Effective molecular Hamiltonian and Hund cases

[Twigg]

Take the Born-Oppenheimer theory expanding the Hamiltonian in the ratio of nuclear to electron mass.

Isn't $\frac{m_{nuc}}{m_e} \geq 1000$? How is that a perturbative power series? If it's a lot of work to type up in TeX, can you provide a citation to the original Born-Oppenheimer paper you mentioned?

 

[Twigg]

@EigenState137 Yeah, sorry, you got invited late to a very long thread with a lot of different questions in it. Right now, we're just looking at the questions in Post #98 (post numbers are in the top right corner of each message), which aren't publication worthy (if that's what your concern is).

Here's a link to Post #98, which we are currently responding to.

From what I've seen in this thread, mostly the OP is just working their way through Brown & Carrington and checking in with folks to sanity check or to fill in the myriad gaps in the logic of B&C. Nothing asked here has seemed novel or publication worthy, so I don't think there's a need to worry about intruding on anything. Further, if the OP worked in a group with diatomic experts, I expect they wouldn't need to post questions here.

 

[EigenState137]

Greetings,

From what I've seen in this thread, mostly the OP is just working their way through Brown & Carrington and checking in with folks to sanity check or to fill in the myriad gaps in the logic of B&C

Not your everyday activity even under Covid19 restrictions. I am not concerned about publishing, I am concerned about a potential research student doint their own research. Old school I admit, but so be it.

Regarding Post#98. I fail to understand the concept of averaging over vibrational states.ES

 

[Twigg]

I fail to understand the concept of averaging over vibrational states.

I think they're just trying to evaluate this mixing Hamiltonian in a given vibrational state. I think the jargon was just a little loose. The OP stated they are only considering one vibrational state, as a simplifying assumption.

I am not concerned about publishing, I am concerned about a potential research student doint their own research. Old school I admit, but so be it.

I get what you mean. I'm sympathetic to the OP's cause because I personally find B&C very difficult to learn from. I had the privilege of working with very smart, very knowledgeable people in my research group who helped me along. Just trying to pay it forward. I respect your position on this

 

[EigenState137]

Greetings,

I get what you mean. I'm sympathetic to the OP's cause because I personally find B&C very difficult to learn from. I had the privilege of working with very smart, very knowledgeable people in my research group who helped me along. Just trying to pay it forward. I respect your position on this

As did I. But that was before the days of online forums. I am just being careful not to step on the toes of some PI--for me it is a policy decision.ES

 

[DrDu]

Isn't $\frac{m_{nuc}}{m_e} \geq 1000$? How is that a perturbative power series? If it's a lot of work to type up in TeX, can you provide a citation to the original Born-Oppenheimer paper you mentioned?

See my post No. 100. The perturbation parameter is evidently $m_\mathrm{e}/m_\mathrm{nuc}$, not the other way round. Try author:Born and author:Oppenheimer in Google Scholar which also yields a link to an english translation:
https://www2.ulb.ac.be/cpm/people/bsutclif/bornopn_corr.pdf

 

[BillKet]

@EigenState137 Yeah, sorry, you got invited late to a very long thread with a lot of different questions in it. Right now, we're just looking at the questions in Post #98 (post numbers are in the top right corner of each message), which aren't publication worthy (if that's what your concern is).

Here's a link to Post #98, which we are currently responding to.

From what I've seen in this thread, mostly the OP is just working their way through Brown & Carrington and checking in with folks to sanity check or to fill in the myriad gaps in the logic of B&C. Nothing asked here has seemed novel or publication worthy, so I don't think there's a need to worry about intruding on anything. Further, if the OP worked in a group with diatomic experts, I expect they wouldn't need to post questions here.

@EigenState137, @Twigg is right. For now I just want to understand diatomic molecules better and while reading B&C (and several other papers meanwhile) I came across different ideas that I asked for help here in order to understand them better. Similarly, the Zeeman and Stark shift came in the same context, me trying to understand diatomic molecules from different perspectives. Also as @Twigg said, right now the question in post #98 is what I am curious about, the other questions I had were well answered in this thread.

