Molecular Physics - Vrije Universiteit Amsterdam

72 Pages·2007·0.84 MB·English

Notes on:
Molecular Physics
2004
Prof. W. Ubachs
Vrije Universiteit Amsterdam
1. Introduction
1.1 Textbooks
There are a number a textbooks to be recommended for those who wish to study molecular
spectroscopy; the best ones are:
1)The series of books by Gerhard Herzberg
Molecular Spectra and Molecular Structure
I. Spectra of Diatomic Molecules
II. Infrared and Raman Spectroscopy of Polyatomic Molecules
III. Electronic Spectra of Polyatomic Molecules
2)Peter F. Bernath
Spectra of Atoms and Molecules
3)Philip R. Bunker and Per Jensen
Molecular Symmetry and Spectroscopy, 2nd edition
4)Hélène Lefebvre-Brion and Robert W. Field
Perturbations in the Spectra of Diatomic Molecules
A recent very good book is that of:
5) John Brown and Alan Carrington
Rotational Spectroscopy of Diatomic Molecules
1.2 Some general remarks on the spectra of molecules
Molecules are different from atoms:
- Apart from electronic transitions, always associated with the spectra of atoms, also purely
vibrational or rotational transitions can occur. These transitions are related to radiation by
multipole moments, similar to the case of atoms. While in atoms a redistribution of the elec-
tronic charge occurs in a molecule the transition can occur through a permanent dipole mo-
ment related to the charges of the nuclei.
- Superimposed on the spectral lines related to electronic transitions, there is always a rovi-
brational structure, that makes the molecular spectra much richer. In the case of polyatomics
threedifferentmomentsofinertiagiverisetorotationalspectra,indiatomicsonlyasinglero-
tational component. Each molecule has 3n-6 vibrational degrees of freedom where n is the
number of atoms.
-Atomscanionizeandionizationcontinuaarecontinuousquantumstatesthatneedtobecon-
sidered.Inmolecules,inaddition,therearecontinuumstatesassociatedwiththedissociation
ofthemolecule.Boundstatescancouple,throughsomeinteraction,tothecontinuaasaresult
of which they (pre)-dissociate.
1.3 Some examples of Molecular Spectra
Thefirstspectrumisthatofiodinevapour.Itshowsresolvedvibrationalbands,recordedby
the classical photographic technique, in the so-called B3P + - X1S + system observed in
0u g
absorption; the light features signify intense absorption. The discrete lines are the resolved
vibrationsintheexcitedstategoingovertothedissociativecontinuumatpointC.Leftward
of point C the spectrum looks like a continuum but this is an effect of the poor resolution.
This spectrum demonstrates that indeed absorption is possible (in this case strong) to the
continuum quantum state.
LatertheabsorptionspectrumwasreinvestigatedbyFourier-transformspectroscopyresult-
ing in the important iodine-atlas covering the range 500-800 nm. There is several lines in
each cm-1 interval and the numbers are well-documented and often used as a reference for
wavelength calibration. Note that the resolution is determined by two effects: (1) Doppler
broadeningand(2)unresolvedhyperfinestructure.Thefigureshowsonlyasmallpartofthe
iodine atlas of Gerstenkorn and Luc.
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The hyperfine structure can be resolved when Doppler-free laser spectroscopic techniques are
invoked. The following spectrum is recorded with saturation spectroscopy. A single rotational
line of a certain band is shown to consist of 21 hyperfine components. These are related to the
angular momentum of the twoI=5/2 nuclei in the I molecule.
2
R17(16-1)
*
-3000 -2000 -1000
Frequency, MHz
Usually molecular spectra appear as regular progressions of lines. In the vibrational bands of
diatomic molecules the rotational lines are in first order at equal separation. If a quantum state
is perturbed that may be clearly visible in the spectrum. This is demonstrated in the spectra of
two bands of the SiO molecule in the H1S +- X1S + system. The upper spectrum pertains to the
(0,0)bandandisunperturbed;theloweroneofthe(1,0)bandclearlyshownperturbationofthe
rotational structure.
