total cross section of 2R . Thus, for very short wavelengths, the total cross section is 2R2. (ii) Classical limit
2RT . 8. Finite repulsive potential
15
Fig. Plot of u(r) as a function of r. (Sakurai). )sin()( 0 krr
CrRout for r>R, with 00 .
For l = 0 (S-wave)
0)()]([)(" 2 rurUkru , (1) where
)(2
)(2
rVrU
, 2
2
2kE
,
and
R
r
V r
V0
16
)()( rrRru . For r<R, we have
0)()()(" 20
2 rukkru ,
where
220020
2kkVU
.
and
220 kk
Noting the boundary condition: 0u at r=0, the inside solution u(r) can be obtained as
)sinh()( rAruin ,
For r>R
0)()(" 2 rukru ,
)sin()( 0 krCruout ,
or
)sin()( 0 krr
CrRout .
((Boundary condition))
We make sure that u is continuous and has a constant first derivative at r = R. The wave function and its derivative are continuous at r = R;
)sinh()sin( 0 RAkRC
)cosh()cos( 0 RAkRCk
Then we get
)tanh()tanh()tan( 0 RR
kRR
kkR
17
with
220
2
220
22
2
kk
k
RkRU
Rk
R
kR
.
For kR<<1, we have
)tanh(0 RR
kRkR
.
The total cross section is given by
222020
22 ]1
)tanh([4
4sin
4
R
RR
kktot .
For 1R , we have
...)(5
2)(
3
11
)tanh( 42 RRR
R
.
or
20
20 )(
3
1)(
3
1]1
)tanh([ RkkRRkR
R
RkR
<0.
Then we get the total cross section as
64
20
264
0 9
16
9
4R
VRktot
.
18
Fig. )4/( 2Ry tot vs Rx .
9. Attractive Square-well potential: low- energy scattering
We consider the spherical square-well potential in three dimensions given by
0
)( 0VrV
Rr
Rr
Fig. Attractive potential. For l = 0 (S-wave)
x R
T 4 R2
0.1 0.2 0.3 0.4 0.5
0.001
0.002
0.003
0.004
0.005
0.006
R r
V r
-V0
19
0)()]([)(" 2 rurUkru , (1) where
)(2
)( 2 rVrU
,
22
2kE
,
and
)()( rrRru . For r<R, we have
0)()()(" 02 ruUkru ,
where
20020
2kVU
.
Noting the boundary condition: 0u at r=0, the inside solution u(r) can be obtained as
)sin()( rAruin ,
where A is an arbitrary constant,
20
20
2 kkUk .
For r>R, we have the free-particle radial equation
0)()(" 2 rukru ,
)sin()( 0 krCruout ,
or
)sin()( 0 krr
CrRout .
where the subscript 0 on denotes l.
20
Fig. Plot of u(r) as a function of r. (Sakurai). )sin()( 0 krr
Cru for r>R, with 00 .
((Boundary condition))
We make sure that u is continuous and has a constant first derivative at r = R. The wave function and its derivative are continuous at r = R;
)sin()sin( 0 RAkRC ,
)cos()cos( 0 RAkRCk ,
Then we get
)tan()tan()tan( 0 RR
kRR
kkR
or
)tan()tan(tan1
tan)tan(
0
0 RR
kR
kR
kR
or
)tan()tan(1
)tan()tan(tan 0
kRRR
kR
kRRR
kR
with
21
20
2
2
20
2
22
2
kk
k
RURk
Rk
R
kR
((Note))
)sin(1
)sin(
])cos(
sin)sin(
[cos
)](sin)([cos)(
0
0
00
0000
0
0
0
krr
C
kr
kre
kr
kr
kr
kre
krnkrjerR
i
i
iout
10. Total cross section for the attractive square-well potential: exact calculation
Here we discuss the exact expression for the total cross section for the S-wave scattering. We consider the change of the phase shift when the kinetic energy is changed while the potential is kept constant. (a) The total cross section tot
Here we start with the expression for 0tan for the S wave, which is given by
)tan()tan(1
)tan()tan(tan 0
kRRR
kR
kRRR
kR
. (exact expression) (1)
The total cross section can be obtained as
22
])tan(1[
)tan()tan(
)(cos4
)tan()tan(1)tan()tan(
)tan()tan(4
)tan()tan(
)tan()tan(11
14
cot1
14
sin4
22
2
22
22
2
2
2
22
022
02
2
RR
kR
kR
kR
R
R
kRR
kRRR
kRkRR
R
kR
kRRR
kR
k
kRRR
kR
kRRR
kRk
k
ktot
(2)
For kR<<1, we get
22
22
1)tan(
4
)tan()tan(4
R
RR
kR
kR
R
RRtot
(3)
tot becomes zero when
1)tan(
R
R
, (4)
or
R 4.49341 (1.430 ), 7.72525 (2.459 ), 10.9041 (= 3.471 ), 14.0662 (= 4.477 ). When )tan( R ( 2/ R , 2/3 , 2/5 ,…),
2
4
ktot
,
which becomes infinity as 0k .
23
((Note)) Taylor expansion For x<<1,
...15
2
3
11
)tan( 42 xxx
x
(b) Phase shift 0
We start with another expression of
)tan(tan)tan(1
tan)tan()tan(
0
00 R
R
kR
kR
kRkR
. (exact expression)
As long as )tan( R is not too large, 1)tan( RR
kR
. Then we have
)tan()tan( 00 RR
kRkRkR
or
]1)tan(
[0 R
RkR
.
Using this value of 0, the total cross section is obtained as
22
2222
02
2
0
22
]1)tan(
[4
]1)tan(
[4
sin4
sin)12(4 max
R
RR
R
RRk
k
k
lk
l
lltot
(5)
which is the same as Eq.(4). We make a plot of this tot as a function of R . This function
becomes zero at R = 4.49341 and 7.72525, 10.9041, 14.0662,…(Ramsauer effect) and becomes infinity at R = /2, 3/2,…. (resonance).
We note that the attractive scattering becomes transparent to the incident beam at
24
1) tan(
R
R
.
Such resonant transparency of an attractive well is experimentally observed in the scattering of low energy electrons by rare gas atoms. The vanishing of the scattering cross-section at a certain low values of the energy is found in a number of wave processes. For example, He or other noble gas atoms are practically transparent to slow electrons of about 0.7 eV energy, while smokes consisting of particles homogeneous in size are virtually transparent to light in a narrow wavelength region. (c) Numerical calculation
We make a plot of )4/( 2Rtot as a function of Rx . This function becomes zero at R =
4.49341 and 7.72525, and becomes infinity at R = /2, 3/2,….
Fig. Plot of 22
]1)tan(
[4
R
R
Rtot
as a function of R . The change of the total cross section
tot as the kinetic energy of the incident particle, where the potential energy is kept
constant. tot becomes zero at R = 4.49341 and 7.72525 (Ramsauer-Townsend effect)
and becomes infinity at R = /2, 3/2,…. (resonance) 11. Ramsauer-Townsend effect ((Discovery))
In a preliminary investigation in 1921 of the free paths of electrons of very low energy (0.75 eV to 1.1 eV) in various gases, Ramsauer found the free paths of these electrons in Ar gas to be very much greater that that calculated from gas-kinetic theory. It was found that the effective
tot
4 R2
R2 4 6 8 10
1
2
3
4
5
25
cross-section (proportional to the reciprocal of the free path) of Ar gas increases with decreasing velocity until the electron energy becomes less than 10 eV. For electron energies below this value, it decreases again to the lowest value found in the preliminary measurements.
Independently, Townsend and Bailey examined the variation of the free path with velocity for electrons with energies between 0.2 and 0.8 eV by a different method, and showed that a maximum of the free path occurs at about 0.39 eV. This was confirmed by much later work of Ramsauer and Kollath. After these classical experiments, the behavior of a large number of gases and vapors has been examined over a wide voltage range. The striking features of the cross-section vs velocity curves are their variation in shape and size and also the marked similarity of behavior of similar atoms, such as those of the heavier rare gases and the alkali metal vapors. At the time of the earlier measurements no satisfactory explanation of the phenomena could be given, but on the introduction of quantum mechanics it was immediately suggested that the effect was a diffraction phenomenon.