I am sorry if "vibrational averaging" was misleading (I am kinda using the B&C terms, but I am not sure how general they are). Basically what I want to understand is this: if I build the full electronic and vibrational Hamiltonian (in my case there are only 2 electronic states with only 1 vibrational level each, but the question can be generalized to realistic cases) and then I diagonalize it, would I get the same results as in B&C, where they first diagonalize the electronic Hamiltonian alone, and then they build a vibrational Hamiltonian for each entry on the diagonal of the electronic Hamiltonian. Intuitively I would say the 2 approaches are the equivalent, but the derivation I wrote above for the simplified case doesn't seem to work. Please let me know if the notation in my derivation in post #98 is confusing, I would be happy to clarify what I meant there.

 

[EigenState137]

Greetings,

For now I just want to understand diatomic molecules better and while reading B&C (and several other papers meanwhile) I came across different ideas that I asked for help here in order to understand them better.

Depending on what you mean by "better" you have articulated a potentially very broad and ambitious aspiration. Most professionals in the field spend their careers focused on one aspect of diatomic molecules, even on one single molecule. It is rare to find those with full breadth knowledge of the field.

Thus I will assume for the moment that your objective is to comprehend a Hamiltonian that describes the structure of a diatomic molecule. Are you comfortable with all of the angular momenta in question? Are you conversant with the corresponding Hamiltonian describing the detailed structure of an atom? If not, I urge you to backtrack and cover those prerequisites.

If you want more than that, then you will ultimately need to master the dynamical properties of diatomics, especially their excited states: photo-ionization, photo-dissociation, pre-dissociation, autoionization and interstate perturbations as examples. That is a life-long quest and frankly one that I simply cannot see a non-practicing professional even attempting to pursue.

Please note that I am not attempting to discourage you. I am attempting to make certain that you understand the breadth and complexity of the field.

On a purely pragmatic note, why have you chosen to follow the development as presented by Brown and Carrington? Our colleague #Twigg has stated

but ho boy do I have a lot of traumatizing books! Brown & Carrington tops that list

There are very complete and high-quality discussions of the structure of diatomic molecules available going back to Herzberg and Townes and Schalow. Lefebvre-Brion and Field, and Carrington, Levy and Miller have published Hamiltonians. Many scholarly treatments are available and I urge you to make use of them rather than to focus too tightly on a single treatment.ES

 

[Twigg]

@DrDu Thanks for bearing with me. With my level of journal access, I was only able to find the original in German. I'm rusty but working my way through it. Is there a particular equation or section that you wanted to highlight, or just the whole paper? Edit: I just re-read your post #99 and I think I've got it.

@EigenState137 Yeah, another user convinced BillKet to branch out into Lefebvre-Brion and Field as well as Brown & Carrington. Also, I called B&C "a traumatizing book" as opposed to "a bad book" because "traumatizing" is subjective and I'm possibly the weak link .

Returning to the questions in post 98, I think your second result (taking the expectation value *before* diagonalizing) is the right approach. Here's my logic for this:

When you say $$\langle \Lambda' | H | \Lambda \rangle = \left( \begin{array} aa(R) & c(R) \\ c(R) & b(R) \\ \end{array} \right)$$, what you are really saying is $$\langle \Lambda'; R | H | \Lambda; R \rangle = \left( \begin{array} aa(R) & c(R) \\ c(R) & b(R) \\ \end{array} \right)$$ where $| \Lambda; R \rangle$ is the state where the $\Lambda$ is well-defined but the vibrational number $\eta_{\Lambda}$ is not defined but instead the vibrational wavefunction has collapsed into the position $R$ (i.e., $\langle R | \psi_{vib} \rangle = \delta (R)$). In other words, if you solve for the eigenvalues of your matrix $\langle \Lambda' | H | \Lambda \rangle$, you are really solving for the energies when the vibrational state is concentrated around a position $R$. It might be a good approximation for coherent states on a dissociating potential in the classical limit (maybe?). However, if you want to talk about the spectrum when the molecule is in the well-defined vibrational states $| \eta_\Lambda \rangle$, then you need to take the vibrational expectation values first like you did in your second approach. If you had more than one vibrational state per electronic manifold, then you would have a block matrix for the Hamiltonian. I'm not betting a kidney on this being correct, but perhaps others can correct me if I'm wrong.

