90
Today’s Outline - April 14, 2015 C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 1 / 12
Today’s Outline - April 14, 2015
• Review problems
• Partial wave phase shifts
• Problem 11.6
• Born approximation
• Problem 11.8
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Midterm Exam #2:Thursday, April 16, 2015, Room 111 Stuart Buildingbring 1 sided page of notes
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 1 / 12
Today’s Outline - April 14, 2015
• Review problems
• Partial wave phase shifts
• Problem 11.6
• Born approximation
• Problem 11.8
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Midterm Exam #2:Thursday, April 16, 2015, Room 111 Stuart Buildingbring 1 sided page of notes
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 1 / 12
Today’s Outline - April 14, 2015
• Review problems
• Partial wave phase shifts
• Problem 11.6
• Born approximation
• Problem 11.8
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Midterm Exam #2:Thursday, April 16, 2015, Room 111 Stuart Buildingbring 1 sided page of notes
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 1 / 12
Today’s Outline - April 14, 2015
• Review problems
• Partial wave phase shifts
• Problem 11.6
• Born approximation
• Problem 11.8
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Midterm Exam #2:Thursday, April 16, 2015, Room 111 Stuart Buildingbring 1 sided page of notes
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 1 / 12
Today’s Outline - April 14, 2015
• Review problems
• Partial wave phase shifts
• Problem 11.6
• Born approximation
• Problem 11.8
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Midterm Exam #2:Thursday, April 16, 2015, Room 111 Stuart Buildingbring 1 sided page of notes
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 1 / 12
Today’s Outline - April 14, 2015
• Review problems
• Partial wave phase shifts
• Problem 11.6
• Born approximation
• Problem 11.8
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Midterm Exam #2:Thursday, April 16, 2015, Room 111 Stuart Buildingbring 1 sided page of notes
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 1 / 12
Today’s Outline - April 14, 2015
• Review problems
• Partial wave phase shifts
• Problem 11.6
• Born approximation
• Problem 11.8
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Midterm Exam #2:Thursday, April 16, 2015, Room 111 Stuart Buildingbring 1 sided page of notes
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 1 / 12
Today’s Outline - April 14, 2015
• Review problems
• Partial wave phase shifts
• Problem 11.6
• Born approximation
• Problem 11.8
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Midterm Exam #2:Thursday, April 16, 2015, Room 111 Stuart Buildingbring 1 sided page of notes
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 1 / 12
Today’s Outline - April 14, 2015
• Review problems
• Partial wave phase shifts
• Problem 11.6
• Born approximation
• Problem 11.8
Homework Assignment #10:Chapter 11:2,4,5,7,9,20due Tuesday, April 28, 2015
Midterm Exam #2:Thursday, April 16, 2015, Room 111 Stuart Buildingbring 1 sided page of notes
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 1 / 12
2015 Lederman Lecture
Anthony J. Leggett
John D. and Catherine T. MacArthurProfessor and Center for AdvancedStudy Professor of Physics
University of Illinois at Urbana-Champaign
Nobel Laureate in Physics, 2003
“for pioneering contributions to thetheory of superconductors and super-fluids”
PHYSICS LEDERMAN LECTURE
Quantum mechanics has been enormously successful in describing nature at the atomic level,and most physicists believe that it is in principle the “whole truth” about the world even at the everyday level. However, such a view prima facie leads to a severe problem: in certain circumstances, the most natural interpretation of the theory implies that no definite outcome of an experiment occurs until the act of “observation.” For many decades this problem was regarded as “merely philosophical,” in the sense that it was thought that it had no consequences which could be tested in experiment. However, in the last 15 or so years, the situation has changed very dramatically in this respect. Leggett will discuss the problem, some popular “resolutions” of it, the current experimental situation, and prospects for the future.
Lederman Lecturer Anthony J. Leggett with Schrödinger’s cat on blackboard.
