segre/phys406/14S/lecture_21.pdf · Phase shifts in 1-D For a 1D system with a solid \wall" at x = 0, we can write theinci-dent andre ectedwaves far from the non-zero potential i(x)

$Page 1: segre/phys406/14S/lecture_21.pdf · Phase shifts in 1-D For a 1D system with a solid \wall" at x = 0, we can write theinci-dent andre ectedwaves far from the non-zero potential i(x)$
Today’s Outline - April 09, 2014

• Partial wave phase shifts

• Problem 11.6

• Born approximation

• Problem 11.8

Homework Assignment #09:Chapter 11:2,4,5,7,9,20due Monday, April 21, 2014

Midterm Exam #2:Monday, April 14, 2014

C. Segre (IIT) PHYS 406 - Spring 2014 April 09, 2014 1 / 13



• Problem 11.6


• Problem 11.8






• Problem 11.6


• Problem 11.8






• Problem 11.6


• Problem 11.8






• Problem 11.6


• Problem 11.8






• Problem 11.6


• Problem 11.8






• Problem 11.6


• Problem 11.8




Phase shifts in 1-D

For a 1D system with a solid “wall”at x = 0, we can write the inci-dent

and reflected waves far fromthe non-zero potential

ψi (x) = Ae+ikx , x < −aψr (x) = Be−ikx , x < −a

if V = 0 for x < 0, we can writethe full solution

ψ0(x) = A(e ikx − e−ikx

)with a V (x) 6= 0, the reflectedwave will gain a phase shift fromtraversing the region −a ≤ x ≤ 0twice and the solution becomes

-a 0x

V

Be-ikx

Ae+ikx

ψ(x) = A(e ikx − e i(2δ−kx)

)


Phase shifts in 1-D

For a 1D system with a solid “wall”at x = 0, we can write the inci-dent

and reflected waves far fromthe non-zero potential

ψi (x) = Ae+ikx , x < −a

ψr (x) = Be−ikx , x < −a




-a 0x

V

Be-ikx

Ae+ikx


)


Phase shifts in 1-D

For a 1D system with a solid “wall”at x = 0, we can write the inci-dent and reflected waves far fromthe non-zero potential

ψi (x) = Ae+ikx , x < −a

ψr (x) = Be−ikx , x < −a




-a 0x

V

Be-ikx

Ae+ikx


)


Phase shifts in 1-D






-a 0x

V

Be-ikx

Ae+ikx


)


Phase shifts in 1-D






-a 0x

V

Be-ikx

Ae+ikx


)


Phase shifts in 1-D





)

with a V (x) 6= 0, the reflectedwave will gain a phase shift fromtraversing the region −a ≤ x ≤ 0twice and the solution becomes

-a 0x

V

Be-ikx

Ae+ikx


)


Phase shifts in 1-D






-a 0x

V

Be-ikx

Ae+ikx


)


Phase shifts in 1-D






-a 0x

V

Be-ikx

Ae+ikx


)C. Segre (IIT) PHYS 406 - Spring 2014 April 09, 2014 2 / 13

Phase shifts in 3-D

Recall that a plane wave can be ex-pressed as a sum of partial waveswith m = 0

each partial wave with a specific to-tal angular momentum scatters in-dependently

for x � 1 and V (r) = 0

ψ(l)0 = Ai l(2l + 1)jl(kr)Pl(cos θ)

jl(x) =1

2

[h(1)l (x) + h

(2)l (x)

]≈ 1

2x

[(−1)l+1e ix + i l+1e−ix

]thus, for the l th partial wave, at large r

ψ(l)0 ≈ A

(2l + 1)

2ikr

[e ikr − (−1)le−ikr

]Pl(cos θ), V (r) = 0

the second term is an incoming spherical wave

and the first is the outgoingwave which is phase-shifted by δl when there is a non-zero potential

ψ(l) ≈ A(2l + 1)

2ikr

[e ikre2iδl − (−1)le−ikr

]Pl(cos θ), V (r) 6= 0


Phase shifts in 3-D



for x � 1 and V (r) = 0


jl(x) =1

2

[h(1)l (x) + h

(2)l (x)

]≈ 1

2x

[(−1)l+1e ix + i l+1e−ix


ψ(l)0 ≈ A

(2l + 1)

2ikr


]Pl(cos θ), V (r) = 0



ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ), V (r) 6= 0


Phase shifts in 3-D



for x � 1 and V (r) = 0


jl(x) =1

2

[h(1)l (x) + h

(2)l (x)

]≈ 1

2x

[(−1)l+1e ix + i l+1e−ix


ψ(l)0 ≈ A

(2l + 1)

2ikr


]Pl(cos θ), V (r) = 0



ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ), V (r) 6= 0


Phase shifts in 3-D



for x � 1 and V (r) = 0


jl(x) =1

2

[h(1)l (x) + h

(2)l (x)

]

≈ 1

2x

[(−1)l+1e ix + i l+1e−ix


ψ(l)0 ≈ A

(2l + 1)

2ikr


]Pl(cos θ), V (r) = 0



ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ), V (r) 6= 0


Phase shifts in 3-D



for x � 1 and V (r) = 0


jl(x) =1

2

[h(1)l (x) + h

(2)l (x)

]

≈ 1

2x

[(−1)l+1e ix + i l+1e−ix


ψ(l)0 ≈ A

(2l + 1)

2ikr


]Pl(cos θ), V (r) = 0



ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ), V (r) 6= 0


Phase shifts in 3-D



for x � 1 and V (r) = 0


jl(x) =1

2

[h(1)l (x) + h

(2)l (x)

]≈ 1

2x

[(−1)l+1e ix + i l+1e−ix

]

thus, for the l th partial wave, at large r

ψ(l)0 ≈ A

(2l + 1)

2ikr


]Pl(cos θ), V (r) = 0



ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ), V (r) 6= 0


Phase shifts in 3-D



for x � 1 and V (r) = 0


jl(x) =1

2

[h(1)l (x) + h

(2)l (x)

]≈ 1

2x

[(−1)l+1e ix + i l+1e−ix


ψ(l)0 ≈ A

(2l + 1)

2ikr


]Pl(cos θ), V (r) = 0



ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ), V (r) 6= 0


Phase shifts in 3-D



for x � 1 and V (r) = 0


jl(x) =1

2

[h(1)l (x) + h

(2)l (x)

]≈ 1

2x

[(−1)l+1e ix + i l+1e−ix


ψ(l)0 ≈ A

(2l + 1)

2ikr


]Pl(cos θ), V (r) = 0



ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ), V (r) 6= 0


Phase shifts in 3-D



for x � 1 and V (r) = 0


jl(x) =1

2

[h(1)l (x) + h

(2)l (x)

]≈ 1

2x

[(−1)l+1e ix + i l+1e−ix


ψ(l)0 ≈ A

(2l + 1)

