# Why do molecules lack the inversion symmetry of the full molecular Hamiltonian?

The nonrelativistic molecular Hamiltonian has inversion symmetry, since the kinetic energy operator and the Coulomb operator have inversion symmetry,

\begin{aligned} \hat p_i^2&\stackrel{i}{\rightarrow} (-\hat p_i)^2 = \hat p_i^2\\ \frac{1}{|\hat r_i- \hat r_j|} &\stackrel{i}{\rightarrow} \frac{1}{|\hat r_j- \hat r_i|}=\frac{1}{|\hat r_i- \hat r_j|}\\ \hat H &= \sum_i \hat p_i^2/2m_i + k\sum_{i>j}\frac{Z_iZ_j}{|\hat r_i- \hat r_j|} \end{aligned} where $$i$$ and $$j$$ run over all particles, i.e., electrons and nuclei.

This means that energy eigenstates should have inversion symmetry as the Hamilton operator commutes with the inversion operator.

This begs the question how molecules without inversion symmetry and constant dipole moments can exist.

What is the connection between molecular structures and the eigenvalues of the molecular Hamiltonian and what causes the symmetry breaking, since we can clearly observe stable molecules that have constant dipole moments.

EDIT:

My reasoning above was based on the assumption that a ground state is also a eigenstate of the parity operator. The discussion in the comments made me realize that this is not the only option. In the case of degeneracy of the parity eigenstates, there is the possibility of a superposition state. Let

\begin{aligned} \hat i|\psi_n^u\rangle &=-1|\psi^u_n\rangle \\ \hat i|\psi_n^g\rangle &=+1|\psi^g_n\rangle \\ \hat H|\psi_n^g\rangle &=E_n|\psi^g_n\rangle \\ \hat H|\psi_n^u\rangle &=E_n|\psi^u_n\rangle \end{aligned}

Then we can form states with energy $$E_n$$ of the form $$|\psi_n\rangle \equiv c_u|\psi_n^u\rangle +c_g|\psi_n^g\rangle$$, where $$g/u$$ stand for even/odd parity.

The expectation value of the dipole operator $$\hat d$$ of such a state contains terms that have overall even symmetry, $$c^*_uc_g\langle \psi_n^u|\hat d| \psi_n^g\rangle + h.c.$$ This means that a ground state with a nonzero dipole moment is possible.

But there is still the question what exactly fixes the coefficients $$c_u,c_g$$ so that we have a ground state with a clearly defined dipole moment. It cannot be energy or minimization of energy, since we started with the assumption of degeneracy, so any combination of $$c_u,c_g$$ will have the same energy.

While I have shown that a ground state with a dipole moment is possible under inversion symmetry, I am still unclear how a particular molecule with a specific dipole moment realizes.

• The symmetry just says that the molecule and its mirror image are eigenvalues of the Hamiltonian of the same energy – not that there are only symmetric solutions. The same holds, e.g. for the rotation symmetry of the hydrogen Hamiltonian. You can choose your solutions as inversion even/inversion odd by linear combination of the usual non-symmetric solutions. Commented Dec 18, 2023 at 16:59
• @SebastianRiese What do you mean with solutions? The observable ground states should be energy eigenstates and these eigenstates are also eigenstates of the inversion operator. This means that they are even or odd. In either case the expectation value of the dipole operator will be zero since the expectation value integral will be odd overall due to the odd parity of the dipole operator. The fact that you can expand states without inversion symmetry into this basis does not resolve this point. Commented Dec 18, 2023 at 18:53
• Possibly useful: Ammonia electric dipole moment according to Philip Warren Anderson and links therein. Commented Dec 18, 2023 at 19:41
• @SebastianRiese I was wrong to assume that the energy eigenstates have to be eigenstates of the parity operator, thank you for pointing that out. You are correct that we can have any mixture in the degenerate case. My mind was still stuck in point group tables where we usually have further symmetry operations that lift the degeneracy with respect to inversion. Commented Dec 18, 2023 at 22:17
• I do not have the time to properly answer now (maybe later), but consider checking Sec. 2.2 of this paper, and its Refs. [19-36] (mainly the work of Woolley). Commented Dec 21, 2023 at 11:40

First of all, there is an incorrect statement in the question "the eigenstates should have the inversion symmetry." This is not correct. Rather, the eigenstates should transform under some representation of the inversion symmetry. This isn't just a nitpick, the answer to the full question partially lies in this misconception. Consider for example the $$2\mathrm{P}$$ state of the hydrogen atom (ignoring electron spin for now), which transforms as a the "triplet representation" under rotations. The three eigenstates of the Hamiltonian with different $$L_z$$ values are NOT symmetric under rotations, as the Hamiltonian is. But they transform like a representation of the rotation group. That is, there is a well-defined way to rewrite a linear combination of $$L_z$$ eigenstates after rotating your coordinate system (the Wigner D matrices I think?).

