MIT Department of Chemistry
5.74, Spring 2004: Introductory Quantum Mechanics II�
Instructor: Prof. Robert Field
15 – 1
5.74 RWF Lecture #15
Visualizing IVR.
[Chem. Phys. Lett. 320, 553 (2000)]
Readings:
Chapter 9.4.9 and 9.4.10, The Spectra and Dynamics of Diatomic Molecules, H. LefebvreBrion and R. Field, 2nd Ed., Academic Press, 2004.
Last time:
Looking for causes of dynamics
d
1
A =
[A, H]
dt
ih
d2
 1
A =
2
 ih 
dt
2
[[A,H],H]
discussed 3 mode, 2 resonance model (H2O)
d
Νj⟩ is transfer rate operators
Νj⟩ found that the cause of dt⟨Ν
derived many operator relationships for ⟨Ν
# of quanta changed
�
e.g.
d
2
N2
=
Ω2 − Ω2†
dt
ih
k11
′ ,33a 12a †3 2
H
(1)
(
) (
k1′,22a 1a †2 2
= Ω1 + Ω1† + Ω2 + Ω2†
)
5.74 RWF Lecture #15
15 – 2
The Two-Level Problem Again
E
H=
0
0   δE
 + 0
E 
EA + EB
2
0 V 
Ω=

0 0
0 
†
 +Ω+Ω
−δE 
EA − EB
2
0
†  0
Ω =

V * 0
E=
δE =
Suppose our pluck at t = 0 is ψA
Ψ(0)⟩ = A⟩ = cos θ/2 +⟩ – sin θ/2 –⟩
verify by plugging in
+ = cosθ 2 A + sinθ 2 B
− = − sinθ 2 A + cosθ 2 B
Ψ (0) = cos2 θ 2 A + cosθ 2sin θ 2 B + sin 2 θ 2 A − cos θ 2sin θ 2 B
(
)
= cos2 θ 2 + sin 2 θ 2 A .
Now put in t-dependence:
Ψ (0) = cos θ 2e −iE + t
h
+ − sinθ 2e −iE − t h −
in the +,– basis and
Ψ ( t) = cosθ 2 e −iE + t
h
− sinθ 2e −iE − t
h
(
(cos θ 2 A + sinθ 2 B )
(− sinθ 2 A + cosθ 2 B )
= cos2 θ 2e −iE + t h + sin 2 θ 2e −iE − t
+ cos θ 2sinθ 2 (e −iE + t h − e −iE − t
h
)A
)B
h
E± = E ± ∆
∆ = [δE 2 + V
]
2 1/ 2
1/ 2
 ∆ + δE 
cos θ 2 = 
 2∆ 
1/ 2
 ∆ − δE 
sin θ 2 = 
 2∆ 
✓checks
We are
going to let
V be real
5.74 RWF Lecture #15
15 – 3
Let us ask several questions for this maximally simple problem that will give us insights into the
relationships between observable properties and the fundamental control parameters in Heff.
H ≡ E is independent of t and E = E A if Ψ (0) = A . This is going to tell
us how to measure the energy of a zero-order state.
1.
2. What is
Hdiag
t
= Ediag ( t)
H res t = Ω + Ω† t = E res
( t) ?
And what do Ediag(t) and Eres(t) tell us?
Evaluate ⟨H⟩.
We can evaluate this in either the A⟩, B⟩, or +⟩, –⟩ basis set.
In the A,B basis set
[
]
+[cos2 θ 2sin 2 θ 2 (2 − 2cosω +− t)] B B
+[cos3 θ 2sin θ 2 (1 − e −iω t ) + cosθ 2sin 3 θ 2 (e +iω
+[cos3 θ 2sin θ 2 (1 − e +iω t ) + cosθ 2sin 3 θ 2 (e −iω
ρ( t) = cos4 θ 2 + sin 4 θ 2 + cos2 θ 2sin 2 θ 2 ( cosω +− t) A A
]
t
− 1)] B
+−
+− t
+−
+−
simplify


V 2
(
)
ρ t = 1 −
1−
t
A A
cosω
(
)

+−
2
2
 2(V + δE )



V2
+
1
t
−
cosω
(
)
 B B
+−
2
2
 2(V + δE )