2ikr


]Pl(cos θ), V (r) = 0



ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ), V (r) 6= 0


Phase shifts in 3-D



for x � 1 and V (r) = 0


jl(x) =1

2

[h(1)l (x) + h

(2)l (x)

]≈ 1

2x

[(−1)l+1e ix + i l+1e−ix


ψ(l)0 ≈ A

(2l + 1)

2ikr


]Pl(cos θ), V (r) = 0

the second term is an incoming spherical wave and the first is the outgoingwave which is phase-shifted by δl when there is a non-zero potential

ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ), V (r) 6= 0


Phase shifts in 3-D



for x � 1 and V (r) = 0


jl(x) =1

2

[h(1)l (x) + h

(2)l (x)

]≈ 1

2x

[(−1)l+1e ix + i l+1e−ix


ψ(l)0 ≈ A

(2l + 1)

2ikr


]Pl(cos θ), V (r) = 0

the second term is an incoming spherical wave and the first is the outgoingwave which is phase-shifted by δl when there is a non-zero potential

ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ), V (r) 6= 0


Connection between al and δl

Comparing this result

with the general solution by partial waves

ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ)

ψ(l) ≈ A

{(2l + 1)

2ikr


]+

(2l + 1)

rale

ikr

}Pl(cos θ)

(2l + 1)

2ikr

(e2iδl − 1

)=

(2l + 1)

rale

ikr

al =1

2ik

(e2iδl − 1

)=

1

ke iδl

e iδl − e−iδl

2i=

1

ke iδl sin(δl)

then, following the partial wave calculation, the scattering factor and totalcross-section become

f (θ) =1

k

∞∑l=0

(2l + 1)e iδl sin(δl)Pl(cos θ)

σ =4π

k2

∞∑l=0

(2l + 1) sin2(δl)



Comparing this result

with the general solution by partial waves

ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ)

ψ(l) ≈ A

{(2l + 1)

2ikr


]+

(2l + 1)

rale

ikr

}Pl(cos θ)

(2l + 1)

2ikr

(e2iδl − 1

)=

(2l + 1)

rale

ikr

al =1

2ik

(e2iδl − 1

)=

1

ke iδl

e iδl − e−iδl

2i=

1

ke iδl sin(δl)


f (θ) =1

k

∞∑l=0


σ =4π

k2

∞∑l=0

(2l + 1) sin2(δl)



Comparing this result with the general solution by partial waves

ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ)

ψ(l) ≈ A

{(2l + 1)

2ikr


]+

(2l + 1)

rale

ikr

}Pl(cos θ)

(2l + 1)

2ikr

(e2iδl − 1

)=

(2l + 1)

rale

ikr

al =1

2ik

(e2iδl − 1

)=

1

ke iδl

e iδl − e−iδl

2i=

1

ke iδl sin(δl)


f (θ) =1

k

∞∑l=0


σ =4π

k2

∞∑l=0

(2l + 1) sin2(δl)




ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ)

ψ(l) ≈ A

{(2l + 1)

2ikr


]+

(2l + 1)

rale

ikr

}Pl(cos θ)

(2l + 1)

2ikr

(e2iδl − 1

)=

(2l + 1)

rale

ikr

al =1

2ik

(e2iδl − 1

)=

1

ke iδl

e iδl − e−iδl

2i=

1

ke iδl sin(δl)


f (θ) =1

k

∞∑l=0


σ =4π

k2

∞∑l=0

(2l + 1) sin2(δl)




ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ)

ψ(l) ≈ A

{(2l + 1)

2ikr


]+

(2l + 1)

rale

ikr

}Pl(cos θ)

(2l + 1)

2ikr

(e2iδl − 1

)=

(2l + 1)

rale

ikr

al =1

2ik

(e2iδl − 1

)=

1

ke iδl

e iδl − e−iδl

2i=

1

ke iδl sin(δl)


f (θ) =1

k

∞∑l=0


σ =4π

k2

∞∑l=0

(2l + 1) sin2(δl)




ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ)

ψ(l) ≈ A

{(2l + 1)

2ikr


]+

(2l + 1)

rale

ikr

}Pl(cos θ)

(2l + 1)

2ikr

(e2iδl − 1

)=

(2l + 1)

rale

ikr

al =1

2ik

(e2iδl − 1

)

=1

ke iδl

e iδl − e−iδl

2i=

1

ke iδl sin(δl)


f (θ) =1

k

∞∑l=0


σ =4π

k2

∞∑l=0

(2l + 1) sin2(δl)




ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ)

ψ(l) ≈ A

{(2l + 1)

2ikr


]+

(2l + 1)

rale

ikr

}Pl(cos θ)

(2l + 1)

2ikr

(e2iδl − 1

)=

(2l + 1)

rale

ikr

al =1

2ik

(e2iδl − 1

)=

1

ke iδl

e iδl − e−iδl

2i

=1

ke iδl sin(δl)


f (θ) =1

k

∞∑l=0


σ =4π

k2

∞∑l=0

(2l + 1) sin2(δl)




ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ)

ψ(l) ≈ A

{(2l + 1)

2ikr


]+

(2l + 1)

rale

ikr

}Pl(cos θ)

(2l + 1)

2ikr

(e2iδl − 1

)=

(2l + 1)

rale

ikr

al =1

2ik

(e2iδl − 1

)=

1

ke iδl

e iδl − e−iδl

2i=

1

ke iδl sin(δl)


f (θ) =1

k

∞∑l=0


σ =4π

k2

∞∑l=0

(2l + 1) sin2(δl)




ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ)

ψ(l) ≈ A

{(2l + 1)

2ikr


]+

(2l + 1)

rale

ikr

}Pl(cos θ)

(2l + 1)

2ikr

(e2iδl − 1

)=

(2l + 1)

rale

ikr

al =1

2ik

(e2iδl − 1

)=

1

ke iδl

e iδl − e−iδl

2i=

1

ke iδl sin(δl)


f (θ) =1

k

∞∑l=0


σ =4π

k2

∞∑l=0

(2l + 1) sin2(δl)




ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ)

ψ(l) ≈ A

{(2l + 1)

2ikr


]+

(2l + 1)

rale

ikr

}Pl(cos θ)

(2l + 1)

2ikr

(e2iδl − 1

)=

(2l + 1)

rale

ikr

al =1

2ik

(e2iδl − 1

)=

1

ke iδl

e iδl − e−iδl

2i=

1

ke iδl sin(δl)


f (θ) =1

k

∞∑l=0


σ =4π

k2

∞∑l=0

(2l + 1) sin2(δl)




ψ(l) ≈ A(2l + 1)

2ikr


]Pl(cos θ)

ψ(l) ≈ A

{(2l + 1)

2ikr


]+

(2l + 1)

rale

ikr

}Pl(cos θ)

(2l + 1)

2ikr

(e2iδl − 1

)=

(2l + 1)

rale

ikr

al =1

2ik

(e2iδl − 1

)=

1

ke iδl

e iδl − e−iδl

2i=

1

ke iδl sin(δl)


f (θ) =1

k

∞∑l=0


σ =4π

k2

∞∑l=0

(2l + 1) sin2(δl)


Problem 11.6

What are the partial wave phase shifts for hard sphere scattering?

