J. Phys. Chem. 1993,97, 2413-2416
2413
Can Molecules Have Permanent Electric Dipole Moments?
W. Klemperer,'++K. K. Lehmann,'**J. K. C. Watson,# and S. C. Wofsyl
Departments of Chemistry and Earth and Planetary Sciences, Harvard University, Cambridge,
Massachusetts 021 38, Department of Chemistry, Princeton University, Princeton, New Jersey, and
Herzberg Institute of Astrophysics, National Research Council, Ottawa, Canada
Received: September 24, 1992; In Final Form: December 22, I992
The character of the first-order Stark effect in the energy levels of real symmetric tops is examined. Group
theory shows that the only allowed levels are nondegenerate, and as a result time reversal symmetry requires
that such eigenstates not have a nonzero orientation required for a permanent electric dipole moment and thus
a first-order Stark effect. The terms responsible for the breaking of the nominal degeneracy of levels with fK
units of angular momentum around the symmetry axis are shown to derive from off-diagonal higher-order
distortion coupling (for ro-vibronic states of nondegenerate symmetry) and from spin-rotation and spin-spin
interactions (for degenerate ro-vibronic states). A semiclassical model demonstrates that both effects can be
viewed as caused by tunneling and as such decrease rapidly for increasing K. For a C3 symmetry symmetric
top, the resulting splittings are a few kilohertz for K = 1 but only on the order of 10-4 H z for K = 2.
Introduction
The Stark effect measures the interaction of a molecular dipole
moment with an electric field, while the Zeeman effect measures
the interaction of a molecular magnetic moment with a magnetic
field. While both electric and magnetic dipoles are rank one
tensor quantities, they have quite different symmetries. For
example, though magnetic moments are common, no elementary
particle can have a nonzero electric dipole moment by both parity
and time reversal symmetries.'J While violation of parity is
common in particle physics, violations of time reversal symmetry
have been established in only one particular case.3 Attempts to
measure a dipole moment for the free neutron, which provide a
test of theories of time reversal symmetry breaking, have so far
only found extremely small upper limitsS4 The same symmetry
arguments which predict that elementary particles should have
zero electric dipole moments apply to nondegenerate states of
molecules as well. In fact, experimentaltests looking for violation
of time reversal symmetry have included attempts to determine
a small static dipole moment in a pure rotational state of TIF.S
Some of the possible mechanisms that can produce such a static
dipole require mixing states of different parity. One would expect
higher sensitivities in molecules with nearly degenerate states of
different parity and time reversal symmetries,since, then, a weak
symmetryviolating interaction could induce a corresponding larger
mixing of the levels and thus a greater nonzero dipole moment.
As such, it may prove useful to these experiments to consider
symmetrictop molecules which are traditionallyviewed as having
true first-order Stark effects. In the discussion, we refer
particularly to field free motion and the energy changes that
Occur upon application of infinitesimal fields which negligibly
perturb the field free motion but provide a convenient coordinate
frame.
Whileall moleculesposessdegeneratelevelsdueto the isotropy
of space (labeled by the magnetic quantum number M), all these
levels have the same parity, and thus this degeneracy does not
in itself allow for an electric dipole moment. Some higher
symmetry is required. For example, the first-order Stark effect
of the hydrogen atom is an important and well-understood
Occurrence that reflects the higher dynamical symmetry of the
Authors for correspondence.
Department of Chemistry, Harvard University.
Princeton University.
f Herzbcrg Institute of Astrophysics.
1 Department of Earth and Planetary Sciences, Harvard University.
+
Coulombic potential. In particular, for the nonrelativistic spinfree hydrogen atom this symmetry is associated with the Lenz
vector, A = (1/2e2me)[L X p - p X L] + r/r being a constant
of thc motion.6 For the relativistic hydrogen atom there exist
degeneracies between states of the same total angular momentum,
j . The first-order Stark effect has been discussed in detail by
Bethe and Salpeter.' However, the Lamb shift demonstrates
that this extra symmetry is lifted when matter-radiation field
interactions are considered. Hyperfine structure would also
remove the exact degeneracies discussed here. In molecules, it
is well-known that rigid spin-free symmetric tops are expected
to have first-order Stark effects due to the degeneracy of the fK
levels. This degeneracy is geometrical in nature, a consequence
of the cylindrical symmetry of the inertial tensor being higher
than that required for the general rigid rotor, namely that of the
four element point group D2.
