Let's say I have the molecule $\ce{N2+}$. Its symmetry point group is clearly $D_{\infty\mathrm h}$. But I'm confused by its irreducible representations. I know there are 8 of them, but as I understand it, they're all representations of the same group, just with different bases.

But today I found out, that they're not completely "equal", i.e. that they mark different states of the system ($\mathrm{A_g}$ is the ground state, $\mathrm{B_{1u}}$ is one of the excited states etc.).

What I'm completely missing is - how are irreducible representations linked with the system states?

Please, try some "low-level" explanation, I'm a beginner both in quantum chemistry and group theory.

  • 1
    $\begingroup$ The point group $D_{\infty h}$ has actually infinitely many IRREPs, the 8 IRREPs you are referring to are from $D_{2h}$ which is the largest abelian subgroup of $D_{\infty h}$. $\endgroup$
    – Feodoran
    Commented Dec 19, 2017 at 8:37
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    $\begingroup$ Agreed that Eenoku is referring to $D_{2h}$. However, $D_{2h}$ is not the largest abelian subgroup of $D_{\infty h}$. There are infinitely many larger abelian subgroups of $D_{\infty h}$, namely the $C_n$ (n>4) and $S_{2n}$ groups (n>2). FWIW, I believe it is the largest non-cyclic, abelian subgroup though. $\endgroup$
    – levineds
    Commented Dec 20, 2017 at 0:55

2 Answers 2


The symmetry of a state is the MO (molecular orbital) product of its occupied molecular orbitals. Let's take the example of water molecule. We can see the symmetry label of each of the MO in the diagram. Each MO is assigned an irreducible symmetry label (associated with the point group of molecule) which can be deduced from the symmetry of MOs in space (or from the LCAOs). The wavefunction associated with the MOs also have the same symmetry species. Now, this arrangement of electrons represents the ground electronic state of water, so the symmetry label of the ground state of water can be obtained as

$(2a_1)^2(1b_2)^2(3a_1)^2(1b_1)^2=(A_1)^2\bigotimes (B_2)^2\bigotimes (A_1)^2\bigotimes (B_1)^2 =A_1$

Follow the same procedure to obtain ground state of $N_2^+$. Let me know if you have any questions.MO diagram of water molecule

  • $\begingroup$ Thank you very much for your answer! I have just one question - how can you multiply two irreducible representations? Up today I understood it just as a row in a character table... $\endgroup$
    – Eenoku
    Commented Dec 20, 2017 at 21:12
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    $\begingroup$ @Eenoku This type of multiplication is called a 'direct product'. It's a general mathematical concept used in group theory and other branches. The following link describes it in the context of molecular symmetry staff.ncl.ac.uk/j.p.goss/symmetry/Products.html $\endgroup$
    – tobiuchiha
    Commented Dec 21, 2017 at 18:46

How are irreducible representations linked with the system states?

In short, the irreducible representation (IRREP) tells you about the symmetry of the electronic state or orbital.

Okay, this seems like a quite obvious answer. Let me try to use some different words. The term "symmetry" has two slightly different meanings here: symmetry of the molecule and symmetry of the wave function.

The symmetry of the molecule, which is the same as the symmetry of the electron density, is always either symmetric or not symmetric with respect to a certain operation. This is because the electron density is positive at all points in space. This is what we use to determine the point group of the system.

On the other hand, the symmetry of the wave function can be symmetric or anti-symmetric. The symmetry operations are the same as for the molecule, but the wave function has nodes and therefore different signs. Therefore a symmetry operation can keep the sign or switch it. This is encoded by the characters in the character table.


\begin{array}{c|cccccccc|cc} \boldsymbol{C_S} & E & \sigma_\mathrm{h} \\ \hline A^\prime & 1 & 1 \\ A^{\prime\prime} & 1 & -1 \\ \end{array}

The $C_S$ point group has 2 symmetry operations ($E$ and $\sigma_h$) and accordingly 2 IRREPs ($A^\prime$ and $A^{\prime\prime}$). The total symmetric IRREP ($A^\prime$) has the character $+1$ for both operations. This can for example be some $s$ orbital. The other IRREP has character $+1$ for $E$ and $-1$ for $\sigma_h$. This could for example be a $p_z$ orbital where $\sigma_h$ is the $xy$ plane. (Note that $p_x$ and $p_y$ would be of IRREP $A^\prime$ here, since their symmetry plane is not included in the point group discussed here.)

The same can be applied to the wave function of an electronic state. @tobiuchiha already explained in his answer how to go from the symmetry of the orbitals to the symmetry of the state (strictly speaking configuration, but the symmetry will be the same anyway).


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