The Ramsauer-Townsend effect can be observed as long as the scattering does not become inelastic by excitation of the first excited state of the atom. This condition is best fulfilled by the closed shell noble gas atoms. Physically, the Ramsauer-Townsend effect may be thought of as a diffraction of the electron around the rare-gas atom, in which the wave function inside the atom is distorted in just such a way that it fits on smoothly to an undistorted wave function outside. The effect is analogues to the perfect transmission found at particular energies in one-dimensional scattering from a square well.
Note that the Born approximation is not applicable to the low-energy collisions of electron with atoms, and the experimental results obtained in this limit clearly show that a more sophisticated theory (in this case, phase shift analysis) is required. We note that the Ramsauer-Townsend experiment as well as the Franck-Hertz experiment (mainly inelastic scattering) is now introduced in the Advance laboratory course of the universities ((Experiment))
26
Fig. Schematic diagram for the apparatus for measuring scattering cross-section. Xenon in
vapor. This apparatus (as one of the Advanced Laboratory in universities) is used to measure the elastic scattering cross-section for low energy electrons (0 - 5 eV) since for high energies inelastic scattering (excitation) dominates (E0 ≈ 10 eV).
27
Fig. Typical results on the Ramusauer-Townsend experiment for Xenon gas. The cross section
times density (
1n ) as a function of V . V is the electron energy. Ionization (Frank-
Hertz effect) occurs at the position denoted by I. (Kukolich). ((Note)) Comparison between Frank-Hertz experiment and Ramsauer-Townsend
experiment The Ramsauer-Townsend experiment is similar to the Frank-Hertz experiment. The
difference between these two experiments is as follows. For the Frank-Hertz experiment, the collision between atoms and electrons is inelastic. The energy of the bombarding electron is lost because of the ionization process of atoms (such as Hg vapor). The electrons in atoms undergo transitions from the lower states to the more excited state. For the Ramsuer-Townsend experiment, on the other hand, the electrons are elastically scattered by atoms (such as Xenon gas). The origin of this effect is the diffraction of electron waves by atoms. REFERENCES
N.F. Mott and H.S.W. Massey, The Theory of Atomic Collisions, 3rd edition (Oxford. 1965). D. Bohm, Quantum Theory (Dover, 1989).
28
S.G. Kukolich, Am. J. Phys. 36 (8) 701, “Demonstration of the Ramsauer-Townsend Effect in a Xenon Thyratron.”
((Note))
The incident particle is electron. m is the mass of electron. The first minimum of the total cross section occurs at the condition
49341.42 2
0222 RV
mRkR
or
(eV)(nm) 769263.049341.42
)2
( 22
20
22
m
RVkm
in the unit of eV for 02
2
2Vk
m
and in the unit of nm for R.
((Note))
In the electron scattering by atoms, R is an atomic radius. It is roughly on the order of several factor of Bohr radius (= 0.53 Å) for complicated atoms. 13. Comment on Ramsauer Effect ((by D. Bohm))
We observe from eq. (1) that if the scattering phase is equal to some integral multiple of for
nonzero k, the cross section vanishes. If is an integral multiple of , then 0tan 0 . For a
square well, we obtain the condition for the vanishing of 0tan from
eq. (2):
kR
kR
R
R )tan()tan(
.
For small k, 1kR . Replacement of kRtan by kR then yields
RR )tan(
For small k, is given approximately by 202 V . If V0 and R are such that the eq. (4) is
satisfied, the scattering cross section will be zero, and if it is nearly satisfied, the cross section will be very small. This vanishing of the scattering cross section for a non-zero potential is peculiar to the wave properties of matter. It would occur, for example, with light waves which were being scattered from small transparent spheres with a high index of refraction, so. chosen that the 0sin corresponding to the scattered wave vanished. This means, essentially, that the
contributions of the various parts of the potential to the scattered wave interfere destructively,
29
leaving only an un-scattered wave. Although this result was derived for a square well, it can easily be extended to any well that has the property that it is fairly localized in space. This is because the vanishing of the phase is determined by the cumulative phase shifts suffered by the wave throughout the entire well, so that it is always possible to obtain a phase shift of n by properly choosing the magnitude and range of the potential. For slow electrons scattered from noble gas atoms, it turns out that 0sin is very small and the cross section for electron-atom
scattering is therefore much smaller than the gas-kinetic cross section. This effect is known as the Ramsauer effect. As the electron energy is increased, the phase of the scattered wave changes, and, eventually, at higher energies above 25 eV the usual gas-kinetic cross section is approached.
The Ramsauer effect is somewhat analogous to the transmission resonances obtained in the one dimensional potential. The analogy, however, is not complete, because the condition for the Ramsauer effect [eq. (4)] is not exactly the same as that for a transmission resonance in a one-dimensional well. The reason for the difference is that in the one-dimensional case we define the transmitted wave as the total wave that comes through the well. In the scattering problems, we have an incident wave that converges on the well. Some of it enters the well and some of it is reflected at the edge of the well. The net effect is to produce an outgoing wave, whose phase depends on what happens to the wave at the well. The question of how much of this outgoing wave corresponds to a scattered wave depends on how large a phase shift it has suffered relative to the outgoing wave which would have been present in the absence of a potential. Thus we see that the intensity of the scattered wave depends on properties of the potential that are somewhat different from those determining the intensity of that part of the wave that is transmitted through the potential and out again on the other side. The vanishing of the cross section in the Ramsauer effect is, as we have already seen, a result of the fact that the contributions of different parts of the potential all add up in such a way as to produce a wave that cannot be distinguished from one which has not been inside a potential at all. 14. Origin of the meta-stable bound states
E0
Vc r
r
30
Suppose that the energy of the particle (E) is a little higher than zero energy. In the case of attractive Coulomb potential, the potential )(rVc is negative and smoothly increases with
decreasing the distance r. There is no drastic change of )(rVc at any r. Thus the particle with the
energy E (>0) (even if E is very small) leak outside the effective range of the potential without ant reflection. Thus it does not become a bound state.
How about the attractive square-well potential? There is a drastic change in the form of the
potential )(rVsq at the wall of r = R. Thus a part of particles with the energy E undergoes
reflections at r = R. The remaining particles leak and transmit outside the potential. The cause of such reflections is due to the drastic change of the wavenumber in the wave function at r = R. Particles having reflections at the wall of the potential attempt to transmit the outside of the potential wall, and finally succeed in leaking the potential. As a result of such a repetition of reflections, particles are temporally bound inside the well of the potential, forming the meta-stable state as a positive eigenvalue. Note that this state is not the same as the bound state of negative energy 15. The phase shift 0 for the S-wave scattering: change of 0 with the variation of V0
changes
We consider the change of the phase shift for the S-wave scattering when the potential is changed while the energy is kept constant. From
)tan(tan)tan(1
tan)tan()tan(
0
00 R
R
kR
kR
kRkR
,
we get the phase shift as
R r
Vs r
V0
E0
31
)tan()tan(1
)tan()tan(tan 0
kRRR
kR
kRRR
kR
and the total scattering cross section is given by
02
20 sin4 k
.
___________________________________________________________________________ Here we define
)tan()tan( qRRR
kR
where q is a wavenumber which is newly introduced. Then we get
)tan()tan()tan(1
)tan()tan(tan 0 kRqR
kRqR
kRqR
or
]1)tan(
[)]tan(arctan[0 R
RkRkRR
R
kRkRqR
(mod )
with the condition
20
22 )()( RkkRR
and
20020
2kVU
Note that 0V gradually increases, also increases.
At very low energies, using xx tan for 1x , we get
)1)tan(
(0 R
RkRkRqR
We imagine that we are slowly deepening the potential well (V0 is increasing slowly)
32
(i) 0R
0 = 0, which means . 00
(ii) R = /2
(the attractive square well just meets the criterion to host a single S-wave bound state)
)tan( R , and
kRkRkRRR
kR
kRRR
kR1
)tan(
1
)tan()tan(1
)tan()tan(tan 0
.
Then 0 goes through /2. In this case a bound state at zero energy is like a resonance. 0 takes
a maximum value (we will discuss later); 20
4
k
, which is dependent on k.
(iii) R ,
kRkR )tan(tan 0 ,
0 is nearly equal to . 00 .
(iv) 2/3 R
(the potential becomes capable of hosting a second bound state, and there is another resonance).
)tan( R ,
and
kRkRkRRR
kR
kRRR
kR1
)tan(
1
)tan()tan(1
)tan()tan(tan 0
.
Then 0 goes through 3/2. 0 takes a maximum value.
