Zdunkowski W., Trautmann T., Bott A. Radiation in the atmosphere: A course in Theoretical Meteorology
Подождите немного. Документ загружается.
8.4 Vibrations and rotations of molecules 305
system is given by
H =
p
2
2m
+
1
2
kx
2
(8.128)
where p is the linear momentum. Using Table 8.1, the Hamiltonian operator is
expressed by
H =−
h
2
8π
2
m
d
2
dx
2
+
1
2
kx
2
(8.129)
and the wave equation by
d
2
ψ
dx
2
+
8π
2
m
h
2
E −
1
2
kx
2
ψ = 0 (8.130)
Introducing the abbreviations
α =
8π
2
m
h
2
E,β=
2π
√
mk
h
(8.131)
(8.130) assumes the form
d
2
ψ
dx
2
+ (α − β x
2
)ψ = 0 (8.132)
Changing the variable according to
ξ =
-
βx,
d
2
dx
2
= β
d
2
dξ
2
(8.133)
we find
d
2
ψ
dξ
2
+
α
β
− ξ
2
ψ = 0 (8.134)
Now we investigate which form ψ must have in order to be an acceptable wave
function for large values of ξ . For sufficiently large values of ξ the ratio α/β can
be neglected in comparison to ξ
2
so that we get the approximate equation
d
2
ψ
dξ
2
− ξ
2
ψ = 0 (8.135)
This equation is approximately satisfied by
ψ = C exp
±
ξ
2
2
with
d
2
dξ
2
exp
±
ξ
2
2
= exp
±
ξ
2
2
(ξ
2
± 1) (8.136)
since ±1 may be neglected in the region of large ξ
2
. For obvious reasons we cannot
use the solution exp(+ξ
2
/2), but exp(−ξ
2
/2) behaves satisfactorily at large values
306 Absorption by gases
of ξ. Thus we are led to the trial solution
ψ(ξ ) = u(ξ)exp
−
ξ
2
2
(8.137)
Substitution of (8.137) into (8.134) yields
d
2
u
dξ
2
− 2ξ
du
dξ
+
α
β
− 1
u = 0 (8.138)
By setting α/β − 1 = 2n, we obtain Hermite’s differential equation whose solu-
tion are the well-known Hermite polynomials H
n
, n = 0, 1,...
H
n
(ξ) = (−1)
n
exp(ξ
2
)
d
n
dξ
n
[exp(−ξ
2
)] (8.139)
A few low-order expressions of these polynomials are listed next
H
0
(ξ) = 1, H
1
(ξ) = 2ξ, H
2
(ξ) = 4ξ
2
− 2 (8.140)
A particular solution to equation (8.138) can be written as
u(ξ ) = H
n
(ξ) (8.141)
so that the wave function is given by
ψ(ξ ) = CH
n
(ξ)exp
−
ξ
2
2
, C = con st (8.142)
Using equation (8.131) we find that the energy E is quantized and must be written
as
E =
h
2π
n +
1
2
%
k
m
=
n +
1
2
hν, n = 0, 1,... (8.143)
where ν is the frequency of the classical harmonic oscillator. The state with n = 0
is the vibrational ground state whose vibrational energy is not zero. The residual
energy is known as the zero point energy. This is in agreement with Heisenberg’s
uncertainty principle which states that one can never precisely know the position
and the momentum of a particle. If the oscillator had zero energy it would have zero
momentum and would be located exactly at the position of the minimum potential
energy.
Finally, we have to find the normalization constant C by considering the condition
(8.142). For the present situation we obtain
∞
−∞
ψ
∗
n
ψ
n
dx =
C
2
√
β
∞
−∞
H
n
(ξ)
2
exp(−ξ
2
) dξ = 1 with C =
β
1/4
-
2
n
n!
√
π
(8.144)
8.4 Vibrations and rotations of molecules 307
Omitting details we find for the normalized wave function the following expression
ψ
n
(ξ) = N
n
H
n
(ξ)exp
−
ξ
2
2
with N
n
=
%
β
π
1
2
n
n!
1/2
(8.145)
with ξ =
√
βx.
The solution of the harmonic oscillator problem outlines the common approach
to solve the wave equation. We convert the wave equation into one of the standard
differential equations whose solutions are known. This technique will also be used
in connection with the rigid rotator problem.
