Download Scattering Phase-Shift Examples - Howard University Physics and

Quantum Mechanics II
Scattering
Basics;
Scattering Phase-Shift;
Scattering Phase-Shift Examples
Tristan Hübsch
Department of Physics and Astronomy, Howard University, Washington DC
http://physics1.howard.edu/~thubsch/
Q
M
II
Scattering
Basics
So far: quantum object (particle or wave)
subject to (classical) agents represented by a surrounding potential
Turn this around: the (classical) scattering target is localized
the quantum object: (plane) wave or a beam of particles
incoming from away, interacts with the target, scatters
…and is detected away from the
⇥ target
=
⇤
g—General Theory
JS µ J I ds
h̄
M
⇥ ⇤S~rY S
Y
=m
⇤
1)
(
Det.
⇥)
= (2 r
d2 W = sin(q )dq df
solid angle
=
f
q
J
S
⇤
⇥ ⇤
⇤
⇥ ⇤
⇤
z
h̄
h̄
y
~
~
r
J I = M =meY
r
Y
J
=
=
m
Y
r
Y
(1)
oI
I
S YI M
S
S
⇤~
⇥
r
h
lT
YI
a
q
r
diff. (2)
cross-section
m
2
e
n
d WGe= sin(q )dq df h̄ = q
h
i
h
i
f
d
M
1 ds
ds
JS
ds
—
=
q
g
d
)function of !:
: 2
:= 2
=
in Determine J SI /JI as (a
=
q
r
e
2
2
dW
rr dW
d W
W
d
JI
r d W
in
att
s
c
|
{z
}
2W=
:S
d
:= s (q, f)
2
Q
M
II
Scattering
Basics
Linear momentum conservation: the target does move
Calculate in CM (center-of-mass)-frame
Compare with Lab-frame measurements
Lab
CM
~v1
~v
~v 0
T
~v MM
+ MT
q1
~v 00
~v2
Must be able to translate between the two
v1 cos(q1 ) = v0 cos(q ) + V
sin(q )
) tan(q1 ) =
0
cos(q ) + vV0
v1 sin(q1 ) = v sin(q )
MT v 2
1
2 M M + MT
M
+
MT v
M + MT
2 ! 1
1
Mv
= 2M
2 MT M + MT
!
Mv
MT M + MT = 0 =
v
0 2
Mv0
3
+
1
2M
v
MT v00
00 2
|
q
{z
}
⇣
⌘
~
~v M+MMT =: V
— velocity of the
CM-frame in Lab
in elastic collisions
M
are unchanged
) T speeds
⇢ 0
v = MM+TMvT
)
v
v00 = MM
+ MT
Q
M
II
Scattering
) tan(q1 ) =
Basics
sin(q )
cos(q ) +
V
v0
Relating the CM-calculation to the Lab-frame measurement:
Lab
CM
~v1
~v
|
~
V
q1
T
:= ~v MM
+ MT
~v 00
~v2
|
~v 0
{z
q
~v M+MMT
}
in particular, for the “differential cross-section”
s1 (q, f) sin(q1 )dq1 df1 = s(q, f) sin(q )dq df
p
⇣ ∂(q, f) ⌘⌘
(1 + 2b cos(q ) + b2 )3
= s(q, f)
s1 (q, f) = s(q, f)
|1 + b cos(q )|
∂(q1 , f1 )
Lab
b :=
CM
V
v0
There is no transformation in ". In fact, for the most part, the target and
so the scattered wave will be "-independent. To first order at least.