(v) 2R
33
0)tan( R , or
kRkR )tan(tan 0 ,
So 0 is nearly equal to 2 . 00 .
Note that when nR , the scattering cross section vanishes identically and the target becomes invisible ( 00 , the Ramsauer-Townsend effect).
We draw the plot of y = 0 vs x= R with kR (= a =0.05, 02
22
kE
) as a parameter by using
Mathematica. For convenience the value of kR is fixed as a small value.
axx
ay )]tan(arctan[00
kRa =fixed
22
0222
022 RkaRkRkRx ,
In the low energy limit, Rkax 00
4
3
2
0
0
0
0
0
0
y
y
y
y
y
y
)2/92/7(
)2/72/5(
)2/52/3(
)2/32/(
)2/0(
x
x
x
x
x
We also calculate the value of 0
2sin as a function of x. The total cross section shows a sharp
peak at 2/ Rx .
34
Fig. The plot of 0
2sin vs R , where k is fixed constant. We can see the change of 02sin
when the potential V0 is varied ((Mathematica))
x=kR
sin2d0
p21.5 1.6 1.7 1.8
0.2
0.4
0.6
0.8
1.0
Clear"Global`"; a 0.05; k1 ArcTana
xTanx a;
y Which0 x 2, k1, 2 x 3 2, k1 ,
3 2 x 5 2, k1 2 , 5 2 x 7 2, k1 3 ,
7 2 x 9 2, k1 4 ;
f1 PlotEvaluatey, x, 0, 7 ,
Ticks Range0, 5 , 2 , Range0, 5 , ,
PlotStyle Red, Thick, PlotPoints 60 ;
f2 GraphicsTextStyle"xR", Black, 12, 8.5 2, 0.8,
TextStyle"y0", Black, 12, 1, 4.2 ;
Showf1, f2, PlotRange All
x=kR
y=d0
0 p
2p
3 p2
2 p5 p2
3 p7 p2
4 p9 p2
p
2 p
3 p
4 p
35
((Summary))
The above results are summarized in two figures. In order to draw these figures, we use the Gaussian distribution function and the Heaviside step function, which are defined by
)2
exp(2
1)( 2
2
1 x
xf and )]2
(1[2
1)(2
xerfxf ,
with = 0.1.
h1 Plot Evaluate Sin y 2, x, 0, 7 , PlotStyle Red, Thick ,
PlotPoints 100,
Ticks Range 0, 5 , 2 , Range 0, 1, 0.2 ,
PlotRange All ;
h2 Graphics Text Style "x R", Black, 12 , 8.5 2, 0.1 ,
Text Style "sin20", Black, 12 , 0.8, 1 ;
Show h1, h2, PlotRange All
x R
sin2 0
2
3
22
5
23
7
24
9
2
0.2
0.4
0.6
0.8
1.
36
Fig. Schematic diagram. The phase shift vs R. shows a drastic change in the vicinity of
)12(2
nR , which means that )12(
200 naRkR in the low energy limit.
p
2p
kR
d0
0p
2p
3 p
2
1
kR
s0
s0max
0p
2p
3 p
2
37
Fig. sin2 vs R. It shows a sharp peak in the vicinity of )12(2
nR .
16. Attractive square-well potential-III: graphical solution (ContourPlot)
We consider the solution of two equations given by
20
22 )()( RUkRR
)tan()tan()tan( 0 RR
kRR
kkR
For simplicity, we have
20
20
20 )( aRkRU or Rka 00
20020
2kVU
, 2
2
2kE
Note that a0 is the depth of attractive potential. The number of bound states strongly depends on the magnitude of a0 (this will be discussed in association with the Levinson’s theorem). We also define as
kRy , Rx .
Rr
V r
V0
Ek0
38
Then we get two equations such that
20
22 ayx (1)
)tan()tan( 0 xyyx . (2)
We note that
yy tan)tan( For n 00 ( 00 )
)tan()tan()tan( 000 ynyy
Thus the ContourPlot of )tan()tan( 0 xyyx is the same as that of
)tan()tan( 0 xynyx .
We solve this problem using the Mathematica graphically. We make a plot of Eqs.(1) and (2) using the ContourPlot for x vs y for various 0 , where a0 is a fixed parameter.
x
y
a0 2 a0 3 2
0 0 0
0 2 0 3 2
2 3 2 5 2
2
3 2
0 2 4 6 8 10
0
2
4
6
8
10
39
Fig. ContourPlot of x vs y. kRy and Rx s . a0 = /2, 3 /2. and 5(black lines).0 =
/2 (red), 0.55, 0.6, 0.65, 0.7, 0.75, 0.8, 0.85, 0.9, 0.95, and . 0 = 3/2 (red), 1.55, 0.6, 1.65, 1.7, 1.75, 1.8, 1.85, 1.9, 1.95, and 2. The horizontal lines between y = /2 and (independent of x) are the trivial solutions derived from the present ContourPlot calculation.
x
y a0 2 a0 3 2
0 0 0
0 2 0 3 2
2 3 2 5 2
2
0 2 4 6 8 10
0.0
0.5
1.0
1.5
2.0
40
x
y
a0 2 a0 3 2
0 0 0
0 2 0 3 2
2 3 2 5 2
2
3 2
0 2 4 6 8 10
0
2
4
6
8
10
41
Fig. ContourPlot of x vs y. kRy and Rx . a0 = /2, 3 /2. and 5(black lines).
0 = (red), 0.05, 0.1, 0.15, 0.2, 0.25, 0.3, 0.35, 0.4, 0.45, and . 0 = (red), 1.05, 1.1, 1.15, 1.2, 1.25, 1.3, 1.35, 1.4, 1.45, and 3/2. The horizontal lines between y = 0 and /2 (independent of x) are the trivial solutions derived from the present ContourPlot calculation.
((Discussion))
There is a significant exception to this independence of the cross section on energy. Suppose that
2022
022 RkRkRkR
In the above figure, this corresponds to the case of [2
x , 0y ,
200
Rka ,
20
],
Then we have
)tan()tan( 0 RR
kRkR
x
y a0 2 a0 3 2
0 0 0
0 2 0 3 2
2 3 2 5 2
2
0 2 4 6 8 10
0.0
0.5
1.0
1.5
2.0
42
or
20
kR
or
20
,
since 1kR . The total cross section takes a maximum as
222
202
20
14
4sin
4
RkR
kkltot ,
leading to the occurrence of the zero-energy resonance. We see a pronounced dependence of the total cross section on energy. The magnitude of the total cross section is much larger than that given before when 0k , 0E . 17. Attractive square-well potential-I: bound states and S-wave resonance
We consider the case when
22
2
bE (<0) for the attractive square-well potential. is
close to zero and real. ((Bound state of a deuteron))
R r
V r
V0
Ek0
43
E coincides with the energy eigenvalue of the bound state. Note that the bound state of a deuteron is E = -2.23 MeV. The value of for a bound state is easily calculated from the fact that outside the potential the wave function is just a decaying exponential (for S waves),
)exp(1
rr
A ,
where
B 2
and B is the binding energy of the deuteron. Thus we obtained for the value of in the bound state.
The wave function inside the well has gone past a maximum and is decreasing with radius to
meet a decaying exponential at r = R. Now, the potential is of the order of 20 MeV deep; hence undergoes only a small change as E is increased from -2.23 MeV to a value of zero or slightly above, simply because the wavelength at any particular point is not changed much by this small fractional increase in kinetic energy. [Bohm D., Quantum Theory]. Here we consider the case for l = 0 (S wave) (i) Schrödinger equation for r<R
uuEuuVdr
ud
2)(
2
22
02
22
or
uumV
dr
ud 22
02
2 2
or
0)(" 20
2 uku
or
0" 2 uu b
where
44
2002
02kU
V
,
2
022 kb .
Then we get the solution as
)sin( rBu b ,
with the boundary condition, 0)( ru . (ii) Schrödinger equation for r>R
0" 2 uu , or
rAeu . The boundary condition at r = R, leads to
Rb AeRB )sin( ,
R
b eARB )cos( .
Then we have
RRR bb )cot( ,
with
20
22 kb .
We determine the constants A and B.
)2sin(2)(sin22
)sin(22 RRRRRRR
RRReA
bbbb
bbR
)2sin(2)(sin22
22 RRRRRRR
RRB
bbbb
b
45
from the normalization of the wave function. Note that the constants A and B are in the units of cm-1/2.
We solve the problem using the Mathematica.
