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?

  • $\begingroup$ hey, but isn't that opposite case? I think that wiki says Λ>0 has two distinct states, symemtric and antisymmetric state. and Λ=0 state has only two states, symmetric or antisymmetric state. I think this is the case because the reflection of plane is same as the letting \phi \rightarrow -\phi in spherical harmonics. With this, the two states must be same in Λ=0 state.. $\endgroup$ – user42298 Jan 28 '15 at 6:31
  • $\begingroup$ I don't get what you mean, but I think Wikipedia's wording isn't very good. What I have found from more reliable sources is that you get symmetric and anti-symmetric states for all $\Lambda$, but I am asking here about the %\Lambda=0$ in particular. $\endgroup$ – guillefix Feb 3 '15 at 15:08
  • $\begingroup$ Related: Is the molecular term ${}^1\Sigma^-$ possible in a molecule?, exploring whether a $\Sigma^-$ term is possible with an exchange-symmetric spatial part and an antisymmetric singlet spin state (answer: yes, it's possible), and the electronic states that are required for it. $\endgroup$ – Emilio Pisanty Sep 6 '18 at 9:40

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 $\Sigma^-$.

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.

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.

* 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.

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This is a very good question.

First of all, you are absolutely correct that, for a single electron, invariance under $\phi\to-\phi$ means invariance under $y\to-y$. This is obvious just from looking at the coordinates $x=\rho\cos(\phi)$ and $y=\rho\sin(\phi)$. It is clear that $\phi\to-\phi$ means the exact same thing as $y\to-y$.

The answer is a little tricky to explain because this a multi-electron effect. The explanation in simplest in terms of Slater determinants.

One other caveat: Remember that in a diatomic molecule the only rotational symmetry is about the $z$ axis. This means that the $m$ label actually labels the different energy levels. This is totally different from monoatomic systems where the $\ell$ label differentiates between energy levels and the $m$ levels for a given $\ell$ are degenerate.

Anyways, hopefully, I can give you an example that clears everything up. The example will be for a single atom with a crystal field applied along the $z$ axis. The purpose of the crystal field is to split the $Y_{1,+1}$ and $Y_{1,-1}$ energies away from the $Y_{1,0}$ energy (just like the analogous orbitals in the diatomic molecule are split because of the symmetry of the potential).

So now, the simplest analog of the $\Sigma^-$ state is a Slater determinant made up of $Y_{1,+1}$ and $Y_{1,-1}$. Remember that $Y_{1,+1}=-e^{+i\phi}$ and $Y_{1,-1}=+e^{-i\phi}$ (ignore the overall normalization constant and theta dependence). The state $Y_{1,+1}$ has z-angular momentum $+1$ and the state $Y_{1,-1}$ has $z$-angular-momentum $-1$. Therefore, I know that application of the $z$-angular-momentum operator does this: $Y_{1,+1}\to +Y_{1,+1}$ and $Y_{1,-1}\to -Y_{1,-1}$.

Remember also (as mentioned above) that $\phi\to-\phi$ means the same thing as $y\to-y$. Therefore, I know that application of reflection in the xz-plane does this: $Y_{1,+1}\to -Y_{1,-1}$ and $Y_{1,-1}\to -Y_{1,+1}$.

Finally, in a multi-electron atom with a symmetric spin wave function, the spatial wave function must be antisymmetric and is most simply approximated as $\Psi(\phi_1,\phi_2)=Y_{1,+1}(\phi_1)Y_{1,-1}(\phi_2)-Y_{1,+1}(\phi_2)Y_{1,-1}(\phi_1),$ where $\phi_1$ is the phi coordinate of electron 1 and $\phi_2$ is the phi coordinate of electron 2.

The total $z$-angular-momnetum operator is: $\hat M=-i\frac{\partial}{\partial\phi_1}+-i\frac{\partial}{\partial\phi_2}$ and, clearly, $\hat M\Psi=0$

On the other hand, explicitly, the reflection about $xz$ affects $\Psi$ as follows: \begin{align} Y_{1,+1}(\phi_1)Y_{1,-1}(\phi_2)-Y_{1,+1}(\phi_2)Y_{1,-1}(\phi_1) & \to Y_{1,+1}(-\phi_1)Y_{1,-1}(-\phi_2)-Y_{1,+1}(-\phi_2)Y_{1,-1}(-\phi_1) \\ & = (-Y_{1,-1}(\phi_1))(-Y_{1,+1}(\phi_2))-(-Y_{1,-1}(\phi_2))(-Y_{1,+1}(\phi_1)) \\ & = Y_{1,-1}(\phi_1)Y_{1,+1}(\phi_2)-Y_{1,-1}(\phi_2)Y_{1,+1}(\phi_1) \\ & = -\Psi(\phi_1,\phi_2) \end{align}

Therefore, it is possible to have $M=0$ and antisymmetry about reflections in the $xz$ plane. Whew!

Let me know if you have any questions. There are a lot of sub-scripts and minus signs running around this answer and I believe I have got them all correct (after a lot of editing...)


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Maybe I am missing the point, but if the article says that the wave functions are "either symmetric or antisymmetric" about this axis, then for the Z=0 case they are (only ever) symmetric which is indeed one of the allowed cases. "Either... or" doesn't mean "it has to be capable of both", but "it can only be one of these".

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  • $\begingroup$ Yeah, I just read Joachain's book and it makes that point clear. In any case, I am not bothered by it being one or the other, but about the possibility of a state with $\Lambda=0$ being anti-symmetric under this symmetry. More specifically, I think that this state satisfies $\frac{\partial \psi}{\partial \phi}=0$, while the symmetry operation is $\phi \rightarrow-\phi$, which I can't see possibily taking $\phi$ to $-\phi$ under the above condition! $\endgroup$ – guillefix Feb 3 '15 at 23:46
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    $\begingroup$ See my answer. It is a multi-electron effect that can not be understood in terms of single particle orbitals. You need the Slater determinant to see how it happens. I.e., you need more that one $\phi$ coordinate... $\endgroup$ – hft Feb 4 '15 at 7:53

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