Recall that the partial wave ampli-tudes for hard sphere scattering are

comparing with the partial wavephases

now h(1)l (x) = jl(x) + inl(x) so

al =jl(ka)

kh(1)l (ka)

=1

ke iδl sin δl

e iδl sin δl =jl(ka)

h(1)l (ka)

e iδl sin δl = ijl

jl + inl= i

1

1 + i [nl/jl ]

1− i [nl/jl ]

1− i [nl/jl ]= i

1− i [nl/jl ]

1 + [nl/jl ]2=

[nl/jl ] + i

1 + [nl/jl ]2

equating the real and imaginary parts

cos δl sin δl =nl/jl

1 + [nl/jl ]2, sin2 δl =

1

1 + [nl/jl ]2

tan δl =1

nl/jl−→ δl = tan−1

[jl(ka)

nl(ka)

]


Problem 11.6





al =jl(ka)

kh(1)l (ka)

=1

ke iδl sin δl


h(1)l (ka)


jl + inl= i

1

1 + i [nl/jl ]

1− i [nl/jl ]

1− i [nl/jl ]= i

1− i [nl/jl ]

1 + [nl/jl ]2=

[nl/jl ] + i

1 + [nl/jl ]2



1 + [nl/jl ]2, sin2 δl =

1

1 + [nl/jl ]2

tan δl =1


[jl(ka)

nl(ka)

]


Problem 11.6





al =jl(ka)

kh(1)l (ka)

=1

ke iδl sin δl


h(1)l (ka)


jl + inl= i

1

1 + i [nl/jl ]

1− i [nl/jl ]

1− i [nl/jl ]= i

1− i [nl/jl ]

1 + [nl/jl ]2=

[nl/jl ] + i

1 + [nl/jl ]2



1 + [nl/jl ]2, sin2 δl =

1

1 + [nl/jl ]2

tan δl =1


[jl(ka)

nl(ka)

]


Problem 11.6





al =jl(ka)

kh(1)l (ka)

=1

ke iδl sin δl


h(1)l (ka)


jl + inl= i

1

1 + i [nl/jl ]

1− i [nl/jl ]

1− i [nl/jl ]= i

1− i [nl/jl ]

1 + [nl/jl ]2=

[nl/jl ] + i

1 + [nl/jl ]2



1 + [nl/jl ]2, sin2 δl =

1

1 + [nl/jl ]2

tan δl =1


[jl(ka)

nl(ka)

]


Problem 11.6





al =jl(ka)

kh(1)l (ka)

=1

ke iδl sin δl


h(1)l (ka)


jl + inl= i

1

1 + i [nl/jl ]

1− i [nl/jl ]

1− i [nl/jl ]= i

1− i [nl/jl ]

1 + [nl/jl ]2=

[nl/jl ] + i

1 + [nl/jl ]2



1 + [nl/jl ]2, sin2 δl =

1

1 + [nl/jl ]2

tan δl =1


[jl(ka)

nl(ka)

]


Problem 11.6





al =jl(ka)

kh(1)l (ka)

=1

ke iδl sin δl


h(1)l (ka)


jl + inl= i

1

1 + i [nl/jl ]

1− i [nl/jl ]

1− i [nl/jl ]= i

1− i [nl/jl ]

1 + [nl/jl ]2=

[nl/jl ] + i

1 + [nl/jl ]2



1 + [nl/jl ]2, sin2 δl =

1

1 + [nl/jl ]2

tan δl =1


[jl(ka)

nl(ka)

]


Problem 11.6





al =jl(ka)

kh(1)l (ka)

=1

ke iδl sin δl


h(1)l (ka)


jl + inl= i

1

1 + i [nl/jl ]

1− i [nl/jl ]

1− i [nl/jl ]= i

1− i [nl/jl ]

1 + [nl/jl ]2=

[nl/jl ] + i

1 + [nl/jl ]2



1 + [nl/jl ]2, sin2 δl =

1

1 + [nl/jl ]2

tan δl =1


[jl(ka)

nl(ka)

]


Problem 11.6





al =jl(ka)

kh(1)l (ka)

=1

ke iδl sin δl


h(1)l (ka)


jl + inl

= i1

1 + i [nl/jl ]

1− i [nl/jl ]

1− i [nl/jl ]= i

1− i [nl/jl ]

1 + [nl/jl ]2=

[nl/jl ] + i

1 + [nl/jl ]2



1 + [nl/jl ]2, sin2 δl =

1

1 + [nl/jl ]2

tan δl =1


[jl(ka)

nl(ka)

]


Problem 11.6





al =jl(ka)

kh(1)l (ka)

=1

ke iδl sin δl


h(1)l (ka)


jl + inl= i

1

1 + i [nl/jl ]

1− i [nl/jl ]

1− i [nl/jl ]= i

1− i [nl/jl ]

1 + [nl/jl ]2=

[nl/jl ] + i

1 + [nl/jl ]2



1 + [nl/jl ]2, sin2 δl =

1

1 + [nl/jl ]2

tan δl =1


[jl(ka)

nl(ka)

]


Problem 11.6





al =jl(ka)

kh(1)l (ka)

=1

ke iδl sin δl


h(1)l (ka)


jl + inl= i

1

1 + i [nl/jl ]

1− i [nl/jl ]

1− i [nl/jl ]

= i1− i [nl/jl ]

1 + [nl/jl ]2=

[nl/jl ] + i

1 + [nl/jl ]2



1 + [nl/jl ]2, sin2 δl =

1

1 + [nl/jl ]2

tan δl =1


[jl(ka)

nl(ka)

]


Problem 11.6





al =jl(ka)

kh(1)l (ka)

=1

ke iδl sin δl


h(1)l (ka)


jl + inl= i

1

1 + i [nl/jl ]

1− i [nl/jl ]

1− i [nl/jl ]= i

1− i [nl/jl ]

1 + [nl/jl ]2

=[nl/jl ] + i

1 + [nl/jl ]2



1 + [nl/jl ]2, sin2 δl =

1

1 + [nl/jl ]2

tan δl =1


[jl(ka)

nl(ka)

]


Problem 11.6





al =jl(ka)

kh(1)l (ka)

=1

ke iδl sin δl


h(1)l (ka)


jl + inl= i

1

1 + i [nl/jl ]

1− i [nl/jl ]

1− i [nl/jl ]= i

1− i [nl/jl ]

