For m.sc. quantum physics

1. JAI NARAIN VYAS UNIVERSITY
JODHPUR ( RAJ. )
PHYSICS DEPARTMENT
ADVANCE QUANTUM MECHANICS
PRESENTEDBY –
RAMNIWAS JAKHAR
SUBMITTEDTO – DR. S. S.
MEENASIR

2. GREEN FUNCTION FOR
SCATTERING PROBLEM AND
LIPPMANN-SCHWINGER
EQUATION

3. • GREEN FUNCTION :-
In mathematical a Green function is the impulse
response of an inhomogeneous linear differential
operator, defined on a domain with specified
initial condition or boundary condition.
If differential operator is L(x) is linear, then one
can superpose them to find the solution.
u(x) = ʃ f(y) G(x , y) dy
Solution of Green function if differential operator
is L than –
L G(x , s) = δn
(s-x)

4. where- δ = Dirac delta function.
n = Dimension
In physics :- the Green function as used in physics is
usually defined with the opposite sign, that is –
L G(s , x) = δ(x-s)
If the operator is translation invariant that is, when
L has constant coefficient with x, the Green
function can be taken to be a convolution kernel,
that is –
G(x , s) = G(x-s)
In this case, Green function is the same as the
impulse of linear time invariant system theory.

5. • GREEN FUNCTION FOR SCATTERING PROBLEM:-
PROPERTY OF GREEN FUNCTION :-
𝛻2 + 𝐾2 G( r, r’)=𝑑3 (r, r’) --------(1)
Schrodinger equation
G( r,r’)=G(r-r’)
 LOWER TRANSFORMATION OF GREEN FUNCTION:-
G(r-r’)= 𝐺 𝑞 iq(r − r′)𝑑3
𝑞 -------(2)
Generate r to q state
𝑑3
𝑟 − 𝑟′
=
1
(2𝜋)3 𝑒𝑞(𝑟−𝑟′)
𝑑3
q
Eq (1),
(q+𝑘2 )G (q)𝑒𝑖𝑞(𝑟−𝑟′)𝑑3q=
1
(2𝜋)3 𝑒𝑖𝑞(𝑟−𝑟′) 𝑑3q

6. (-𝑞2
+𝑘2
) 𝐺(𝑞) 𝑒𝑖𝑞(𝑟−𝑟′)
𝑑3
q=
1
(2𝜋)3 𝑒𝑖𝑞(𝑟−𝑟′)
𝑑3
q
[(𝑞2 − 𝑘2)G(q)+
1
(2𝜋)3]𝑒𝑖𝑞(𝑟−𝑟′)𝑑3𝑞 = 0
𝑒𝑖𝑞(𝑟−𝑟′)
𝑑3
q≠0
(𝑞2
− 𝑘2
)G(q)+
1
(2𝜋)3 = 0
G(q)= -
1
(2𝜋)3
1
( 𝑞2−𝑘2)
--------(4)
By equation (2) and(4)
G(r-r’)= 𝐺(𝑞)𝑒𝑖𝑞(𝑟−𝑟′)𝑑3𝑞
G(r-r’)=-
1
(2𝜋)3
𝑒𝑖𝑞(𝑟−𝑟′)
(𝑞2−𝑘2)
𝑑3q
Value of green function

7. :𝑑3
q=𝑞2
dqsin𝜃𝑑𝜃d∅
G(r-r’)=-
1
(2𝜋)3 0
∞
0
𝜋
0
2𝜋 𝑒𝑖𝑞(𝑟−𝑟′)
(𝑞2−𝑘2)
𝑞2
𝑑𝑞sin𝜃𝑑𝜃𝑑∅
If ∞ = (𝑟 − 𝑟′
)
G(r-r’)= -
1
(2𝜋)2 0
∞
0
𝜋 𝑒𝑖𝑅𝑐𝑜𝑠𝜃
(𝑞2−𝑘2)
𝑞2
sin𝜃d𝜃d∅
= -
1
(2𝜋)2 0
𝜋
𝑒𝑖𝑅𝑐𝑜𝑠𝜃
𝑠𝑖𝑛𝜃𝑑𝜃 0
∞ 𝑞2𝑑𝑞
(𝑞2−𝑘2)
=(𝐼1 + 𝐼2)
𝐼1 =
0
𝜋
𝑒𝑖𝑅𝐶𝑂𝑆𝜃 sinθdθ
COS𝜃 = 𝑡
-SIN𝜃𝑑𝜃 = 𝑑𝑡
𝐼1 = − 1
−1
𝑒𝑖𝑅𝑞𝑡
dt