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2. Energy levels in molecules; the quantum structure
2.1. The Born-Oppenheimer approximation
The Hamiltonian for a system of nuclei and electrons can be written as:
2 2
H = – -(cid:34)------(cid:229) (cid:209) 2–(cid:229) ----(cid:34)-------(cid:209) 2 + V(R,r)
2m i 2M A
i A A
wherethesummationireferstotheelectronsandAtothenuclei.Thefirsttermontheright
corresponds to the kinetic energy of the electrons, the second term to the kinetic energy of
the nuclei and the third term to the Coulomb energy, due to the electrostatic attraction and
repulsion between the electrons and nuclei. The potential energy term is equal to:
(cid:229) ZAe2 (cid:229) ZAZBe2 (cid:229) e2
V(R,r) = – ------------------- + ----------------------------------- + -----------------
4p e r 4p e R – R 4p e r
A,i 0 Ai A>B 0 A B i> j 0 ij
The negative terms represent attraction, while the positive terms represent Coulomb-repul-
sion. Note that a treatment with this Hamiltonian gives a non-relativistic description of the
molecule, in which also all spin-effects have been ignored.
Now assume that the wave function of the entire molecular system is separable and can
be written as:
Y (r ,R ) = y (r ;R)c (R)
mol i A el i nuc
wherey representstheelectronicwavefunctionandc thewavefunctionofthenuclear
el nuc
motion.Inthisdescriptionitisassumedthattheelectronicwavefunctioncanbecalculated
for a particular nuclear distanceR. Then:
(cid:209) 2y c = c (cid:209) 2y
i el nuc nuc i el
(cid:209) 2y c = y (cid:209) 2c +2((cid:209) y )((cid:209) c ) +c (cid:209) 2y
A el nuc el A nuc A el A nuc nuc A el
TheBorn-Oppenheimerapproximationnowentailsthatthederivativeoftheelectronicwave
function with respect to the nuclear coordinates is small, so (cid:209) y is negligibly small. In
A el
words this means that the nuclei can be considered stationary, and the electrons adapt their
positions instantaneously to the potential field of the nuclei. The justification for this origi-
natesinthefactthatthemassoftheelectronsisseveralthousandtimessmallerthanthemass
of the nuclei. Indeed the BO-approximation is the least appropriate for the light H -mole-
2
cule.
If we insert the separable wave function in the wave equation:
HY = EY
then it follows:
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HY = c (cid:237)(cid:236) – -(cid:34)----2--(cid:229) (cid:209) 2+ (cid:229) -------e---2------- – (cid:229) ----Z----A---e---2-----(cid:253)(cid:252) y +
mol nuc(cid:238) 2m i 4p e r 4p e r (cid:254) el
i i> j 0 ij A,i 0 Ai
+ y (cid:237)(cid:236) (cid:229) --------Z----A----Z----B---e---2--------- –(cid:229) ----(cid:34)---2----(cid:209) 2(cid:253)(cid:252) c = E Y
el(cid:238) 4p e R – R 2M A(cid:254) nuc total mol
A>B 0 A B A A
The wave equation for the electronic part can be written separately and solved:
(cid:237)(cid:236) – -(cid:34)----2--(cid:229) (cid:209) 2 + (cid:229) -------e---2------- – (cid:229) ----Z----A---e---2-----(cid:253)(cid:252) y (r ;R) = E (R)y (r ;R)
(cid:238) 2m i 4p e r 4p e r (cid:254) el i el el i
i i> j 0 ij A,i 0 Ai
for each value of R. The resulting electronic energy can then be inserted in the wave equation
describing the nuclear motion:
(cid:237)(cid:236) –(cid:229) ----(cid:34)---2----(cid:209) 2 + (cid:229) --------Z----A----Z----B---e---2---------(cid:253)(cid:252) c (R)+ E (R)c (R) = E c (R)
(cid:238) 2M A 4p e R – R (cid:254) nuc el nuc total nuc
A A A>B 0 A B
Wehavenowinacertainsensetwoseparateproblemsrelatedtotwowaveequations.Thefirst
relates to the electronic part, where the goal is to find the electronic wave function y (r ;R)
el i
and an energy E (R). This energy is related to the electronic structure of the molecule analo-
el
gouslytothatofatoms.Notethatherewedealwithan(infinite)seriesofenergylevels,aground
stateandexcitedstates,dependentontheconfigurationsofallelectrons.Bysearchingtheeigen
valuesoftheelectronicwaveequationforeachvalueofRwefindafunctionfortheelectronic
energy, rather than a single value.