The present definition of a dipole moment is in conflict with
the traditional %hemicall definition of a dipole moment which
is based upon the point group symmetry of a rigid molecule. The
point group symmetry determines whether at the equilibrium
geometry a molecule can have a nonzero dipole moment in the
molecular axis system. However, the molecular axis system is
not static but rotates with the molecule. Externalfieldsareapplied
in the laboratory axis system, and it is the possible existence of
a static projection of the dipole moment in this coordinate system
with whichweareconcerncdin this paper. Ifonewantstoconsider
the effects of rotation, even for a rigid molecule, one must go
beyond the point group of a molecule and consider the molecular
symmetry group, which is a subgroup of the complete permutation-inversion group which includes all 'feasible" operation^.^
If one considers the full permutation-inversion group for a
molecule, it can be shown that the only energy levels allowed by
quantum statistics belong to one of two nondegenerate representations (i.e., there are no exact degeneraciesdue to symmetry
as discussed above for atomic hydrogen).*,9 Further, since these
two representations are of opposite parity symmetryL0for all
achiral molecules, the true molecular eigenstatesshould not have
first-order Stark effects. For a chiral molecule the states allowed
by the full permutation-inversion symmetry group are split into
levelsof opposite parity due to tunneling that converts the molecule
into its stereoisomer. The splitting due to this inversion can be
estimated and is believed to be typically smaller than parityviolating terms in the molecular Hamiltonian." Thus, by
considerationsof parity alone, one may not rule out the existence
0022-3654/93/2091-2413$04.00/00 1993 American Chemical Society
Klemperer et al.
2414 The Journal of Physical Chemistry, Vol. 97, No. 10, 1993
of a dipole moment in a chiral symmetric top (such as a molecule
with symmetry C3 for example).
When one considers the requirements of time reversal symmetry, however, even this possible exception can be ruled out.
Even if we consider only a permutation subgroup of the full
permutation-inversion group, the nuclear spin-ro-vibronic states
allowed by quantum statistics will still be nondegenerate representations. Time reversal symmetry will still be applicable
(though parity will not be for a chiral molecule), and by the same
argument that applies to elementary particles,2we can conclude
that such a nondegenerate state cannot have a dipole moment or
thus a first-order Stark effect.
The proof of this statement is as follows. Consider a molecule
in a state
(The index a labels degrees of freedom other than
angular momentum.) To have a nonvanishing expectation value
for the dipole moment operator requires that this dipole be along
a conserved vector. The standard argument examines the angular
momentum vector J , since rotation will average out any perpendicular components. Thus, an average dipole moment, given
by the expectation value of p in state J,a, must be proportional
to J.
= C(a)( # J a l J k J a )
Consider the effect of the time reversal operator. T-IpT = p
(because p is defined in terms of particle positions and charges
both of which are invariant under time reversal) while T-IJT =
-J. If the state JIJa is nondegenerare, the time reversal symmetry
of the Hamiltonian requires T$ja = eisJIJo. This equation will
still hold if two levels of stereoisomers are degenerate since time
reversal symmetry will not interconvert isomers (unlike parity).
The above equation relating ( p ) and ( J ) is consistent with ( p )
invariant under time reversal and ( J ) changing sign only if C(a)
is pure imaginary. Since both J and p are Hermitian, C(a)must
be real; thus C(a) = 0.
While the above arguments based upon symmetry are very
powerful, because they transcend any specific molecular model,
they are also very limited for the same reason. Thus, while they
prove that no molecule can have a true first-order Stark effect,
they do not give any indication as to what real physical effects
remove all degeneracies. Another way of interpreting the lack
of any dipole moment for an eigenstate is that, time averaged,
the molecule must spend equal time pointing up, K M positive,
or down, K M negative. But this gives no indication as to how
long, if we prepare the molecule with an on average up orientation,
it will stay that way. It is the magnitude of the splitting of this
IKM degeneracy that must be examined. For this purpose, we
need to explicitly look at a real molecular model. Consider the
prototypical polar symmetrical top, noncoplanar XY3. First we
will present a quantum mechanical treatment of the problem.
This will be followed by a semiclassical treatment, which we
believe gives a better physical model for why these effects occur.
OkaIoexplicitly stated that the spin-rotation interactions lead
to a lifting of all residual degeneracies, but to the best of our
knowledge no one has given an analysis of the exact terms
responsible for the nondegeneracy of the levels and how the
splitting patten depends upon rotational quantum numbers. Below
we show that, for a CH3Y molecules, the K = f l levels are split
by terms coming from both the anisotropy of the IH spin-rotation
interaction as well as from IH-Y spin-spin coupling. Both of
there terms are on the order of a few kilohertz. For states with
K = f 2 the coupling terms are second order in hyperfine
interactions and totally negligible. Thus, in this case (and higher
K = 3N f 1) the splittings predicted by symmetry are of no
dynamical consequence and molecular orientation will be preserved, even without external fields, for time scales longer than
any experiments currently feasible.
