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Femtochemistry: A theoretical overviewFemtochemistry: A theoretical overview
Mario [email protected]
VI – Transition probabilities
This lecture can be downloaded athttp://homepage.univie.ac.at/mario.barbatti/femtochem.html lecture6.ppt
2
Fermi’s golden rule
3
Fermi’s Golden RuleFermi’s Golden Rule
kpk kHiW 22
Transition rate:
i
kk ,
Quantum levels of the non-perturbed
system
Perturbation is applied
Transition is induced
4
Derivation of Fermi’s Golden RuleDerivation of Fermi’s Golden Rule
• See Fermi‘s Golden Rule paper at the course homepage
Time-dependent formulation
pHHH 0H0 – Non-perturbated Hamiltonian
Hp – Perturbation Hamiltonian
0
Ht
i
n which solves: 00 nEH n
ijji and ijji EE
5
Derivation of Fermi’s Golden RuleDerivation of Fermi’s Golden Rule
n
tiEn neta n
pHHH 0
0
Ht
i
Prove it!
Multiply by at left and integrate
n
tinp
k knetanHktta
i
k
Note that the non-perturbated Hamiltonian is supposed non-dependent on time.
6
An approximate way to solve the differential equationAn approximate way to solve the differential equation
n
tinp
k knetanHktta
i
Guess the “0-order” solution: tan)0(
Use this guess to solve the equation and to get the 1st-order approximation: tak)1(
n
tinp
k knetanHkdt
tdai )0(
)1(
Use the 1st-order to get the 2nd-order approximation and so on.
n
tipnp
pk knetanHkdt
tdai )1(
)(
7
First order approximationFirst order approximation
Guess the “0-order” solution: nin ta )0(
n
tinp
k knetanHkdt
tdai )0(
)1(
Suppose the simplified perturbation:
0
'kn
p
HnHk
Constant between 0 and Otherwise
t
0
'knH
0
nHk p
8
First order approximationFirst order approximation
0
')1(
n
tininmk
kneH
dt
tdai
Between 0 and Otherwise
ki
kiiik
tiikk
knki eHdteHai
2/sin2 2/'
0
')1(
It was used:
a ikxikx
kkx
edxe0
2/ 2/sin2
9
Transition probabilityTransition probability
22
22'2 2/sin
4ki
kiikkik HaP
0
0.0
0.2
0.4
0.6
0.8
1.0
sin(
)2 /
2
In this derivation for constant perturbation, only transitions with ~ 0 take place. If the perturbation oscillates harmonically (like a photon), ≠ 0 can occur.
The final result for the Fermi’s Golden Rule is still the same.
10
Physically meaningful quantityPhysically meaningful quantity
kneark
ikk PW'
'
1
i
kk ,
Near k: density of statesdEdn
k
kikikkkikkneark
ikk dPdEPPW
11
''
11
Physically meaningful quantityPhysically meaningful quantity
kikikkkikkneark
ikk dPdEPPW
11
''
Using
22
22' 2/sin
4ki
kiikik HP
kikk HW 2'2
kiki
kikk dHWki
2
22' 2/sin1
41
2/
12
Fermi’s Golden Rule: photons and moleculesFermi’s Golden Rule: photons and molecules
kti
k dtekiW ik
2
0
2 Ed
Transition rate:
i
kk ,
Ed 0HHH0 – Non-perturbated molecular Hamiltonian
– Light-matter perturbation HamiltonianEd
13
Transition dipole momentTransition dipole moment
kti
k dtekiW ik
2
0
2 Ed
i
kk ,
kikZeiki eikN
N
n
N
mmnn
at el
ddrRd
1 1
0
Electronic transition dipole moment
14
Einstein coefficientsEinstein coefficients
i
k
ik
iN
Rate of absorption i → k iika
ik NBW
Einstein coefficient B for absorption
2
2061
kig
gB e
i
kik d
ik
ng - degeneracy of state n
(see Einstein coefficients text in the course homepage)
15
Einstein coefficientsEinstein coefficients
i
k
ki
kN
Rate of stimulated emission k → i kkist
ki NBW
Einstein coefficient B for stimulated emission
kik
iik B
g
gB
ki
16
Einstein coefficientsEinstein coefficients
i
k
ki
kN
Rate spontaneous decay k → i 2NAW kisp
ki
Einstein coefficient A for spontaneous emission
kiki
ki Bc
A 32
3
ki
17
Einstein coefficient and oscillator strengthEinstein coefficient and oscillator strength
kiki
ki Ae
mcf 22
302
In atomic units:
kiki
ki AE
cf 2
3
2
18
Einstein coefficient and lifetimeEinstein coefficient and lifetime
kikiki fE
cA 2
3
21
1
2
R
E
If E21 = 4.65 eV and f21 = -0.015, what is the lifetime of the excited state?
au
eVE
.1708840
211396.27/65.4
65.421
au10
2
3
21
1029.0
)015.0()170884.0(2)137(
ns
s
70
104189.21029.0 171021
Converting to nanoseconds:
19
non-adiabatic transition probabilities
20
Non-adiabatic transitionsNon-adiabatic transitions
11,
22 ,
21,
12 ,
0
x
E
0 H11
H22 E2
E1
Problem: if the molecule prepared in state 2 at x = ∞ moves through a region of crossing, what is the probability of ending in state 1 at x = ∞?