Rx b , Ry
In this case,
xxy cot , 20
220
22 aRkyx
Fig. Plot of xxy cot (red line) and 20
20
22 )( Rkayx . a0 = /2, a0 = 3/2 (light green
line), a0 = 5/2 (green line), and a0 = 7/2 (blue line), Rx b . Ry . One bound state
for a0 = /2. Two bound states for a0 = 3/2. Three bound states for a0 = 5/2. As a0
increases, the potential well becomes deep. The vertical axis; 2
2
RRy . The
radius; RV
a2
00
2
.
2 3 2 5 2 7 2
x bR
y R
0 2 4 6 8 10 12
0
2
4
6
8
10
12
46
The solution is the intersection of xxy cot (red line) and 20
22 ayx (a0 = /2; blue line).
2
x , y = 0,
20
a .
implying that
200
Rka ,
2
200 2
RkU
,
The energy eigenvalue is
02
22
m
Eb
(quasi-bound state)
When the radius a0 is slightly larger than /2, the intersection of two curves is seen at y>0. Since
Ry , comes to take a small positive value. In this case (-) slightly becomes negative,
forming the bound state. For 2
0 0
a , there is no bound state.
(i) 20
a . One bound state with 0bE .
2
Rx b , 0 Ry
(ii) 2
3
2 0
a . One bound state with 2
2
2
mEb
<0
(iii) 2
30
a . One bound state with 0bE . One bound state with 2
2
2
mEb
<0
56582.2x , 95262.3y 2/x , 0y
47
Fig. Plot of u(r) vs r/R for the bound states with 56582.2 Rx b , 95262.3 Ry .
2
30
a .
(iv) 2
5
2
30
a . Two bound state with 2
2
2
mEb
<0.
(v) 2
50
a . One bound state with 0 . Two bound state with 2
2
2
mEb
<0
77982.2x , 34559.7y 50627.5x , 60053.5y
2/5x , 0y
Fig. Plot of u(r) vs r/R for the bound states with 77982.2x , 34559.7y (red), and
50627.5 Rx b , 60053.5 Ry (blue). 2
50
a
ur
rR0.5 1.0 1.5 2.0
0.2
0.4
0.6
0.8
1.0
1.2
1.4
ur
rR0.5 1.0 1.5 2.0
-1.0
-0.5
0.5
1.0
1.5
48
Fig. 2
50
a . Bound states with E1 and E2. Quasi bound state with 0bE . The particle with
0bE will go outside from the inside of the square-well potential through a tunneling
effect.
(vi) 2
70
a . One bound state with 0 . Three bound state with 2
2
2
mEb
<0
87687.2x , 6126.10y 73454.5x , 38187.9y 53599.8x , 93196.6y
2/7x , 0y
Fig. Plot of u(r) vs r/R for the bound states with. 87687.2 Rx b , 6126.10 Ry (red),
73454.5x , 38187.9y (blue), and 53599.8x , 93196.6y (green). 2
50
a
RO r
Vr
-V0
E2
E1
E0
ur
rR0.5 1.0 1.5 2.0
-1.0
-0.5
0.5
1.0
1.5
49
16. Connection between scattering amplitude and binding energy We start with the total cross section for the S wave scattering,
22 ]1
)tan([4
R
RR
s
stot
,
where
22
2kEs
, 2
020
2
Vk
.
Suppose that a positive energy
sE of the particle (scattering) shifts to a negative energy bE
(bound state)
22
2kEs
→ 2
2
2
bE ,
where s denotes the scattering problem and b denotes the bound state problem. Effectively this means that the replacement of the wavenumber occurs as
ik in this process. Correspondingly the wavenumber changes as
220 kk , → 22
0 kb .
50
Fig. The complex k-plane with bound-state pole at ik . Region of physical scattering is
denoted by real k (>0) (scattering state). After this replacement, the scattering problem is reduced to the bound-state problem which is discussed above. The boundary condition of the bound-state problem is given by
RRR bb )cot( .
Then we have
RR
R
R
R
boundb
b
scatt
1)tan()tan(
.
For 1R , we get
)21(4
)1(4
)1
1(4
1)tan(
4
2
22
22
22
R
R
RR
R
RRtot
.
Real k
k= ia
O
51
Such a method (the analytical continuity) allows one to bypass the problem of determining the potential and then calculating the cross section. This method only works when is very small. ((Summary)) Attractive potential
202
0
2
Vk
Scattering state Bound state k i
02
22
k
Ek
0
2
22
bE
20
2 kk 20
2 kb
,...2
3,
2
R (resonance) ,...2
3,
2
Rb (bound state)
)cot()cot( 0 kRkRRR RRR bb )cot( 2
2 1)tan(
4
R
RRtot
)21(4
2 Rtot
18. Effective potential range-I for attractive square-well potential R.G. Sachs, Nuclear theory (Addison-Wesley, 1953)
We consider the S wave (l = 0). The Schrödinger equation is given by
0)()]([)( 12
112
2
rurUkrudr
d. (1a)
with the energy 21
2
1 2kE
and 0)0(1 u .
0)()]([)( 12
222
2
rurUkrudr
d. (1b)
where the energy 22
2
2 2kE
and 0)0(2 u . Multiplying Eq.(1a) by u2 and Eq.(1b) by u1,
subtract and integrate, getting
0
212
12
202112 )(|]''[ druukkuuuu (2a)
52
Now we perform the same procedure, using the solution at distances large enough compared to the range of forces.
012
121
2
wkr
w
1
111 sin
)sin()(
rk
rw
Similarly, we have
022
222
2
wkr
w
2
222 sin
)sin()(
rk
rw
where
0][lim ii
rwu (i = 1, 2)
In other words, two functions },{ ii wu have the same asymptotic forms. Then we have
0
212
12
202112 )(|]''[ drwwkkwwww (2b)
Subtraction of Eq.(2a) from Eq.(2b) leads to
0
21212
12
20211202112 )()(|]''[|]''[ druuwwkkuuuuwwww
or
0
21212
12
221122112 )()()]0(')0()0(')0([)]0(')0()0(')0([ druuwwkkuuuuwwww
or
0
21212
12
22112 )()()0(')0()0(')0( druuwwkkwwww
53
We note that
1)0( iw , iii kw cot)0('
Then we get
0
21212
12
21122 )()(cotcot druuwwkkkk
Now we take E1 and E2 to be very close to E. Then we have
0
222
)(cot)(
druwkkd
d
where w and u are solutions associated with energy E. From this, we get
...2
1)cot(cot 2
00 krkk k
where
0
022
0 )(2
1druwr k
Here we need to note that l (= 0) is the angular momentum (the S-wave), and k is the wavenumber. In the present case the phase shift means 0 , Nevertheless, we use instead of
0 , in order to avoid confusion between k = 0 and l = 0.
We now consider the differential equation with 0k . The wave function )(0 rw satisfies the
differential equation given by
0)(02
2
rwdr
d.
The solution of this equation is obtained as
)1()(0 a
rrw
where
)0(0 rw , a
rw
)0('0 (3a)
54
We now return to the function )(rwk ;
sin
)sin()(
krrwk ,
sin
)cos()('
krkrwk .
In the limit of 0k , we assume that
1)0(lim0
rwkk
, (3a)
cotlim)0('lim
00krw
kk
k (3b)
leading to
1 , a
kk
cotlim0
using Eq.(3a) and 3(b). Thus we have
02
2
11cot rk
ak . (effective potential range)
The parameter cotk depends only on two parameters and k. Then the total cross section is given by
r
w0 r
0a
55
22
22
22
4cot1
14
sin4
k
k
k
Note that the total scattering cross section has a Breit-Wigner form. We define the scattering length as
ak
k
1cotlim
0
,
or
a
1 .
Then the total scattering cross section is
22
2
22 1
41
4
ak
a
ak
tot
.
which means that tot depends on the energy. The scattering length can now be characterized by
the fact that
2
04lim atot
k
.
19. Effective potential with 0l and quasi bound state
56
Fig. The centrifugal barrier combines with the potential well to form an effective potential,
which can produce a metastable state.
