Question about chiral molecules

[BillKet]

Hello! I started reading a bit about chiral molecules, and I have some questions. For reference, I will use this article for my questions. In Figure 1 in that article they show the energy levels for a typical chiral molecule (my question is for the case in which the left/right enantiomers are long lived i.e. according to their notation $\Delta_{pv} E^0 \gg \Delta E_\pm$). As far as I understand, normally one has positive and negative parity levels, but when the condition above is met, what we see in the lab are actually linear combinations of positive and negative parities (as the tunneling from left to right is basically zero during an actual experiment). What I am not sure I understand is their definition of parity. In Figure 1 they have (using nomenclature from diatomic molecules) a potential energy curve, and the horizontal lines are vibrational levels built upon this curve. However a vibrational level doesn't have a well defined parity (at least not in the case of diatomic molecules). The parity is given by different angular momenta and each vibrational level has many rotational levels, hence changing the parity with every J. So what exactly do they mean by alternating positive/negative parity levels of vibrational states (for example in the excited electronic level in that figure)?

 

[TeethWhitener]

I haven't looked at the article in depth, but I think you can justify most of your concerns if you think of Fig 1 as analogous to the ammonia inversion potential energy surface, if you're familiar with that. As you note, the eigenstates of the chiral molecule in Fig 1 are not the enantiomers, but rather linear combinations of the enantiomers. The + and - indicate whether those linear combinations are symmetric or antisymmetric. Symmetric wavefunctions have positive parity ($\Psi(x)=\Psi(-x)$) and antisymmetric wavefunctions have negative parity ($\Psi(-x)=-\Psi(x)$). Individual enantiomers do not have a well-defined parity in this model.

Also, as far as diatomic molecules (or any molecular vibration) is concerned, if the potential is harmonic, then the eigenfunctions have well-defined parity (symmetric or antisymmetric). Of course, the potential isn't exactly harmonic, so YMMV. But that seems to be what they're doing in the excited electronic state of Figure 1.

 

[BillKet]

I haven't looked at the article in depth, but I think you can justify most of your concerns if you think of Fig 1 as analogous to the ammonia inversion potential energy surface, if you're familiar with that. As you note, the eigenstates of the chiral molecule in Fig 1 are not the enantiomers, but rather linear combinations of the enantiomers. The + and - indicate whether those linear combinations are symmetric or antisymmetric. Symmetric wavefunctions have positive parity ($\Psi(x)=\Psi(-x)$) and antisymmetric wavefunctions have negative parity ($\Psi(-x)=-\Psi(x)$). Individual enantiomers do not have a well-defined parity in this model.

Also, as far as diatomic molecules (or any molecular vibration) is concerned, if the potential is harmonic, then the eigenfunctions have well-defined parity (symmetric or antisymmetric). Of course, the potential isn't exactly harmonic, so YMMV. But that seems to be what they're doing in the excited electronic state of Figure 1.

Thanks for this. I agree with what you said, but that doesn't really answer my question. For example, assuming that the potential is perfectly harmonic (and say a diatomic molecule for simplicity), a given vibrational level doesn't have a well defined parity. For example, if I am in the $\nu=1$ vibrational state in the J=3 rotational level (assuming for simplicity I have no other angular momenta involved), the parity of my state is $(-1)^{J=3}=-1$. However, if I am in the exactly same $\nu=1$ vibrational state but in the J=4 rotational state, the parity is $(-1)^{J=4}=1$ (and the same is true for any value of $\nu$). So it doesn't make sense to talk about the parity of a vibrational level, as the parity depends on the rotational level you are in, and a vibrational level has both positive and negative parity states. Am I missing something?

 

[TeethWhitener]

What I meant is that parity at its most basic just means reversing the sign on the coordinates in the wavefunction. The harmonic oscillator hamiltonian commutes with the parity operator, with eigenvalues of $\pm1$. For the even-numbered eigenstates of the harmonic oscillator, the parity eigenvalue is $+1$, and for the odd-numbered eigenstates, the parity eigenvalue is $-1$. It looks like what they're doing in the excited electronic state of figure 1 is assigning those parity eigenvalues to the vibrational eigenstates of the harmonic oscillator.

 

[BillKet]

What I meant is that parity at its most basic just means reversing the sign on the coordinates in the wavefunction. The harmonic oscillator hamiltonian commutes with the parity operator, with eigenvalues of $\pm1$. For the even-numbered eigenstates of the harmonic oscillator, the parity eigenvalue is $+1$, and for the odd-numbered eigenstates, the parity eigenvalue is $-1$. It looks like what they're doing in the excited electronic state of figure 1 is assigning those parity eigenvalues to the vibrational eigenstates of the harmonic oscillator.

But vibrational levels don't have fixed parity, that depends on the rotational level. In all experiments where they search for parity violation they mix 2 levels of opposite parity, but they always do that in the same vibrational level. If a given vibrational level would have fixed parity that wouldn't make sense.