Now for the bigger, more important question - how can molecules have permanent dipole moments. This is critical for your understanding of quantum mechanics: In free space, the groundstate of a molecule DOES NOT have a dipole moment which is oriented in a specific direction in space. However, in the sense that I'll describe in this paragraph, it still has a dipole moment. In the absence of an external electric field, the grounstate of a molecule will indeed be symmetric under rotations. And its energy will increase quadratically with the electric field strength. Note that except with a different central force potential, different charges, and much different masses, a negatively charged ion orbiting a positively charged one is the same thing as a hydrogen atom. However, rotational excited states will have energies that change linearly with the applied electric field. This comes from a term in the Hamiltonian of the form $$p\cdot \mathbf{E}$$, where $$p=q(r_1-r_2)$$, and through degenerate perturbation theory, rotationally excited states will have linear combinations which do have a polarization. But the energy difference between the rotational groundstate and the excited states is much smaller than the energy level difference between the $$1\mathrm{S}$$ state of hydrogen and the $$2\mathrm{P}$$ state. And mixing between the groundstate and the excited states is what gives rise to a regime where the energy starts to change linearly $$|\mathbf{E}|$$. So how strong the electric field needs to be before the quadratic behavior ends is much lower for the molecule than for the hydrogen atom, and the electric field needed to start observing a roughly constant dipole moment is much lower. Therefore, with any reasonable electric field you almost immediately see a linear response in the molecule, indicating the presence of a permanent dipole moment. And unlike in hydrogen, that dipole moment changes very weakly with the strength of the applied electric field, because its magnitude is mostly set by the bond length which is mostly independent of the rotational quantum numbers. In the hydrogen atom groundstate, the dipole moment needs to be first created by the electric field ($$p=\alpha \mathbf{E}$$ for the hydrogen $$1\mathrm{S}$$ state up to very high electric fields).

You may be surprised to find that a Hydrogen atom in the $$2\mathrm{P}$$ state can also have a polarization, even though the $$2\mathrm{P}$$, $$L_z=0$$ state says the electron is equally likely to be in any direction, and the two other $$L_z$$ eigenstates alone do not make an electric dipole moment nonzero expectation value. This is to say that the energy of the three $$2\mathrm{P}$$ states separate linearly with the magnitude of the applied electric field. This is because there are linear combinations of the three degenerate $$L_z$$ eigenvalues which do have a charge distribution (the electron is more likely to be on one side of the proton than the other). So when you apply an electric field, with degenerate perturbation theory, you find that the electric field determines which linear combinations are still eigenvalues of the Hamiltonian including the electric field.