ANTHONY J. LEGGETT
Nobel Laureate in Physics John D. and Catherine T. MacArthur Professor
Center for Advanced Study Professor of Physics at the University of Illinois at Urbana -Champaign
The event is free. All are welcome. Public reception after the lecture For more information and RSVP contact [email protected]
DOES THE EVERYDAY WORLD REALLY OBEY QUANTUM MECHANICS?
April 20, 2015, 3:30 p.m.Perlstein Hall Auditorium
Physics
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 2 / 12
Problem 9.7
The first term in the equation
cb ∼= −Vba
2~
[e i(ω0+ω)t − 1
ω0 + ω+
e i(ω0−ω)t − 1
ω0 − ω
]
comes from the e iωt/2 part of cos(ωt). and the second from e−iωt/2.Thus droppingthe first term is formally equivalent to writingH ′ = (V /q)e−iωt , which is to say,
H ′ba =Vba
2e−iωt , H ′ab =
Vab
2e iωt
Rabi noticed that if you make the rotating wave approximation at thebeginning of the calculation, the time dependent coefficient equations canbe solved exactly with no need for perturbation theory, and no assumptionof field strength.
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 3 / 12
Problem 9.7
(a) Solve for the time dependent coefficients with the usual startingconditions: ca(0) = 1, cb(0) = 0. Express your results in terms of theRabi flopping frequency,
ωr ≡ 12
√(ω − ω0)2 + (|Vab|/~)2
(b) Determine the transition probability, Pa→b(t), and show that it neverexceeds 1. Confirm |ca(t)|2 + |cb(t)|2 = 1.
(c) Check that Pa→b(t) reduces to perturbation theory result when theperturbation “small,” and state precisely what small means in thiscontext, as a constraint on V .
(d) At what time does the system first return to its initial state?
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 4 / 12
Problem 9.15
Develop time-dependent perturbation theory for a multilevelsystem starting with the generalization of
H0ψn = Enψn, 〈ψn|ψm〉 = δnm
At time t = 0 we turn on a perturbation H ′(t), so that thetotal Hamiltonian is
H = H0 + H ′(t)
(a) If the generalized wavefuction is given by
Ψ(t) =∑
cn(t)ψne−iEnt/~
derive the expression for cm
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 5 / 12
Problem 9.15
(b) If the system starts out in the state ψN , derive the firstorder expression for cN and cm
(c) If the perturbing Hamiltonian is constant but turned onat t = 0 and off again at t, find the transitionprobability from state N to state M (M 6= M)
(d) Suppose H ′ = V cos(ωt), show that transitions onlyoccur to states with energy EM = EN ± ~ω
(e) Show that the transition rate for stimulated emission isthe same as that for a two level system
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 6 / 12
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)
)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 7 / 12
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
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)
)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 7 / 12
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
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)
)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 7 / 12
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)
)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 7 / 12
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)
)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 7 / 12
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)
)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 7 / 12
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)
)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 7 / 12
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)
)C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 7 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 8 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 8 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 8 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 8 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 8 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 8 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 8 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 8 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 8 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 8 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 8 / 12
Connection between al and δl
Comparing this result
with the general solution by partial waves
ψ(l) ≈ A(2l + 1)
2ikr
[e ikre2iδl − (−1)le−ikr
]Pl(cos θ)
ψ(l) ≈ A
{(2l + 1)
2ikr
[e ikr − (−1)le−ikr
]+
(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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 9 / 12
Connection between al and δl
Comparing this result
with the general solution by partial waves
ψ(l) ≈ A(2l + 1)
2ikr
[e ikre2iδl − (−1)le−ikr
]Pl(cos θ)
ψ(l) ≈ A
{(2l + 1)
2ikr
[e ikr − (−1)le−ikr
]+
(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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 9 / 12
Connection between al and δl
Comparing this result with the general solution by partial waves
ψ(l) ≈ A(2l + 1)
2ikr
[e ikre2iδl − (−1)le−ikr
]Pl(cos θ)
ψ(l) ≈ A
{(2l + 1)
2ikr
[e ikr − (−1)le−ikr
]+
(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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 9 / 12
Connection between al and δl
Comparing this result with the general solution by partial waves
ψ(l) ≈ A(2l + 1)
2ikr
[e ikre2iδl − (−1)le−ikr
]Pl(cos θ)
ψ(l) ≈ A
{(2l + 1)
2ikr
[e ikr − (−1)le−ikr
]+
(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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 9 / 12
Connection between al and δl
Comparing this result with the general solution by partial waves
ψ(l) ≈ A(2l + 1)
2ikr
[e ikre2iδl − (−1)le−ikr
]Pl(cos θ)
ψ(l) ≈ A
{(2l + 1)
2ikr
[e ikr − (−1)le−ikr
]+
(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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 9 / 12
Connection between al and δl
Comparing this result with the general solution by partial waves
ψ(l) ≈ A(2l + 1)
2ikr
[e ikre2iδl − (−1)le−ikr
]Pl(cos θ)
ψ(l) ≈ A
{(2l + 1)
2ikr
[e ikr − (−1)le−ikr
]+
(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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 9 / 12
Connection between al and δl
Comparing this result with the general solution by partial waves
