Timeline for Why do molecules lack the inversion symmetry of the full molecular Hamiltonian?
Current License: CC BY-SA 4.0
16 events
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| Jan 11 at 18:17 | comment | added | my2cts | Downvoting in general is silly. To downvote a correct answer is worse than silly. It is stupid. | |
| Dec 21, 2023 at 22:39 | comment | added | my2cts | @dennismoore94 Good point. I would argue that in the case of HCl the inversion probability is incredibly much smaller than in the NH$_3$ case. NH$_3$ is a pet example because both time scales are accessible. Another example: a phosphorous donor atom in a silicon lattice has no lattice symmetry at any reasonable time scale. Yet the hopping hamiltonian element is never strictly zero. Therefor the stationary solution will be a linear combination of P at every Si site with phase factors corresponding to a crystal wave function. It is a matter of time scale. | |
| Dec 21, 2023 at 19:28 | comment | added | dennismoore94 | Right, every subgroup of $O(3)$ can be called a point group, that was a slip-up on my side, sorry. But your answer and comments still do not resolve how dipole moment arises when the full molecular Hamiltonian is considered. The abelian subgroup $C_i$ (inversion of electron and nuclear coordinates) has only one-dimensional irreps leading to even/odd parity states, both of which have zero average dipole moment. For example, what about the dipole moment of HCl (as a system of 18 electrons and 2 nuclei)? The "zero due to fast umbrella inversion" argument certainly does not work here... | |
| Dec 21, 2023 at 19:18 | comment | added | Hans Wurst | It sounded to me that you were talking about the inversion process like a dynamic process, since you mention a time scale. Looking at your link, I assumed you were thinking about a superposition state of the ground state of the double well potential and the first excited state of the double well. Such a superposition state would have an amplitude that oscillates between the two minima, corresponding to an inversion motion. But that is a superposition state and not an energy eigenstate. I think we are talking past each other at this point. | |
| Dec 21, 2023 at 19:11 | history | edited | my2cts | CC BY-SA 4.0 |
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| Dec 21, 2023 at 18:02 | comment | added | my2cts | @HansWurst How do you mean? The solutions I am talking about have no time dependency. They are eigenfunctions of the hamiltonian. That is, at a time scale exceeding 25 GHz. | |
| Dec 21, 2023 at 17:49 | comment | added | Hans Wurst | That does not resolve the problem. The expectation value of an eigenstate has no time-dependency. We should be able to average at any point in time over an ensemble in a pure state and get zero-dipole moment. Why does that not happen? | |
| Dec 21, 2023 at 16:26 | comment | added | my2cts | @HansWurst Take the example of NH$_3$. This molecule has a permanent electric dipole moment. However ammonia may invert spontaneously. At much longer than electronic time scale it does not have a permanent dipole moment. The time scale is close to 25 GHz. demonstrations.wolfram.com/… | |
| Dec 21, 2023 at 16:15 | comment | added | Hans Wurst | So you agree that any non degenerate ground state cannot have a permanent dipole moment? How do you resolve this with the observation of molecules with permanent dipole moment? | |
| Dec 21, 2023 at 14:35 | history | edited | my2cts | CC BY-SA 4.0 |
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| Dec 21, 2023 at 14:33 | comment | added | my2cts | @dennismoore Point groups only require a fixed position of the entire molecule. Use of a point group also does not imply the Born-Oppenheimer approximation. See my updated answer. Which other group operations would you like to include? | |
| Dec 21, 2023 at 13:18 | comment | added | Hans Wurst | Molecular point groups only come into play once the nuclei are assume to have fixed positions. But we have no fixed positions as long as we treat the nuclei on the same footing as the electrons. And the fundamental Hamiltonian clearly leads to an expectation value of zero for the permanent dipole moment for non-degenerate ground states. This is somewhat disconcerting, given that real molecules have permanent dipoles. So how is it, that we end up with molecules with permanent dipoles despite the properties of the principle Hamiltonian? It is not obvious in my opinion. | |
| Dec 21, 2023 at 12:06 | comment | added | dennismoore94 | The complete group of symmetries is actually much bigger than $C_i$. But even if you only consider inversions, that already points to OP's problem: how to reconcile the experimental fact of molecules having a dipole moment with the exact electron-nucleus energy eigenstates being also parity eigenstates. Referencing point groups is not helpful, since point groups already imply the treatment of nuclei as point charges in fixed positions. Solving the electronic Schrodinger equation for H$_2$O will give you a dipole moment, while solving the full e-N equation will seemingly not. | |
| Dec 21, 2023 at 11:54 | comment | added | my2cts | If you would solve the full hamiltonian you would find superpositions of combined nuclear and electronic states that transform according to the representations of the point group $C_i$. | |
| Dec 21, 2023 at 11:34 | comment | added | dennismoore94 | But the question is exactly about those situations when you cannot rely on the point group concept, since you do not work in the Born-Oppenheimer separation. If nuclei are treated as active quantum particles, then you only have more general symmetries (inversion/rotation of all coordinates, permutation of identical particles, charge conjugation, etc.). | |
| Dec 21, 2023 at 11:25 | history | answered | my2cts | CC BY-SA 4.0 |