The effective potential for the l-th partial wave ( 0l ) is given by
2
2
0 2
)1(
r
llVVeff
(r<R)
rR
V r
-V0
Veff
E
57
where V0>0 and is the reduced mass. As shown in the above figure, the effective potential has an attractive well followed by a repulsive barrier at larger distances. The particle can be trapped inside., but cannot be trapped forever. Such a trapped state has a finite lifetime as a consequence of quantum-mechanical tunneling. The particle leaks through the barrier to the outside region. Such a state is called a quasi-bound state. ((Townsend))
A particle with energy E greater than zero but less than the height of the barrier can tunnel through the barrier and form a metastable bound state in the wall. This state is metastable (and not stable) because a particle “trapped” inside the well can also tunnel out.
We consider the resonance scattering from a potential well
0)(])1(
)([)("2
2
rur
llrUkru , (1)
where
)()( rrRru . For r<R
0)(])1(
[)("20
2
rur
llUru
with
22
2
mE
022 U
)()(
)( rAjr
rurR l
klkl
For r>R
0)())1(
()("2
2
rur
llkru
)](sin)([cos)(
)( krnkrjBer
rurR llll
iklkl
l
The continuity of )(rukl and its derivative at r = R:
58
)(tan)(
)('tan)('
)(sin)(cos
)]('sin)('[cos
)(
)('
kRnkRj
kRnkkRkj
kRnkRj
kRnkRjk
Rj
Rj
lll
lll
llll
llll
l
l
.
Thus we get
)(')()()('
)(')()()('tan
RjkRnRjkRkn
RjkRjRjkRkj
llll
lllll
For lkRx , we use the approximation
ll x
lxj
)12(7531
1)(
)1()12(7531)( l
l xlxn ,
1
)12(7531)('
l
l xl
lxj
)2()1)(12(7531)(' l
l xllxn
(i) Now we investigate the resonance scattering in detail for small energies and a very deep potential well;
RlkR .
)(
)('1
)(
)('
)(]!)!12[(
)12(tan 12
2
Rj
RjRl
Rj
RjRl
kRl
l
l
l
l
l
ll
For 0kR , we have
12)( ll kR
The total cross section is
lllll kkR
kkk424
2
2
22
2)(
14sin
4 .
59
For sufficiently small energy, the partial waves with 1l therefore do not contribute.
(ii) We note that 2
l , when
0)(')()()(' RjkRnRjkRkn llll .
Then we have
0)(')()()1( 1 RjRjkRkl ll
or
0)(
)('1
Rj
RjRl
l
l
.
Using the asymptotic form
)2
sin(1
)( lxx
xjl
, )
2cos(
1)(' lx
xxjl
,
)2
cos(1
)( lxx
xnl
, )
2sin(
1)(' lx
xxnl
.
So we have
0)2
cos()2
sin(1
lRlRR
l ,
or
R
llR
1
)2
cot(
,
or
R
llR
1
)22
tan(
.
Since the right hand side is very small, we get
R
ln
lR
)
2
1(
2.
60
where n is an positive integer. The resonant scattering occurs when the incident energy is just such as to match an energy level. 20. Connection between resonance and binding energy (a) S-matrix element for the low-energy scattering
We start with
)cot()cot( 0 kRkR (scattering due to the attractive potential))
or
)cot()cot( 0 kRR
k
for the low-energy S-wave scattering, where
220
2222 RkRkR .
Here we determine the S-matrix element
02 ieS . using the formula
1
1cot
2
2
ix
ix
e
eix .
Note that for 0 kRx , we have
)cot(1
1)cot( )(2
)(2
0 0
0
Rke
eikR kRi
kRi
or
)cot(1
1)(2
)(2
0
0
Rk
ie
ekRi
kRi
or
61
)sin()cos(
)sin()cos(
)cos()sin(
)cos()sin(
)cot(1
)cot(1)(2 0
Rk
iR
Rk
iR
Rk
iR
Rk
iR
Rk
i
Rk
ie kRi
Thus we get
)sin()cos(
)sin()cos(22 0
Rk
iR
Rk
iReeS ikRi
As is expected, the expression for S has unit modulus; 1S . There is a remarkable relation
between the S-matrix and bound states. We set
ik , b
Then the function S has a pole for
0)sin()cos( RR bb
b
or
)cot( Rbb (bound state)
which is exactly the same expression derived from the approach from the wave function of the bound state. In general, the poles of )(kSl for ik give the position of the bound states in the
l-th partial wave. (c) Scattering length from the effective range expansion
We start with the relation
0
2
0 2cot r
kk (the effective range expansion)
62
The scattering length a is defined by
ar
kk
kk
1)
2(limcotlim 0
2
00
0
or
a
1.
______________________________________________________________________________ ((Note))
In the early days of nuclear physics, many attempts were made to fit experimental data on low–energy scattering phase shifts for the nucleon-nucleon system. It was found that there was a peculiar insensitivity to the precise potential shape and the data could be fit by almost any shape. The essential result was that the function 0cotk is to excellent approximation a linear function
of k2,
02
0 2
1cot rkk ,
The parameters a is called the scattering length, whereas r0 is called the effective range, and the approximation is referred to as the effective-range approximation. _____________________________________________________________________________
Using this scattering length, the total cross section can be rewritten as
22
2
22
0222
022
1
4
14
cot
4
cot1
14
ka
a
ak
kk
ktot
The low-energy form of the scattering amplitude is given by
63
a
ik
iika
a
kik
i
ii
k
ik
ek
kf i
1
cot
1
)sin(cos
)sin)(cossin(cossin
1
)sin(cossin1
sin1
)(
0
00
00000
000
00
,
using the effective-range approximation. Then )(kf has a simple pole at
ia
ik
We note that in the low energy limit the optical theorem is
222
2
22
41
14
]1
)1(Im[
4
]1
Im[4
)](Im[4
aak
a
ak
ikaa
k
ika
a
k
kfktot
((Note)) Deuteron as an example
s = 1 3S1 ((Definition of the scattering length ))
aRR
kk
kk
1
2)
2(limcotlim
222
00
0
or
64
Ra 2
1 2 . (1)
In the spin triplet channel, the scattering length a and effective range R are
a = 5.42 fm, R = 1.75 fm,
respectively, while the binding energy is 2.23 MeV. fm = 10-15 m = 10-13 cm. ((Note)) The femtometre (American spelling femtometer, symbol fm Ancient Greek: μέτρον, metrοn, "unit of measurement") is an SI unit of length equal to 10−15 metres. This distance can also be called fermi and was so named in honor of physicist Enrico Fermi, as it is a typical length-scale of nuclear physics. https://en.wikipedia.org/wiki/Femtometre This means that
22
2
bE = 2.23 MeV, 232.0 fm-1.
where
np
np
mm
mm 8.36887 x 10-25 g
With these values both sides of Eq.(1) have the value 0.185 fm-1;
1845.01
a fm-1,
which is in good agreement with
1848.02
2
R fm-1.
21. Breit-Wigner resonance formula
The cross section for pure elastic scattering for the l-th partial wave is
l
llel
lk
lk
22
22
cot1
1)12(
4
sin)12(4
65
This has a maximum when 2
)( 0
El at E = E0.
2)( 0
El
We expand )](cot[ El around E = E0 using the Taylor expansion as
...)(sin
1
....)(cot)](cot[)](cot[
02
00
0
0
EEdE
d
EEdE
dEE
EE
l
l
EElll
Here we define
2
0EE
l
dE
d.
Then we get
...)(2
)](cot[ 0
EEEl
0
)(cos)(12)(l
ll Pkflf ,
with
2)(
21
)(2
11
)sin()cos(
)sin(1
)sin(1
)(
0
0
iEEk
iEEk
ik
ek
kf
ll
l
li
ll
,
The total cross section is
66
20
2
2
2
22
22
)(4
4)12(4
cot1
1)12(
4
sin)12(4
EEl
k
lk
lk
l
ll
The factor
20
2
2
)(4
4)(EE
EP
,
is called as a Breit-Wigner factor. 22. Definition of scattering length a
In the limit of 0k ,
)('
)tan
lim('
]tan)cos()[sin(cos
)()(
0
0
00
arCk
rC
krkrC
rrRru
k
outout
Here we define the scattering length a as
ka
k
0
0
tanlim
.
or
ak
k
1cotlim 0
0
.
Thus we have the wave function as
)(')( arCruout .
The total cross section in the limit of 0k is given by
67
2
00
02
24
cot
1lim4sin
4a
ikkk ktot
.
(a) Repulsive potential. The scattering length a is positive (a>0). For the infinite potential
height, the scattering length a is equal to R, since 0)( Rru .
0)(3
1]1
)tanh([ 2
00 RkkRR
RkR
.
