Allowed permutation symmetry in atomic and molecular wavefunctions. Simple examples

TL;DR

This paper explores how permutation symmetry in the symmetric group relates to the antisymmetry requirement of wavefunctions in multi-electron systems, using simple models to illustrate the compatibility of symmetry species with antisymmetry.

Contribution

It analyzes the compatibility of permutation symmetry species with the antisymmetry principle in atomic and molecular wavefunctions, providing simple illustrative examples.

Findings

Certain symmetry species are compatible with antisymmetry

Hamiltonian invariance under permutations influences wavefunction symmetry

Simple models demonstrate symmetry-antisymmetry relationships

Abstract

It is well known that the allowed wavefunctions for an $N$-electron system should be antisymmetric with respect to the permutation of any pair of electron labels. On the other hand, the Hamiltonian for such system is invariant under any permutation of electron labels and, consequently, its eigenfunctions are basis for the irreducible representations of the symmetric group $S_N$.Here, we investigate which symmetry species of the $S_N$ group are compatible with the antisymmetry principle. We illustrate the conclusions by means of simple $N$-particle one-dimensional models with harmonic interactions.

Equations

$$P_{ij}=\left[\begin{array}[]{l}1,2,\ldots,i,\ldots,j,\ldots N\\ 1,2,\ldots,j,\ldots i,\ldots N\end{array}\right]$$

$$P_{[i]}$$

$$i_k$$

$$|{\chi }_{i_1}{\chi }_{i_2}\dots {\chi }_{i_N}⟩$$

$$\Phi =\sum _{i_1,i_2,\dots ,i_N}{C}_{i_1,i_2,\dots ,i_N}|{\chi }_{i_1}{\chi }_{i_2}\dots {\chi }_{i_N}⟩$$

$$\begin{array}[]{c|ccc|}C_{3v}&\hat{E}&2C_{3}&3\sigma_{v}\\ \hline\cr A_{1}&1&1&1\\ A_{2}&1&1&-1\\ E&2&-1&0\end{array}$$

$$\hat{E}$$

$${\sigma }_{v_1}$$

$${\mathcal{P}}_{A_1}$$

$${\mathcal{P}}_{A_2}$$

$${\mathcal{P}}_E$$

$$H=-\frac{1}{2}\left(\frac{{\partial }^2}{\partial x_1^2}+\frac{{\partial }^2}{\partial x_2^2}+\frac{{\partial }^2}{\partial x_3^2}\right)+\frac{1}{2}\left(x_1^2+x_2^2+x_3^2\right)+\xi \left(x_1{x}_2+x_1{x}_3+x_2{x}_3\right),$$

$$y_1=\frac{\sqrt{2}{x}_2}{2}-\frac{\sqrt{2}{x}_3}{2},y_2=\frac{\sqrt{6}\left(2x_1-x_2-x_3\right)}{6},y_3=\frac{\sqrt{3}\left(x_1+x_2+x_3\right)}{3},$$

$$H$$

$$k$$

$${\psi }_{n_1{n}_2{n}_3}(x_1,x_2,x_3)={\varphi }_{n_1}(k,y_1){\varphi }_{n_2}(k,y_2){\varphi }_{n_3}(k^{\prime },y_3)$$

$$E_{n_1{n}_2{n}_3}=\sqrt{k}\left(n_1+n_2+1\right)+\sqrt{k^{\prime }}\left(n_3+\frac{1}{2}\right),$$

$$n_1,n_2,n_3=0,1,2,\dots .$$

$$H_{HO}{\varphi }_n(k,q)$$

$$H_{HO}$$

$${\mathcal{P}}_{A_2}{\omega }_i(x_1){\omega }_j(x_2){\omega }_k(x_3){\mathcal{P}}_{u}f(x_1,x_2,x_3)$$

$$H$$

$$y_1$$

$$y_4$$

$$H$$

$${\psi }_{n_1{n}_2{n}_3{n}_4}\left(x_1,x_2,x_3,x_4\right)={\varphi }_{n_1}(k,y_1){\varphi }_{n_2}(k,y_2){\varphi }_{n_3}(k,y_3){\varphi }_{n_4}(k^{\prime },y_4)$$

$$k=1-\xi ,k^{\prime }=1+3\xi$$

$$E_{n_1{n}_2{n}_3{n}_4}=\sqrt{1-\xi }\left(n_1+n_2+n_3+\frac{3}{2}\right)+\sqrt{1+3\xi }\left(n_4+\frac{1}{2}\right)$$

Full text

1 Introduction

There has recently been a controversy about the permutation symmetry of atomic and molecular Hamiltonians and the approaches commonly used to obtain their eigenvalues and eigenfunctions. On one hand it has been stated that Hartree-Fock and related methods do not take into account the permutation symmetry of the non-relativistic Hamiltonian, which “leads to false concepts, misinterpretations and unjustifiable approximations when dealing with many-electron systems”[1]. On the other hand, it has been shown that the arguments put forward in that paper violate well known mathematical theorems and that the analysis is based on an incorrect application of the permutation operators of the symmetric group $S_{N}$[2]. Such interesting discussion motivated the analysis of the permutation symmetry of electronic systems carried out below in this paper.