($Jobl$Ja)
Quantum Treatment
If we consider the rotational wave functions for XY3, we have
in the rigid rotor picture a degeneracy between levels with f K . I 2
Such states will have an average dipole moment proportional to
KM. If we can find any interaction that couples, directly or
indirectly, the states K,Mand -K,M, then the eigenstates will be
Wang functions ( l / ~ / Z ) ~ I J , K , M ) f I J , - K , MSuch
) ~ . states do
not havedipole moments. Thesplittingof the two Wang functions,
in hertz, can be interpreted as the rate at which the molecule
“flips” from pointing up to down and back up again, just as the
splitting of the symmetric and antisymmetric tunneling levels in
NH3 gives the rate of molecular inversion.
The simplest case to consider is where atoms Y have zero nuclear
spin, since in this case we do not have toconsider hyperfine effects.
However, for Zy = 0, the only states that are allowed by BoseEinstein statistics are those with K = 3N, N = an integer.g Those
rotational levels are nondegenerate, however, in the C3, point
group. As first shown by Nielsen and DennisonI3 in their
discussion of NH3, there exist off-diagonal sextic distortion
interactions (1 /2)h3(J46 which cause a lifting of the degeneracy.
In particular, 1
4 = 3 levels are directly connected by this
interaction and thus have a splitting first order in h3. h3 is on
the order of B S / w 4 my-3, where my is the mass of Y. For NH3,
the value of h3 = 350 Hz.I4 1
4 = 6 levels are split by second4 = 9 levels are
order coupling through the K = 0 level, while 1
split in third order by this interaction, etc. Thus, the splitting
goes up as a high power of J but decreases exponentially with K ,
a sign that the physical origin of the coupling is some type of
tunneling, much like the asymmetry splitting of an asymmetric
top. Thus, it is a quantum effect which disappears rapidly as the
mass or quantum numbers of the molecule increase. If one looks
at symmetric tops with a higher-order axis (C4,CS,C6, ...), the
same situation holds, namely, that the only states that are allowed
by BoseEinstein statistics are split by centrifugal distortion terms.
The higher the symmetry axis, the higher order the coupling, and
thus the smaller the characteristicsize of the coupling constant.
If we consider molecules for which ZY = I/*or higher, all values
of K are allowed by quantum statistics. We will focus on the
common case of Zy = ’ / 2 , but our conclusions will hold for higher
spin as well. The levels for K = 3N f 1 will remain degenerate
to all ,,den in the rotation-vibration Hamiltonian, since these
levels form an E symmetry representation in the molecular
symmetry group of the m ~ l e c u l e .These
~
levels are allowed by
quantum statistics, however, because they can be combined with
E symmetry nuclear spin functions to produce states with total
nondegenerate symmetry. Since both the ro-vibrational and
nuclear wave functions are E symmetry, it is natural to suspect
that it is the interaction of these degrees of freedom that removes
the degeneracy. In fact, the Hamiltonian for CH3D was worked
out by Wofsy, Muenter, and KlempererlSunder the assumption
of tetrahedral geometry. It was found that in a basis set where
only the three H nuclear spins are coupled together that there is
a coupling between levels with AK = f 2 ,
-
.
In this equation c = (2/9)(c, - cg)[B(CH,D)/B(CH4)] = 3.0
kHz, and c, and cg are the two spin-rotation interaction constants
for methane. c, is the spin-rotation interaction produced by
rotation with Jparallel to a given C-H bond, and cg is for rotation
perpendicular to that bond. MJ, M H ,and M D are the Z-axis
projections of J and the two nuclear spins angular momenta; p~
and p~ are the magnetic moments of H and D nuclei. Wofsy,
Muenter, and Klemperer used esu units and thus do not include
the factor of 4 ~ ~ Thus,
0 .
even without atom D, there is a splitting
of the K = f l levels due to the spin-rotation interaction. The
Can Molecules Have Permanent Electric Dipole Moments?