ψ(l) ≈ A(2l + 1)
2ikr
[e ikre2iδl − (−1)le−ikr
]Pl(cos θ)
ψ(l) ≈ A
{(2l + 1)
2ikr
[e ikr − (−1)le−ikr
]+
(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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 9 / 12
Connection between al and δl
Comparing this result with the general solution by partial waves
ψ(l) ≈ A(2l + 1)
2ikr
[e ikre2iδl − (−1)le−ikr
]Pl(cos θ)
ψ(l) ≈ A
{(2l + 1)
2ikr
[e ikr − (−1)le−ikr
]+
(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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 9 / 12
Connection between al and δl
Comparing this result with the general solution by partial waves
ψ(l) ≈ A(2l + 1)
2ikr
[e ikre2iδl − (−1)le−ikr
]Pl(cos θ)
ψ(l) ≈ A
{(2l + 1)
2ikr
[e ikr − (−1)le−ikr
]+
(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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 9 / 12
Connection between al and δl
Comparing this result with the general solution by partial waves
ψ(l) ≈ A(2l + 1)
2ikr
[e ikre2iδl − (−1)le−ikr
]Pl(cos θ)
ψ(l) ≈ A
{(2l + 1)
2ikr
[e ikr − (−1)le−ikr
]+
(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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 9 / 12
Connection between al and δl
Comparing this result with the general solution by partial waves
ψ(l) ≈ A(2l + 1)
2ikr
[e ikre2iδl − (−1)le−ikr
]Pl(cos θ)
ψ(l) ≈ A
{(2l + 1)
2ikr
[e ikr − (−1)le−ikr
]+
(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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 9 / 12
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)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 10 / 12
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)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 10 / 12
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)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 10 / 12
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)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 10 / 12
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)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 10 / 12
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)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 10 / 12
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)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 10 / 12
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
= 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
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)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 10 / 12
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)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 10 / 12
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 ]
= i1− 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)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 10 / 12
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)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 10 / 12
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)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 10 / 12
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)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 10 / 12
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)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 10 / 12
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)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 10 / 12
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)
]
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 10 / 12
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)
]C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 10 / 12
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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 11 / 12
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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 11 / 12
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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 11 / 12
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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 11 / 12
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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 11 / 12
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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 11 / 12
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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 11 / 12
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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 11 / 12
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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 11 / 12
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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 11 / 12
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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 11 / 12
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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 11 / 12
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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 11 / 12
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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 11 / 12
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)
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 11 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 12 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 12 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 12 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 12 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 12 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 12 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 12 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 12 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 12 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 12 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 12 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 12 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 12 / 12
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
C. Segre (IIT) PHYS 406 - Spring 2015 April 14, 2015 12 / 12