4
Q
M
II
Scattering
Scattering Phase-Shift
Scattering is a 2-body interaction
b
b=
H
~R :=
h̄2 ~
h̄2 ~
r2 + W (~r1
2M1 r1
2Mr
2
M
M1rr~1 + M
M2~rr2
~r := ~r1 ~r2
M1 + M2
CM-position
b=
H
Thus
~r2 )
µ := MM11+MM22
b
relative position
2
2
h̄
~~
r
2( M1 + M2 ) R
h̄
2µ
reduced mass
+ W (~r )
motion as a whole relative dynamics
⇥(free particle) ⇤
~J I :=
Choose:
h̄
µ
⇥ ⇤
⇤ ⇤
⇥
⇥
~ yI (~r ) ⇤ ~JS :=
=m yI (~r )r
r
yI (~r ) = A e
i~k·~r
1
r
h̄2
2µ
⇤
+ W (~r ) y(~r ) = Ey(~r )
y(~r ) = yI (~r ) + yS (~r )
incident scattered
h̄
µ
⇥
⇤
~ yS (~r )
=m yS (~r )r
⇤
~J I = | A|2 h̄~k~k ⇠ | A|2~v uniform plane-wave
µ
yS (~r ) = A f (q, f) e
I
⇥
i |~k||~r |
~JS = | A|2 | f (q, f)|2 h̄|~k| 12 êr
µ rI
5
spherical wave
S2
d2~s · ~JS = const.
Q
M
II
Scattering
Scattering Phase-Shift
Simple example: spherical potential p
2
⇥
~ y(~r ) + k2
r
y(~r ) =
Â
`m
⇥
⇤
⇤
U (r ) y(~r ) = 0
u ` (r )
m
a` m Y` (q, f) r
k :=
u00` (r ) +
⇥
⇥
2µE
h̄
k2
2µ
U (r ) := 2 W (r )
h̄
`(`+1) ⇤
U (r )
u ` (r )
r2
=0
Now: for very large r, ignore U(r) and the centrifugal barrier
u00` (r ) + k2 u` (r ) ⇠ 0
for r ! •
)
u` (r ) ⇠ e±ikr
BTW, the Coulomb potential cannot be ignored for large r!
⇠
)
⇠
!
Restrict to “short-range” potentials
i~k·~r
yI = e = Â(2`+1) i` j` (kr ) P` (cos (q ))
Know:
`
Try:
y = y + y = (2`+1) i` A R (r ) P (cos (q ))
satisfying
I
⇥
1 d2
r dr2 (rR` )
S
⇤
Â
⇥
`
2
+ k
`
U
6
`
`(`+1) ⇤
R`
r2
`
U = 0,
= 0 where
spherical Bessel eq.
Q
M
II
Scattering
⇥
~
⇤
⇥
⇤
y)
= Â(2`+1⇠
) i` A` R` (r ) P`!
(cos (q ))
⇠
Scattering Phase-Shift
⇥
⇤ ⇥
⇤
yI = eik·~r =
`
Â(2`+1) i` j` (kr) P` (cos (q ))
`
As the solution of the spherical Bessel equation,
a`2 + b`2 = 1
R` = a` j` (kr ) + b` n` (kr )
= cos(d` ) j` (kr ) + sin(d` ) n` (kr )
⇠
⇠
sin(kr 21 p `)
cos(d` )
kr
sin(kr
1
2 p`
kr
+ d` )
+ sin(d` )
cos(kr
kr
phase-shift
1
2 p `)
for kr
1
at the detector!
The phase-shifts are not multiples of #/2 and n$(kr) is not ruled out
…because the origin (where the n$(kr) diverge) is excluded by the target
⇠
Equating
y(~r ) = yI (~r ) + yS (~r ) = yI (~r ) + f (q, f)
h
`
(
2
`+
1
)
i
P` (cos (q )) A`
Â
`
sin(kr
1
2
p `+d` )
kr
7
eikr
r
sin(kr 12 p `) i
eikr
= f (q, f) r
kr
Q
M
II
Scattering
Scattering Phase-Shift Â
`
h
h
`
(2`+1)i P (cos (q )) A
`
`
sin(kr
1
2
p `+d` )
kr
i
sin(kr 12 p `)i
ikr
= f (q, f) e r
kr
i
Equating coefficients hof e –ikr (projecting
w/e –ikr):
`
(
2
`+
1
)
i
P` (cos (q )) A` exp(kr
Â
1
2
p `+d` )
exp(kr
1
2
`
Projecting with P$(cos(!)): A` = eid`
Equating coefficients of e +ikr (projecting w/e +ikr):
f (q, f) =
1
2ik
`
(
2
`+
1
)
i
e
Â
`
Â(2`+1) e
id`
Thus: s(q, f) = f (q, f)
2
=
1
k
`
s :=
Z
⇥⇥
p `/2
e2id`
4p
k2
p `) = 0
⇤⇤
1 P` (cos (q ))
sin(d` ) P` (cos (q ))
d2 W s(q, f)) =
=
i
This is indeed
independent of "
2
(
2
`+
1
)
sin
(d` )
Â
`
So, calculating &(!,") and & reduces to calculating '$.