The apparent contradiction is the following. The general non-relativistic Hamiltonian \begin{aligned} {\cal{H}}&= -\sum_{a=1}^{\cal{N}}\frac{1}{2m_a}\nabla_{\rho,a}^2+\sum_{\substack{a,b=1 \\ a is invariant under inversions $$\vec{\rho}_a\rightarrow-\vec{\rho}_a$$ (that is, $$(\vec{r}_i,\vec{R}_A)\rightarrow(-\vec{r}_i,-\vec{R}_A)$$), meaning its non-degenerate energy eigenstates are also parity eigenstates which cannot carry a permanent dipole moment: $$\vec{\mu}=\langle\Psi|\sum_{A=1}^{N_{\text{nuc}}}Z_A\vec{R}_A-e\sum_{i=1}^{N_{\text{e}}}\vec{r}_i|\Psi\rangle=0 \ .$$ However, at the same time, the clamped nucleus approximation of $$\cal{H}$$ (esentially the first half of the Born-Oppenheimer separation: letting $$M_A\rightarrow\infty$$ and treating $$\vec{R}$$ as parameters) leads to \begin{aligned} H(\vec{R})= -\sum_{i=1}^{N_{\text{e}}}\frac{1}{2m}\nabla_i^2+\sum_{\substack{i,j=1 \\ i which does give rise to a dipole moment: $$\vec{\mu}=\sum_{A=1}^{N_{\text{nuc}}}Z_A\vec{R}_A-e\langle\Phi(\vec{R})|\sum_{i=1}^{N_{\text{e}}}\vec{r}_i|\Phi(\vec{R})\rangle=\sum_{A=1}^{N_{\text{nuc}}}Z_A\vec{R}_A-e \int\mathrm{d}^3r\vec{r}\rho(\vec{r};\vec{R}) \ .$$ In the above, $$\vec{R}$$ refers to the (parametric) dependence on all the nuclear coordinates, and $$\rho$$ is just the electronic density that can be calculated from $$\Phi$$. Neglecting any vibrational effects and simply using the equilibrium (minimal energy) nuclear configurations gives us permanent dipole moments that are in rather good agreement with experiment. For example[1][2]: \begin{aligned} \mu_{\text{calc}}(\text{HCl})\approx 1.084 \, \text{D} \ \ \ &\leftrightarrow \mu_{\text{exp}}(\text{HCl})\approx 1.093 \, \text{D} \ , \\ \mu_{\text{calc}}(\text{H_2O})\approx 1.840 \, \text{D} \ \ \ &\leftrightarrow \mu_{\text{exp}}(\text{H_2O})\approx 1.857 \, \text{D} \ . \end{aligned} How could we generate a non-zero dipole moment from the clamped nucleus approximation, and how can such a value be extracted from the general formalism? The punchline is that dipole moment is only zero in a laboratory-fixed frame due to free rotations/inversions. Clamping the nuclei, however, picks out a molecule-fixed frame (position and orientation being fixed by the point charges of the nuclei), in which a non-zero dipole moment can be found; this corresponds to the fact that a molecule of $$N_\text{nuc}$$ nuclei has only $$3N_\text{nuc}-6$$ internal degrees of freedom ($$3N_\text{nuc}-5$$ when linear). The question is whether such a molecule-fixed frame can be found without clamping the nuclei.

The first thing to realize is that the kinetic energy associated with center-of-mass motion must be excluded from $${\cal{H}}$$, otherwise the spectrum would be continuous, and no internal motion could be described: $${\cal{H}}=-\frac{1}{2M_{\text{tot}}}\nabla^2_{\text{COM}}+H_\text{in} \ .$$ There are, however, an infinite number of coordinate transformations that can achieve this[3][4]! Any invertible linear transformation $$\vec{\rho}'_a=\sum_{b=1}^{\cal{N}}t_{ab}\vec{\rho}_b$$ satisfying the additional conditions $$t_{1b}=\frac{m_b}{M_\text{tot}} \ \ \ , \ \ \ \sum_{b=1}^{\cal{N}}t_{ab}=\delta_{a1}$$ is sufficient to completely decouple the motion of $$\vec{\rho}'_1=\vec{\cal{R}}_{\text{COM}}$$.