1 + [nl/jl ]2=

[nl/jl ] + i

1 + [nl/jl ]2



1 + [nl/jl ]2, sin2 δl =

1

1 + [nl/jl ]2

tan δl =1


[jl(ka)

nl(ka)

]


Problem 11.6





al =jl(ka)

kh(1)l (ka)

=1

ke iδl sin δl


h(1)l (ka)


jl + inl= i

1

1 + i [nl/jl ]

1− i [nl/jl ]

1− i [nl/jl ]= i

1− i [nl/jl ]

1 + [nl/jl ]2=

[nl/jl ] + i

1 + [nl/jl ]2



1 + [nl/jl ]2, sin2 δl =

1

1 + [nl/jl ]2

tan δl =1


[jl(ka)

nl(ka)

]


Problem 11.6





al =jl(ka)

kh(1)l (ka)

=1

ke iδl sin δl


h(1)l (ka)


jl + inl= i

1

1 + i [nl/jl ]

1− i [nl/jl ]

1− i [nl/jl ]= i

1− i [nl/jl ]

1 + [nl/jl ]2=

[nl/jl ] + i

1 + [nl/jl ]2



1 + [nl/jl ]2,

sin2 δl =1

1 + [nl/jl ]2

tan δl =1


[jl(ka)

nl(ka)

]


Problem 11.6





al =jl(ka)

kh(1)l (ka)

=1

ke iδl sin δl


h(1)l (ka)


jl + inl= i

1

1 + i [nl/jl ]

1− i [nl/jl ]

1− i [nl/jl ]= i

1− i [nl/jl ]

1 + [nl/jl ]2=

[nl/jl ] + i

1 + [nl/jl ]2



1 + [nl/jl ]2, sin2 δl =

1

1 + [nl/jl ]2

tan δl =1


[jl(ka)

nl(ka)

]


Problem 11.6





al =jl(ka)

kh(1)l (ka)

=1

ke iδl sin δl


h(1)l (ka)


jl + inl= i

1

1 + i [nl/jl ]

1− i [nl/jl ]

1− i [nl/jl ]= i

1− i [nl/jl ]

1 + [nl/jl ]2=

[nl/jl ] + i

1 + [nl/jl ]2



1 + [nl/jl ]2, sin2 δl =

1

1 + [nl/jl ]2

tan δl =1

nl/jl

−→ δl = tan−1[jl(ka)

nl(ka)

]


Problem 11.6





al =jl(ka)

kh(1)l (ka)

=1

ke iδl sin δl


h(1)l (ka)


jl + inl= i

1

1 + i [nl/jl ]

1− i [nl/jl ]

1− i [nl/jl ]= i

1− i [nl/jl ]

1 + [nl/jl ]2=

[nl/jl ] + i

1 + [nl/jl ]2



1 + [nl/jl ]2, sin2 δl =

1

1 + [nl/jl ]2

tan δl =1


[jl(ka)

nl(ka)

]C. Segre (IIT) PHYS 406 - Spring 2014 April 09, 2014 5 / 13

Integral form of the Schrodinger equation

Starting with the time-independentSchrodinger equation

and rewritingit in a more compact form using

k ≡√

2mE

~, Q ≡ 2m

~2Vψ

if we can find a solution of thisequation, G (~r), for a delta functionsource then the solution to the ac-tual source, Q, becomes

Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0

and this satisfies the Schrodinger equation(∇2 + k2

)ψ(~r) =

∫ [ (∇2 + k2

)G (~r −~r0)

]Q(~r0) d3~r0

=

∫δ3(~r)Q(~r0) d3~r0 = Q(~r) =

2m

~2Vψ(~r)



Starting with the time-independentSchrodinger equation

and rewritingit in a more compact form using

k ≡√

2mE

~, Q ≡ 2m

~2Vψ


Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0


)ψ(~r) =

∫ [ (∇2 + k2

)G (~r −~r0)

]Q(~r0) d3~r0

=

∫δ3(~r)Q(~r0) d3~r0 = Q(~r) =

2m

~2Vψ(~r)



Starting with the time-independentSchrodinger equation and rewritingit in a more compact form using

k ≡√

2mE

~, Q ≡ 2m

~2Vψ


Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0


)ψ(~r) =

∫ [ (∇2 + k2

)G (~r −~r0)

]Q(~r0) d3~r0

=

∫δ3(~r)Q(~r0) d3~r0 = Q(~r) =

2m

~2Vψ(~r)




k ≡√

2mE

~,

Q ≡ 2m

~2Vψ


Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0


)ψ(~r) =

∫ [ (∇2 + k2

)G (~r −~r0)

]Q(~r0) d3~r0

=

∫δ3(~r)Q(~r0) d3~r0 = Q(~r) =

2m

~2Vψ(~r)




k ≡√

2mE

~, Q ≡ 2m

~2Vψ


Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0


)ψ(~r) =

∫ [ (∇2 + k2

)G (~r −~r0)

]Q(~r0) d3~r0

=

∫δ3(~r)Q(~r0) d3~r0 = Q(~r) =

2m

~2Vψ(~r)




k ≡√

2mE

~, Q ≡ 2m

~2Vψ


Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0


)ψ(~r) =

∫ [ (∇2 + k2

)G (~r −~r0)

]Q(~r0) d3~r0

=

∫δ3(~r)Q(~r0) d3~r0 = Q(~r) =

2m

~2Vψ(~r)




k ≡√

2mE

~, Q ≡ 2m

~2Vψ

if we can find a solution of thisequation, G (~r), for a delta functionsource

then the solution to the ac-tual source, Q, becomes

Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0


)ψ(~r) =

∫ [ (∇2 + k2

)G (~r −~r0)

]Q(~r0) d3~r0

=

∫δ3(~r)Q(~r0) d3~r0 = Q(~r) =

2m

~2Vψ(~r)




k ≡√

2mE

~, Q ≡ 2m

~2Vψ

if we can find a solution of thisequation, G (~r), for a delta functionsource

then the solution to the ac-tual source, Q, becomes

Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0


)ψ(~r) =

∫ [ (∇2 + k2

)G (~r −~r0)

]Q(~r0) d3~r0

=

∫δ3(~r)Q(~r0) d3~r0 = Q(~r) =

2m

~2Vψ(~r)




k ≡√

2mE

~, Q ≡ 2m

~2Vψ


Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0


)ψ(~r) =

∫ [ (∇2 + k2

)G (~r −~r0)

]Q(~r0) d3~r0

=

∫δ3(~r)Q(~r0) d3~r0 = Q(~r) =

2m

~2Vψ(~r)




k ≡√

2mE

~, Q ≡ 2m

~2Vψ


Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0


)ψ(~r) =

∫ [ (∇2 + k2

)G (~r −~r0)