VδE
iV sin ω +− t 

+
1 − cosω +− t) +
(
1/ 2  A B
2
2
2[V 2 + δE 2 ] 
 2(V + δE )
iV sinω +− t 
VδE

+
1 − cosω +− t) −
(
1/ 2  B A
2
2
2[V 2 + δE 2 ] 
 2(V + δE )
− 1) A B
A
5.74 RWF Lecture #15
15 – 4
in the +,– basis
ρ( t) = cos
2 θ 2 + + + sin 2 θ 2 − −
− sin θ 2cosθ 2e −iω +− t + −
− sin θ 2cosθ 2e +iω +− t − +
=
 ∆ + δE 
 ∆ − δE 
+ + +
− −
 2∆ 
 2∆ 
1/
2
 ∆ 2 − δE 2 
−

 4∆ 2 
[e −iω
+− t
+ − + e +iω +− t − + ]
Want H t = Trace(ρH ) = ∑ ρ jk H kj
j ,k
= ρ11H11 + ρ12H 21 + ρ21H12 + ρ22H 22
= 0 in +,– basis
because H21 = H12 = 0
in the +,– basis
Ht=
 ∆ + δE 
 ∆ − δE 
E +
E
 2∆  +  2∆  −
E ± = E ± ∆
Ht=
2∆
2δE∆
= E + δE = E A !
E+
2∆
2∆
So we have our first really important result. If you know your pluck perfectly, then, for any H,
⟨H⟩t = Epluck
independent of time.
To determine the energy of the plucked zero-order state, simply measure the intensity weighted
average energy in the absorption spectrum.
5.74 RWF Lecture #15
15 – 5
Now let us divide H into Hdiag and Hres = Ω + Ω †
Ediag ( t) = Hdiag
t
E res ( t) = Ω + Ω
Ω
†
t
of course E = Ediag ( t) + E res
( t) must be independent of t, but Ediag and Eres are certainly time­
dependent.
For our Ψ (0) = A pluck of a two-level system, we want to express both H and ρ in the A, B Basis
set.


V2
ρ( t) = 1 −
1 − cosω +− t)  A A
(
2
2
 2(V + δE )



V2
+
1 − cosω +− t) 
B B
(
2
2
 2(V + δE )

VδE
iV sinω +− t 

+
−
+
t
1
A B cosω
(
+− )
1
2
/

2
2
2[V 2 + δE 2 ] 
 2(V + δE )
iV sinω +− t 
VδE

−
+
−
1
cosω
t
B A (
)
+−
1
2
/

2
2
2
2
2[V + δE ] 
 2(V + δE )
5.74 RWF Lecture #15
15 – 6
Thus
diag
+
Hdiag t = Trace ρHdiag = ρAA Hdiag
ρ
H
BB BB
AA
E + δE
E − δE
= ρAA E A + ρBB E B = (ρAA + ρBB ) E + (ρAA − ρBB )δE
V 2δE
=E− 2
1− cosω +− t) + δE
2(
V + δE
at t = 0
Hdiag = E + δE = E A
H res t = ρAB H BA + ρBA H AB
V 2δ
E
= 2
(1− cosω +− t)
V + δE
at t = 0
H res t = 0
Hdiag t + H res t = E + δE = E A
In a more general situation, when there are several resonances, we have a polyad Heff that is larger
than 2 × 2, we need help sorting out which resonance is most important for the post-pluck dynamics
of each conceivable pluck.
See HLB-RWF Chapter 9.4.10 and Chem. Phys. Lett. 320, 553 (2000).
5.74 RWF Lecture #15
15 – 7
Let E res ( t) = ∑ Ω j + Ω†j
j
sum over
resonances
we still have E = Ediag ( t) + ∑ E res, j ( t
)
j
two useful quantities:
E res, j
(
1 T
= lim ∫0 dt Ωk + Ωk†
T →∞ T
)
and fractional importance, for a specific pluck, of the j-th resonance
fj =
E res, j
E res
What do we do with E res, j , E res, j , and fj?
We attempt to correlate what we see in the spectrum to the various resonance terms that could cause
the observed effects.
We look for a pluck which is maximally sensitive to one particular resonance term.
In order to do this, we need a working model for Heff. We then use insights guided by the resonance
operator dynamics.
See Figure 9.11. The survival probability of the plucked state is directly observable because it is the
F. T. of the pluck-related piece of the spectrum. The dynamical features in the survival probability
encode Fourier components and amplitudes associated with each of the dynamically important
resonance operators. We want to know what causes each of the dynamical features present in any
do-able experiment.
Strategy: use computed simple dynamical quantities to sharpen up correspondences with observable
dynamical features.
5.74 RWF Lecture #15
15 – 8
Figures removed due to copyright reasons.