8.4.2 Vibration of diatomic molecules
We consider the vibrations of the two atoms relative to each other. The simplest
form of vibrations in a diatomic molecule is that each atom moves toward or away
from the other in simple harmonic motion. For a molecule consisting of two like
atoms such as N
2
,O
2
(homonuclear) the dipole moment is zero, and therefore no
transitions between the different vibrational levels will be observed. This means that
no infrared absorption or emission occurs. In contrast to this, diatomic molecules
such as HCl (heteronuclear) do have a permanent dipole moment.
The harmonic oscillator potential energy curve is not particularly accurate when
considering actual molecules. A more satisfactory procedure is to assume some
appropriate analytical expression for the potential energy curve such as
V (r) = D
(
1 − exp
[
−β(r − r
e
)
]
)
2
(8.146)
This is the so-called Morse function. Here D is the dissociation energy of the
molecule which is obtained if r →∞. The quantity β is a constant which differs
from molecule to molecule. If the distance r is equal to the equilibrium value r
e
the potential energy has the minimum value V = 0. For r = 0 the potential energy
approaches a large but finite value. (8.146) is satisfactory for many situations.
However, for reasons of mathematical simplicity we will presently continue to use
the harmonic oscillator approximation.
To determine the vibrational energy states it will be useful to introduce the
concept of the reduced mass. If r
1
and r
2
are the respective distances of the atoms
from the center of mass of the molecule, we obtain
m
1
r
1
= m
2
r
2
(8.147)
If r = r
1
+r
2
then r
1
and r
2
are given by
r
1
=
m
2
m
1
+ m
2
r, r
2
=
m
1
m
1
+ m
2
r (8.148)
308 Absorption by gases
The kinetic energy K of the diatomic system is expressed by
K =
1
2
m
1
˙
r
2
1
+
1
2
m
2
˙
r
2
2
=
1
2
µ
˙
r
2
with µ =
m
1
m
2
m
1
+ m
2
(8.149)
where µ is the reduced mass. Thus the classical Hamilton function is given by
H =
1
2
µ
˙
r
2
+ V (r) with V (r) =
1
2
k(r − r
e
)
2
(8.150)
so that Schr¨odinger’s equation follows immediately
d
2
ψ
dr
2
+
8π
2
µ
h
2
[
E − V (r)
]
ψ = 0 (8.151)
Setting r − r
e
= x, the latter equation has the form (8.130). Thus, according to
(8.145) and (8.143), we find that the wave function and the energy of the diatomic
system are given by
ψ
n
(x) =
√
β/π
2
n
n!
1/2
H
n
(
-
βx)e
−
1
2
βx
2
, β = 2π
-
µk/ h
(8.152)
and
E =
h
2π
v +
1
2
&
k
µ
=
v +
1
2
hν
(8.153)
As is customary in spectroscopy, we have used the symbol v for the vibrational
quantum number. It will be observed that the vibrational spectrum of a diatomic
molecule considered as a harmonic oscillator consists of one frequency. The selec-
tion rule v =±1 will be derived later.
8.4.3 Vibration of polyatomic molecules
As stated before, a molecule consisting of n atoms has 3n degrees of freedom. Three
coordinates are required to describe the translational motion of the entire system
considered as concentrated at the center of mass, and three degrees are required, in
general, to describe the rotational motion of the system about its center of mass.
Thus 3n − 6 degrees of freedom are left to describe the vibrational motion of the
nuclei of the atoms relative to the axes with origin at the center of mass. For the
linear CO
2
molecule (3n − 5 vibrational degrees of freedom) and for the water
vapor molecule we have shown how to find the normal coordinates and normal
frequencies. Any vibrational motion of the molecule may be constructed from the
superposition of the normal vibrations.