]Q(~r0) d3~r0

=

∫δ3(~r)Q(~r0) d3~r0 = Q(~r) =

2m

~2Vψ(~r)




k ≡√

2mE

~, Q ≡ 2m

~2Vψ


Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0

and this satisfies the Schrodinger equation

(∇2 + k2

)ψ(~r) =

∫ [ (∇2 + k2

)G (~r −~r0)

]Q(~r0) d3~r0

=

∫δ3(~r)Q(~r0) d3~r0 = Q(~r) =

2m

~2Vψ(~r)




k ≡√

2mE

~, Q ≡ 2m

~2Vψ


Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0


)ψ(~r) =

∫ [ (∇2 + k2

)G (~r −~r0)

]Q(~r0) d3~r0

=

∫δ3(~r)Q(~r0) d3~r0 = Q(~r) =

2m

~2Vψ(~r)




k ≡√

2mE

~, Q ≡ 2m

~2Vψ


Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0


)ψ(~r) =

∫ [ (∇2 + k2

)G (~r −~r0)

]Q(~r0) d3~r0

=

∫δ3(~r)Q(~r0) d3~r0

= Q(~r) =2m

~2Vψ(~r)




k ≡√

2mE

~, Q ≡ 2m

~2Vψ


Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0


)ψ(~r) =

∫ [ (∇2 + k2

)G (~r −~r0)

]Q(~r0) d3~r0

=

∫δ3(~r)Q(~r0) d3~r0 = Q(~r)

=2m

~2Vψ(~r)




k ≡√

2mE

~, Q ≡ 2m

~2Vψ


Eψ = − ~2

2m∇2ψ + Vψ

Q =(∇2 + k2

)ψ

δ3(~r) =(∇2 + k2

)G (~r)

ψ(~r) =

∫G (~r −~r0)Q(~r0) d3~r0


)ψ(~r) =

∫ [ (∇2 + k2

)G (~r −~r0)

]Q(~r0) d3~r0

=

∫δ3(~r)Q(~r0) d3~r0 = Q(~r) =

2m

~2Vψ(~r)


Green’s functions

G (~r) is a Green’s function and represents the response of a lineardifferential equation to a delta function source

by determining the Green’s function, we can solve the differentialequation’s response to an arbitrary source using a simple integral equation

the task is to solve the delta func-tion source equation for the Green’sfunction which can be done by tak-ing a Fourier transform

δ3(~r) =(∇2 + k2

)G (~r)

G (~r) =1

(2π)3/2

∫e i~s·~rg(~s) d3~s

(∇2 + k2

)G (~r) =

1

(2π)3/2

∫ [(∇2 + k2

)e i~s·~r

]g(~s) d3~s

1

(2π)3

∫e i~s·~r d3~s =

δ3(~r) =

1

(2π)3/2

∫ (−s2 + k2

)e i~s·~rg(~s) d3~s

g(~s) =1

(2π)3/2(k2 − s2)−→ G (~r) =

1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s


Green’s functions




δ3(~r) =(∇2 + k2

)G (~r)

G (~r) =1

(2π)3/2


(∇2 + k2

)G (~r) =

1

(2π)3/2

∫ [(∇2 + k2

)e i~s·~r

]g(~s) d3~s

1

(2π)3

∫e i~s·~r d3~s =

δ3(~r) =

1

(2π)3/2

∫ (−s2 + k2


g(~s) =1

(2π)3/2(k2 − s2)−→ G (~r) =

1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s


Green’s functions



the task is to solve the delta func-tion source equation for the Green’sfunction

which can be done by tak-ing a Fourier transform

δ3(~r) =(∇2 + k2

)G (~r)

G (~r) =1

(2π)3/2


(∇2 + k2

)G (~r) =

1

(2π)3/2

∫ [(∇2 + k2

)e i~s·~r

]g(~s) d3~s

1

(2π)3

∫e i~s·~r d3~s =

δ3(~r) =

1

(2π)3/2

∫ (−s2 + k2


g(~s) =1

(2π)3/2(k2 − s2)−→ G (~r) =

1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s


Green’s functions



the task is to solve the delta func-tion source equation for the Green’sfunction

which can be done by tak-ing a Fourier transform

δ3(~r) =(∇2 + k2

)G (~r)

G (~r) =1

(2π)3/2


(∇2 + k2

)G (~r) =

1

(2π)3/2

∫ [(∇2 + k2

)e i~s·~r

]g(~s) d3~s

1

(2π)3

∫e i~s·~r d3~s =

δ3(~r) =

1

(2π)3/2

∫ (−s2 + k2


g(~s) =1

(2π)3/2(k2 − s2)−→ G (~r) =

1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s


Green’s functions




δ3(~r) =(∇2 + k2

)G (~r)

G (~r) =1

(2π)3/2


(∇2 + k2

)G (~r) =

1

(2π)3/2

∫ [(∇2 + k2

)e i~s·~r

]g(~s) d3~s

1

(2π)3

∫e i~s·~r d3~s =

δ3(~r) =

1

(2π)3/2

∫ (−s2 + k2


g(~s) =1

(2π)3/2(k2 − s2)−→ G (~r) =

1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s


Green’s functions




δ3(~r) =(∇2 + k2

)G (~r)

G (~r) =1

(2π)3/2


(∇2 + k2

)G (~r) =

1

(2π)3/2

∫ [(∇2 + k2

)e i~s·~r

]g(~s) d3~s

1

(2π)3

∫e i~s·~r d3~s =

δ3(~r) =

1

(2π)3/2

∫ (−s2 + k2


g(~s) =1

(2π)3/2(k2 − s2)−→ G (~r) =

1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s


Green’s functions




δ3(~r) =(∇2 + k2

)G (~r)

G (~r) =1

(2π)3/2


(∇2 + k2

)G (~r) =

1

(2π)3/2

∫ [(∇2 + k2

)e i~s·~r

]g(~s) d3~s

1

(2π)3

∫e i~s·~r d3~s =

δ3(~r) =

1

(2π)3/2

∫ (−s2 + k2


g(~s) =1

(2π)3/2(k2 − s2)−→ G (~r) =

1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s


Green’s functions




δ3(~r) =(∇2 + k2

)G (~r)

G (~r) =1

(2π)3/2


(∇2 + k2

)G (~r) =

1

(2π)3/2

∫ [(∇2 + k2

)e i~s·~r

]g(~s) d3~s

1

(2π)3

∫e i~s·~r d3~s =

δ3(~r) =

1

(2π)3/2

∫ (−s2 + k2


g(~s) =1

(2π)3/2(k2 − s2)−→ G (~r) =

1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s


Green’s functions




δ3(~r) =(∇2 + k2

)G (~r)

G (~r) =1

(2π)3/2


(∇2 + k2

)G (~r) =

1

(2π)3/2

∫ [(∇2 + k2

)e i~s·~r

]g(~s) d3~s

1

(2π)3

∫e i~s·~r d3~s =

δ3(~r) =

1

(2π)3/2

∫ (−s2 + k2


g(~s) =1

(2π)3/2(k2 − s2)−→ G (~r) =

1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s


Green’s functions




δ3(~r) =(∇2 + k2

)G (~r)