Edit: I thought of a better example in which your first approach is valid. Atomic collisions! So long as the change in interatomic potentials over one de Broglie wavelength is small, then you can approximate the atoms as classical particles. There's a name for this approximation and it eludes me.

 

[EigenState137]

Greetings,

However, if you want to talk about the spectrum when the molecule is in the well-defined vibrational states , then you need to take the vibrational expectation values first like you did in your second approach. If you had more than one vibrational state per electronic manifold, then you would have a block matrix for the Hamiltonian.

I believe that is correct, and any real molecule has more than one vibrational state, thus the general treatment is better. Also, I would treat the vibrational states as anharmonic in general.

I thought of a better example in which your first approach is valid. Atomic collisions! So long as the change in interatomic potentials over one de Broglie wavelength is small, then you can approximate the atoms as classical particles. There's a name for this approximation and it eludes me.

Are you thinking of van der Waals broadening?

Yeah, another user convinced BillKet to branch out into Lefebvre-Brion and Field as well as Brown & Carrington.

Levy and Miller should not be overlooked. ``Electron Resonance of Gaseous Diatomic Molecules,'' A. Carrington, D. H. Levy, and T. A. Miller, Adv. Chem. Phys. 18, 149 (1970). http://dx.doi.org/10.1002/9780470143650.ch4 I think that is the right presentation.ES

 

[Twigg]

Are you thinking of van der Waals broadening?

I just found it in Metcalf's book, and what I was thinking of was the Gallagher-Pritchard model (see pg. 18 of this review) for atomic collisions in the presence of a near-resonant laser field. Similar ideas to vdW broadening. It's mainly relevant for collisions of laser-cooled, trapped alkali atoms. It contributes to loss of atoms out of traps and it is relevant to the field of photoassociation/magnetoassociation of alkali dimer molecules at ultracold temperatures (like the fermi-degenerate KRb gas they reported at JILA a few years back).

 

[EigenState137]

Greetings,

More homework! I'll read it over breakfast in the morning.

Sorry for being slow--I was posting a belated introduction to the community.ES

 

[Twigg]

Ah, no need to put a lot of time into it. It's a bit niche haha. I just threw out the link so people would have something to follow if they were interested.

 

[EigenState137]

Ah mon ami, breakfast is for reading!

 

[BillKet]

Greetings,

Depending on what you mean by "better" you have articulated a potentially very broad and ambitious aspiration. Most professionals in the field spend their careers focused on one aspect of diatomic molecules, even on one single molecule. It is rare to find those with full breadth knowledge of the field.

Thus I will assume for the moment that your objective is to comprehend a Hamiltonian that describes the structure of a diatomic molecule. Are you comfortable with all of the angular momenta in question? Are you conversant with the corresponding Hamiltonian describing the detailed structure of an atom? If not, I urge you to backtrack and cover those prerequisites.

If you want more than that, then you will ultimately need to master the dynamical properties of diatomics, especially their excited states: photo-ionization, photo-dissociation, pre-dissociation, autoionization and interstate perturbations as examples. That is a life-long quest and frankly one that I simply cannot see a non-practicing professional even attempting to pursue.

Please note that I am not attempting to discourage you. I am attempting to make certain that you understand the breadth and complexity of the field.

On a purely pragmatic note, why have you chosen to follow the development as presented by Brown and Carrington? Our colleague #Twigg has stated

There are very complete and high-quality discussions of the structure of diatomic molecules available going back to Herzberg and Townes and Schalow. Lefebvre-Brion and Field, and Carrington, Levy and Miller have published Hamiltonians. Many scholarly treatments are available and I urge you to make use of them rather than to focus too tightly on a single treatment.ES

Thank you for your reply. I totally agree with you about the fact that trying to understand *all* (or most of) diatomic molecular physics would be basically impossible. But my questions are quite basics. I guess I just want to understand a basic Hamiltonian better and this vibrational averaging would be an important step before aiming to understand mode advanced topics. Also, I am not using just B&C. @Twigg adn @amoforum directed me towards Lefebvre-Brion and Field and I am also using Herzberg and Demtroder. However, I didn't find an answer to this particular question in their books (or maybe I missed/missunderstood it).