Solution of the nuclear part then gives the eigen functions c (R) and eigen energies:
nuc
E = E – E (R) = E + E
nuc total el vib rot
IntheBO-approximationthenucleiaretreatedasbeinginfinitelyheavy.Asaconsequencethe
possibleisotopicspecies(HClandDCl)havethesamepotentialintheBO-picture.Alsoallcou-
plings between electronic and rotational motion is neglected (e.g.L -doubling).
2.2. Potential energy curves
The electrostatic repulsion between the positively charged nuclei:
2
Z Z e
V (R) = (cid:229) ------------A--------B---------------
N 4p e R – R
A>B 0 A B
is a function of the internuclear distance(s) just as the electronic energy. These two terms can
be taken together in a single function representing the potential energy of the nuclear motion:
V(R) = V (R)+ E (R)
nuc el
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In the case of a diatom the vector-character can be removed; there is only a single internu-
clear distance between two atomic nuclei.
In the figure below a few potential energy curves are displayed, for ground and excited
states. Note that:
-atsmallinternuclearseparationtheenergyisalwayslarge,duetotheedominantroleof
the nuclear repulsion
- it is not always so that de electronic ground state corresponds to a bound state
- electronically excited states can be bound.
V
V
R R
Electronictransitionscantakeplace,justasintheatom,iftheelectronicconfigurationinthe
molecule changes. In that case there is a transition form one potential energy curve in the
moleculetoanotherpotentialenergycurve.Suchatransitionisaccompaniedbyabsorption
oremissionofradiation;itdoesnotmakeadifferencewhetherornotthestateisbound.The
binding (chemical binding) refers to the motion of the nuclei.
2.3. Rotational motion in a diatomic molecule
Staring point is de wave equation for the nuclear motion in de Born-Oppenheimer approxi-
mation:
2
(cid:34)
– ------D +V(R) c (R) = Ec (R)
2m R nuc nuc
where,justasinthecaseofthehydrogenatomtheproblemistransferredtooneofareduced
mass. Note that m represents now the reduced mass of the nuclear motion:
M M
m = ---------A---------B-----
M + M
A B
Before searching for solutions it is interesting to consider the similarity between this wave
equation and that of the hydrogen atom. If a 1/R potential is inserted then the solutions (ei-
genvaluesandeigenfunctions)ofthehydrogenatomwouldfollow.Onlythewavefunction
c (R) has a different meaning: it represents the motion of the nuclei in a diatomic mole-
nuc
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cule. In general we do not know the precise form of the potential function V(R) and also it is
not infinitely deep as in the hydrogen atom.
AnalogouslytothetreatmentofthehydrogenatomwecanproceedbywritingtheLaplacian
in spherical coordinates:
D = --1---- ¶ (cid:230) R2 ¶ (cid:246) + --------1--------- ¶ (cid:230) sinq ¶ (cid:246) + --------1--------- ¶ 2
R R2¶ RŁ ¶ Rł R2sinq ¶ q Ł ¶ q ł R2sinq ¶ f 2
Now a vector-operatorN can be defined with the properties of an angular momentum:
(cid:230) ¶ ¶ (cid:246) (cid:230) ¶ ¶ (cid:246)
N = –i(cid:34) y – z = i(cid:34) sinf + cotq cosf
x Ł ¶ z ¶ xł Ł ¶ q ¶ f ł
(cid:230) ¶ ¶ (cid:246) (cid:230) ¶ ¶ (cid:246)
N = –i(cid:34) z – x = i(cid:34) (–cosf ) + cotq sinf
y Ł ¶ x ¶ zł Ł ¶ q ¶ f ł
(cid:230) ¶ ¶ (cid:246) ¶
N = –i(cid:34) x – y = i(cid:34)
z Ł ¶ y ¶ xł ¶ f
The Laplacian can then be written as:
2
D = --1---- ¶ (cid:230) R2 ¶ (cid:246) – ---N---------
R R2¶ RŁ ¶ Rł (cid:34)2R2
The Hamiltonian can then be reduced to:
– ----(cid:34)----2----- ¶ (cid:230) R2 ¶ (cid:246) + ------1-------N2+ V(R) c (R) = Ec (R)
2m R2¶ RŁ ¶ Rł 2m R2 nuc nuc
BecausethispotentialisonlyafunctionofinternuclearseparationR,theonlyoperatorwithan-
gular dependence is the angular momentum N2, analogously to L2 in the hydrogen atom. The
angular dependent part can again be separated and we know the solutions:
N2|N, Mæ = (cid:34)2N(N + 1)|N, Mæ with N = 0,1,2,3,etc
N |N, Mæ = (cid:34)M|N, Mæ with M = –N,– N + 1, N
z
Theeigenfunctionsfortheseparatedangularpartarethusrepresentedbythewell-knownspher-
ical harmonics:
|N, Mæ = Y (q ,f )
NM
and the wave function for the molecular Hamiltonian:
c (R) = ´ (R)Y (q ,f )
nuc NM
Inserting this function gives us an equation for the radial part:
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2
– ----(cid:34)--------- d (cid:230) R2 d (cid:246) + N(N + 1)+ V(R) ´ (R) = E ´ (R)
2m R2dRŁ dRł vN
Now the wave equation has no partial derivatives, only one variable R is left.