G (~r) =1

(2π)3/2


(∇2 + k2

)G (~r) =

1

(2π)3/2

∫ [(∇2 + k2

)e i~s·~r

]g(~s) d3~s

1

(2π)3

∫e i~s·~r d3~s =

δ3(~r) =

1

(2π)3/2

∫ (−s2 + k2


g(~s) =1

(2π)3/2(k2 − s2)−→ G (~r) =

1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s


Green’s functions




δ3(~r) =(∇2 + k2

)G (~r)

G (~r) =1

(2π)3/2


(∇2 + k2

)G (~r) =

1

(2π)3/2

∫ [(∇2 + k2

)e i~s·~r

]g(~s) d3~s

1

(2π)3

∫e i~s·~r d3~s =

δ3(~r) =1

(2π)3/2

∫ (−s2 + k2


g(~s) =1

(2π)3/2(k2 − s2)−→ G (~r) =

1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s


Green’s functions




δ3(~r) =(∇2 + k2

)G (~r)

G (~r) =1

(2π)3/2


(∇2 + k2

)G (~r) =

1

(2π)3/2

∫ [(∇2 + k2

)e i~s·~r

]g(~s) d3~s

1

(2π)3

∫e i~s·~r d3~s = δ3(~r) =

1

(2π)3/2

∫ (−s2 + k2


g(~s) =1

(2π)3/2(k2 − s2)−→ G (~r) =

1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s


Green’s functions




δ3(~r) =(∇2 + k2

)G (~r)

G (~r) =1

(2π)3/2


(∇2 + k2

)G (~r) =

1

(2π)3/2

∫ [(∇2 + k2

)e i~s·~r

]g(~s) d3~s

1

(2π)3

∫e i~s·~r d3~s = δ3(~r) =

1

(2π)3/2

∫ (−s2 + k2


g(~s) =1

(2π)3/2(k2 − s2)

−→ G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s


Green’s functions




δ3(~r) =(∇2 + k2

)G (~r)

G (~r) =1

(2π)3/2


(∇2 + k2

)G (~r) =

1

(2π)3/2

∫ [(∇2 + k2

)e i~s·~r

]g(~s) d3~s

1

(2π)3

∫e i~s·~r d3~s = δ3(~r) =

1

(2π)3/2

∫ (−s2 + k2


g(~s) =1

(2π)3/2(k2 − s2)−→ G (~r) =

1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s


Integrating the Green’s function

r

s

θ

φ

choose spherical coordinateswith the polar axis fixedalong ~r for the integrationover ~s

thus, ~s ·~r = sr cos θ and theφ integral is equal to 2π

G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s

=1

(2π)2

∫ ∞0

∫ π

0

e isr cos θ

(k2 − s2)s2sin θ dθ ds

the θ integral is

Iθ = − e isr cos θ

isr

∣∣∣∣π0

=2 sin(sr)

sr

G (~r) =1

(2π)22

r

∫ ∞0

s sin(sr)

(k2 − s2)ds

=1

4π2r

∫ ∞−∞

s sin(sr)

(k2 − s2)ds

this integral needs to be perfomed usingCauchy’s formula



r

s

θ

φ



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s

=1

(2π)2

∫ ∞0

∫ π

0

e isr cos θ


the θ integral is


isr

∣∣∣∣π0

=2 sin(sr)

sr

G (~r) =1

(2π)22

r

∫ ∞0

s sin(sr)

(k2 − s2)ds

=1

4π2r

∫ ∞−∞

s sin(sr)

(k2 − s2)ds




r

s

θ

φ



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s

=1

(2π)2

∫ ∞0

∫ π

0

e isr cos θ


the θ integral is


isr

∣∣∣∣π0

=2 sin(sr)

sr

G (~r) =1

(2π)22

r

∫ ∞0

s sin(sr)

(k2 − s2)ds

=1

4π2r

∫ ∞−∞

s sin(sr)

(k2 − s2)ds




r

s

θ

φ



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s

=1

(2π)2

∫ ∞0

∫ π

0

e isr cos θ


the θ integral is


isr

∣∣∣∣π0

=2 sin(sr)

sr

G (~r) =1

(2π)22

r

∫ ∞0

s sin(sr)

(k2 − s2)ds

=1

4π2r

∫ ∞−∞

s sin(sr)

(k2 − s2)ds




r

s

θ

φ



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s

=1

(2π)2

∫ ∞0

∫ π

0

e isr cos θ


the θ integral is


isr

∣∣∣∣π0

=2 sin(sr)

sr

G (~r) =1

(2π)22

r

∫ ∞0

s sin(sr)

(k2 − s2)ds

=1

4π2r

∫ ∞−∞

s sin(sr)

(k2 − s2)ds




r

s

θ

φ



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s

=1

(2π)2

∫ ∞0

∫ π

0

e isr cos θ


the θ integral is


isr

∣∣∣∣π0

=2 sin(sr)

sr

G (~r) =1

(2π)22

r

∫ ∞0

s sin(sr)

(k2 − s2)ds

=1

4π2r

∫ ∞−∞

s sin(sr)

(k2 − s2)ds




r

s

θ

φ



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s

=1

(2π)2

∫ ∞0

∫ π

0

e isr cos θ


the θ integral is


isr

∣∣∣∣π0

=2 sin(sr)

sr

G (~r) =1

(2π)22

r

∫ ∞0

s sin(sr)

(k2 − s2)ds

=1

4π2r

∫ ∞−∞

s sin(sr)

(k2 − s2)ds




r

s

θ

φ



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s

=1

(2π)2

∫ ∞0

∫ π

0

e isr cos θ


the θ integral is


isr

∣∣∣∣π0

=2 sin(sr)

sr

G (~r) =1

(2π)22

r

∫ ∞0

s sin(sr)

(k2 − s2)ds

=1

4π2r

∫ ∞−∞

s sin(sr)

(k2 − s2)ds




r

s

θ

φ



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s

=1

(2π)2

∫ ∞0

∫ π

0

e isr cos θ


the θ integral is


isr

∣∣∣∣π0

=2 sin(sr)

sr

G (~r) =1

(2π)22

r

∫ ∞0

s sin(sr)

(k2 − s2)ds

=1

4π2r

∫ ∞−∞

s sin(sr)

(k2 − s2)ds




r

s

θ

φ



G (~r) =1

(2π)3

∫e i~s·~r

1

(k2 − s2)d3~s

=1

(2π)2

∫ ∞0

∫ π

0

e isr cos θ


the θ integral is


isr

∣∣∣∣π0

=2 sin(sr)

sr

G (~r) =1

(2π)22

r

∫ ∞0

s sin(sr)