The Journal of Physical Chemistry, Vol. 97, No. 10, 1993 2415
second term reflects spin-spin interaction between the H and D
nuclear spins (R is the distancebetween the nuclei in the molecule),
with 3 p ” p ~ / 4 a p a 3 h= 9.28 kHz. Thus we see that the K = f l
levels will be split by a few kilohertz, which implies that the
orientation of such a molecule in zero field will only be preserved
for a fraction of a millisecond. This is a time scale which can
in principle be observed in the laboratory but requires that stray
fields be reduced below -1 mV/cm. If we consider higher K
values, they have no direct coupling by either the spin-spin or
spin-rotation operators (which are second-rank tensors), but they
will be coupled in higher order through order K levels. For
example, K = 2 will be coupled through the K = 0 levels. This
will produce a splitting that is smaller by the ratio of the K = 1
splitting to 4(A - B) (the K = 0 and 2 splitting). For CH3D, this
ratio is on the order of
and thus it will take about 1 h for
K = 2 molecules to flip their orientation! Thus, this effect is
totally negligible for all but K = 1 levels.
Semiclassical Interpretation of K-Doubling Effects
While the above discussion provides a correct quantummechanical description as to how the apparent time-independent
molecular orientation of a symmetric top is lost due to centrifugal
and hyperfine interactions, it is appealing to have a semiclassical
interpretation of this effect as well. In this section, we provide
such a picture in the frameworkof motion on the rotational energy
surface (RES), which has been used by Harter and Patterson16
to provide a physical visualization of rotational dynamics. As
suggested by the quantum-mechanical treatment, the effects can
be understood as due to tunneling.
Consider a symmetric top which will be treated as prolate,
although nothing below depends upon this choice. The classical
rotational energy can be written as
+
E(J,O,X) = J2[Acos2((?) B sin2(0)]
where 8,x are the polar coordinates that determine the direction
of the instantaneous angular momentum vector in the molecular
axis system. For a given magnitude of angular momentum, in
each direction 8,x, we draw a point at a distance E(J,O,X). This
traces out the RES, which is a surface of revolution in the present
case. Classically, the rotational motion of the symmetric top is
represented by precession that scribesa circle on the RES. (More
generally, the curve is the intersection of the constant angular
momentum RES and a constant energy sphere.) Quantum states
with a specific are a pair of quantizing circlesthat make angles
given by cos(8*K) = fK/[J(J + 1)] as shown in Figure 1. All
points on the RES surface have the same magnitude of angular
momentum but differ in the direction of the angular momentum
in the molecular axis system.
In general, we can expect tunneling between these two
degenerate states, as shown in Figure 1. For a given value of x ,
the minimum tunneling path will take us to the same x , Le., be
a curve of constant longitude on Figure 1. For each x , we expect
a tunneling amplitude t(X) proportional to [Harter and Paterson,
eqs 4.2 and 4.41
In this formula, Jr is the angular momentum conjugate to the
tunneling path, which will be on the equator for the minimal
tunneling path y. Note that along the tunneling path y is real
while Jr and the other angle used to describe the direction of J
( p in the notation of Harter and Patterson) are imaginary. JT
is found by solving eq 4.6 of Harter and Patterson, which is
E =A(J~
- J,’) cos2 7 + B ( J ~- J,’) sin2 7 + CJ,’
-
where E is the rotational energy of the state J,K and B = C for
a symmetrictop. As y a/2(which is when the tunneling path
crosses the equator), the value of .
I
,
goes to i times infinity and
Figure 1. Rotational energy surface (RES)for a prolate symmetric top.
The quantum states, J,K are shown for J = 8. 6 K = cos-’ [ K / ( J ( J
1))’/*] is shown for K = 7. Tunneling paths along arcs of constant x are
shown for both K = f3 and f7. This figure was kindly provided by Prof.
William G. Harter.
+
we get a logarithmic divergence of the tunneling integral, and
thus the tunneling rate is zero. The reason for this divergence
is simply that coefficient of Jr is zero in the above equation for
y = a/2. It can be shown that addition of terms diagonal in J
and Kwill not change this situation; i.e., the fKlevels will remain
exactly degenerate.
If we consider a rigid asymmetric top, C < B, then the value
of Jr stays finite for y = a/2 if we consider tunneling that takes
the rotational angular momentum over the B rotational axis. We
have a tunneling path of equal amplitude if we go over the -B
rotation1 axis as well (see Harter and Patterson, Figure 4). We
must add the amplitude of these two paths, but it can be shown
that they will constructively interfere for any K.
Now let us consider a distortable C3 rotor with the (l/2)h3(Jk)6 terms added to the rotational Hamiltonian. Now the
classical“trajectories”of each K will dip down toward the equator
six times on each precession around the symmetry axis. This will
lead to six equivalent tunneling paths on each cycle of the motion.
