Femtochemistry: A theoretical overview Mario Barbatti mario.barbatti@univie.ac.at VI – Transition...

Femtochemistry: A theoretical overviewFemtochemistry: A theoretical overview

Mario Barbattimario.barbatti@univie.ac.at

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

Contactmario.barbatti@univie.ac.at

This lecture can be downloaded athttp://homepage.univie.ac.at/mario.barbatti/femtochem.html lecture6.ppt