(k2 − s2)ds

=1

4π2r

∫ ∞−∞

s sin(sr)

(k2 − s2)ds



Radial integral of G (~r)

G (~r) =1

4π2r

∫ ∞−∞

s sin(sr)

(k2 − s2)ds

using eix − e−ix = 2i sin x

=i

8π2r

{∫ ∞−∞

s e isr

(s − k)(s + k)ds −

∫ ∞−∞

s e−isr

(s − k)(s + k)ds

}=

i

8π2r[I1 − I2]

both integrals are of the form towhich we can apply Cauchy’s inte-gral formula

∮f (z)

(z − z0)dz = 2πif (z0)

if z0 lies within the contour, other-wise 0

in this case, the pole singularities lie along the path of integration so weneed to avoid the poles to use Cauchy’s formula



G (~r) =1

4π2r

∫ ∞−∞

s sin(sr)

(k2 − s2)ds using eix − e−ix = 2i sin x

=i

8π2r

{∫ ∞−∞

s e isr

(s − k)(s + k)ds −

∫ ∞−∞

s e−isr

(s − k)(s + k)ds

}=

i

8π2r[I1 − I2]


∮f (z)

(z − z0)dz = 2πif (z0)





G (~r) =1

4π2r

∫ ∞−∞

s sin(sr)


=i

8π2r

{∫ ∞−∞

s e isr

(s − k)(s + k)ds −

∫ ∞−∞

s e−isr

(s − k)(s + k)ds

}

=i

8π2r[I1 − I2]


∮f (z)

(z − z0)dz = 2πif (z0)





G (~r) =1

4π2r

∫ ∞−∞

s sin(sr)


=i

8π2r

{∫ ∞−∞

s e isr

(s − k)(s + k)ds −

∫ ∞−∞

s e−isr

(s − k)(s + k)ds

}=

i

8π2r[I1 − I2]


∮f (z)

(z − z0)dz = 2πif (z0)





G (~r) =1

4π2r

∫ ∞−∞

s sin(sr)


=i

8π2r

{∫ ∞−∞

s e isr

(s − k)(s + k)ds −

∫ ∞−∞

s e−isr

(s − k)(s + k)ds

}=

i

8π2r[I1 − I2]


∮f (z)

(z − z0)dz = 2πif (z0)





G (~r) =1

4π2r

∫ ∞−∞

s sin(sr)


=i

8π2r

{∫ ∞−∞

s e isr

(s − k)(s + k)ds −

∫ ∞−∞

s e−isr

(s − k)(s + k)ds

}=

i

8π2r[I1 − I2]


∮f (z)

(z − z0)dz = 2πif (z0)





G (~r) =1

4π2r

∫ ∞−∞

s sin(sr)


=i

8π2r

{∫ ∞−∞

s e isr

(s − k)(s + k)ds −

∫ ∞−∞

s e−isr

(s − k)(s + k)ds

}=

i

8π2r[I1 − I2]


∮f (z)

(z − z0)dz = 2πif (z0)




Contour integration

Re{s}

Im{s}

s=-k

s=+k

deform the path to loop aroundthe negative pole in the positive di-rection by an infinitesimal amount,and the positive pole in the nega-tive direction

close the contour at Re{s} → ±∞in a semi-circle such that |s| → ∞

∮f (z)

(z − z0)dz = 2πif (z0)

I1 =

∮ [s e isr

s + k

]1

s − kds

= 2πi

[s e isr

s + k

∣∣∣∣s=k

= iπ e ikr

I2 = −∮ [

s e−isr

s − k

]1

s + kds

= −2πi

[s e−isr

s − k

∣∣∣∣s=−k

= −iπ e ikr

G (~r) =i

8π2r

[(iπ e ikr

)−(−iπ e ikr

)]= − e ikr

4πr


Contour integration

Re{s}

Im{s}

s=-k

s=+k



∮f (z)

(z − z0)dz = 2πif (z0)

I1 =

∮ [s e isr

s + k

]1

s − kds

= 2πi

[s e isr

s + k

∣∣∣∣s=k

= iπ e ikr

I2 = −∮ [

s e−isr

s − k

]1

s + kds

= −2πi

[s e−isr

s − k

∣∣∣∣s=−k

= −iπ e ikr

G (~r) =i

8π2r

[(iπ e ikr

)−(−iπ e ikr

)]= − e ikr

4πr


Contour integration

Re{s}

Im{s}

s=-k

s=+k



∮f (z)

(z − z0)dz = 2πif (z0)

I1 =

∮ [s e isr

s + k

]1

s − kds

= 2πi

[s e isr

s + k

∣∣∣∣s=k

= iπ e ikr

I2 = −∮ [

s e−isr

s − k

]1

s + kds

= −2πi

[s e−isr

s − k

∣∣∣∣s=−k

= −iπ e ikr

G (~r) =i

8π2r

[(iπ e ikr

)−(−iπ e ikr

)]= − e ikr

4πr


Contour integration

Re{s}

Im{s}

s=-k

s=+k



∮f (z)

(z − z0)dz = 2πif (z0)

I1 =

∮ [s e isr

s + k

]1

s − kds

= 2πi

[s e isr

s + k

∣∣∣∣s=k

= iπ e ikr

I2 = −∮ [

s e−isr

s − k

]1

s + kds

= −2πi

[s e−isr

s − k

∣∣∣∣s=−k

= −iπ e ikr

G (~r) =i

8π2r

[(iπ e ikr

)−(−iπ e ikr

)]= − e ikr

4πr


Contour integration

Re{s}

Im{s}

s=-k

s=+k



∮f (z)

(z − z0)dz = 2πif (z0)

I1 =

∮ [s e isr

s + k

]1

s − kds

= 2πi

[s e isr

s + k

∣∣∣∣s=k

= iπ e ikr

I2 = −∮ [

s e−isr

s − k

]1

s + kds

= −2πi

[s e−isr

s − k

∣∣∣∣s=−k

= −iπ e ikr

G (~r) =i

8π2r

[(iπ e ikr

)−(−iπ e ikr

)]= − e ikr

4πr


Contour integration

Re{s}

Im{s}

s=-k

s=+k



∮f (z)

(z − z0)dz = 2πif (z0)

I1 =

∮ [s e isr

s + k

]1

s − kds

= 2πi

[s e isr

s + k

∣∣∣∣s=k

= iπ e ikr

I2 = −∮ [

s e−isr

s − k

]1

s + kds

= −2πi

[s e−isr

s − k

∣∣∣∣s=−k

= −iπ e ikr

G (~r) =i

8π2r

[(iπ e ikr

)−(−iπ e ikr

)]= − e ikr

4πr


Contour integration

Re{s}

Im{s}

s=-k

s=+k



∮f (z)