8.4 Vibrations and rotations of molecules 309
In textbooks on theoretical mechanics, see also the previous examples, it is shown
that in terms of the normal coordinates the kinetic and the potential vibrational
energy assume the simple form
K =
1
2
3n−6
k=1
˙
q
2
k
, V =
1
2
3n−6
k=1
λ
k
q
2
k
(8.154)
For the linear molecule the summation extends to 3n − 5. In V there are no
terms present involving the cross-products of the coordinates q
k
. In the Hooke’s
law approximation the λ
k
are constants. By naively treating the q
k
as if they were
ordinary Cartesian coordinates, ignoring any interaction with rotational motion, the
wave equation can be written as
3n−6
k=1
∂
2
ψ
∂q
2
k
+
8π
2
h
2
E −
1
2
3n−6
k=1
λ
k
q
2
k
ψ = 0 (8.155)
This equation is separable into 3n − 6 (or 3n − 5) equations by using the
substitution
ψ = ψ
1
(q
1
)ψ
2
(q
2
) ···ψ
3n−6
(q
3n−6
) (8.156)
This leads to wave equations of the type
d
2
ψ
k
(q
k
)
dq
2
k
+
8π
2
h
2
E
k
−
1
2
λ
k
q
k
ψ
k
(q
k
) = 0 (8.157)
The total energy E is the sum of the energies E
k
associated with each normal
coordinate, i.e.
E =
3n−6
k=1
E
k
(8.158)
Here each equation is an ordinary differential equation in one variable. From a
comparison with the wave equation for the linear harmonic oscillator (8.151) and
the energy relation (8.153) we find that the eigenvalues for this problem are given
by
E
k
= hν
k
v
k
+
1
2
, v
k
= 0, 1,... (8.159)
and the ν
k
follow from
ν
k
=
1
2π
-
λ
k
(8.160)
This is the classical oscillation frequency of the normal vibration k and v
k
is
the vibrational quantum number. The total vibrational energy of the system of n
310 Absorption by gases
x
y
z
x
2
x
1
y
1
y
2
z
2
z
1
m
1
m
2
R
R sin
θ
ϕ
θ
Fig. 8.9 Coordinates of the rigid rotator.
particles is found by introducing (8.159) into (8.158). For further details see, for
example, Eyring et al. (1965).
8.4.4 Rotation of diatomic molecules
As an idealization of a diatomic molecule, we assume that the molecule consists of
two atoms rigidly connected by a weightless link of constant length R. In place of
the space coordinates (x
1
, y
1
, z
1
) and (x
2
, y
2
, z
2
) of the two point masses m
1
and
m
2
we are going to introduce the center of mass coordinates (x, y, z) of the system
and the spherical coordinates (r,θ,ϕ) of one particle referred to the other as origin.
The spherical coordinates are given by
x
2
− x
1
= R sin θ cos ϕ, y
2
− y
1
= R sin θ sin ϕ, z
2
− z
1
= R cos θ
(8.161)
were θ is the polar angle and ϕ the azimuth angle of the system, see also
Figure 8.9. The center of mass coordinates of the diatomic system are given by
x =
m
1
x
1
+ m
2
x
2
m
1
+ m
2
, y =
m
1
y
1
+ m
2
y
2
m
1
+ m
2
, z =
m
1
z
1
+ m
2
z
2
m
1
+ m
2
(8.162)
For brevity of notation we introduce the symbols a and b
a =
m
2
m
1
+ m
2
R, b =
m
1
m
1
+ m
2
R (8.163)
By eliminating the coordinates (x
2
, y
2
, z
2
) and then (x
1
, y
1
, z
1
) from (8.161) and
(8.162), we find
x
1
= x − a sin θ cos ϕ, y
1
= y − a sin θ sin ϕ, z
1
= z − a cos θ
x
2
= x + b sin θ cos ϕ, y
2
= y + b sin θ sin ϕ, z
2
= z + b cos θ
(8.164)
8.4 Vibrations and rotations of molecules 311
Thus the center of mass coordinates have been separated from the spherical coor-
dinates. Since we are not interested in the translational motion of the molecule in
space, we may regard the center of mass as fixed. In terms of the original coordinates
(x
