Ultrafast processes in molecules

Mario Barbattibarbatti@kofo.mpg.de

VI – Transition probabilities

2

Fermi’s golden rule

3

Fermi’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 Rule

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 Rule

n

tiEn neta n

pHHH 0

0

Ht

i

Prove it!

Multiply by at left and integrate

n

tinp

k knetanHkttai

k

Note that the non-perturbated Hamiltonian is supposed non-dependent on time.

6

An approximate way to solve the differential equation

n

tinp

k knetanHkttai

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 approximation

Guess the “0-order” solution: nin ta )0(

n

tinp

k knetanHkdt

tdai )0()1(

Suppose the simplified perturbation:

0

'kn

pH

nHk Constant between 0 and tOtherwise

t0 t

'knH

0

nHk p

8

First order approximation

0

')1(

n

tininmk

kneH

dttdai

Between 0 and tOtherwise

ki

kiiik

tiikk

knki eHdteHai

tt tt 2/sin2 2/'

0

')1(

It was used:

a ikxikx

kkxedxe

0

2/ 2/sin2

9

Transition probability

22

22'2 2/sin4ki

kiikkik HaP

tt

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 quantity

kneark

ikk PW'

'1t

i

kk ,

Near k: density of states dEdn

k

kikikkkikkneark

ikk dPdEPPW t

tt

11'

'

11

Physically meaningful quantity

kikikkkikkneark

ikk dPdEPPW t

tt

11'

'

Using

22

22' 2/sin4ki

kiikik HP

t

kikk HW 2'2

kiki

kikk dHWki

tt

2

22' 2/sin141

2/

12

Fermi’s Golden Rule: photons and molecules

kti

k dtekiW ik t

2

0

2 Ed

Transition rate:

i

kk ,

Ed 0HH H0 – Non-perturbated molecular Hamiltonian – Light-matter perturbation HamiltonianEd

13

Transition dipole moment

kti

k dtekiW ik t

2

0

2 Ed

i

kk ,

kikZeiki eikN

N

n

N

mmnn

at el

ddrRd

1 1

0

Electronic transition dipole moment

14

Einstein coefficients

i

k

ik

iN

Rate of absorption i → k iika

ik NBW

Einstein coefficient B for absorption

22

061 ki

ggB e

i

kik d

ik

ng - degeneracy of state n

15

Einstein coefficients

i

k

ki

kN

Rate of stimulated emission k → i kkist

ki NBW

Einstein coefficient B for stimulated emission

kik

iik B

ggB

ki

16

Einstein 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 strength

kiki

ki Aemcf 22

302

In atomic units:

kiki

ki AE

cf 2

3

2

18

Einstein coefficient and lifetime

kikiki fE

cA 2

3

21

t

1

2

R

E

If E21 = 4.65 eV and f21 = -0.015, what is the lifetime of the excited state?

au

eVE

.1708840211396.27/65.4

65.421

au10

2

3

21

1029.0

)015.0()170884.0(2)137(

t

nss

70104189.21029.0 1710

21

t

Converting to nanoseconds:

19

non-adiabatic transition probabilities

20

Non-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 transitions

xFHH 122211 constant, 1212 FH

1. Landau-Zener

12

2122expF

Hv

P

2. Demkov / Rosen-Zener

constant21 EE /sech1212 xhh x

xvhEEP

12

212

4sech

3. Nikitin

xAH

xAHH

expsin2

expcos

012

02211

02

02

02

cottan21exp

12/cosexp

v

P

4. Bradauk; 5. Delos-Thorson; 6. …

22

Derivation of Landau-Zener formula

nn

t

nnn dttHita

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 etaHidt

tda

12121

2 etaHidt

tda

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):

011

2122

122112

12

aHdtdaHHi

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

011

2122

122112

12

aHdtdaHHi

dtad

25

01

01

22

1222

1222

2

12

1221

1221

2

aHdt

davtFidt

ad

aHdtdavtFi

dtad

Problem: Find a2(+∞) subject to the initial condition a2(∞) = 1.

The solution is:

0x

E

0

12 a

01 a ?2 a

?1 a

12

212

2 expF

Hv

a

26

The probability of finding the system in state 1 is:

12

2122

22exp

FH

vaPnad

The probability of finding the system in state 2 is:

12

2122exp11F

Hv

PP nadad

0

0

1

Pro

babi

lity

|H12

|2

Pnad

Pad

0

0

1

|F12

| or v

Pad

Pnad

27

212

** 2expH

P

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

011577.02exp2

**

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

2122expF

Hv

Pnad

H12

30

0

x

E

0

0

x

E

0

For the same , Rosen-Zener predicts:

Non-adiabatic (but not diabatic!) Adiabatic

vv

xnad vhEEP

12

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

cottan21exp

12/cosexp

v

Pnad

32

Marcus Theory

AB → A+B 2022 1 exp

44ABBB

l Gk H

k Tk T

Ener

gy

Reaction coordinateA-B A+-B

2HAB

G0

33

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.

34

Next lecture

Organic photovoltaics

Contactbarbatti@kofo.mpg.de