MIT Department of Chemistry
5.74, Spring 2004: Introductory Quantum Mechanics II
Instructor: Prof. Andrei Tokmakoff
p. Supp. 6-1
Slowly Applied (Adiabatic) Perturbation
All of our perturbations so far have been applied suddenly at t > t0 (step function)
V (t ) = θ (t − t 0 )V (t )
This leads to unphysical consequences—you generally can’t turn on a perturbation fast enough to
appear instantaneous. Since first-order P.T. says that the transition amplitude is related to the
Fourier Transform of the perturbation, this leads to additional Fourier components in the spectral
dependence of the perturbation—even for a monochromatic perturbation!
So, let’s apply a perturbation slowly . . .
V (t ) = V eη t
η: small and positive
V(t)
V
t
t0
at t = −∞ . Find Pk (t ).
The system is prepared in state
bk = k UI
bk =
=
−i
t
∫−∞ dτ e
iω k τ
eητ
kV
− iVk exp[η t + iω k t ]
η + iωk
= Vk
exp[ηt + i (Ek − E )t /
Ek − E + iη
2
Pk = bk =
Vk
2
2
exp[2η t ]
=
η 2 + ω 2k
]
Vk
2
exp[2ηt ]
(Ek − E )
2
+ (η
)2
p. Supp. 6-2
This is a Lorentzian lineshape in ω k with width 2η .
Step Response Perturbation
Gradually Applied Perturbation
2π
2η
ωkl
0
/ t
ωkl
The gradually turned on perturbation has a width dependent on the turn-on rate, and is independent
of time. (The amplitude grows exponentially in time.) Notice, there are no nodes in Pk .
η−1 is the effective turn-on time of the perturbation:
Now, let’s calculate the transition rate:
V
∂P
w kl = k = k2
∂t
2
2η e 2 η t
η2 + ωk2
Look at the adiabatic limit; η → 0 .
setting e2 ηt → 1; and using
wk =
2π
2
Vk
2
δ ( ωk ) =
2π
lim
η
= πδ ( ωk
2
η → 0 η + ω2k
Vk
2
δ ( Ek − E
)
)
We get Fermi’s Golden Rule—independent of how perturbation is introduced!
p. Supp. 6-3
If we gradually apply the Harmonic Perturbation,
V (t ) = V eηt cosωt
i ωτ
+ e −i ωτ 
i ω k τ + ητ  e
d
τ
V
e


∫−∞ k
2


i ω +ω t
i(ω k −ω )t

Vk η t  e ( k )
e
= e 
+

2
 − (ω k + ω ) + iη − (ω k − ω ) + iη 
bk =
−i
t
Again, we have a resonant and anti-resonant term, which are now broadened by η .
If we only consider absorption:
Pk = bk
2
V
= k
4
2
2
e 2η t
1
(ω k
− ω) + η2
2
which is the Lorentzian lineshape centered at ω k = ω with width ∆ ω = 2η .
Again, we can calculate the adiabatic limit, setting η → 0 . We will calculate the rate of transitions
ω k = ∂Pk / ∂t . But let’s restrict ourselves to long enough times that the harmonic perturbation has
cycled a few times (this allows us to neglect cross terms) → resonances sharpen.
wk =
π
Vk
2 2
2
 δ ( ωk − ω) + δ ( ωk + ω) 