1
, y
1
, z
1
) and (x
2
, y
2
, z
2
) the kinetic energy of the system is given by
K =
m
1
2
˙
x
2
1
+
˙
y
2
1
+
˙
z
2
1
+
m
2
2
˙
x
2
2
+
˙
y
2
2
+
˙
z
2
2
(8.165)
Using (8.164) the kinetic energy can also be written as
K =
m
1
a
2
+ m
2
b
2
2
dθ
dt
2
+ sin
2
θ
dϕ
dt
2
(8.166)
Setting the fixed center of mass coordinates (x, y, z) equal to zero, the moment of
inertia I about an axis through the center of mass and perpendicular to the molecular
axis is I = m
1
a
2
+ m
2
b
2
so that the kinetic energy can be written as
K =
I
2
dθ
dt
2
+ sin
2
θ
dϕ
dt
2
(8.167)
The quantum mechanical operator for the kinetic energy, see Table 8.1, can be
written down by replacing the Laplacian in rectangular coordinates by the Laplacian
in spherical coordinates. Thus the quantum mechanical Hamiltonian is given by
H =−
h
2
8π
2
m
1
r
2
∂
∂r
r
2
∂
∂r
+
1
r
2
sin θ
∂
∂θ
sin θ
∂
∂θ
+
1
r
2
sin
2
θ
∂
2
∂ϕ
2
+ V
(8.168)
Since no external forces are acting on the rotator we may set the potential energy
V = 0. Moreover, setting r = 1, mr
2
= m = I , we find that the Schr¨odinger equa-
tion is given by
1
sin θ
∂
∂θ
sin θ
∂ψ
∂θ
+
1
sin
2
θ
∂
2
ψ
∂ϕ
2
+
8π
2
IE
h
2
ψ = 0 (8.169)
which is a partial differential equation with two independent variables. We attempt
to solve this equation by separating the variables, i.e. we are looking for a solution
in the form
ψ = (θ)(ϕ) (8.170)
Substituting (8.170) into (8.169) gives
sin θ
∂
∂θ
sin θ
∂ψ
∂θ
+
8π
2
IE
h
2
sin
2
θ =−
1
∂
2
∂ϕ
2
(8.171)
312 Absorption by gases
Since the left-hand side of this equation depends on the variable θ , the right-hand
side on the variable ϕ, both sides must be equal to a constant, say M
2
. Thus we
obtain the two differential equations
d
2
dϕ
2
=−M
2
(8.172)
and
1
sin θ
∂
∂θ
sin θ
∂
∂θ
−
M
2
sin
2
θ
+
8π
2
IE
h
2
= 0 (8.173)
From (8.172) we immediately obtain the solution
(ϕ) = C exp(±iMϕ) (8.174)
where C is an integration constant. This is an acceptable solution provided that M
is an integer. This condition arises because the function (ϕ) must be single-valued
which implies
(ϕ) = (ϕ + 2π) or exp(iMϕ) = exp[iM(ϕ + 2π)] (8.175)
This requires that exp(i2π M) is unity which is possible only if M is an integer. It
is easy to show that the normalized function
M
is given by
(ϕ) =
M
(ϕ) =
1
√
2π
exp(±iMϕ), M = 0, 1,... (8.176)
In order to solve equation (8.173) we set x = cos θ and introduce the derivatives
d/dθ and d
2
/dθ
2
according to
x = cos θ,
d
dθ
=−sin θ
d
dx
,
d
2
dθ
2
= sin
2
θ
d
2
dx
2
− cos θ
d
dx
(8.177)
After some simple rearrangements we obtain
(1 − x
2
)
d
2
dx
2
− 2x
d
dx
+
8π
2
IE
h
2
−
M
2
1 − x
2
= 0 (8.178)
This equation has the form
(1 − x
2
)
d
2
u
dx
2
− 2x
du
dx
+
J (J + 1) −
M
2
1 − x
2
u = 0 (8.179)
which is known as the associated Legendre equation having the solution
u = P
M
J
(x) where P
M
J
(x) = (1 − x
2
)
M/2
d
M
dx
M
P
J
(x) (8.180)
8.4 Vibrations and rotations of molecules 313
The P
M
J
(x) are the associated Legendre polynomials and the P
J
(x) the ordinary
Legendre polynomials. Whenever M > J the function P
M
J
(x) = 0. We have dis-
cussed the orthogonality properties of the Legendre polynomials in Section 2.4
when treating the scattering problem. Equations (8.178) and (8.179) are identical
if
8π
2
IE
h
2
= J (J + 1) or E =
h
2
8π
2
I
J (J + 1), J = 0, 1,... (8.181)
Since M enters equation (8.178) only as M
2
we must have
J,M
=
J,−M
.
