reflection - How to understand the symmetry $Sigma^-$ in the wavefunction of a diatomic molecule?

In Wikipedia (and elsewhere), a particular symmetry of the quantum system of a diatomic molecule is mentioned: symmetry under reflection along a plane containing the internuclear axis. The wavefunctions may be either symmetric or antisymmetric under the this symmetry.

It is then asserted that, in particular, for the $\Sigma$ wavefunctions, which have z-angular momentum $0$, both kinds exist, so that there are two distinct wavefunctions labelled $\Sigma^+$ and $\Sigma^-$ for the symmetric and antisymmetric case, respectively.

My question is, how can a wavefunction with a vanishing $z$ component of the angular momentum be antisymmetric under reflection through such a plane. In particular and w.l.o.g. take the reflection to be $y \rightarrow -y$, how can the wavefunction change sign under the action of this, when if the z-angular momentum is $0$, we know that the wavefunction is uniform in any cross-section parallel to the $xy$ plane?

Answer

The answer is that the term symbol refers to the entire, multielectronic, molecule. You are indeed correct that a single electron in a $\Sigma$ state must be symmetric under reflection about a plane that contains the internuclear axis. If you have multiple electrons, however, you can still have a global antisymmetry under such reflections, and get their angular momenta to cancel out to give a global $\Sigma$ state. Wikipedia is correct in its statement but not particularly clear; this is a better resource.

The simplest example of a $\Sigma^-$ state comes from two identical $\pi$ electrons. To get the antisymmetric spatial state, you need the spin state to be spatially symmetric, which requires you to combine the spins into the triplet representation, and this essentially means that you can assume both spins to be parallel to each other and the internuclear axis. In any case, the bottom line is that you need the electrons to occupy orthogonal orbitals. These can be a $\pi_x$ and a $\pi_y$, or a $\pi_+$ and a $\pi_-$; it doesn't matter because the state must be antisymmetrized w.r.t. electron exchange, and the final two-electron wavefunction, $$ |\Psi⟩= \tfrac1{\sqrt{2}}\left[|\pi_x\pi_y⟩-|\pi_y\pi_x⟩\right]= \tfrac1{\sqrt{2}}\left[|\pi_+\pi_-⟩-|\pi_-\pi_+⟩\right], $$ turns out to be exactly equal in both cases.*

This representation is quite useful, and from it you can see quite clearly that:

1. 
   

   The state has zero total angular momentum, with $m_z=+1-1=0$, so it's in a $\Sigma$ state.

   
   


2. 
   

   The state is completely rotationally symmetric, because each $\pi_\pm$ state is transforms under a rotation by $\theta$ around the internuclear axis as $U(R(\theta))|\pi_\pm⟩=e^{\pm i\theta}|\pi_\pm⟩$, so the phases cancel out and each product state $|\pi_\pm\pi_\mp⟩$ is rotationally symmetric.

   
   


3. 
   
   

   The state is antisymmetric under reflections about the $yz$ plane, which sends $x\mapsto -x$, because $|\pi_x⟩\mapsto-|\pi_x⟩$ and $|\pi_y⟩\mapsto|\pi_y⟩$.

   
   


4. 
   

   (Alternatively, you can do this for a reflection about the $xz$ plane, which flips the circular orbitals, i.e. $|\pi_+⟩\mapsto|\pi_-⟩$ and $|\pi_-⟩\mapsto|\pi_+⟩$, and therefore introduces a global minus sign to the state.)

   
   


5. 
   

   Because of the rotational symmetry, the state is antisymmetric under any reflection about a plane which includes the internuclear axis.

   
   

That about sums it up: here is an electronic state of a (multielectronic) molecule which has zero angular momentum about the internuclear axis, but which has odd symmetry about such reflection planes; its term symbol must therefore be $\Sigma^-$.

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Now, from the above, it's easy to get the impression that this is just one of those highly-correlated multi-electronic states that is just impossible to fully understand, but that's not really the case - this $\Sigma^-$ state is actually rather simple in structure.

To see this in practice, set up your $\pi$ orbitals such that $$ ⟨\varphi,\vec \rho|\pi_\pm⟩ = e^{\pm i \varphi} \psi(\vec \rho), $$ where $\vec\rho$ represents the non-azimuthal dependence (i.e. $(r,\theta)$ in spherical coordinates or $(\rho,z)$ in a cylindrical system). Then, for the two-electron $\Sigma^-$ state above, the full wavefunction reads \begin{align} ⟨\varphi_1,\vec\rho_1,\varphi_2,\vec\rho_2|\Psi⟩ & = \tfrac1{\sqrt{2}}\left[ ⟨\varphi_1,\vec\rho_1|\pi_+⟩⟨\varphi_2,\vec\rho_2|\pi_-⟩ -⟨\varphi_1,\vec\rho_1|\pi_-⟩⟨\varphi_2,\vec\rho_2|\pi_+⟩ \right] \\ & = \tfrac1{\sqrt{2}}\left[ e^{i(\varphi_1-\varphi_2)} -e^{-i(\varphi_1-\varphi_2)} \right] \psi(\vec \rho_1)\psi(\vec \rho_2) \\ & = \sqrt{2}i \ \sin(\varphi_1-\varphi_2) \ \psi(\vec \rho_1)\psi(\vec \rho_2) , \end{align} or, in other words, the azimuthal and radial/longitudinal dependences factor out, and we can just look at the azimuthal dependence, which is a rather simple two-electron one-dimension wavefunction, $$ ⟨\varphi_1,\varphi_2|\Psi⟩ \sim \sin(\varphi_1-\varphi_2), $$ which lays out that the electrons prefer to be at an angle $|\varphi_1-\varphi_2| \approx \pi/2$ from each other, but:

• with no preferred global direction and with no phase gained under a global rotation, i.e. a wavefunction invariant under $\varphi_j \mapsto \varphi_j +\theta$ for a global rotation angle $\theta$, which then guarantees that $L_z=0$;


• with negative parity under spatial inversions, which here read $\varphi_j \mapsto -\varphi_j$; and


• in an antisymmetric combination under electron exchange.

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Finally, let me expand a bit on the spin arguments that I touched on above. To get this term, you essentially need to demand the spin part of the wavefunction to be in the triplet representation (so the full term is $^3\Sigma^-$). This is a triply-degenerate, spin-1 representation that contains the states $|\uparrow\uparrow⟩$, $\tfrac{1}{\sqrt{2}} \left[ |\uparrow\downarrow⟩ +|\downarrow\uparrow⟩\right]$ and $|\downarrow\downarrow⟩$, and you don't care which one you're in. These are all symmetric under electron exchange; since the global wavefunction must be antisymmetric, this requires the spatial wavefunction to be antisymmetric. In turn, this requires that the spatial orbitals occupied by the two electrons be orthogonal, since otherwise they would be unable to support an antisymmetric state. This then feeds into the above.

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^{* This is a general principle: Slater determinants are invariant under any unit-determinant linear transformation amongst their orbitals. This is why chemists rightfully insist that molecular orbitals are not meaningful physical concepts.}