0)(3
1)tanh(1)(lim 2
00
00
0
Rk
Rk
Rk
ka
k
.
68
(b) Attractive potential without a bound state. The scattering length a is negative (a<0).
u(r=0) = 0. u(r) is proportional to )(')( arCruout with a<0.
]1)tan(
[0 R
RkR
s
s
,
0]1)tan(
[)(lim0
00
0
Rk
RkR
ka
k
.
69
(c) Deeper attractive potential with a single bound state. a is the scattering length. a>0.
u(r=0) = 0. u(r) has a peak for r<R, indicating the existence of the bound state. For r>R, u(r) exponentially decays with increasing r.
0cotk , or
1tan 0 k
,
1tan
lim 0
0
ka
k>0.
23. Levinson's theorem
The Levinson’s theorem relates the phase shift as zero and infinite energy to the number of bound states.
There is a remarkable theorem due to Norman Levinson, which relates the behavior of the phase shift for E>0 to the number of bound states with E<0. We already show that the number of bound states as a function of R is closely related to that of the phase shift as a function of R .
(i) 2
0 R 0
There is no bound state
(ii) 2
3
2
R
There is one bound state
70
(iii) 2
5
2
3 R 2
There is two bound states
(iv) 2
7
2
5 R 3
There is three bound states. In other words, we have the relation
N where N is the number of bound states. This relation is called the Levinson's theorem. In general we have
ll N
for any l.
Fig. Graphical solution for the number of bound states.
p2 3p2 5p2 7p2x=kR
y=aR
0 2 4 6 8 10 12
0
2
4
6
8
10
12
71
Fig. Schematic diagram for the phase shift vs R. 24. How to determine the depth of the square-well potential.
20
22 )()( RkkRR
p
2p
kR
d0
0p
2p
3 p
2
kR
stot
0 p2
p3 p2
2 p5 p2
3 p7 p2
4 p9 p2
5 p11 p
26 p
13 p2
7 p
2
4
6
8
72
and
20020
2kV
mU
Suppose that R is a littler smaller than 2
12 n. We slightly increase 2)(kR (which is
proportional to the kinetic energy). When
2
12
nR
the total cross section increases drastically. Since 1kR , we can evaluate the depth of the potential as
2
122020
nRV
mRk
.
27 Scattering length in neutron scattering
73
Figure shows the geometry of a scattering experiment. An incident neutron, specified by its wavevector k, is scattered into a new state having a wavevector kf. The origin of coordinates is at the position of the nucleus and the neutron is scattered to a point r. The direction of scattering is defined by the azimuthal angle f and by the angle 2q between the incident and scattered beams. Scattering occurs in an elementary cone of solid angle d . If the scattering is elastic, then the magnitude of the wavevector is unchanged on scattering, i.e., fi kk .
The total cross section is defined by
0
1
Itotal (number of neutrons scattered per sec in all directions .
74
The incident flux 0I is the number of neutrons striking unit are of the sample in unit time, where
the area is taken to be perpendicular to the incident neutron beam. Let us consider scattering by a single nucleus. An incident plane wave of neutrons travelling in the z direction is expressed by
ikzi eA0 ,
where 0A is the normalization factor, and /2k is the wavenumber. The probability of
finding a neutron in a volume V is
12
0
2 VAdVi ,
leading to the value of 0A as
VA
10 ,
V
vvAvI i 2
0
2
0 . (1)
The wavelength of slow neutrons vastly exceeds the nuclear radius, and so there is S wave
scattering which is isotropic with no dependence on direction. The wave scattered by an isolated nucleus, is of the form
r
eAb
ikr
f , (2)
which r is the distance from the scattering nucleus and b is known as the scattering length of the nucleus. We assume that the nucleus is fixed so that the scattering is elastic. The minus sing in Eq.(1) is adopted to ensure that the most values of b for the elements are positive. In the absence of an appropriate theory of nuclear forces, the scattering length is treated as a parameter to be determined experimentally for each kind of nucleus. The scattering length for the elements are listed in the textbooks on the neutron scattering. In the thermal neutron region b is independent of wavelength. The scattered flux is
vr
bAf 2
222
velocity , (3)
and the number of neutrons scattered per second is flux times area
vbArvr
bAI f )4()4( 222
2
22 .
Hence from Eqs.(1) and (3), we get
75
2
0
4 bI
I ftot ,
which is the effective area of the nucleus viewed by the neutron. The units used for cross-sections are known as barns, where 1 barn = 10-28 m2, and the units used for scattering lengths are fermis, where 1 fermi = 10-15 m.
The differential cross section is obtained as
0
1
Id (number of neutrons scattered per sec into a solid angle d
or
dbdrvA
vr
bA
dvrvA
d f22
2
2
22
22
2
1 ,
or
2bd
d
.
28. One dimensional transmission-reflection problem (Lipmann-Schwinger equation)
76
The Lippmann-Schwinger formalism can also be applied to a one-dimensional transmission-reflection problem with a finite-range potential, 0)( xV for ax only.
(a)
)(
0
)( ˆˆ1
V
iHE
Show that the Green function )',( xxG is given by
'
0
2
2'
ˆ1
2)',( xxike
k
ix
iHEx
mxxG
(b) The Lippmann-Schwinger equation for )(x can be expressed by
)('
2)( ')'(
2'
2
2
1
xxVek
idx
mex xxikikx
(c) We consider the special case of an attractive function potential
)(2
2
xm
V ( 0 ).
Solve the integral equation to obtain the transmission and reflection amplitudes.
(e) The one-dimensional function potential with 0 admits one (and only one) bound
state for any value of . Show that the transmission and reflection amplitudes you computed have bound-state poles at the expected positions when k is regarded as a complex variable.
(f) Make a plot of the probability of transmission and reflection as a function of )2/( k . __________________________________________________________________________ ((Solution)) (a), (b) We use the Lippmann-Schwinger equation
)(
0
)( ˆˆ1
V
iHE
Green function:
77
im
pE
edpm
eppi
m
pE
edpdpm
xppiHE
ppxdpdpm
xiHE
xm
xxG
xxip
xipxip
2
'1
'2
1
2
"'
2
'1
"'2
1
2
'""ˆ1
''"'2
'ˆ1
2)',(
2
)'('2
'"
2
'2
0
2
0
2
where
xip
epx'
2
1'
, ",'"' pppp
Here we put
m
kE
2
22 , '' kp
ikk
edkxxG
xxik
22
)'(
''
2
1)',(
78
)(
)2
(
)2
1(
)1(
)('
2
2/12
2/12
ikk
ik
k
ik
k
ik
ikk
((Jordan’s lemma, residue theorem)) The integrand has poles in the complex k-plane at
ikk ' , and )(' ikk When 'xx , we take the path in the upper plane. When 'xx , we take the path in the lower plane.
Re
Im
k i
k i
O
79
(i) For 'xx ,
k
ei
ikksixxG
xxik
2
)'(Re22
1)',(
)'(
(ii) For 'xx ,
k
ei
ikksixxG
xxik
2
)'(Re)2(2
1)',(
)'(
Combining Eqs.(1) and (2),
'
2)',( xxike
k
ixxG
The Lippmann-Schwinger equation for )(x ;
)('
2)( ')'(
2'
2
2
1
xxVek
idx
mex xxikikx
(c), (d)
)(2
2
xm
V ( 0 ).
)(
)('
)(2
'
2)(
022
1
')]'(2
'2
1
')]'(2
[2
'2
2
1
xikikx
xxikikx
xxikikx
ek
ie
xxek
idxe
xxm
ek
idx
mex
When x = 0,
)()( 022
10
k
i
80
or
k
i
21
1
2
10 )(
Then we get
xikikx e
k
ik
i
ex
21
22
1
2
1)(
For x>0
ikxe
k
ix )
21
1(
2
1)(
For x<0
]
21
2[2
1)( ikxikx e
k
ik
i
ex
T: probability for the transmission R: probability for the reflection
2
21
1
k
T
,
2
2
21
21
k
kTR
((Bound state)) The wave function of bound state Schrodinger equation
81
)()()()(2 2
22
xExxVxdx
d
m
with
)(2
)(2
xm
xV
For 0x , 0)( xV .
)()(2 2
22
xExdx
d
m
For 0 EE (bound state)
)()(2
)( 222
2
xxEm
xdx
d
The solution of this equation is
xAex )( where
mE
2
22
Note that )(x is continuous at x = 0, but dxxd /)( is not continuous.
dxxm
dxxxm
dxxdx
d
m)(
2)()(
2)(
2
222
2
22
or
)0()]([ x
dx
d
leading to the relation
2
.
