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Physics 452
Quantum mechanics II
Winter 2012
Karine Chesnel
Phys 452
Class- evaluation
Please take some time to submit your
class evaluation on line by April 13
http://studentratings.byu.edu
Quiz on Wed April 11
5 points
Click if you have submitted evaluation
or commit to submit the evaluation by Friday
Phys 452
Quiz scores
Quiz 1 – 10: part I (January)
Quiz 11 – 19: part II (February)
Quiz 20:
test 2 review
Quiz 21- 30: part III (March-April)
Review your quiz score, and if you see
missing entries, let me know by Friday
Phys 452
Class scores W 2012
Tentative
grade scale
Homework (40%)
92.0
Quizzes
(10%)
89.9
Midterms (30%)
76.9
Final
(20%)
Class average
86.3
Complete all your homework by Friday!!
Prepare well for the final!!
A
AB+
B
BC+
C
CD+
D
D-
93
89
85
80
75
70
60
55
50
45
40
Phys 452
Final test
Saturday, April 14
2:30pm – 5:30 pm
Where: classroom SWKT 337
Time – limited: 3 hours
Comprehensive
Closed book, closed notes
Bring a calculator
Phys 452
Final test
Choose 5 out of 6 problems
Sample Test (2011)
1. Two particle & perturbation theory
Ch 5 & 6
2. Hydrogen fine structure
Ch 6
3. Variational principle
Ch 7
4. WKB approximation
Ch 8
5. Two level & magnetic resonance
Ch 9
6. Scattering
Ch 11
Phys 452
Final test
Choose 5 out of 6 problems
Planned test (2012)
1. Ch 6: perturbation theory
2. Ch 7: variational principle
3. Ch 8: WKB approximation
4. Ch 9: Time dependant pertubation
5. Ch 10: Adiabatic approx.
6. Ch 11: Scattering
Phys 452
Techniques to solve
for the allowed energies
1. The perturbation theory
2. The variational principle
3. The WKB approximation
Quiz 32 a
For which of these techniques
you need to have a first
estimate of the eigenstates?
A.
B.
C.
D.
E.
1
2
3
1&2
All of them
Phys 452
Perturbation theory
H H H'
0
Unperturbed
states
Building the
true states and
true energies
to some order
H E
0
0
n
0
n
0
n
 n      ...
0
n
1
n
2
n
En  E  E  E  ...
0
n
1
n
zeroorder
firstorder
2
n
secondorder
Phys 452
Non-degenerate Perturbation theory
First-order correction
Energy
State
En1   n0 H '  n0
 
1
n
nm

H'
0
m
E
0
n
E
0
m
0
n


0
m
Phys 452
Non-degenerate Perturbation theory
Second-order correction
Energy
E  
2
n
0
n
E 
2
n
m n
H'

0
m
1
n
H'
0
n
2
En0  Em0
Only works if the energies are non-degenerate
Phys 452
Degenerate perturbation theory
General method
• Start with an ortho-normal basis of the unperturbed states

0
1
, 20 , 30 ,...

• If the state is non-degenerate:
En1   n0 H '  n0
• If the state is degenerate: build
W  Wij   i0 H '  0j 
• Diagonalize W : the eigenvalues are
En1,4
En1,3
1
n,k
E
0
n
E
En1,2
En1,1
l
Phys 452
The fine structure of hydrogen
2
p
H 
 V (r )
2m
0
Motion of
the electron
Bohr’s
energies
Coulomb interaction
between e- and nucleus
2
 mc 2  e2 2  1
2
1

mc 2 1
0
En   


  2 
2
2
2 n2
2ma n
 2  4 0 c   n
Phys 452
Quiz 32 b
What kind of interaction is at the origin
of the spin-orbit coupling effect?
A. An interaction between the spins of two electrons located at different orbits
B. The spin of an electron interacting with the spin of the nucleus
C. The spin of an electron interacting with field created by its angular momentum
D. The spin of an electron interacting with the field created
by another electron’s angular momentum
E. An interaction between the spins of two electrons located in the same orbit
Phys 452
The fine structure of hydrogen
Fine structure
Bohr’s
energy
E=
E 
0
n
+ Relativistic
correction
2
2
coupling
B
 mc 1
2
+ Spin-orbit
H '  Trel  Tclassic
n2
e
S
-
p4
E  H'   3 2
8m c
1
r
“Classical
e
+
view”
Emag  M S .BL
Phys 452
The fine structure of hydrogen
Fine structure
E=
E 
0
n
Bohr’s
energy
+ Relativistic
correction
 mc 1
2
E 