8
Q
M
II
Scattering
u00` (r ) +
Phase-Shift Examples
A spherically limited potential
W (r ) =
nU ( r )
0
⇥
k
2
U (r )
`(`+1) ⇤
u ` (r )
r2
=0
for r < a
for r > a
The inside solution R$(r) = u$(r)/r must be regular at r = 0,
so u$(r) ~ r b, we must have b ≥ 1
@r ⇠ 0
⇥
u00` (r ) +
b 2
b ( b 1)r
`(`+1) ⇤
u ` (r )
r2
b 2
`(`+1)r
⇠0
⇠0
b=
n `+1
`
regular
singular
As long asnr2 U(r) < ∞ for r ~ 0, one solution is regular, one singular
Turns out, if r4 U(r) ~ ∞ for r ~ 0, both solutions are singular @ r ~ 0.
Assume that r2 U(r) < ∞ for r ~ 0, solve the radial equation , compute
(in)
g`
h d ln ru(in) (r ) i
`
and match this to the logarithmic
:=
dr
r ! a derivative computed from the outside
Matching u$(r) itself will match the amplitudes & the normalization
9
Q
M
II
Scattering
Phase-Shift Examples
h
i
⇥
00 n `+1 2
` (r ) + k
bu=
`
b ( b 1)r b
U (r )
2
`(`+1)r b
`(`+1) ⇤
u ` (r )
r2
2
=0
The outside solution is R$(r)!= cos('$) j$(kr) – sin('$) n$(kr)
hhd ln R(out) (r ) ii
h
(in)
(out)
`
h
i
:
d
ln
ru
(
r
)
g`
=
(in)
`
dr
:
r
!
a
g
=
dr
r`! a
⇥
⇤
⇥
dr
r!a
0
hh
0
0
i
k cos(d` ) j` (z) sin
(d` ) n` (z) (out)
!
(in)
⇥
h
i
=
g
=
h
d ln R` (r )
`
(out)
cos(d` ) j` (z)g sin:=
(d` ) n` (z) z!ka
`
dr
r!a
⇥
⇤i
This is the equation that determines hthe
phase-shifts,
'
$
0
0
k cos(in)(d` ) j` (z) sin(d` ) n` (z)
0
k j`=
(ka) g` j` (ka)
cos
(
d
)
j
(
z
)
sin
(
d
)
n
(
z
)
z!ka
tan d` =
`
`
`
`
(in)
k n0` (ka) g` n` (ka)
k j`0 (ka) g`(in) j` (ka)
(in)
Quick example: impenetrable
inside
solution
=
0
tansphere.
d` = The
,
g
!•
(in)
0
`
• )
k n` ``(ka
n2s` (ka) z`
sg
`
(
1
)
(
s
+`)
!
j` (ka)
j` (z) = 2 z  s!(2s+2`+1)! z ⇡ (2`+1)!!
tan d` =
s =0
z⌧1
n` (ka)
•
`+
1
s
( 1)
( 1) (s `)!
(2` 1)!!
n` (z) = 2` z`+1 Â s!(2s 2`)! z2s ⇡
z`+1
so
tan d` ⇡
(2`+1)
2`+1
(
ka
)
[(2`+1)!!]2
s =0
ka ⌧ 1 Lowest-ℓ terms contribute most.