82
Therefore the wave function of the bound state is
2/
2)( xex
The value of k corresponding to the bound state is 2
ik .
)2
)(2
(
1
21
12
ki
ki
k
T
)2
)(2
(
2
21
2
2
2
2
ki
ki
k
k
kR
T and R has a pole of 2
ik
We make a plot of T and R as a function of k
x2
.
________________________________________________________________________ REFERENCES
R
Tx 2k
1 2 3 4 5
0.2
0.4
0.6
0.8
1.0
83
J.J. Sakurai and J. Napolitano, Modern Quantum Mechanics, second edition (Addison-Wesley, New York, 2011).
A. Das and A.C. Melissinos, Quantum Mechanics A Modern Introduction (Gordon and Breach Science Publishers, New York, 1986).
R.G. Sachs, Nuclear Theory (Addison-Wesley, 1953). David Park, Introduction to the Quantum Theory, 3rd edition (McGraw-Hill, Inc., New York,
1974). M.L. Bellac, Quantum Physics (Cambridge University Press, 2006). Stephen Gasiorowicz, Quantum Mechanics, third edition (John Wiley & Sons, Inc., New York,
2003). F. Schwabl, Quantum Mechanics, 4-th edition (Springer Verlag, Berlin, 2007). ISBN 978-3-540-
71932-8 Richard L. Liboff, Introductory Quantum Mechanics, 4th edition (Addison Wesley, New York,
2003). John S. Townsend , A Modern Approach to Quantum Mechanics, second edition (University
Science Books, 2012). A.G. Sitenko, Lectures in Scattering Theory (Pergamon Press, Oxford, 1971). H.A. Bethe and P. Morrison, Elementary Nuclear Theory (John Wiley & Sons, New York, 1947). D. Bohm, Quantum Theory (Dover, 1979). B.T.M. Willis and C.J. Carlile, Experimental neutron scattering (Oxford, 2000). S.W. Lovesey, Theory of Neutron Scattering from Condensed Matter (Oxford, 1986). ______________________________________________________________________________ APPENDIX I What the Ramsauer-Townsend effect meant for Bohr in 1920’s
Before I read the book of H. Kragh, I do not understand why this effect is so often discussed in the scattering in the quantum mechanics. But after I read the book, I realized that this effect is very important as well as the Frank-Hetz experiment. The Ramsuaer effect is the phenomenon of elastic scattering for electrons, while the Frank-Hertz experiment is the phenomenon of inelastic scattering for electrons. Here I put the following sentences in the book written by H. Kragh. H. Kragh, Niels Bohr and the Quantum Atom: The Bohr Model of Atomic Structure 1913-1925 (Oxford, 2012) p.261. ((Kragh))
There were other anomalies that played a similarly minor role in the crisis that eventually led to the fall of the Bohr–Sommerfeld theory. One of them was the Ramsauer effect, so named after the Heidelberg physicist Carl Ramsauer who, at a meeting of German scientists in Jena in September 1921, reported some startling results concerning the penetrability of slow electrons in an argon gas. A few earlier physicists had observed that slow cathode rays move more freely through a gas than fast ones, but it was only with Ramsauer’s work that the effect became generally known and aroused widespread attention in the physics community. Franck (James Franck, known as the Frank-Hertz experiment), who had participated in the Jena meeting,
84
reported to Bohr: ‘In Jena I was particularly interested in a paper of Ramsauer that I am not able to believe, though I cannot show any mistake in the experiment. Ramsauer obtained the result that in argon the free path lengths are tremendously large at very low velocity of electrons. If this result is right, it seems to me fundamental. Bohr replied that he was very interested in the new result and wanted more information about it. He thought that the question was probably ‘very closely connected with the general views of atomic structure. Franck and Bohr were not the only physicists who found Ramsauer’s experiments puzzling. In November Born wrote to Einstein about ‘Ramsauer’s quite crazy assertion (in Jena) that in argon the path length of the electrons tends to infinity with decreasing velocity (slow electrons pass freely through atoms!)’. He added that ‘This we would like to refute!’ The initial skepticism with regard to the Ramsauer effect evaporated after it was confirmed by experiments carried out by, among others, Gustav Hertz in Eindhoven, Hans Mayer in Heidelberg, and Rudolph Minkowski and Hertha Sponer in Göttingen. Not only was the effect real, it also turned out that it was not limited to argon but appeared in the other noble gases as well and possibly was a general property of matter in the gaseous state. The phenomenon defied theoretical explanation, whether in terms of classical theory or quantum theory. The first to take up the challenge was young Friedrich Hund, who at the time was a doctoral student under Born in Göttingen. Inspired by Franck, he developed a theory based on quantum conditions and the correspondence principle, from which followed that slow electrons would not be influenced by collisions with gas molecules.126 Bohr was keenly interested in Hund’s theory, which he knew in outline from his correspondence with Franck. Although the theory was somewhat unorthodox, he thought it agreed with the spirit of quantum theory. ‘I see no other simple explanation of the Ramsauer experiment’, he wrote to Franck, ‘and am so skeptical about the established principles of physics that I do not feel justified in rejecting your [and Hund’s] ideas as total nonsense’.127 However, Hund’s theory turned out to be untenable and it was not replaced by better theories. The Ramsauer effect thus remained unexplained, without the lack of explanation causing much concern at the time. Although physicists in Copenhagen and Göttingen were convinced that it was a quantum effect, other physicists thought of it in terms of classical gas theory or simply avoided attempts of explanation. At any rate, by far the most work on the Ramsauer effect was experimental, an autonomous line of research that was uninfluenced by quantum theory. The Ramsauer effect was anomalous, but it was not obvious that the anomaly belonged to the domain of quantum theory. This may explain the limited role it played during the last years of the old quantum theory, when it was no more significant than the Paschen–Back effect. It is noteworthy that the Ramsauer anomaly did not appear in any of the editions of Sommerfeld’s Atombau or in Born’s Atommechanik. It only received a partial explanation in 1925, when Walter Elsasser used Louis de Broglie’s new ideas of matter-waves to explain how slow electrons can penetrate almost freely through gases because of their very large wavelength (as given by the de Broglie formula λ = h/mv).128 At about the same time Bohr returned to the effect and how to understand it in a broader physical context. He revealed some of his thoughts in a letter to Geiger of April 1925. Recently I have also felt that an explanation of collision phenomena, especially Ramsauer’s results on the penetration of slow electrons through atoms, presents difficulties to our ordinary space-time description of nature similar in kind to those presented by the simultaneous understanding of interference phenomena and a coupling of changes of state of separated atoms by radiation. I believe that these difficulties exclude the retention of the ordinary space-time description of phenomena to such an extent that, in spite of the existence of coupling, conclusions about a possible corpuscular nature of radiation lack a sufficient basis.
85
______________________________________________________________________________ APPENDIX II Summary on the resonance scattering at low energies
We consider the attractive potential with a depth V0.
202
0
2
Vk
2
22kEk
,
2
22bE
22
0 kk 220 kb ,
We assume that the depth V0 is changed as a parameter. (a) The bound state
The number of the bounds states increases with increasing 0V . When )2
1( nRb , the
number of bound states is n. The highest level of the bound states is zero. The other bound states are well below the zero energy.
86
For 2
Rb , 0R . The energy level of the bound state is equal to zero ( 0R ).
For 2
3 Rb , 0R . The energy level of the bound state is equal to zero ( 0R ). There is
one bound state well below zero energy.
For 2
5 Rb , The energy level of the bound state is equal to zero ( 0R ) (E0 =0). There are
two bound state well below zero (E1 and E2)
2 3 2 5 2 7 2
x bR
y R
0 2 4 6 8 10 12
0
2
4
6
8
10
12
87
For 2
7 Rb , The energy level of the bound state is equal to zero ( 0R ). There are three
bound state well below zero. From the condition that 0R
)2
1(0
22220 nRkRRkRb
we have
2
222
0 2)
2
1(
RnV
.
(b) Scattering
The total cross section (scattering side) takes a peak when the depth V0 satisfies the condition
that )2
1( nR . We note that the kinetic energy of the particle is kept small while the depth
of the potential is changed as a parameter.
)1)tan(
(0 R
RkR
when 1kR .