(z − z0)dz = 2πif (z0)

I1 =

∮ [s e isr

s + k

]1

s − kds

= 2πi

[s e isr

s + k

∣∣∣∣s=k

= iπ e ikr

I2 = −∮ [

s e−isr

s − k

]1

s + kds

= −2πi

[s e−isr

s − k

∣∣∣∣s=−k

= −iπ e ikr

G (~r) =i

8π2r

[(iπ e ikr

)−(−iπ e ikr

)]= − e ikr

4πr


Contour integration

Re{s}

Im{s}

s=-k

s=+k



∮f (z)

(z − z0)dz = 2πif (z0)

I1 =

∮ [s e isr

s + k

]1

s − kds

= 2πi

[s e isr

s + k

∣∣∣∣s=k

= iπ e ikr

I2 = −∮ [

s e−isr

s − k

]1

s + kds

= −2πi

[s e−isr

s − k

∣∣∣∣s=−k

= −iπ e ikr

G (~r) =i

8π2r

[(iπ e ikr

)−(−iπ e ikr

)]= − e ikr

4πr


Contour integration

Re{s}

Im{s}

s=-k

s=+k



∮f (z)

(z − z0)dz = 2πif (z0)

I1 =

∮ [s e isr

s + k

]1

s − kds

= 2πi

[s e isr

s + k

∣∣∣∣s=k

= iπ e ikr

I2 = −∮ [

s e−isr

s − k

]1

s + kds

= −2πi

[s e−isr

s − k

∣∣∣∣s=−k

= −iπ e ikr

G (~r) =i

8π2r

[(iπ e ikr

)−(−iπ e ikr

)]= − e ikr

4πr


Contour integration

Re{s}

Im{s}

s=-k

s=+k



∮f (z)

(z − z0)dz = 2πif (z0)

I1 =

∮ [s e isr

s + k

]1

s − kds

= 2πi

[s e isr

s + k

∣∣∣∣s=k

= iπ e ikr

I2 = −∮ [

s e−isr

s − k

]1

s + kds

= −2πi

[s e−isr

s − k

∣∣∣∣s=−k

= −iπ e ikr

G (~r) =i

8π2r

[(iπ e ikr

)−(−iπ e ikr

)]= − e ikr

4πr


Contour integration

Re{s}

Im{s}

s=-k

s=+k



∮f (z)

(z − z0)dz = 2πif (z0)

I1 =

∮ [s e isr

s + k

]1

s − kds

= 2πi

[s e isr

s + k

∣∣∣∣s=k

= iπ e ikr

I2 = −∮ [

s e−isr

s − k

]1

s + kds

= −2πi

[s e−isr

s − k

∣∣∣∣s=−k

= −iπ e ikr

G (~r) =i

8π2r

[(iπ e ikr

)−(−iπ e ikr

)]= − e ikr

4πr


Contour integration

Re{s}

Im{s}

s=-k

s=+k



∮f (z)

(z − z0)dz = 2πif (z0)

I1 =

∮ [s e isr

s + k

]1

s − kds

= 2πi

[s e isr

s + k

∣∣∣∣s=k

= iπ e ikr

I2 = −∮ [

s e−isr

s − k

]1

s + kds

= −2πi

[s e−isr

s − k

∣∣∣∣s=−k

= −iπ e ikr

G (~r) =i

8π2r

[(iπ e ikr

)−(−iπ e ikr

)]= − e ikr

4πr


Contour integration

Re{s}

Im{s}

s=-k

s=+k



∮f (z)

(z − z0)dz = 2πif (z0)

I1 =

∮ [s e isr

s + k

]1

s − kds

= 2πi

[s e isr

s + k

∣∣∣∣s=k

= iπ e ikr

I2 = −∮ [

s e−isr

s − k

]1

s + kds

= −2πi

[s e−isr

s − k

∣∣∣∣s=−k

= −iπ e ikr

G (~r) =i

8π2r

[(iπ e ikr

)−(−iπ e ikr

)]

= − e ikr

4πr


Contour integration

Re{s}

Im{s}

s=-k

s=+k



∮f (z)

(z − z0)dz = 2πif (z0)

I1 =

∮ [s e isr

s + k

]1

s − kds

= 2πi

[s e isr

s + k

∣∣∣∣s=k

= iπ e ikr

I2 = −∮ [

s e−isr

s − k

]1

s + kds

= −2πi

[s e−isr

s − k

∣∣∣∣s=−k

= −iπ e ikr

G (~r) =i

8π2r

[(iπ e ikr

)−(−iπ e ikr

)]= − e ikr

4πr


General solution

The general solution to the Schrodinger equation is thus

ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d3~r0

with ψ0(~r) being the solution of the free particle Schrodinger equation(∇2 + k2

)ψ0(~r) = 0


General solution


ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d3~r0


)ψ0(~r) = 0


General solution


ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d3~r0


)ψ0(~r) = 0


General solution


ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d3~r0


)ψ0(~r) = 0


Problem 11.8

Show that

G (~r) = − e ikr

4πr

satisfies (∇2 + k2

)G (~r) = δ3(~r)


First Born approximation

Suppose we have a potential localized about ~r0 = 0

we want to solve the integral Schrodinger equation

ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d3~r0

at points far away from the scattering center, we have |~r | � |~r0| for thepoints where the scattering potential is significant

|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr

e ik|~r−~r0| ∼= e ikre−i~k·~r0 −→ e ik|~r−~r0|

|~r −~r0|∼=

e ikr

re−i

~k·~r0

ψ0(~r) = Ae ikz , ψ(~r) ∼= Ae ikz − m

2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d3~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d3~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d3~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d3~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d3~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d3~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d3~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d3~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0

∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d3~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d3~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)

−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d3~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d3~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d3~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d3~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d3~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d3~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr

e ik|~r−~r0| ∼= e ikre−i~k·~r0

−→ e ik|~r−~r0|

|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d3~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d3~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d3~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d3~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0

ψ0(~r) = Ae ikz ,

ψ(~r) ∼= Ae ikz − m

2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d3~r0





ψ(~r) = ψ0(~r)− m

2π~2

∫e ik|~r−~r0|

|~r −~r0|V (~r0)ψ(~r0) d3~r0


|~r −~r0|2 = r2 − r20 − 2~r ·~r0 ∼= r2(

1− 2~r ·~r0r2

)−→ |~r −~r0| ∼= r − r ·~r0

if we let ~k ≡ kr


|~r −~r0|∼=

e ikr

re−i

~k·~r0


2π~2e ikr

r

∫e−i

~k·~r0V (~r0)ψ(~r0) d3~r0


Documents

segre/phys406/14S/lecture_21.pdf · Phase shifts in 1-D For a 1D system with a solid \wall" at x = 0, we can write theinci-dent andre ectedwaves far from the non-zero potential i(x)