The postulates of quantum mechanics state that the wavefunction for a system of particles should be symmetric or antisymmetric under the permutation of the variables of identical particles if they are either bosons or fermions, respectively. In the particular case of an $N$-electron system the wavefunction should be antisymmetric with respect to the transposition of the coordinates of any pair of electrons. For this reason approximate calculations of the electronic structure of atoms and molecules is commonly based on Slater determinants constructed from suitably chosen spin-orbitals. The configuration interaction (CI) method is known to provide accurate atomic and molecular electronic energies[3]. On the other hand, it is well known that the non-relativistic Hamiltonian for a system of $N$ electrons is invariant under the $N!$ permutations of the electronic variables. For this reason its eigenfunctions are basis for the irreducible representations (irreps) of the symmetric group $S_{N}$. Since the Schrödinger equation for an $N$-electron system is not exactly solvable for $N>1$ there are no available comparisons between the exact solutions of the non-relativistic system and sufficiently accurate results provided by widely used methods like CI, except for some exactly-solvable models[4]. It would be interesting, for example, to know to which non-relativistic energy levels converges a CI calculation based on a Slater-determinant basis set.

The purpose of this paper is to fill this gap by means of exactly-solvable non-relativistic models with $S_{N}$ symmetry that can be easily treated by means of CI to a great degree of accuracy. In section 2 we outline the concepts of permutation symmetry that are relevant for present discussion. In section 3 we solve the Schrödinger equation for a simple non-relativistic model with $N=3$ identical particles. In section 4 we carry out a similar analysis for $N=4$. Finally, in section 5 we summarize the main results and draw conclusions.

2 Permutation symmetry

The Hamiltonian operator $H$ for an $N$-electron system is invariant under the transposition $P_{ij}$ of the variables of any pair of electrons $i$,$j$; that is to say: $P_{ij}HP_{ij}^{-1}=H$. There are $N(N-1)/2$ such transpositions that satisfy $P_{ij}=P_{ji}=P_{ij}^{-1}$. Since $P_{ij}^{2}=\hat{E}$ (the identity operator) then the eigenvalues of every transposition operator are $\pm 1$. The invariance of $H$ under transpositions can also be written in terms of vanishing commutators $[H,P_{ij}]=0$. Since the transpositions do not commute, then we cannot obtain a complete set of eigenfunctions common to $H$ and all $P_{ij}$. We can write a transposition as

[TABLE]

which means to substitute the electron variables $\mathbf{r}_{j},\mathbf{r}_{i}$ for $\mathbf{r}_{i},\mathbf{r}_{j}$ (it may also include spin variables when necessary)

The Hamiltonian $H$ is also invariant under any permutation

[TABLE]

which means to substitute $\mathbf{r}_{i_{1}}$, $\mathbf{r}_{i_{2}}$,…, $\mathbf{r}_{i_{N}}$ for $\mathbf{r}_{1}$, $\mathbf{r}_{2}$,…,$\mathbf{r}_{N}$. There are $N!$ such permutations of the variables of the $N$ electrons that can be split into $N!/2$ even and $N!/2$ odd permutations. Any permutation can be written as a non-unique product of a finite number of transpositions[5]. However, given a permutation, the number of such factors is either even or odd and we commonly say that the permutation is even or odd, respectively. The set of all $N!$ permutations of the $N$ electrons form the symmetric group $S_{N}$. The invariance of the Hamiltonian may be expressed either as $P_{[i]}HP_{[i]}^{-1}=H$ or $\left[H,P_{[i]}\right]=0$ for any of the $N!$ permutations.

If $\psi$ is an eigenfunction of $H$ with eigenvalue $E$, then $P_{ij}H\psi=HP_{ij}\psi=EP_{ij}\psi$. Therefore, if $\psi$ is non-degenerate then $P_{ij}\psi=\pm\psi$ for all $i$, $j$. In the case of a degenerate energy level both $\psi$ and $P_{ij}\psi$ may be linearly independent eigenfunctions of $H$. In fact, since $\left[P_{ij},P_{kl}\right]\neq 0$ then the non-degenerate states are not, in general, eigenfunctions of all the permutation operators $P_{[i]}$. Despite of this fact it has been stated that $\left[H,P_{[i]}\right]=0$ implies that any eigenfunction of $H$ is an eigenfunction of $P_{[i]}$[1, 6, 7].