For example the states $|\nu=1,J=3>$ has parity - and the state $|\nu=1,J=4>$ has parity +. Are you saying that the state $|\nu=1>$ has parity - regardless of the J level? I don't see how defining the parity of a given vibrational level without specifying the rotational makes any sense.

 

[TeethWhitener]

But vibrational levels don't have fixed parity, that depends on the rotational level. In all experiments where they search for parity violation they mix 2 levels of opposite parity, but they always do that in the same vibrational level. If a given vibrational level would have fixed parity that wouldn't make sense.

For example the states $|\nu=1,J=3>$ has parity - and the state $|\nu=1,J=4>$ has parity +. Are you saying that the state $|\nu=1>$ has parity - regardless of the J level? I don't see how defining the parity of a given vibrational level without specifying the rotational makes any sense.

I may be wrong, but I think you’re overthinking it. Here's a site with probably a clearer explanation of what I'm trying to say:
https://quantummechanics.ucsd.edu/ph130a/130_notes/node153.html
(Ctrl+F for parity--sorry I'm too tired right now to write everything out directly, but the site does a good job explaining parity in the harmonic oscillator eigenfunctions.)

Edit: parity literally just means, given $\Psi(x,y,z)$, consider the operation $\mathcal{P}\Psi(x,y,z)=\Psi(-x,-y,-z)$. If $\mathcal{P}\Psi(x,y,z)=\Psi(x,y,z)$, then the eigenvalue for the parity operator is 1 and the parity is even (sometimes denoted with a +). If $\mathcal{P}\Psi(x,y,z)=-\Psi(x,y,z)$, then the eigenvalue for the parity operator is -1 and the parity is odd (sometimes denoted with a -). Applying the parity operator to harmonic oscillator eigenfunctions gives the result that the ground, 2nd, and 2n-th excited states are even parity, whereas the 1st, 3rd, and 2n+1-th excited states are odd parity.

 

[BillKet]

I may be wrong, but I think you’re overthinking it. Here's a site with probably a clearer explanation of what I'm trying to say:
https://quantummechanics.ucsd.edu/ph130a/130_notes/node153.html
(Ctrl+F for parity--sorry I'm too tired right now to write everything out directly, but the site does a good job explaining parity in the harmonic oscillator eigenfunctions.)

Edit: parity literally just means, given $\Psi(x,y,z)$, consider the operation $\mathcal{P}\Psi(x,y,z)=\Psi(-x,-y,-z)$. If $\mathcal{P}\Psi(x,y,z)=\Psi(x,y,z)$, then the eigenvalue for the parity operator is 1 and the parity is even (sometimes denoted with a +). If $\mathcal{P}\Psi(x,y,z)=-\Psi(x,y,z)$, then the eigenvalue for the parity operator is -1 and the parity is odd (sometimes denoted with a -).

Hmmm, but that is in the lab frame, no? Vibrational modes in a molecule are defined in the molecule frame. For example, if in a diatomic molecule we denote by z the coordinate between the 2 nuclei in the molecule frame, in the molecular frame that will have a parity (as in the article you sent), but in the lab frame that will average out (same way the dipole moment averages out in the lab frame). And given that we define parity in the lab frame, doesn't it mean that the parity of the vibrational motion in the LAB frame is always positive?

 

[TeethWhitener]

Hmmm, but that is in the lab frame, no? Vibrational modes in a molecule are defined in the molecule frame. For example, if in a diatomic molecule we denote by z the coordinate between the 2 nuclei in the molecule frame, in the molecular frame that will have a parity (as in the article you sent), but in the lab frame that will average out (same way the dipole moment averages out in the lab frame). And given that we define parity in the lab frame, doesn't it mean that the parity of the vibrational motion in the LAB frame is always positive?

I'll have to look at this more closely when I'm not so tired, but I'm not sure this is how parity is defined when talking about chiral molecules and the coordinate that reflects through a mirror plane. Even in the lab frame, you can tell the difference between enantiomers (with circular dichroism, for example), and I think that's why the states wrt that reaction coordinate can have well-defined parity.

 

[DrDu]

I think Bunker discusses this in detail in his book on molecular symmetry.
The relevant symmetries of a molecule are given by the CNPI group, the group of "complete nuclear permutations and inversion" group. Specifically, the inversion is the parity operation.
The parity operation will lead to an equivalent rotation which acts on the rotational wavefunction, which will be either symmetric or antisymmetric under this operation.
The point is that you can factor it out and consider the symmetry of the electronic- vibrational-nuclear spin part of the wavefunction separately.

PS: The book by Bunker and Jensen is online here: https://www.theochem.ru.nl/files/local/molecular-symmetry-and-spectroscopy-bunker-jensen.pdf