2
2
n
+ Spin-orbit
2
0
n
coupling
2
 4n

 3
2 
2mc  l  1/ 2 
1
n
E
1
n
E
E  n  j ( j  1)  l (l  1)  s(s  1) 


0
n
mc 2
2
l (l  1/ 2)(l  1)
Phys 452
The fine structure of hydrogen
E=
En0  
Bohr’s
energy
 2 mc 2 1
2
n2
+
Fine structure
+

En0 
4n  
E fs  E 1 
3


2 
j  1/ 2  
 2mc 
0
n
New relevant quantum numbers:
n, l, s, j and mj
Zeeman effect
?
Phys 452
Zeeman effect
Bext
S
L
e-
“Classical view”
Weak-field
Fine structure
dominates
e
H 
( L  2S ).Bext
2m
'
Z
• Comparing:
Bext
Intermediate
field
'
H '  H SO
 H Z'
and
Bint
Strong field
Zeeman effect
dominates
Phys 452
Zeeman effect
Bext
S
Weak -field
L
Good eigenstates:
e-
Bext  Bint
n, l , j , m j
E  B g j m j Bext
1
Z
j ( j  1)  l (l  1)  3 / 4
with Lande factor: g j  1 
2 j ( j  1)
2



0
Etot  En 1  2 cnj   B g j m j Bext
 n

Phys 452
Zeeman effect
Bext
S
Strong -field
L
e-
Bext  Bint
Good eigenstates:
n, l , ml , s, ms
E  B  ml  2mS  Bext
1
Z
0
l (l  1)  ml ms 
E
1
2  3
1
E fs   3   

n
 4n l (l  1/ 2)(l  1) 
Phys 452
Variational principle
???
Hamiltonian
Schrödinger
Equation…
… very hard
to solve!
Many
particles
Egs   H 
Ground
state
Expectation value
on any normalized
function 
Phys 452
Quiz 32 c / d
With the variational principle,
we are guaranteed to find out the ground state
A. TRUE
B. FALSE
What are we basically adjusting
in the variational principle?
A. The Hamiltonian
B. The wave function
C. Both
Phys 452
Variational principle
The method:
• Define your system, and the Hamiltonian H
• Pick a normalized wave function 
• Calculate
 H
• Minimize
 H
• You get an estimate
of ground state energy
Egs   H 
min
Phys 452
The ground state of Helium
2 particles system
r1
p12 p22
2e2  1 1  e2  1
H



  
2m 2m 4 0  r1 r2  4 0  r1  r2



r2
Kinetic
energy
Interaction
with proton
He atom
Electron- electron
interaction
Zero-order Hamiltonian H0
Perturbation
Vee
Phys 452
The ground state of Helium
• Use the variational principle to account for screening effects
Z3
 Z  r1  r2  / a
  r1 , r2   3 Ae
a
r1
r2
Z 2
Vee
e2
2 3
1
 
  r1 , r2  d r2 d 3r1
4 0 r1  r2
He atom
Vee  
5Z
E1
4
Phys 452
The ground state of Helium
• Use the variational principle to account for screening effects
r1
r2
Z 2
p12 p22
2e2  1 1  e2  1
H



  
2m 2m 4 0  r1 r2  4 0  r1  r2
He atom
p12 p22 Ze 2  1 1  e 2  Z  2 Z  2
1
H





  
2m 2m 4 0  r1 r2  4 0  r1
r2
r1  r2
H
2

 1
e
2
 2Z E1  2( Z  2) 

4

r
0 

 Vee






Phys 452
Hydrogen molecule ion H2+
 0  r2 
 0  r1 
Energy Minimization
H
 E1
R
Equilibrium separation distance:

Req  2.4a  1.3 
x
Presence of a minimum:
Evidence of bonding
R
a
Phys 452
The WKB approximation
V(x)
Turning points
E
Non-classical
region (E<V)
Non-classical
region (E<V)
Classical region (E>V)
Phys 452
The WKB approximation
Excluding the turning points:
E  V  x
E  V  x
i
 WKB  x  
p ( x ) dx
C

e
p( x)
 WKB  x  
p ( x ) dx

e


C
1
p( x)
where
p  x   2m( E  V  x )
Phys 452
Tunneling trough a barrier
V(x)
V0
A
F
B
-a
x
+a
2
Transmission
coefficient
F
2
T
e
A

1
a
 p  x  dx
a
Phys 452
Quiz 34 e
A particle with an incident energy E is approaching a barrier
of potential V and width a.
Which one of these statements does NOT apply?
A. The transmission coefficient through the barrier depends on E, V and a
B. The transmission coefficient increases when a decreases for a given E and V
C. The transmission coefficient increases when V decreases for a given E and a
D. The transmission coefficient increases when E decreases for a given V and a
E. The particle has some chances to be reflected by the barrier if V>E
Phys 452
The WKB approximation
Patching – upward slope
V(x)
Patching region
Linear approximation
Overlap 1
Overlap 2
E
Classical region (E>V)
X=0
Non-classical
region (E<V)
Phys 452
The WKB approximation
Patching – upward slope
• General expression for the wave function
 2D

