MIT Department of Chemistry�
5.74, Spring 2004: Introductory Quantum Mechanics II
Instructor: Prof. Robert Field
5.74 RWF Lecture #11
11 – 1
Complex Energy Heff: Nondegenerate Perturbation Theory
Reading:
Chapter 9.3, The Spectra and Dynamics of Diatomic Molecules, H. Lefebvre-Brion and R.
Field, 2nd Ed., Academic Press, 2004.


Last time:
vibration - rotation
 spectra I (ω)
electronic - vibration - rotation
pure rotation
genuinely localized plucks
grand recurrences
Supplements: 1.
2.
Today:
Stationary Phase for Vibration and Electronic Transitions (nature of pluck)
Heller’s Fractionation Measure (inversely proportional to the fraction of state- or
phase-space accessed.)
sharp bright state
broad doorway state
(broad because it couples to the quasi-continuum bath)
reduce many-state interactions to a simple two-level problem.
vs.
bright
doorway
continuum
bright
continuum
“Lumpy
continuum”
As the bright state is systematically tuned (perhaps by changing rotational level), it can mix either indirectly
or directly with the continuous, dark bath.
Our goal is to find a shortcut for dealing with the interaction between a bright state and a bath, especially via
a doorway state.
5.74 RWF Lecture #11
11 – 2
We can develop an intuitive understanding for a two- or three-level problem much more easily than for a
large-dimension random matrix problem. It is also computationally much quicker.
The basis for the idea is, if H is time-independent,
Ψ j (Q, t) = ψ j (Q)e
−iE j t / h
Hψ j = E j ψ j
Suppose we allow Ej to be complex?
E j = ε j − iγ j / 2
e
−iE j t / h
=e
=e
−iε j t / h ( −i )( −i )( γ
e
j
2) t / h
−iε j t / h − γ j t /2 h
e
exponential
decay
− γ j t / h
(
)
P j t = Ψ j ( t) Ψ j ( t) = e
(an exponentially decaying population! Should come in handy.)
h
γ has units of energy
γ
= τ or γ =
h
τ
Width of Pj ( t) = Ψ j ( t) Ψ j ( t) in energy or cm–1?
Lorentzian L( ν; ν0 , ∆ν) =
*
( ∆ν / 2)
1
π ( ∆ν / 2) 2 + ( ν − ν0 ) 2
∆ν is FWHM
* ∫ dνL( ν; ν0 ,∆ν) = 1
1
* ∆ν[cm−1 ] =
2πcτ
normalized to 1
FWHM of a Lorentzian line
associated with exp. decay of
probability with lifetime τ
!
5.74 RWF Lecture #11
11 – 3
P ( t) = e − γt / h
for our case:
τ=
FWHM[ E ] =
h
τ
h
γ
units of E
=γ
We can think of decay rate as a property of a “quasi-eigenstate”. We can just tack the decay rate onto the
normal Ψj(Q,t) without changing anything. Seems to be too good to be true!
One bad thing about allowing Heff to have complex energies along the diagonal is that Heff is no longer
Hermitian. We will soon see the difficulties this causes and how to patch them up.
We would like to use non-degenerate and quasi-degenerate perturbation theory to describe the effects of
interactions between states with finite lifetimes. [Decay rates is better concept than lifetimes.
Why?]
Are decay rates shared between interacting states the way other properties are shared? Let’s review what
perturbation theory tells us about sharing of properties. Use the Zeeman tuning coefficient (g-value)
as the shared property.
Non-degenerate perturbation theory
H=H
( 0)
()
+H1
( )
()
( )
E j = E j0 + E j1 + E j2
( ) ( )
( ) ( )
H 0 ψ j0 = E j0 ψ j0
( 0)
(1) 2
(1)
E j = H jj + H jj + ∑
k≠ j
( 0)
(1)
( )
( )
E j0 − E k0
()
( 0)
ψj =ψj +ψj =ψj + ∑
k≠ j
Not normalized — must at least consider renormalization
H jk
H jk1
( 0)
( 0)
E j − Ek
( )
ψ k0
5.74 RWF Lecture #11
11 – 4
H Zeeman = µ Bohr Bz (L z + 2Sz )
E Zeeman
= ψ j H Zeeman ψ j
j
ψj ψj

