1000 Solved Problems in Modern Physics

(Grace) #1

226 3 Quantum Mechanics – II


Exchange ofx →−x, y →−y, z →−zimpliesθ →π−θand
φ→π+φ, so that
P 1 m(cosθ)→(−1)l+m
andeimφ→(−1)meimφ,ThusPρ=(−1)^2 m(−1)l=(−1)lρ
wheremis an integer. Thusρis symmetrical for evenland antisymmetrical
for oddl.
First consider zero nuclear spin. The total wave functionψis antisymmet-
rical for oddland symmetrical for evenl. As the nuclei must obey either
Fermi or Bose statistics, either only thel=odd states must exist or only the
l=even states must exist. It turns out that for nuclei with zero spin only the
even rotational states exist and odd rotational states are missing.
Next consider the case of non-zero spin. A nucleus of total angular momen-
tumIcan have a componentMin any prescribed direction taking 2I+1 values
in all (I,I− 1 ,...−I), that is 2I+1 states exist. For the two identical nuclei
(2I+1)^2 wave functions of the formψM 1 (A)ψM 2 (B) can be constructed. If
the two nuclei are identical, these simple products must be replaced by lin-
ear combination of those products which are symmetric or antisymmetric for
interchange of nuclei. IfM 1 = M 2 , the products themselves are (2I+1)
symmetric wave functions, the remaining (2I+1)^2 −(2I+1)= 2 I(2I+1)
functions withM 1
=M 2 have the formψM 1 (A)ψM 2 (B)andψM 2 (A)ψM 1 (B).
Each such pair can be replaced by one symmetric and one antisymmetric
wave function of the formψM 1 (A)ψM 2 (B)±ψM 2 (A)ψM 1 (B). Thus, half of
2 I(2I+1) functions, that isI(2I+1) are symmetric and an equal num-
ber antisymmetric. Therefore, total number of symmetric wave functions
=(2I+1)+I(2I+1)=(2I+1)(I+1). Total number of antisymmetric
wave functions=I(2I+1). Therefore, the ratio of the number of symmetric
and antisymmetric functions is (I+1)/I.
From the previous discussion it was shown that for the symmetric elec-
tronic wave function of the molecule the interchange of nuclei produces a
factor (−1)lin the molecular wave function. Thus, for nuclei obeying Bose
statistics symmetric nuclear spin functions must combine with evenl. Because
of the statistical weight attached to spin states, the intensity of even rota-
tional lines will be (I+1)/Ias great as that of neighboring odd rotational
lines.
For nuclei obeying Fermi statistics, the spin and rotational states combine
in a manner opposite to the previously described and the odd rotational lines
are more intense in the ratio (I+1)/I.
Thus, by determining which lines are more intense, even or odd, the nuclear
statistics is determined and by measuring the ratio of intensities of adjacent
lines the nuclear spin is obtained. The reason for comparing the intensity of
neighboring lines is that the intensity of rotational lines varies according to
the occupation number of rotational states governed by the Boltzmann distri-
bution.
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