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390 11MOLECULAR SPECTROSCOPY
�e Franck–Condon factor is given by [11F.5–464] and involves the square of
integral of the product of the two wavefunctions
I = N1N0 ∫
+∞
−∞
e−ax
2/2(x − x0)e−a(x−x0)
2/2 dx
= (a/π)1/4(4a3/π)1/4 ∫
+∞
−∞
(x − x0)e−a[x
2/2+(x−x0)2/2] dx
= (2a2/π)1/2 ∫
+∞
−∞
(x − x0)e−a[x
2/2+x2/2−xx0+x20/2] dx
= (2a2/π)1/2 ∫
+∞
−∞
(x − x0)e−a[x
2−xx0+x20/2] dx
= (2a2/π)1/2 ∫
+∞
−∞
(x − x0)e−a(x−x0/2)
2
e−ax
2
0/4 dx
the �nal equality above is veri�ed by expanding out the square and recombining
the terms.�e next step is to take out the constant factors and split the integral
into two using (x − x0) = (x − x0/2) − x0/2
I = (2a2/π)1/2e−ax
2
0/4 ∫
+∞
−∞
(x − x0)e−a(x−x0/2)
2
dx
= (2a2/π)1/2e−ax
2
0/4
× [∫
+∞
−∞
(x − x0/2)e−a(x−x0/2)
2
dx − (x0/2)∫
+∞
−∞
e−a(x−x0/2)
2
dx]
�e �rst integral is of an odd function over a symmetric interval and so is zero.
�e second integral is of the form of Integral G.1 with k = a and evaluates to
(π/a)1/2 (it does not matter that it is centred at x0 rather than 0). Hence
I = −(2a2/π)1/2e−ax
2
0/4(x0/2)(π/a)1/2 = −(2a)1/2(x0/2)e−ax
2
0/4
�e Franck–Condon factor is I2 = (ax20/2)e−ax
2
0/2 . As expected, this factor is
zero when x0 = 0 as then the two wavefunctions align and have zero overlap on
account of one being even andone being odd. By taking the derivative of I2 with
respect to x0 it is easily shown that the Franck–Condon factor is a maximum
at x0 = ±(2/a)1/2.
E11F.6(b) �e Franck–Condon factor is given by [11F.5–464] and involves the square of
integral of the product of the two wavefunctions. �e region over which both
wavefunctions are non-zero is from L/2 to L: this is the domain of integration
I = (2/L)∫
L
L/2
sin(πx/L) sin(π[x − L/2]/L)dx
Note that sin(π[x − L/2]/L) = sin(πx/L − π/2) = − cos(πx/L), this gives a
useful simpli�cation
I = −(2/L)∫
L
L/2
sin(πx/L) cos(πx/L)dx