 ψ ( 0) L + 2S ψ ( 0) +
=
∑
z
j
2
 j z
(1)
≠j
k


H jk

1 + ∑  ( 0)
( 0) 

k ≠ j  E j − Ek 
µ Bohr Bz
()


H jk1
 ( 0)
( 0) 
E
E
−
 j
k 
2
( 0)
( 0)
ψ k L z + 2Sz ψ k




mixing fraction
renormalization
The sum of a renormalization-corrected zero-order term (nominal character) plus expectation values of
HZeeman weighted by the fractional character of each “remote perturber” state (borrowed character). This is
exactly what we would expect for sharing of properties other than energy.
What about sharing of decay rates?
H =
H
H
( 0)
()
+H1
( 0) ( 0)
ψj
( )
( )
( )
= E j0 ψ j0 = ( E j − iΓ j / 2) ψ j0
H(0) is not Hermitian! H(1) is Hermitian. It has only real diagonal matrix elements.
The energy denominators from perturbation theory are complex.
1
( )
( )
E j0 − Ek0
=
1
(ε j − ε k ) − i(Γ j − Γk ) 2
(ε j − ε k )
(ε j − ε k )2 + (Γ j − Γk )2
i( Γ j − Γk ) 2
+
(ε j − ε k )2 + (Γ j − Γk )2
=
rationalize
denominator
real part
4
4
imaginary
part
This means that the borrowing and lending of decay rate follows the usual perturbation theory form, except
for two subtle modifications, one in the numerator and one in the denominator. There is also a
hidden surprise that can only be explored when ordinary, non-degenerate perturbation theory is
()
invalid:
H jk1
( )
( )
E j0 − E k0
> 1.
Return to non-degenerate perturbation theory.
5.74 RWF Lecture #11
11 – 5
Real part:
( )2
Level shift
δε j = ∑
k≠ j
(εj − ε k )
(ε j − ε k )2 + (Γ j − Γk )2
H jk1
4
* if εj > εk j-th level gets pushed up as expected
()
1
* if εj = εk (and, to stay in the non-degenerate perturbation theory limit, H jk
(Γ j − Γk ) 2 << 1),
then there is no level shift at all (surprise!)
* if | εj – εk | >> |(Γj – Γk)/2|, then a Taylor series expansion gives
(1) 2
  Γ − Γ 2  2 
(
j
k)
δε j =
1 − 
 .
ε j − ε k   ε j − ε k  


H jk
Imaginary part:
( )2
Decay rate change
δΓ j = − ∑
k≠ j
(Γ j − Γk ) 2
(ε j − ε k ) + (Γ j − Γk )2
H jk1
2
4
* if Γj > Γk (level-k decays more slowly than level-j) the decay rate of level-j is reduced (it lends
decay rate to level-k). The decay rates of levels j and k become more nearly equal.
()
* if Γj = Γk (and
H jk1
ε j − εk
<< 1), then there is no change of decay rate (surprise!)
* if Γj – Γk/2 >> εj – εk
()
δΓ j = − ∑
k≠ j
1
H jk
(Γ j − Γk )
  ε j − εk  2 
1 −
.


2  ( Γ j − Γk ) 2  


reduced decay
rate attraction
5.74 RWF Lecture #11
11 – 6
So part of this non-degenerate perturbation theory model seems reasonable, but there are some surprises.
The surprises are most prominent when
()
H jk1
( 0)
( 0)
E j − Ek
>1
for which we must use quasi-degenerate perturbation theory. This is an algebraic nightmare, even in the
Γ = 0 situation.
Recall
 EA
H= *
 H AB
H AB   E
=
E B   0
E=
 EA − EB
0 
2
 +
E   H*
AB



