Essentials of Computational Chemistry

14.5 TRANSITION PROBABILITIES 509
If we cancel the equivalent sums on the left and right we are left with
− ¯h ∂ck
i ∂t
k
e−(iEkt/¯h) k = e0r sin(2πνt)
k
cke −(iEkt/¯h) k
(14.27)
We now multiply on the left by m and integrate, where m indexes the stationary state
for which we are interested in measuring the probability of transition. This gives
− ¯h ∂ck
i ∂t
k
e−(iEkt/¯h) 〈m|k〉 =e0 sin(2πνt)
cke
k
−(iEkt/¯h)
〈m|r|k〉 (14.28)
Note that the expectation value on the l.h.s. of Eq. (14.28) is simply δmk, because of the
orthogonality of the stationary-state eigenfunctions. Thus, only the term k = m survives, and
we may rearrange the equation to
∂cm
∂t =−i
¯h e0 sin(2πνt)
cke [i(Em−Ek)t/¯h]
〈m|r|k〉 (14.29)
k
If we assume that our perturbation was small, and applied for only a short time, we may
further assume that the expansion coefficients on the r.h.s. of Eq. (14.29) have their initial
(ground-state) values. This leads to the further simplification
∂cm
∂t =−i
¯h e0 sin(2πνt)e [i(Em−E0)t/¯h] 〈m|r|0〉 (14.30)
In order to determine cm at (and after) time τ, we must integrate t from 0 to τ, giving
where
and
cm(τ) =− i
¯h e0
τ
= 1
2i¯h e0
sin(2πνt)e
0
[i(Em−E0)t/¯h]
〈m|r|0〉dt
i(ωm0+ω)τ e − 1
ωm0 + ω − ei(wm0−ω)τ
− 1
〈m|r|0〉 (14.31)
ωm0 − ω
ω = 2πν (14.32)
ωm0 = Em − E0
¯h
(14.33)
We now ask the question, for what values of m is |cm| 2 large? Given a particular frequency
of radiation ω, the magnitude of cm will be large if ωm0 is close to ω, thereby making the
denominator in the second term in brackets very small (note that even when ωm0 is equal to
ω, the expansion coefficient is well behaved because of the way the numerator approaches
zero, cf. Section 10.5.2). This result is consistent with the notion that a photon of energy hν