Let us find the appropriate molecule-fixed frame for diatomic molecules; the rest of my answer is limited to this case. The internal coordinates are specifically \begin{aligned} \vec{\rho}'_1&=\vec{R}'_1=\vec{\cal{R}}_{\text{COM}} \ , \\ \vec{\rho}'_2&=\vec{R}'_2=\vec{R}_2-\vec{R}_1 \ , \\ \vec{\rho}'_{i+2}&=\vec{r}'_i=\vec{r}_i-\frac{1}{2}(\vec{R}_1+\vec{R}_2) \ , \end{aligned} so that an internuclear vector $$\vec{R}=\vec{R}'_2-\vec{R}'_1$$ emerges, and all electron coordinates are measured from the geometric center of the molecule (it is easy to check that this is a special case of the above general transformation). This transformation leads to $$H_\text{in}=-\frac{1}{2M_+}\nabla_R^2-\frac{1}{2m}\sum_{i=1}^N{\nabla'}_i^2 -\frac{1}{8M_+}\sum_{i,j=1}^N{\nabla'}_i\cdot{\nabla'}_j+\frac{1}{2M_-}\nabla_R\cdot\sum_{i=1}^N{\nabla'}_i+V \ ,$$ where we introduced $$\frac{1}{M_\pm}=\frac{1}{M_2}\pm\frac{1}{M_1} \ .$$ Note that this frame is space-fixed in the sense that its orientation is still tied to that of the original laboratory-fixed frame. To finally have the molecule-fixed frame, we write $$\vec{R}=R\cos(\phi)\sin(\theta)\vec{e}'_x+R\sin(\phi)\sin(\theta)\vec{e}'_y+R\cos(\theta)\vec{e}'_z \ ,$$ and switch to a new coordinate system whose "$$z$$" axis coincides with $$\vec{R}$$ in the primed coordinate system: \begin{aligned} \vec{e}''_x&=\frac{\partial_\theta\vec{R}}{|\partial_\theta\vec{R}|}=\cos(\phi)\cos(\theta)\vec{e}'_x+\sin(\phi)\cos(\theta)\vec{e}'_y-\sin(\theta)\vec{e}'_z \ , \\ \vec{e}''_y&=\frac{\partial_\phi\vec{R}}{|\partial_\phi\vec{R}|}=-\sin(\phi)\vec{e}'_x+\cos(\phi)\vec{e}'_y \ , \\ \vec{e}''_z&=\frac{\vec{R}}{R}=\cos(\phi)\sin(\theta)\vec{e}'_x+\sin(\phi)\sin(\theta)\vec{e}'_y+\cos(\theta)\vec{e}'_z \ . \end{aligned} The notation is somewhat confusing, since these are actually the $$\vec{e}_\theta,\vec{e}_\phi,\vec{e}_R$$ unit vectors of the spherical coordinate system, but this is the convention set by Kolos and Wolniewicz[5] that everyone seems to follow. Using $$\vec{r}'_i=x'_{i}\vec{e}'_x+y'_{i}\vec{e}'_y+z'_{i}\vec{e}'_z= x''_{i}\vec{e}''_x+y''_{i}\vec{e}''_y+z''_{i}\vec{e}''_z$$ the components in the space-fixed (primed) and molecule-fixed (double-primed) systems can be seen to related: $$\begin{bmatrix} x_i'' \\ y_i'' \\ z_i'' \end{bmatrix} = \begin{bmatrix} \cos(\phi)\cos(\theta) & \sin(\phi)\cos(\theta) & -\sin(\theta) \\ -\sin(\phi) & \cos(\phi) & 0 \\ \cos(\phi)\sin(\theta) & \sin(\phi)\sin(\theta) & \cos(\theta) \end{bmatrix} \begin{bmatrix} x_i' \\ y_i' \\ z_i' \end{bmatrix} \ ,$$ and with this in hand, the Hamiltonian can be transformed to the molecule-fixed coordinates. The dipole moment operator is defined in this very frame: $$\hat{\vec{\mu}}=\frac{Z_2-Z_1}{2}R\vec{e}''_z-e\sum_{i=1}^n\vec{r}''_i \ ,$$ with the most important property that it does not flip sign upon inversion in the space-fixed coordinate system. The angles change as $$\theta\rightarrow\pi-\theta$$, $$\phi\rightarrow\pi+\phi$$ upon inversion, which means that[5][6] $$(x_i',y_i',z_i')\rightarrow(-x_i',-y_i',-z_i') \ \ \ \Leftrightarrow \ \ \ (x_i'',y_i'',z_i'')\rightarrow(-x_i'',+y_i'',+z_i'') \ .$$ The dipole moment thus does not vanish by symmetry in the molecule-fixed frame and it can in principle be calculated as an expectation value with the eigenstates of $$H_\text{in}$$ (one can show that it does vanish for the homonuclear case $$Z_1=Z_2$$, $$M_1=M_2$$, as it should).

The main (or basically only) application of this formalism was the perturbative calculation of the dipole moment for the hydrogen deuteride (HD) molecule. The clamped nucleus approximation could only predict $$\vec{\mu}(\text{HD})=\vec{\mu}(\text{H}_2)=\vec{\emptyset}$$, since it is insensitive to isotopic mass differences; in reality, HD does have a tiny dipole moment of roughly $${\mu}_\text{exp}(\text{HD})\approx8.78\cdot10^{-4} \, \text{D}$$. Blinder, Kolos, Wolniewicz and many other workers of the 60s-80s calculated this to be roughly $${\mu}_\text{calc}(\text{HD})\approx8\cdot10^{-4} \, \text{D}$$ with the above approach, although the results did not (and to my knowledge, still do not) agree about the second significant digit.