RO r
Vr
-V0
E2
E1
E0
88
We make a plot of )4/( 2Rtot as a function of Rx . This function becomes infinity at
R = /2, 3/2,….
x=kR
sin2d0
p21.5 1.6 1.7 1.8
0.2
0.4
0.6
0.8
1.0
p
2p
kR
d0
0p
2p
3 p
2
89
Fig. Plot of 22
]1)tan(
[4
R
R
Rtot
as a function of R . The change of the total cross section
tot as the kinetic energy of the incident particle, where the potential energy is kept
constant. tot becomes zero at R = 4.49341 and 7.72525 (Ramsauer-Townsend effect)
and becomes infinity at R = /2, 3/2,…. (resonance).
(c) Resonance
)2
1( nRb , with 0
)2
1( nR with 0k
Note that ik in the limit of 0k . The resonance occurs when a part of particle is at the top of the bound state. These particles stray at the bound state for finite times and go out of the range of potential. There is some resonance
tot
4 R2
R2 4 6 8 10
1
2
3
4
5
90
(d) Fabry-Perot etalon experiment
The heart of the Fabry–Pérot interferometer is a pair of partially reflective glass optical flats
spaced micrometers to centimeters apart, with the reflective surfaces facing each other. (Alternatively, a Fabry–Pérot etalon uses a single plate with two parallel reflecting surfaces.)
p
2p
kR
d0
0p
2p
3 p
2
91
The flats in an interferometer are often made in a wedge shape to prevent the rear surfaces from producing interference fringes; the rear surfaces often also have an anti-reflective coating. (e) Physical interpretation
The wave of particles (with very small positive energy) inside the well potential undergo reflections at the boundary r = R. A part of waves goes out side of the well potential as a form of transmission. The cause of such reflections is due to the drastic change of wave vectors. Both the reflection and transmission at the boundary are similar to the phenomenon of the Fabry-Perot etalon.
Note that particles stay inside the well potential for a short time, forming a so-called meta stable state.
220)
2
1( kkRnR
or
02
2222
2
2)
2
1(
2V
RnkEk
When
2
222
0 2)
2
1(
RnV
there is a meta- stable state at E = 0, leading to the resonant scattering. APPENDIX III Selected Problems and Solutions (Capri Quantum mechanics) (a) Low energy-S-wave amplitude (Capri, Problem 19-9)) Show that the scattering amplitude for low energy S waves may be written as
arkika
a
ikrka
kf
02
02
0
2
11
2
111
)(
as well as in the form
92
ikkkf
cot
1)(0
where a is the scattering length and r0 is the effective range. Also verify that both versions of the amplitude satisfy the optical theorem. ((Solution))
kik
i
ii
k
ik
ek
kf i
0
00
00000
000
00
cot
1
)sin(cos
)sin)(cossin(cossin
1
)sin(cossin1
sin1
)( 0
(1)
02
0 sin1
)(Im k
kf
On the other hand, for low energies the phase shift is given in terms of the scattering length a and effective range r0 by
02
0 2
11cot rk
ak . (2)
Substituting this into Eq.(1), we get
02
02
0
2
11
2
111
)(arkiak
a
kirka
kf
. (3)
The optical theorem states that the total cross section tot is given by
kaark
af
ktot2
02
2
)21
1(
4)]0(Im[
4
Here we note that )(0 kf is independent of . On the other hand, the total cross section can be
directly calculated as
93
02
2
2
0 sin4
)( k
kfddtot
So, in this case, the optical theorem is verified.
If we start with Eq.(3) we have
2220
2
2
2220
2
2
0
)21
1(
4
)21
1(
4)]0(Im[
4
kaark
a
kaark
ka
kf
ktot
since
2220
2
2
0
)2
11(
)](Im[kaark
kakf
On the other hand by direct calculation we see that
2220
2
2
2220
2
2
2
0
)2
11(
4
)2
11(
)(
kaark
a
kaark
ad
kfdtot
Thus, in this case we have also verified the optical theorem. (b) Phase shifts for Yukawa potential (Capri, quantum mechanics, Problem 19-8)
Compute approximate l = 0and l = 1 phase shift for scattering a high energy particle of mass m by a short range Yukawa potential
r
eVrV
r
0)( ,
with 00 V and 0 . Use whatever seems to be an appropriate approximation.
((Solution)) Here we use the Born approximation. The scattering amplitude is evaluated as
94
220
2
220
2
)cos1(2
12
12)(
k
Vm
Q
Vmf
.
since
)cos1(2)2
(sin4 2222 kkQ
The partial wave expansion is given by
0
0
)(cos)](1[122
)(cos)(12)(
lll
lll
PkSlk
i
Pkflf
,
with
)](1[2
)1(2
)( 2 kSk
ie
k
ikf l
il
l .
However, the Born approximation assumes that the scattering amplitude is small. Thus we must have small phase shifts. In this case
lli iie l 2)21(1)1( 2
Thus we get
0
0
0
)(cos121
)(cos)2(122
)(cos)(12)(
lll
lll
lll
Plk
Pilk
i
Pkflf
95
l
mlmm
mlmml
k
lm
k
dPPmk
dPf
2
12
212
1
sin)(cos)(cos121
sin)(cos)(
0,
0 00
or
dPfk
ll sin)(cos)(2 0 .
where
',
0
' 12
2sin)(cos)(cos llll l
dPP
For l = 0 (S-wave)
)4
1ln(2
)cos1(2
sin
sin)(cos)(2
2
2
220
0222
0
0
00
k
k
mV
k
dmV
dPfk
For l = 1 (P-wave)
)]4
1ln()2(4
11[
)cos1(2
cossin
sin)(cos)(2
2
2
2
2
220
0222
0
0
11
k
kk
mV
k
dmV
dPfk
We use the Mathematica for the calculation of integrals.
96
(c) Effective range, scattering length for an attractive potential (Capri Quantum Mechanics, 19-12)
Given the attractive potential
0
)( 0VrV
ar
ar
find the effective range and the scattering length for the S-wave ( 0l ) ((Solution)) Suppose that
'00
]2
)2sin([
)]2cos(1[
)(
)(sin2
)]()[(2
'sinsin
220
02
0
02
22
02
0
20
22
000
kaka
k
mV
krdrk
mV
kr
krdrrkV
m
krjrVdrrkm
e
a
a
i
or
0)]2sin(2
1['sin
220
0 kakak
mV
For 0<x<<1, we have an Taylor expansion as
)(15
2
3
2)2sin(
2
1 753
xOxx
xx
Thus we have
])(5
1[
3
1
])(15
2)(
3
2['sin
3
53220
0
kaka
kakak
mV
97
where
2
202
amV
Using the result of 'sin 0 , we find
22
02
2/10
2
0
00
)(18
11
)sin2
11(
)sin1(
'cos
)'cos(cos
ka
Thus we get
2
22
3
22
0
0
0
00
)(51
1
)(18
113
])(51
[31
)(18
11
'sin
'cos
)'sin(
)'cos(cot
ka
ka
akaka
kak
k
kk
or
02
0
20 2
11)
5
6
3(
2
13cot rk
ak
ak
So we have
aa3
10 ,
5
6
30 r
(d) Effective range, scattering length for an attractive Yukawa potential (Capri
Quantum Mechanics, 19-13) For a Yukawa potential,
98
r
eVrV
r
0)(
We start with the approximate equation
)4
1ln(2
)(sin12
)(
)(sin)(
2
)]()[(2
sin
2
2
20
0
22
0
02
2
02
2
0
20
220
0
k
k
mV
krer
drk
mV
kr
kr
r
eVdrrk
m
krjrVdrrkm
e
r
r
i
Suppose that
'00
)4
1ln(2
'sinsin
2
2
20
000
k
k
mV
ei
or
)21(
)2(2
)84(2
)4
1ln(2
'sin
2
2
3
3
220
4
4
2
2
20
2
2
20
0
kk
kkmV
kk
k
mV
k
k
mV
or
)21('sin2
2
0 kk
99
where
2202
mV (dimensionless)
...84)41ln( 422 xxx for 1x .
We note that
22
2
2
22
2/10
2
0
00
)21(2
11
)'sin1(
'cos
)'cos(cos
kk
Thus we have
)21(
)21(21
1
'sin
'cos
sin
coscot
2
2
22
2
2
22
0
0
0
00
kk
kk
k
k
kk
or
20
20 )
4(
2
1
2
11cot krk
ak
with
a ,
40 r