EA − EB 

−

2
H AB
EA + EB
2
1/ 2
 E A − E B  2

2
E± = E ± 
+ H AB 

2


= E ± δE
1/ 2
 E A − E B  2

2
+ H AB 
δE ≡ 

2


(CTDL, page 407)
H AB = H AB e −iφ
ψ + = cos θ 2e −iφ/2 ψ A + sin θ 2eiφ/2 ψ B
ψ − = − sin θ 2e −iφ/2 ψ A + cos θ 2eiφ/2 ψ B
tanθ =
2 H AB
EA − EB
 2 H AB 
θ = tan −1 

 EA − EB 
0≤θ< π
5.74 RWF Lecture #11
11 – 7
more explicitly (Herzberg, page 283)
 δE + ( E A − E B ) 2 
cosθ 2 = 

2δE

1/ 2 δE − ( E A − E B ) 2 
sinθ 2 = 

2δE


1/ 2
verify that ψ+⟩ belongs to E+ > E– (you should always do this when using borrowed, rather than derived,
equations).
ψ + H ψ + = cos2 θ 2 H AA + sin 2 θ 2 H BB + eiθH AB cosθ 2sinθ 2 + e −iθH BA cosθ 2sinθ 2
= cos2 θ 2 E A + sin 2 θ 2 E B + 2 H AB cosθ 2sinθ 2
=
=
E A [δE + ( E A − E B ) 2]
2δE
δE ( E A + E B ) +
E A2
+
E B [δE − ( E A − E B ) 2]
2 − EA EB +
2δE
2δE
E B2
2 + 2 H AB
δE ( E A + E B ) + ( E A − E B ) 2 + 2 H AB
=
2δE
+ 2 H AB
(δE 2 − ( E A − E B )2 4)1/2
2δE
2
2
δE ( E A + E B ) + 2δE 2
E A − E B
=
=
+ δE = E +
2δE
2
insert
definition
of δE2 here
Confirmed. Regardless of sign of HAB and (EA – EB), H ψ + = E + ψ + .
Now we are ready to try to solve for E±, ψ±⟩ using an H that has complex diagonal elements.
But we also have to worry about a big problem!
( )
()
If we use perturbation theory to define ψ j = ψ j0 + ψ j1 , then ψj and ψk are not orthogonal!
Consider a 2-state basis set, ψA⟩ and ψB⟩, and let HAB be real.
�
5.74 RWF Lecture #11
11 – 8
H AB
 ( )
( ) 
ψ A = N  ψ A0 +
ψ B0 
−
Γ
−
Γ
2
ε
−
ε
i
( A B) ( A B)


2

H AB
2
ψ A ψ A = N A 1 +

2
2
 (ε A − ε B ) − i (ΓA − ΓB ) 4 
1/ 2

 (ε A − ε B ) 2 + (ΓA − ΓB ) 2 4

NA =  2
2
2
 H AB + (ε A − ε B ) + (ΓA − ΓB ) 4 
2

H AB
2
ψ B ψ B = N B 1 +

2
2
 (ε B − ε A ) + (ΓB − ΓA ) 4 
|NA| = |NB|
H BA
H BA

2
ψA ψB = NA 
+

 (ε B − ε A ) − i(ΓB − ΓA ) 2 (ε A − ε B ) + i(ΓA − ΓB ) 2 
1
1


2
= N A H BA 
−

 (ε B − ε A ) − i(ΓB − ΓA ) 2 (ε B − ε A ) + i(ΓB − ΓA ) 2 
i(ΓB − ΓA )
2
= N A H BA
≠ 0!
(ε B − ε A ) 2 + (ΓB − ΓA ) 2 4
The usual orthogonality integral is not 0 and is pure imaginary. Something is wrong! Cannot use either
()
orthogonality or completeness to expand ψ j1 . So we cannot use the results of ordinary non-degenerate
perturbation theory without a correction.
The concept of “biorthogonality” saves us.
When H is not Hermitian, ψi⟩ are eigenfunctions of H in the sense Hψi⟩ = Eiψi⟩
∼
∼
but ⟨ψ
i are the conjugate transpose of ψ i ⟩, which are eigenfunctions of H* in the sense
H * ψ˜ i = Ei* ψ˜ i !