See Ref. [4] about the details of this technique and for the references to the aforementioned original works. See also Ref. [7] for current experimental and theoretical values, and for an alternative post-Born-Oppenheimer calculation.

This answer is incomplete in the sense that I only dealt with diatomic molecules. Molecule-fixed frames can be found for more than two nuclei as well, but I do not know of any dipole moment calculation for such systems (I doubt anyone actually tried it, if nothing else, due to the severe computational cost). Also, for polyatomic molecules, questions like this one are closely related to the question of whether molecular structure is a meaningful concept beyond the Born-Oppenheimer framework, which is not settled, to say the least.

References

 [1]: NIST database

 [5]: Kolos, Wolniewicz: Rev. Mod. Phys. 35 3 473 1963

 [6]: Landau, Lifshitz: Quantum Mechanics $$-$$ Non-relativistic Theory; Sec. 86.

• Great answer. It is strange that the subtleties of the separation of coordinates are often swept under the rug when the BO approximation is taught, although it is such an essential part for the theoretical framework of chemistry and the concept of a molecule in general. Commented Jan 12 at 9:25
• There is no mention of $\psi^{g}$ and $\psi^{u}$!!! the basis of the question... Commented Jan 12 at 17:12
• @TheTiler What do you mean? I explicitly stated that non-degenerate eigenfunctions of ${\cal{H}}$ are also parity eigenfunctions (hence either of $g$ or of $u$ symmetry)... Commented Jan 12 at 17:25
• "The expectation value of the dipole operator $\hat d$ of "such a state contains terms that have overall even symmetry, $c^*_uc_g\langle \psi_n^u|\hat d| \psi_n^g\rangle + h.c.$ This means that a ground state with a nonzero dipole moment is possible. "the question concerns the invariance of the Hamiltonian by inversion and the cause of the dipole for $\psi^{g}....$ Commented Jan 12 at 17:39
• I have nothing against your answer... :-) Commented Jan 12 at 18:06

Generally, permutations of particle coordinates of particles of different kind don't qualify as a symmetry. The Hamiltonian has subdivided into symmetric sub-sums of 1-particle operators, symmetric pair operators in its special tensor product of identical particles, and the sum over pairs of such subgroups.

The conventional corpus of quantum theory books avoids the fact, that all operators in such a mixed system have to be completed with a tensorial map of identities in all other particle coordinates, e.g. the Schrödinger operator of $$NH_4$$ has th full Form

$$\mathbb H = \frac{P_N^2}{2M_N} \otimes 1^{\otimes 3} + 1\otimes \frac{p_{H\, 1}^2}{2m_H} \otimes 1^{\otimes 3 } \dots + V_N(X_N-x_{H,1})\otimes 1^{\otimes 3} +\dots + 1\otimes V_{H,H}(x_1-x_2)\otimes 1^{\otimes 2}\dots$$

This notation is suited to work on a tensor product of 5 Hilbert spaces, the four electron subspace represented as the antisymmetric subspace of the free 4-tensor product. Spin complicates the situation, notationally, too.

The basis sets of states of the one particle spaces are taken from an eigenvalue problem, wrt. symmetries somehow near the actual problem. With respect to accuracy of approximations, without much success beyond the classical solvalble 3-body problems with a high symmetry.

The state of a system has to transform as a representation of the symmetry group of the Hamiltonian. That presentation is not necessarily the fully symmetric one. For molecules you need to consider representations of point groups. See e.g. https://www.chem.uci.edu/~lawm/10-2.pdf

Note after reading comments: If you would solve the full hamiltonian you would find superpositions of combined nuclear and electronic states that transform according to the representations of the point group $$C_i$$. Point groups only require a fixed position of the entire molecule. Use of point groups does not require the Born-Oppenheimer approximation.