The Hamiltonians of some systems of identical particles are also invariant under coordinate inversion $\hat{\imath}f(\mathbf{x})=f(-\mathbf{x})$ about the origin. Since $[H,\hat{\imath}]=0$ and $[P_{ij},\hat{\imath}]=0$ then the eigenfunctions of $H$ are either even or odd with respect to inversion: $\hat{\imath}\psi=\pm\psi$.

The results above apply to any system of $N$ identical particles but we restrict ourselves to electrons because we are interested in the electronic structure of atoms (with the nucleus clamped at origin) and molecules (under the Born-Oppenheimer approximation). Since the Schrödinger equation for such systems cannot be solved exactly we resort to approximate methods. In order to obtain a suitable basis set for such calculations we commonly construct the required antisymmetric functions as Slater determinants[3]

[TABLE]

where $p_{[i]}$ reflects the parity (even or odd) of $P_{[i]}$ and $\chi_{j}$ is a spin-orbital given by the product of a space orbital factor $\varphi_{i}$ and a spin one $\omega_{k}$ that equals either $\alpha$ ($m_{s}=1/2$) or $\beta$ ($m_{s}=-1/2$). The CI method is a Rayleigh-Ritz variational approach with the ansatz

[TABLE]

commonly chosen to be an eigenfunction of the total spin operators $S^{2}$ and $S_{z}$ when $H$ is the non-relativistic (spin-free) Hamiltonian[3].

It is not possible to compare the approximate variational calculation based on the trial function (7) with an exact result because the Schrödinger equation for any atomic or molecular system with $N>1$ cannot be solved exactly. For this reason in the following sections we consider two exactly solvable models with $S_{N}$ symmetry.

3 Exactly-solvable three-particle model

The case $N=2$ is trivial because the only permutation operators are $\hat{E}$ and $P_{12}$[2, 5]. Therefore, any eigenfunction $\psi$ of the non-relativistic Hamiltonian $H$ satisfies $P_{12}\psi=\pm\psi$ and a symmetric spatial function is multiplied by an antisymmetric spin one (singlet state), whereas an antisymmetric spatial function is multiplied by a symmetric spin one (triplet) in order to obtain an antisymmetric total wavefunction $\Phi$. Obviously, in this particular case we can easily omit the spin part in the construction of an approximate wavefunction. Therefore, the first non-trivial case is $N=3$.

The symmetric group $S_{3}$ is isomorphic to $C_{3v}$ (and also to $D_{3}$)[8]; its character table being

[TABLE]

There is a one-to-one correspondence between the permutation operators and the $C_{3v}$ ones given by

[TABLE]

The well known projection operators

[TABLE]

will be most useful for present analysis.

One can easily verify that the Hamiltonian

[TABLE]

exhibits $S_{3}$ permutation symmetry and is parity invariant. It describes a system of three identical particles in a one-dimensional space that interact with a different one clamped at origin by means of the terms $x_{j}^{2}/2$ and between them by means of the terms $\xi x_{i}x_{j}$. It resembles, for example, the Lithium atom with the nucleus clamped at origin. One may reasonably argue that this one-dimensional toy model is unsuitable for the study of atomic systems but the point is that here we are merely interested in the permutation symmetry of the Hamiltonian operator. The great advantage of this simple model is that the Schrödinger equation is separable and exactly solvable. It is a simplified version of the oscillator models widely used by Moshinsky[4].

By means of the change of variables[2]

[TABLE]

the Hamiltonian becomes

[TABLE]

We appreciate that there are bound states provided that $-1/2<\xi<1$. Under this condition the eigenfunctions and eigenvalues are given by

[TABLE]

where $\phi(k,q)$ is a normalized eigenfunction of the dimensionless Hamiltonian $H_{HO}$ for the harmonic oscillator

[TABLE]