8. =
1
𝑖𝑞𝑅
(𝑒𝑖𝑞𝑟
− 𝑒−𝑖𝑞𝑅
)
G(r-r’)= -
1
(2𝜋)2 0
∞ 𝑞2
(𝑞2+𝑘2)𝑖𝑞𝑡
𝑒𝑖𝑞𝑅 − 𝑒−𝑖𝑞𝑅 𝑑𝑞
= -
1
(2𝜋)2 −∞
∞ 𝑞𝑒𝑖𝑞𝑡
(𝑞2−𝑘2)𝑖𝑞𝑅
𝑑𝑞
= -
1
4𝜋2 −∞
∞ 𝑞𝑒𝑖𝑞𝑅
(𝑞+𝑘)(𝑞−𝑘)
dq
In equation(5) we can define
q=±k Because value of
𝑞𝑒𝑖𝑞𝑡
(𝑞+𝑘)(𝑞−𝑘)
(i) become=∞
Can be solved by
𝑓 𝑧 𝑑𝑧 = 2𝜋𝑖(𝑠𝑢𝑚 𝑜𝑓 𝑟𝑒𝑠𝑖𝑑𝑢𝑒)
q ±𝑘 (solve)
Solve at q +k
cos

9. Residue=𝑠𝑖𝑛𝑧 +𝑎 (2-a)f(2)
=𝑠𝑖𝑛𝑧 −𝑎 (2-a)f(2)
(1) For q +k (2) For q -k
Residue=𝑠𝑖𝑛𝑞 𝑘 (q-k)
𝑞
(𝑞2−𝑘2)
𝑒𝑖𝑞𝑅
or
=𝑠𝑖𝑛𝑞 𝑘
𝑞−𝑘 𝑞𝑒𝑖𝑞𝑅
(𝑞+𝑘)(𝑞−𝑘)
=
𝑘
𝑘+𝑘
𝑒𝑖𝑞𝑅
Residue
𝑒𝑖𝑘𝑅
2
So,
I= 2𝜋𝑖 −
𝑒𝑖𝑘𝑅
2
=𝜋𝑖𝑒𝑖𝑘𝑅
G(r-r’)=
−1
4𝜋2𝑖𝑅
𝜋𝑖𝑒𝑖𝑘𝑅
= -
1
4𝜋𝑅
𝑒𝑖𝑘𝑅
G(r-r’)= -
1
4𝜋(𝑟−𝑟′)
𝑒𝑖𝑘(𝑟−𝑟′)
Residue=𝑠𝑖𝑛𝑞 𝑘
𝑞𝑒𝑖𝑞𝑅
(𝑞2−𝑘2)
Residue=-
−𝑘𝑒𝑖𝑞𝑅
−2𝐾
I=2𝜋𝑖
1
2
𝑒−𝑖𝑘𝑅
=𝜋𝑖𝑒−𝑖𝑘𝑅
G(r-r’)= -
1
2
𝜋𝑖𝑒−𝑖𝑘𝑅
=-
1
4𝜋𝑅
𝑒−𝑖𝑘𝑅
Or,
G(r-r’)= -
1
4𝜋(𝑟−𝑟′)
𝑒−𝑖𝑘(𝑟−𝑟′)
=
𝑒−𝑖𝑘𝑅
2

10. G(r-r’)= -
1
4𝜋(𝑟−𝑟′)
𝑒−𝑖𝑘(𝑟−𝑟′)
+ 𝑒𝑖𝑘(𝑟−𝑟′)
For scattering ,we want to study about outgoing wave. So we cross restarted green
function than,
G(r-r’)= -
1
4𝜋(𝑟−𝑟′)
𝑒𝑖𝑘(𝑟−𝑟′)
ψ = 𝜑𝑖𝑛 + 𝜑𝑠𝑐
𝜑𝑠𝑐 =
2𝜋
ħ2
ℎ(𝑟𝑟′)𝜑(𝑟′)𝑑3𝑟′
𝜑𝑠𝑐 =
2
ħ2
−
1
4𝜋 𝑟 − 𝑟′
𝑒𝑖𝑘 𝑟−𝑟′
𝜑(𝑟′
)𝑑3
(𝑟′
)
𝜑 = 𝑒𝑖𝑘𝑟
−
2𝜋ħ2
𝑒𝑖𝑘 𝑟−𝑟′
𝑟−𝑟′ 𝜑 𝑟 𝑑3
r’
General solution of scattering problem of green function.

11. • Lippmann-Schwinger Equation :-
: Whenever large no. of component of
scattered particles are appreciable then we
use lippmann-schwinger equation. It give an
integral solution of scattering problem.
: By Green function, we analyze the partial
wave or particle scattering and it useful for
lower energy particle scattering.
: But Lippmann-Schwinger equation is useful
for higher energy scattering particles.

12. to find lippmann-schwinger equation we use
Helm-Holtz operator which is known as –
( 2 + K2 )
and we use the Green function –
2G(r , r’) = - 4πδ3[r - r’]
by solving these we can write Helm-Holtz
operator in form of Green function –
Ѱ(r) Ѱin + Ѱsc
r→ꚙ
by these equations we got a solution -

13. ( 2 + K2 ) Ѱ(r) = 2μ/ћ2 { μ(r’) Ѱ(r’) }
this is not a unique solution.
For unique solution we use Ѱ(r) in form of Ѱ0(r), r
and r’.
where - r = stationary vector / constant
r’ = variable vector
( both are position vector )
therefore we found a unique solution –
( 2 + K2 ) Ѱ(r) = eiK.r + (2μ/ћ2) ʃ d3r’ δ3(r-r’) g(r’)
Ѱ(r’)
This equation is known as Lippmann-Schwinger
equation.

14. THANK YOU