Solution functions that are finite, have integrable squares and are single valued,
exist only for conditions that J is zero or a positive integer and J ≥|M|. Thus the
correct normalized solution function is given by
(θ) =
|M|
J
(θ) =
&
(2J + 1)
2
(J −|M|)!
(J +|M|)!
P
|M|
J
(cos θ ) (8.182)
and the complete wave function by
ψ
J,M
=
M
(ϕ)
|M|
J
(θ) (8.183)
The allowed wave functions depend on the quantum numbers J and M which
are known as the rotational and magnetic quantum numbers, respectively. For every
value of J , there will be 2J + 1 values of M. For example, if J = 2, M can have
the values 0, ±1, ±2. This is called a (2J + 1) degeneracy. In the presence of an
electric or magnetic field, this degeneracy is removed if the molecule has an electric
or magnetic dipole moment, and the energy of the state will depend on M also. We
will return to this topic later.
In order that radiation may interact with the molecule to produce rotation or
that a rotating molecule may emit or absorb radiation, it is necessary that the
molecule possesses an electric moment implying that the molecule must have a
dipole moment. For this reason homonuclear molecules (having a symmetrical
charge distribution about their center of mass) do not have a pure rotation spectrum.
We will show later that the selection rule governing rotational transitions is given
by
J = J
− J
=±1 (8.184)
Thus the frequencies ν
R
absorbed or emitted by a rotating molecule correspond to
energy differences between adjacent energy levels. Denoting the rotational quantum
numbers J
and J
, with J
> J
, we find from equation (8.181) the following
relation
ν
R
=
h
8π
2
I
[J
(J
+ 1) − J
(J
+ 1)] =
h
4π
2
I
(J
+ 1), J
> J
(8.185)
314 Absorption by gases
or ˜ν
R
= ν
R
/c, that is division of the frequency ν by the speed of light c gives the
wave number ˜ν.
8.4.5 Vibration–rotation of diatomic molecules
If a molecule absorbs electromagnetic energy of sufficiently high frequency, both
vibration and rotation may occur simultaneously. While the rigid rotator consists
of two mass points connected by a massless bar, the nonrigid (vibrating) rotator
consists of two mass points which are connected by a massless spring. In order
to describe the nonrigid rotator mathematically, we make use of the Hamiltonian
operator (8.168) and find the following equation
1
r
2
∂
∂r
r
2
∂ψ
∂r
+
1
r
2
sin θ
∂
∂θ
sin θ
∂ψ
∂θ
+
1
r
2
sin
2
θ
∂
2
ψ
∂ϕ
2
+
8π
2
µ
h
2
[E − V (r)]ψ = 0
(8.186)
where µ is the reduced mass. By writing the wave function as the product
ψ = ψ
r
(r) (θ) (ϕ), (8.186) may be separated into three ordinary differential
equations. In the interest of brevity we omit mathematical details which can be
found in Eyring et al. (1965), Pauling and Wilson (1935), and in many other mod-
ern textbooks on quantum mechanics.
As before, solving the and equations leads to the introduction of the rota-
tional quantum number J. This is the Schr¨odinger equation for the r -component
of the wave function
1
r
2
∂
∂r
r
2
∂ψ
r
∂r
+
8π
2
µ
h
2
[E − V (r)] −
J (J + 1)
r
2
ψ
r
= 0 (8.187)
Introducing into this equation the potential energy function V (r) of the harmonic
oscillator, we find that the energy of the system is given by
E =
v +
1
2
hν
e
+ J(J + 1)
h
2
8π
2
I
−
J
2
(J + 1)
2
h
4
128π
6
ν
2
e
I
2
with ν
e
=
1
2π
&
k
µ
(8.188)
where µ, as before, is the reduced mass. The subscript e on the frequency symbol
refers to the equilibrium position of the molecule. Inspection of equation (8.188)
shows that the first term describes the vibrational energy of the molecule. The
second term is the energy of rotation, assuming that the molecule is rigid, while the
third term introduces a correction taking into account the stretching of the actual
nonrigid molecule due to the rotation.