 1 x2
sin   p  x ' dx ' 

x
4


p
x



  x  
x
D
1



exp    p  x ' dx '
x2
 p  x



x  x2
x  x2
Phys 452
The WKB approximation
Connection formulas
• Potential with 2 walls
a
 p  x dx  n
0
x2
• Potential with 1 wall

0
x2
• Potential with no walls
x2

x1
1

p  x dx   n   
4

1

p  x dx   n   
2

Phys 452
Dynamical systems
d
H  i
dt
V depends on time
General solution:
   cn  t  n eiEnt /
n
Probability to measure the energy En:
cn  t 
2
Phys 452
Two- level systems
Time- dependent perturbation
Sinusoidal perturbation
Eb
Ea
H '  V cos t 
2
Vab sin 0    t / 2 
2
2
Probability of transition: Pa b  t  



 0 
2
P()
for a given time t
Resonance
effect
0

Phys 452
Emission and absorption
of a radiation
z
Eb
x
Ea
y
E  E0 cos t  nz
Hab '   pab E0 cos t 
Transition
rate
Rba  t  
with
 pab
3 0
2
pab  q  a r  b
2
 0   B  0 
Phys 452
Quiz 32f
What can we say about the transition rate
of a stimulated emission ?
A. It is the same as the transition rate for absorption
B. It is opposite to the transition rate for absorption
C. It is inverse of the transition rate for absorption
D. It adds up with the spontaneous emission rate
to cancel the absorption’s one
Phys 452
Emission and absorption:
Einstein coefficients
Eb
Nb
Ea
Na
Bab : stimulated absorption rate
Bab : stimulated emission rate
A : spontaneous emission rate
Bab Bba A
Thermal
equilibrium
 0  
dNb
  Nb A  Nb Bba  0   N a Bab  0   0
dt
A
 Na / Nb  Bab  Bba
03
2
A
p
3
3 0 c
Boltzman distribution of particles
Analogy with Planck’s blackbody formula
Excited state lifetime
  1/ A
Phys 452
Emission and absorption
Selection rules
E
l 0
n4
n3
n2
n 1
l 1
l 2
l 3
Electric Dipole
transitions
l  1
m  0, 1
Phys 452
Adiabatic approximation
  t   cn  0  ei t ei t  n  t 
General solution
Adiabatic approx
with
n
 m t   i
t

0
m
 m
dt '
t

m
 R m
n  t   
1
t
 E t 'dt '
n
0
Geometric phase
 m t   i
n
dR
Dynamic phase
Berry’s phase
Phys 452
Scattering
Develop the solution in
terms of spherical harmonics,
Solution to Coulomb potential
  r, ,   R  r  Yl m  , 
2
d 2u 
l (l  1) 


V
r




 u  Eu
2
2
2m dr
2m r 

2
kr
1
V 0
Radiation zone
V 0
V 0
intermediate zone
Scattering zone
Phys 452
Scattering
Partial wave analysis
kr
1
Connecting all three regions and expressing the
Global wave function in spherical coordinates
 ikz
eikr 
  r ,   A e  f   
r 

V 0
V 0
Rayleigh’s formula
Scattered waves


  r ,   A i l  2l  1 jl  kr   ikal hl1  kr  Pl  cos 
l
Total
cross-section
   D   d  4   2l  1 al
l
2
To be determined
by the type of potential
+ boundary conditions
Phys 452
Scattering
Phase - shifts
al 
Scattering
factor
Scattering
Cross-section


1 2il
1
e  1  eil sin  l 
2ik
k
1 
f      2l  1eil sin  l  Pl  cos  
k l 0
4
 2
k

2
2
l

1
sin


 l 

l 0
Phys 452
Born approximation
k'
Scattering vector
q
k
Spherical wave
Plane wave
f  ,    
m
2
2
iq .r
3
e
V
r
d
  r

Phys 452
Born approximation
• Low energy approximation
f  ,    
q.r
m
2
2
1
3
V
r
d
   r
• Case of spherical potential

2m
f  ,     2  rV  r  sin  qr  dr
q0
Phys 452
Compton scattering
Quantum theory
• We retrieve the conservation laws:
p ' k '  p  k
E '  '  E  
• Furthermore, we can evaluate
the cross-section:
2
 q2
  k'
d
2


.

'
  
d  k '  4 0 mc 2   k 
2