As each tunneling path goes across the equator, Jr has a finite
value given approximately by i[(A - B)E/h3]1/6. As a result,
the tunneling integral is convergent, and we have a nonzero
tunneling amplitude for each of the six equivalentpaths. However,
when one adds up the contributions to each path, it can be shown
that for K = 3N the contributions constructively interfer and we
get a nonzero tunneling rate, while for K = 3N f 1 there is exact
destructive interference of the six contributions and the total
tunneling rate is exactly zero.
Let us now considerthe effect of includinghyperfine interactions
in the rotational Hamiltonian. If all spins point in the same
direction, then the sum of the hyperfine interactions give
contributions that depend upon cos2 y but are independent of x .
These terms are equivalent to a slight change in the A and B
rotational constants but do not change the cylindrical symmetry
of the RES and thus do not change the picture given earlier for
a symmetric top. If, instead, one of the proton spins points up
and the other two point down (or visa versa), then both the spin
rotation and H-D spin-spin interactions lead to a contribution
2416
Klemperer et al.
The Journal of Physical Chemistry, Vol. 97,No. 10, I993
to the energy proportional to sin2 y cos2 x . These terms have the
same functional form as making B - C nonzero. The direction
of the B and C axes will be such that one passes through the
proton with unique spin. As a result of this effective asymmetry,
just as for the true asymmetric top treated earlier, one will get
a finite tunneling integral, and constructive interference of the
two equivalenttunneling amplitudes. This will allow for a splitting
of the otherwise exactly degenerate K = 3N f 1 levels. It is
exactly for theserotational levels that quantum statistics demands
that the allowed nuclear spin functions have one spin pointing
opposite to the other two.
The conclusionof this section is that the spectroscopicsplittings
that prevent a truly static orientation, and thus dipole moment,
have a straightforward semiclassical interpretation. Because of
spin-rotation and spin-spin interactions, a state with the spins
of equivalent atoms pointing in different directions will behave
like a very slightly asymmetric top.
Conclusions
The present paper has presented an analysis of whether a true
first-order Stark effect, or equivalently a static orientation, is
expected in a symmetric top molecule. Group theory provides
a rigorous answer of "no". But such an answer makes no reference
to the time scales for the loss of orientation; if they exceed the
time scales of an experiment that depends upon orientation, the
molecular orientation will appear "static". Likewise, once the
Stark coupling matrix element between nearby states exceeds
their energy separation, the spontaneous reorientation will be
quenched and a quasi-first-order Stark effect will be observed.
The loss of orientation is due to a quantum tunneling phenomenon
and as such is important only for the very lowest levels. It is
interesting, however, that for CH3Xmolecules,even with extreme
cooling of the rotational degrees of freedom in a supersonic jet,
about one-half the molecules are in the J = K = 1 levels. These
molecules can be readily oriented by an electrostatic hexapole
focuser. Such molecules can then be extracted and used for
orientational studies of scattering or reactions. It is clear that,
for the fields that exist within the hexapole focusers,the hyperfine
splittings discussed above will be negligibly small. The present
analysis shows, however, that in a field-free environment the
orientation of such states will be averaged out in a time scale of
a fraction of 1 ms. If K = 2 levels are populated in the expansion,
in a true field-free region, these molecules should retain their
initial orientation for time scales much longer than any practical
experiment. In recent experiments,Gandhi and Bernstein showed
that they could maintain molecular orientation of CHJ only in
the presence of a static field of at least 300 mV/cm.17 This is
much higher than would be estimated by the above predicted
splittings. Further, they observed a loss of orientation for both
the (J,K,M) = ( l , l , l ) and (2,2,2) states at about the same field.
Based upon these facts, it is likely that the loss of orientation they
observed reflects not a true spontaneous reorientation but a
reorientation induced by precession about stray electric fields or
by nonadiabatic transitions as molecules leave or enter regions
of significant fields.
Note Added in hoof. The matrix elements of the hyperfine
operator, off-diagonal in K, have been discussed by I. Ozier and
W.L.Meerts(Can.J. Phys. 1981,59,15O)inthecontextoftheir
Stark crossing experiments on symmetric tops. We thank Prof.
Oka for bringing this reference toour attention. We have received
a manuscript from R. J. Butcher, Ch. Chardonnet, and Ch. J.
Borde who have reached a conclusion similar to the present work.
Acknowledgment. We wish to thank Professors Ian Mills and
Takeshi Oka for valuablesuggestionsand Prof. William G. Harter
for helpful discussions and for providing the figure presented in
this paper. This work was supported by the National Science
Foundation and the donors of the Petroleum Research Fund,
administered by the American Chemical Society.
References and Notes
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