10
Q
M
II
Scattering
tan d` ⇡
Phase-Shift Examples
For very low energies, ka ≪ 1,
s=
⇡
2
Â(2`+1) sin
4p
k2
2
(
2
`+
1
)
tan
(d` )) =
=
Â
=
(d` )) =
`
⇥
2
= 4pa 1 +
4
1
(
ka
)
3
+
tan2 (d` )
(2`+1) 1+tan2 (d )
`
4p
k2
Â
4p
k2
Â(2`+1)
`
1
405 ( ka )
`
8
+...
rapid convergence
⇤
4 × the
⇥ geometric cross-section ⇤
For very high energies,
j` (ka)
⇡
tan(d` ) =
n` (ka)
ka ⌧ 1
4p
k2
`
(2`+1)
2`+1
(
ka
)
2
[(2`+1)!!]
1
ka sin( ka
1
ka cos( ka
⇣
(2`+1)
2`+1
(
ka
)
[(2`+1)!!]2
`p/2)
= tan ka ` p2
`p/2)) must sum over all ℓ
Summation (Sakuri, 1982) produces & = 2$ a2
…still 2 × the geometric cross-section
11
⌘2
Q
M
II
Scattering
⇥
u00` (r ) +
Phase-Shift Examples
k
2
U (r )
`(`+1) ⇤
u ` (r )
r2
=0
Consider a finite-strength spherical potential barrier
U(r) = V0 > 0 for 0 ≤ r ≤ a (inside), U(r) = 0 outside.
K0 : =
Focus on $ = 0
u(in)
0 (r ) =
0r  a
p
2MV0
h̄
k :=
p
p
{ := pK02 k2
2ME
h̄
n A sinh({ r )
u(out)
0 (r ) = sin( kr + d0 )
A0 sin({ r )
r a
Equating at r = a the outside and the inside logarithmic derivative
!
1 k
{ tanh({ a )
!
ka cot(ka + d0 ) = { 0 a cot({ 0 a) E > V0 d0 = tan 1 {k0 tan({ 0 a)
⌧ 0 ka{⌧
⇡ 1 k ⌧ K0 { ⇡ K0
However, for⌧low energies,
⌧
⌧
2`+1
⇠
tan(d` ) ⇠ (ka)
⌧ 1 ka + d0 ⌧ 1 cot(ka+d0 ) ⇡ ka+1 d0
tanh(K0 a)
K0 a
ka
1
) d0 ⇡ ka
K0 a
ka+d0 ⇡ tanh(K0 a)
ka cot(ka + d0 ) = { a coth({ a)
E < V0
12
d0 = tan
s0 ⇡ 4pa2
tanh(K0 a)
K0 a
ka
ka
1
2
Q
M
II
Scattering
u00` (r ) +
Phase-Shift Examples
⇥
k
2
U (r )
Consider a spherical '-function bubble
aZ+e
Focus on $ = 0
00
0
0
`(`+1) ⇤
u ` (r )
2
r
⇥
=0
dr u0 (r ) = u0 ( a+e) u0 (r e) = la u0 ( a)
h h 0
i i
0
au0 ( a+e)
au0 (r e)
a e
lim u (a)
=l
u
(
a
)
0
i
!e!0 h 0
The solution is
n A sin(kr )
for r < a
n
u0 (r ) =
sin(kr +d0 ) for r > a
…for which the boundary condition is
⇥
⇤
k cot(ka+d0 ) cot(ka) = l
and produces:
⇥
⇣
⌘⇤
tan(ka)
1
l
d0 = tan
ka
tan(ka)
1
+
l
ka
while continuity yields: s
tan2 (ka) + 1
A=
tan(ka) 2
2
tan (ka) + 1 + l ka
13
just like a
particle in
a hard box
En ⇡
n2 p 2 h̄2
2µa2
Resonances!
tan(k n a) ⇡ 0
k n ⇡ n pa
Quantum Mechanics II
Now, go forth and
calculate!!
Tristan Hübsch
Department of Physics and Astronomy, Howard University, Washington DC
http://physics1.howard.edu/~thubsch/