Since the symmetric group $S_{3}$ is isomorphic to $C_{3v}$ we can label the irreps as $A_{1}$, $A_{2}$ (both one-dimensional) and $E$ (two-dimensional). If we added the inversion, then the suitable group would be $D_{3h}$ (among others) with irreps $A_{1}^{\prime}$, $A_{2}^{\prime}$, $E^{\prime}$, $A_{1}^{\prime\prime}$, $A_{2}^{\prime\prime}$ and $E^{\prime\prime}$, but we will restrict ourselves to the permutation symmetry. The Hamiltonian (10) exhibits also dynamical symmetry because it commutes with a set of five operators that depend on the coordinates and conjugate momenta. Consequently, the degeneracy of the energy levels given by $n_{1}+n_{2}+1$ is considerably greater than the one predicted even by $D_{3h}$. However, for present purposes it will suffice to consider just $C_{3v}$ because we are interested only in the permutation symmetry. Note that the Hamiltonian operator for Lithium (under the clamped-nucleus approximation) commutes with the total angular momentum of the electrons $L^{2}$ and $L_{z}$; therefore, the symmetric group $S_{3}$ will be insufficient in this realistic case too. Since the dynamical symmetry is model-dependent we will omit it from now on.

The variables $y_{1}$ and $y_{2}$ are basis for the irrep $E$ while $y_{3}$ is basis for $A_{1}$. For this reason all the states $\psi_{00j}$ are basis for $A_{1}$ and the symmetry of the states of the non-relativistic Hamiltonian is completely determined by the quantum numbers $n_{1}$ and $n_{2}$. For example, $\psi_{10j}$ and $\psi_{01j}$ are basis for $E$ and the three degenerate functions with $n_{1}+n_{2}=2$ are basis for both $A_{1}$ and $E$. The state $\psi_{11j}$ is basis for $E$ and by means of the projection operators we easily verify that the linear combinations $\psi_{20j}+\psi_{02j}$ and $\psi_{20j}-\psi_{02j}$ are basis for $A_{1}$ and $E$, respectively. We can carry out this analysis for every energy level; for example the four states with $n_{1}+n_{2}=3$ are basis for $A_{1}$, $A_{2}$ and $A_{3}$. A straightforward calculation shows that $3\psi_{21j}-\psi_{03j}$, $\psi_{30j}-3\psi_{12j}$ and $\left(\psi_{30j}+\psi_{12j},\psi_{21j}+\psi_{03j}\right)$ are basis for $A_{1}$, $A_{2}$ and $E$, respectively. In this case only the basis functions for the irreps $A_{1}$ and $A_{2}$ are eigenfunctions of all the permutation operators.

Let us now turn to the construction of antisymmetric spatial-spin functions. Since $\mathcal{P}_{A_{2}}=\sqrt{6}\mathcal{A}$ we will resort to this projection operator for the construction of antisymmetric functions. In the case of three electrons we expect one quadruplet and two doublets. In order to determine which non-relativistic functions will appear in a standard CI calculation we choose an arbitrary function $f(x_{1},x_{2},x_{3})$ and construct antisymmetric functions according to the following expression

[TABLE]

where $u=A_{1},A_{2},E$. The procedure is quite simple: on inserting a product of three monoelectronic spin states the result may be zero or a valid Slater determinant. This straightforward calculation shows that the non-relativistic states that are basis for $A_{1}$ are not allowed by the principles of quantum mechanics. In other words, the non-degenerate energy levels $E_{00j}$ will not appear in a CI calculation. The states that are basis for $A_{2}$ appear in the quadruplet, and those belonging to $E$ in the doublets. It is worth noting that the conclusions based only on the permutation of the electron variables are model independent and therefore apply to more realistic models. For example, in the case of Lithium we expect approximate antisymmetric spatial-spin functions with spatial parts that are basis for the irreps $A_{2}$ ($S=3/2$) and $E$ ($S=1/2$). In other words, we would obtain meaningful results with spin-free basis-set functions belonging to the symmetry species just mentioned. Also notice that equation (15) can be easily generalized to any number of electrons for which we only need the projection operators for the corresponding symmetric group.

It is not difficult to take into account that the Hamiltonian is also parity-invariant. We simply apply equation (15) with the projection operators $\mathcal{P}_{u}$ for the symmetry point group $D_{3h}$ ($u=A_{1}^{\prime},A_{2}^{\prime},E^{\prime},A_{1}^{\prime\prime},A_{2}^{\prime\prime},E^{\prime\prime}$). The result is that $A_{1}^{\prime}$ and $A_{1}^{\prime\prime}$ do not appear in the Slater determinants, $A_{2}^{\prime}$ and $A_{2}^{\prime\prime}$ appear in the quadruplet, $E^{\prime}$ and $E^{\prime\prime}$ appear in the doublets. This result agrees with the analysis of the permutation symmetry of the hydrogen atoms in $H_{3}^{+}$ carried out, for example, by Bunker and Jensen[10] and is called missing levels.