I think I am now more comfortable with the coupling cases. I am not sure exactly what you mean by the atomic case. Do you mean the electronic part of the Hamiltonian only?

 

[BillKet]

@DrDu Thanks for bearing with me. With my level of journal access, I was only able to find the original in German. I'm rusty but working my way through it. Is there a particular equation or section that you wanted to highlight, or just the whole paper? Edit: I just re-read your post #99 and I think I've got it.

@EigenState137 Yeah, another user convinced BillKet to branch out into Lefebvre-Brion and Field as well as Brown & Carrington. Also, I called B&C "a traumatizing book" as opposed to "a bad book" because "traumatizing" is subjective and I'm possibly the weak link .

Returning to the questions in post 98, I think your second result (taking the expectation value *before* diagonalizing) is the right approach. Here's my logic for this:

When you say $$\langle \Lambda' | H | \Lambda \rangle = \left( \begin{array} aa(R) & c(R) \\ c(R) & b(R) \\ \end{array} \right)$$, what you are really saying is $$\langle \Lambda'; R | H | \Lambda; R \rangle = \left( \begin{array} aa(R) & c(R) \\ c(R) & b(R) \\ \end{array} \right)$$ where $| \Lambda; R \rangle$ is the state where the $\Lambda$ is well-defined but the vibrational number $\eta_{\Lambda}$ is not defined but instead the vibrational wavefunction has collapsed into the position $R$ (i.e., $\langle R | \psi_{vib} \rangle = \delta (R)$). In other words, if you solve for the eigenvalues of your matrix $\langle \Lambda' | H | \Lambda \rangle$, you are really solving for the energies when the vibrational state is concentrated around a position $R$. It might be a good approximation for coherent states on a dissociating potential in the classical limit (maybe?). However, if you want to talk about the spectrum when the molecule is in the well-defined vibrational states $| \eta_\Lambda \rangle$, then you need to take the vibrational expectation values first like you did in your second approach. If you had more than one vibrational state per electronic manifold, then you would have a block matrix for the Hamiltonian. I'm not betting a kidney on this being correct, but perhaps others can correct me if I'm wrong.

Edit: I thought of a better example in which your first approach is valid. Atomic collisions! So long as the change in interatomic potentials over one de Broglie wavelength is small, then you can approximate the atoms as classical particles. There's a name for this approximation and it eludes me.

Thanks a lot for this. I agree that doing the vibrational averaging first would make more sense. In principle calculating the matrix elements for the electronic+vib+rot wavefunctions (in the Hund case basis I choose) should be done before doing any diagonalization. However, unless I missunderstand it, B&C don't do that. For example for the rotational term, they calculate $B^{(1)}(R)$ and $B^{(2)}(R)$ first, in equation 7.85, and only much later, in equation 7.171 they do the vibrational averaging. And given that $B^{(2)}(R)$ comes from the electronic diagonalization, it looks like they actually diagonalize the electronic Hamiltonian first, then they do vibrational averaging. Am I missing something?

 

[EigenState137]

Greetings,

I think I am now more comfortable with the coupling cases. I am not sure exactly what you mean by the atomic case. Do you mean the electronic part of the Hamiltonian only?

I meant the Hamiltonian for atoms. They have all the angular momentum without the complications of the additional degrees of freedom of vibration and rotation. Thus, they can be taken as being less complex. If an atomic Hamiltonian proves troublesome, you are not prepared for diatomic molecules.

it looks like they actually diagonalize the electronic Hamiltonian first, then they do vibrational averaging. Am I missing something?

Given Carrington's reputation, you can be certain that his approach works. How do Lefebvre-Brion and Field and the others treat this exact issue?ES

 

[BillKet]

Greetings,I meant the Hamiltonian for atoms. They have all the angular momentum without the complications of the additional degrees of freedom of vibration and rotation. Thus, they can be taken as being less complex. If an atomic Hamiltonian proves troublesome, you are not prepared for diatomic molecules.Given Carrington's reputation, you can be certain that his approach works. How do Lefebvre-Brion and Field and the others treat this exact issue?ES

Lefebvre-Brion and Field use the same approach as B&C, which is (if I understand it correctly), the other way compared to what @Twigg described and I am not sure I understand why they follow that approach, as @Twigg's approach seems to make more sense physically (unless the 2 approaches are actually equivalent, but I don't see it right now). Basically my question is why do they first diagonalize the electronic Hamiltonian and only after that they define the vibrational levels.