As an example, consider NH$$_3$$. It is well known to have an electric dipole moment. However, it can spontaneously invert its structure and by this its dipole moment. The ground state for a model with sufficient flexibility is a pair of states, one symmetric and one asymmetric, that are separated by an energy corresponding to about 25 GHz. Both states have zero electric dipole moment. This shows that for a sufficiently flexible model the solutions transform as representations of the symmetry group of the hamiltonian. Only at time scales shorter than 40 ps there is an electric dipole moment. https://demonstrations.wolfram.com/AmmoniaInversionClassicalAndQuantumModels/

• But the question is exactly about those situations when you cannot rely on the point group concept, since you do not work in the Born-Oppenheimer separation. If nuclei are treated as active quantum particles, then you only have more general symmetries (inversion/rotation of all coordinates, permutation of identical particles, charge conjugation, etc.). Commented Dec 21, 2023 at 11:34
• The complete group of symmetries is actually much bigger than $C_i$. But even if you only consider inversions, that already points to OP's problem: how to reconcile the experimental fact of molecules having a dipole moment with the exact electron-nucleus energy eigenstates being also parity eigenstates. Referencing point groups is not helpful, since point groups already imply the treatment of nuclei as point charges in fixed positions. Solving the electronic Schrodinger equation for H$_2$O will give you a dipole moment, while solving the full e-N equation will seemingly not. Commented Dec 21, 2023 at 12:06
• So you agree that any non degenerate ground state cannot have a permanent dipole moment? How do you resolve this with the observation of molecules with permanent dipole moment? Commented Dec 21, 2023 at 16:15
• That does not resolve the problem. The expectation value of an eigenstate has no time-dependency. We should be able to average at any point in time over an ensemble in a pure state and get zero-dipole moment. Why does that not happen? Commented Dec 21, 2023 at 17:49
• It sounded to me that you were talking about the inversion process like a dynamic process, since you mention a time scale. Looking at your link, I assumed you were thinking about a superposition state of the ground state of the double well potential and the first excited state of the double well. Such a superposition state would have an amplitude that oscillates between the two minima, corresponding to an inversion motion. But that is a superposition state and not an energy eigenstate. I think we are talking past each other at this point. Commented Dec 21, 2023 at 19:18

Suppose that:$$\;\;|\psi\rangle=c_{g}|\psi^{g}\rangle+c_{u}|\psi^{u}\rangle\;\;$$

By reporting in the eigenvalue equation: $$H|\psi\rangle=E|\psi\rangle\;,$$by multiplying on the left by $$\langle\psi^{g}|$$ and by $$\langle\psi^{u}|$$, we obtain the system:$$\begin{cases} \langle \psi^{g}|H|\psi^{g}\rangle c_{g}+\langle \psi^{g}|H|\psi^{u}\rangle c_{u}=E(c_{g}+c_{u}S)\\ \langle \psi^{u}|H|\psi^{g}\rangle c_{g}+\langle \psi^{u}|H|\psi^{u}\rangle c_{u}=E(Sc_{g}+c_{u}) \end{cases}$$ Assuming that: $$\langle \psi^{g}|H|\psi^{g}\rangle= \psi^{u}|H|\psi^{u}\rangle =E\;,\;\langle \psi^{u}|H|\psi^{g}\rangle=\langle \psi^{g}|H|\psi^{u}\rangle=0\;\;(3)\;,\;\int_{\mathbb{R^{3}}}\psi^{u}\psi^{g}d^{3}r=0=S$$ if we assume that: $$\langle \psi^{g}|H|\psi^{g}\rangle= \psi^{u}|H|\psi^{u}\rangle =e\;,\;\langle \psi^{u}|H|\psi^{g}\rangle=\langle \psi^{g}|H|\psi^{u}\rangle=v\;,\;\int_{\mathbb{R^{3}}}\psi^{u}\psi^{g}d^{3}r=S$$ we obtain: $$E=\frac{e \pm v }{1\pm S}=E_{\pm}$$

as in

so $$|\psi\rangle=c_{g}|\psi^{g}\rangle$$ or $$\psi\rangle=c_{u}|\psi^{u}\rangle$$, and in addition, the symbols $$g,u$$ (see (1)) are used when there is a cetre of symmetry that is to say when we have for example two identical atom .

Moreover, if the molecule is homonuclear (A = B), the system is symmetrical in inversion with respect to the middle of the bond. The only invariant vector in this operation is the zero vector: no dipole moment. the existence of a permanent dipole moment d, of greater or lesser magnitude, results physically from the fact that, given different nuclear charges, electrons are more attracted to one nucleus than to the other (the more electronegative one), giving an asymmetrical electron density that is not invariant to reflection (or parity)(2).

(2) Claude Aslangul: Mécanique quantique 2, Développements et applications à basse énergie.