2.4. The rigid rotor
Now assume that the molecule consists of two atoms rigidly connected to each other. That
means that the internuclear separation remains constant, e.g. at a value R . Since the zero
e
pointofapotentialenergycanbearbitrarilychosenwechooseV(R )=0.Thewaveequation
e
reduces to:
1 2
---------------N c (R ) = E c (R )
2m R 2 nuc e rot nuc e
e
The eigenvalues follow immediately:
2
(cid:34)
E = ---------------N(N +1) = BN(N +1)
N 2m R 2
e
whereBisdefinedastherotationalconstant.Hencealadderofrotationalenergylevelsap-
pears in a diatom. Note that the separation between the levels is not constant, but increases
with the rotational quantum number N.
ForanHClmoleculetheinternuclearseparationisR =0.129nm;thisfollowsfromtheanal-
e
ysis of energy levels. Deduce that the rotational constants 10.34 cm-1.
N
This analysis gives also the isotopic scaling for the rotational levels of an isotope:
1
B(cid:181) ---
m
2.5 The elastic rotor; centrifugal distortion
In an elastic rotor R is no longer constant but increases with increasing amount of rotation
as a result of centrifugal forces. This effect is known as centrifugal distortion. An estimate
of this effect can be obtained from a simple classical picture. As the molecule stretches the
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centrifugal force F is, at some new equilibrium distance R ’, balanced by the elastic binding
c e
forceF , which is harmonic. The centripetal and elastic forces are:
e
2
F = m w 2R ¢ @ ----N---------- F = k(R ¢ – R )
c e m R ¢ 3 e e e
e
By equatingF =F and by assuming R ¢ » R it follows:
c e e e
2
N
R ¢ – R = -------------
e e m kR3
e
The expression for the rotational energy including the centrifugal effect is obtained from:
2
N 1
E = ----------------- + ---(R ¢ – R )
2m R ¢ 2 2 e e
e
Now use R ’ for the above equations and expanding the first term of the energy expression it
e
follows:
2 (cid:230) 2 (cid:246) –2 4 2 4
N N 1 N N N
E = -------------(cid:231) 1+ ------------(cid:247) + ------------------ @ ------------- – ------------------ +
2m R2Ł m kR4ł 2m 2kR6 2m R2 2m 2kR6
e e e e e
The quantum mechanical Hamiltonian is obtained by replacing N by the quantum mechanical
operator N. It is clear that the spherical harmonics Y (W ) are also solutions of that Hamilto-
NM
nian. the result for the rotational energy can be expressed as:
E = BN(N +1) – DN2(N + 1)2
N
where:
3
4B
e
D = ---------
w 2
e
isthecentrifugaldistortionconstant.Thisconstant isquitesmall,e.g. 5.32x10-4cm-1 inHCl,
but its effect can be quite large for high rotational angular momentum states (N4 dependence).
Selection rules for the elastic rotor are the same as for the rigid rotor (see later).
2.6. Vibrational motion in a non-rotating diatomic molecule
IfwesettheangularmomentumNequalto0intheSchrödingerequationfortheradialpartand
introduceafunctionQ(R)with ´ (R) = Q(R)⁄ R thanasomewhatsimplerexpressionresults:
2 2
(cid:34) d
– ------ +V(R) Q(R) = E Q(R)
2m 2 vib
dR
ThisequationcannotbesolvedstraightforwardlybecausetheexactshapeofthepotentialV(R)
isnotknown.Forboundstatesofamoleculethepotentialfunctioncanbeapproximatedwitha
quadraticfunction.Particularlynearthebottomofthepotentialwellthatapproximationisvalid
(see figure).
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