4 Exactly-solvable four-particle model

The symmetric group $S_{4}$ is isomorphic to $O$ and $T_{d}$ and we will choose the former point-group symmetry here. In this case we apply a somewhat different strategy. First, we derive the $24$ matrices that produce all the permutations of the elements of a four-dimensional column vector $\mathbf{x}$. Second, we collect the matrices into their respective group classes and determine the order (also called period length) of each of them[8]. In this way we derive a one-to-one correspondence between the matrix classes and those appearing in the character table of the group $O$. Third, with each matrix $\mathbf{M}_{i}$ we build the corresponding operator $\hat{M}_{i}$ by means of the well known expression $\hat{M}_{i}f(\mathbf{x})=f\left(\mathbf{M}_{i}^{-1}\mathbf{x}\right)$. In this case we will show neither the character table nor the projection operators that can be easily constructed by means of well known expressions[8]. We will just discuss the results.

As an illustrative example we resort to the oscillator model

[TABLE]

that exhibits the appropriate symmetry. By means of the change of variables

[TABLE]

the resulting Hamiltonian is separable an exactly solvable

[TABLE]

It exhibits bound states when $-1/3<\xi<1$ and its eigenfunctions and eigenvalues are given by

[TABLE]

The degeneracy of the states of this oscillator is considerably greater than the one for the preceding example: $\left(n_{1}+n_{2}+n_{3}+1\right)\left(n_{1}+n_{2}+n_{3}+2\right)/2$.

The variables $y_{1}$, $y_{2}$, $y_{3}$ are basis for the irrep $T_{2}$ and $y_{4}$ is basis for $A_{1}$. For this reason all the states of the form $\psi_{000j}$ are basis for $A_{1}$ and the symmetry of the non-relativistic states is determined by the quantum numbers $n_{1}$, $n_{2}$ and $n_{3}$. For each value of $n_{4}$ the three degenerate states with $n_{1}+n_{2}+n_{3}=1$ are basis for $T_{2}$. The six degenerate states with $n_{1}+n_{2}+n_{3}=2$ are basis for $A_{1}$, $E$ and $T_{2}$. The ten degenerate states with $n_{1}+n_{2}+n_{3}=3$ are basis for $A_{1}$, $E$, $T_{1}$ and $T_{2}$. The basis functions for the irrep $A_{2}$ appear in a much higher energy level with $n_{1}+n_{2}+n_{3}=6$. As in the preceding example only the basis functions for the irreps $A_{1}$ and $A_{2}$ are eigenfunctions of all the permutation operators.

In order to determine which non-relativistic spatial functions contribute to the antisymmetric spatial-spin ones we proceed as in equation (15) adding an additional electron to that expression and choosing the projection operators for the symmetry point-group $O$. In the case of four electrons we expect one quintuplet, three triplets and two singlets. Our results show that the spatial functions that are basis for $A_{1}$ and $T_{2}$ are not allowed by the antisymmetry principle. The basis functions for $A_{2}$, $T_{1}$ and $E$ are responsible for the quintuplets, triplets and singlets, respectively. This conclusion is not model dependent because it is based entirely on the symmetry of the problem and applies, for example, to Beryllium under the clamped-nucleus approximation.

5 Further comments and conclusions

Throughout this paper we have analyzed the connection between the antisymmetric spatial-spin functions given in terms of Slater determinants and the eigenfunctions of the non-relativistic Hamiltonian that are basis for the irreps of the symmetric group $S_{N}$. We restricted ourselves to the particular cases of $N=3$ and $N=4$ electrons because they can be analyzed by means of the character tables of the point groups $C_{3v}$ and $O$, respectively, that are well known for most chemists and physical chemists. The exactly solvable models chosen here are suitable for illustration but are not necessary for obtaining the main theoretical results. From the point of view of symmetry they are identical to the Lithium and Beryllium atoms and the particular forms of the Hamiltonians mimic those atoms in the clamped-nucleus approximation. However, the main conclusions about the symmetry of the spatial parts of the Slater determinants also applies to the case of finite nuclear mass. If we remove the motion of the center of mass and place the coordinate origin at the nucleus the resulting Hamiltonian exhibits the same symmetry $S_{N}$. The reason is that the coupling terms that appear in the kinetic-energy operator (the so called mass polarization terms) do not change the permutational symmetry of the Hamiltonians[9].

That the main theoretical results derived in this paper are not model dependent is clearly illustrated by the fact that present analysis of three $1/2$-spin identical particles by means of the $D_{3h}$ point group agrees with the results derived by Bunker and Jensen[10] for the hydrogen nuclei of the $H_{3}^{+}$ molecule.

Acknowledgments

The author is indebted to Diego R. Alcoba and Ofelia Oña for their full CI calculations that confirmed present theoretical results.