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

Question

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.

Answer 1

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.

Answer 2

The apparent contradiction is the following. The general non-relativistic Hamiltonian $$ \begin{aligned} {\cal{H}}&= -\sum_{a=1}^{\cal{N}}\frac{1}{2m_a}\nabla_{\rho,a}^2+\sum_{\substack{a,b=1 \\ a<b}}^{\cal{N}}\frac{\alpha z_az_b}{|\vec{\rho}_a-\vec{\rho}_b|} \\ &= -\sum_{i=1}^{N_{\text{e}}}\frac{1}{2m}\nabla_i^2+\sum_{\substack{i,j=1 \\ i<j}}^{N_{\text{e}}}\frac{\alpha}{|\vec{r}_i-\vec{r}_j|} -\sum_{i}^{N_{\text{e}}}\sum_{A=1}^{N_{\text{nuc}}}\frac{\alpha Z_A}{|\vec{r}_i-\vec{R}_A|} -\sum_{A=1}^{N_{\text{nuc}}}\frac{1}{2M_A}\nabla_A^2+\sum_{\substack{A,B=1 \\ A<B}}^{N_{\text{nuc}}}\frac{\alpha Z_AZ_B}{|\vec{R}_A-\vec{R}_B|} \end{aligned} $$ 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 $$ \begin{aligned} H(\vec{R})= -\sum_{i=1}^{N_{\text{e}}}\frac{1}{2m}\nabla_i^2+\sum_{\substack{i,j=1 \\ i<j}}^{N_{\text{e}}}\frac{\alpha}{|\vec{r}_i-\vec{r}_j|} -\sum_{i=1}^{N_{\text{e}}}\sum_{A=1}^{N_{\text{nuc}}}\frac{\alpha Z_A}{|\vec{r}_i-\vec{R}_A|} +\sum_{\substack{A,B=1 \\ A<B}}^{N_{\text{nuc}}}\frac{\alpha Z_AZ_B}{|\vec{R}_A-\vec{R}_B|} \ , \end{aligned} $$ 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]: $$ \begin{aligned} \mu_{\text{calc}}(\text{HCl})\approx 1.084 \, \text{D} \ \ \ &\leftrightarrow \mu_{\text{exp}}(\text{HCl})\approx 1.093 \, \text{D} \ , \\ \mu_{\text{calc}}(\text{H$_2$O})\approx 1.840 \, \text{D} \ \ \ &\leftrightarrow \mu_{\text{exp}}(\text{H$_2$O})\approx 1.857 \, \text{D} \ . \end{aligned} $$ 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

$ $ [2]: Damour, Quintero-Monsebaiz, Caffarel, Jacquemin, Kossoski, Scemama, Loos: J. Chem. Theory Comput. 19 1 221 (2022)

$ $ [3]: Sutcliffe, Woolley: Phys. Chem. Chem. Phys. 7 3664 (2005)

$ $ [4]: Fernandez, Echave: Chapter 6. of Computational Spectroscopy (editor: Jorg Grunnenberg; 2010) (arXiv link)

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

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

$ $ [7]: Hobson, Valeev, Csaszar, Stanton: Mol. Phys. 107 8 1153 2009

Answer 3

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.

Answer 4

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/

Answer 5

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).

(1) non-relativistic quantum mechanics landau

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

(3) Landau,Lifchitz