Fermi Golden Rule

2.5

FERMI’S GOLDEN RULE

The  transition  rate  and  probability  of  observing  the  system  in  a  state

after  applying  a

k

perturbation  to  from  the  constant  first-order  perturbation  doesn’t  allow  for  the  feedback

between quantum states, so it turns out to be most useful in cases where we are interested just the

rate  of  leaving  a  state.This question  shows up  commonly  when  we  calculate  the transition

probability  not  to  an  individual  eigenstate,  but a  distribution  of  eigenstates.    Often  the  set  of

eigenstates  form  a  continuum  of  accepting  states,  f or  instance,  vibrational  relaxation  or

ionization.

Transfer to a set of continuum (or bath) states forms the basis for a describing irreversible

relaxation.  You can think of  the material Hamiltonian for our problem being partitioned into two

( )

portions,

=

H

+

H

+

V

t

,  where  you  are  interested in  the  loss  of  amplitude  in  the

H

H

S

B

SB

S

states  as it leaks into

. Qualitatively, you expect deterministic, oscillatory feedback between

H

B

discrete quantum  states.   However, the amplitude of one discrete state coupled to a continuum

will decay due to destructive interferences between the oscillating frequencies for each member

of the continuum.

So, using the same ideas as before, let’s calculate the transition probability from    to a

distribution of final states:

.

P

k

Probability of  obser ving amplitude in discrete eigenstate of

2

=

b

H

P

k

k

0

( )

:  Density  of   states—units  in

,  describes  distribution  of  final

1

E

E

k

k

states—all eigenstates of

H

0

If we start in a state  , the total transition probability is a sum of probabilities

P

=

P

.  (2.161)

k

k

k

We are just inter ested in the rate of  leaving    and occupying any state

or for a continuous

k

distribution:

2-43

( )

=

dE

E

P

(2.162)

P

k

k

k

k

For a constant perturbation:

( )

( )

sin

E

E

t

/ 2

( )

2

P

=

dE

E

4

V

2

(2.163)

k

k

k

k

k

E

E

2

k

Now, let’s make two assumptions to evaluate this expression:

( )

1)

varies slowly with frequency and there is a

E

k

continuum of f inal states. (By slow what we are saying is

that the observation point

t

is relatively long) .

2)  The matrix element

is invariant across the f inal

V

k

states.

These assumptions allow those variables to be factored out of integral

( )

E

E

t

/ 2

2

P

=

V

+

dE

4 sin

(2.164)

2

k

( )

k

k

k

E

E

2

k

( )

Here,  we  have  chosen  the  limits

since

is  broad  relative  to

.    Using  the

+

P

E

k

k

identity

sin

a

2

+

d

=

a

(2.165)

2

with

we have

=

t

/

a

2

P

=

V

2

t

(2.166)

k

k

The total transition probability is linearly propor tional to time.  For relaxation processes, we will

be concerned with the transition rate,

:

w

k

2-44

P

w

=

k

t

k

(2.167)

2

2

w

=

V

k

k

Remember that

is centered sharply at

.  So although   is a constant, we usually write

=

E

P

E

k

k

( )

( )

eq. (2.167) in terms of

=

E

or more commonly in ter ms of

E

:

E

E

k

k

( )

2

w

=

E

=

E

V

(2.168)

2

k

k

k

( )

( )

2

w

=

V

E

E

=

dE

E

w

(2.169)

2

w

k

k

k

k

k

k

k

This expression is known as Fermi’s Golden Rule.  Note the rates are independent of time.  As

we will see going forward,  this first-order perturbation theory expression involving the matrix

element  squared  and the  density of  states  is  very  common in the  calculation of  chemical  rate

processes.

Range of validit y

<<

For discrete states we  saw  that the first order expression  held  for

,  and for

V

k

k

times such that

never varies fr om initial values.

P

k

( )

1

=

w

t t

(2.170)

t

<<

P

w

k

k

0

k

( )

However, transition probability must also be sharp compared to

, which implies

E

k

(2.171)

>>

/

E

t

k

So, this expr ession is useful where

2-45

E

>>

w

k