H12
21
Models for non-adiabatic transitionsModels for non-adiabatic transitions
xFHH 122211 constant, 1212 FH
1. Landau-Zener
12
2
122exp
F
H
vP
2. Demkov / Rosen-Zener
constant21 EE
/sech1212 xhh x
xvhEE
P12
212
4sech
3. Nikitin
xA
H
xAHH
expsin2
expcos
012
02211
02
02
02
cottan2
1exp
12/cosexp
v
P
4. Bradauk; 5. Delos-Thorson; 6. …
22
Derivation of Landau-Zener formulaDerivation of Landau-Zener formula
nn
t
nnn dttHi
ta
exp
Multiply by at left and integrate
kn
nknk netaHdt
tdai
k
nkkn HH
t
nnn dtHi
0
Ht
i
In the deduction it was used:
nn
t
nn HdtHdtd
23
kn
nknk netaHdt
tdai
Since there are only two states:
21212
1 etaHi
dttda
12121
2 etaHi
dttda
jiij
(i)
(ii)
Solving (i) for a2 and taking the derivative:
dt
tdadt
tadHei
dtda
HeHH 2
21
2
12
1
12
22111212
(iii)
Substituting (iii) in (ii):
01
1
2
1221
221121
2
aHdtda
HHi
dtad
24
Zener approximation: tHH 2211
21,
x
0
E
0 11,
22 ,
12 ,
222 F
dxdH
111 F
dxdH
vtFxFH 1111
vtFxFH 2222
vtFFHH 212211
vF12
01
1
2
1221
221121
2
aHdtda
HHi
dtad
25
01
01
2
2
1222
1222
2
1
2
1221
1221
2
aHdt
davtF
idt
ad
aHdtda
vtFi
dtad
Problem: Find a2(+∞) subject to the initial condition a2(∞) = 1.
The solution is:
(The complete derivation is in the paper on Landau-Zener in the course homepage)
0x
E
0
12 a
01 a ?2 a
?1 a
12
2
122 exp
F
H
va
26
The probability of finding the system in state 1 is:
12
2
122
2
2exp
F
H
vaPnad
The probability of finding the system in state 2 is:
12
2
122exp11
F
H
vPP nadad
0
0
1
Pro
babi
lity
|H12
|2
Pnad
Pad
0
0
1
|F12
| or v
Pad
Pnad
27
2
12** 2exp
HP
Example: In trajectory in the graph, what are the probability of the molecule to remain in the * state or to change to the closed shell state?
0 2 4 6 8 10 12 14
-224.85
-224.80
-224.75
-224.70
-224.65
*
cs
H11
= t
Ene
rgy
(au)
Time (fs)
H22
= t
H12
= au
cs
*
au0.001au/f0435.0
01126.003224.0
s
fs104.2au1 2
time
tHH 2211
vF12 1
28
Example: In trajectory in the graph, what are the probability of the molecule to remain in the * state or to change to the closed shell state?
43.0001.0
011577.02exp
2
**
P
57.01 *** PP cs
0 2 4 6 8 10 12 14
-224.85
-224.80
-224.75
-224.70
-224.65
*
cs
Ene
rgy
(au)
Time (fs)
cs
*
0.43
0.57
29
0
x
E
0
0
x
E
0
vv
For the same H12, Landau-Zener predicts:
Non-adiabatic (diabatic) Adiabatic
12
2
122exp
F
H
vPnad
H12
30
0
x
E
0
0
x
E
0
For the same , Rosen-Zener predicts:
Non-adiabatic (but not diabatic!) Adiabatic
vv
xnad vhEE
P12
212
4sech
xh12
31
0
x
E
0
0
x
E
0
For the same 0 (H12), Nikitin predicts:
Non-adiabatic (diabatic) Adiabatic
v v
02
02
02
cottan2
1exp
12/cosexp
v
Pnad
32
The problem with the previous formulations is that they only predict the total probability at the end of the process.
If we want to perform dynamics, it is necessary to have the instantaneous probability.
33
Next lectureNext lecture
Quantum dynamics methods
This lecture can be downloaded athttp://homepage.univie.ac.at/mario.barbatti/femtochem.html lecture6.ppt