 

[Twigg]

Just a thought, but it could be that they're working under the approximation that $\langle 0_\Pi | b(R) | 0_\Pi \rangle \approx b(R_{eq,\Pi})$ where $R_{eq,\Pi}$ is the equilibrium internuclear distance in the $\Pi$ manifold: $\langle 0_\Pi | R | 0_\Pi \rangle$. This would seem applicable if $\frac{db}{dR}(R_{eq}) \Delta R$ where $\Delta R$ is some measure of the width of the vibrational wavefunction (standard deviation, maybe?). Where this gets confusing is talking about $c(R)$. I'm not sure how that works, since presumably the equilibrium distances are different in the $\Sigma$ and $\Pi$ states.

 

[BillKet]

Just a thought, but it could be that they're working under the approximation that $\langle 0_\Pi | b(R) | 0_\Pi \rangle \approx b(R_{eq,\Pi})$ where $R_{eq,\Pi}$ is the equilibrium internuclear distance in the $\Pi$ manifold: $\langle 0_\Pi | R | 0_\Pi \rangle$. This would seem applicable if $\frac{db}{dR}(R_{eq}) \Delta R$ where $\Delta R$ is some measure of the width of the vibrational wavefunction (standard deviation, maybe?). Where this gets confusing is talking about $c(R)$. I'm not sure how that works, since presumably the equilibrium distances are different in the $\Sigma$ and $\Pi$ states.

I mean, that might be a reasonable approximation, but I don't see why would they do that, given that they aim at a pretty complete treatment, not just an approximation of a PT approach. Also, doing the vibrational averaging first would give the exact result without any extra work, so no approximation needed, no? Also, why would they do the vibrational averaging part at all, if we assume $R=R_{eq}$?

 

[EigenState137]

Greetings,

First of all we just have to acknowledge that Carrington, Lefebvre-Brion, and Field are masters of this field. Thus from a purely functional perspective their approaches work and they work in the general case not just in some contrived pedagogical example such as only one vibrational state per electronic state. My group has utilized Field's Hamiltonian as well as his fitting code to analyze one spectrum with well over 300 rovibronic transitions including perturbations (interstate couplings) observed under Doppler-free resolution.

Second, I still fail to comprehend the vibrational averaging.

I realize none of those comments directly address the question posed, nor is it obvious to me that it should matter physically which way the vibrational states are treated. However, this is becoming a tempest in a tea pot.ES

 

[EigenState137]

Van Vleck transformation

 

[Twigg]

I hear you ES137. We're not questioning the Hamiltonian, we're trying to uncover the logic in deriving it. It sounds like you don't have a copy on hand, so let me summarize. What B&C do is they derive an expression for the rotational constant B(R) under the oversimplified assumption that the ground and excited electronic manifolds have only one vibrational state. They (apparently) do this by solving the spectrum as a function of internuclear separation, and then integrating w.r.t. the vibrational wavefunction ("vibrational averaging"), as in the first method discussed in post 98.

This is why I find B&C traumatizing. You know the results are correct, but at times it feels like the derivations are missing key info for you to replicate them (and sometimes they really are missing key info, so you always end up wondering).

 

[EigenState137]

Greetings,

I am suggesting that the confusion is that they utilize the Van Vleck transformation without explicitly stating that.

The Van Vleck transformation allows each Born-Oppenheimer state to be treated individually while still allowing any number of energetically remote states. Thus using the Van Vleck transformation accounts for interactions between different electronic states as well as interactions between different vibrational states within any given electronic state. It supports the construction of the rotational constant as described above as a function of internuclear separation.

I admit to not having had my coffee yet this morning, but that seems as if it might explain the confusion regarding the Brown and Carrington treatment.ES

 

[BillKet]

Greetings,

I am suggesting that the confusion is that they utilize the Van Vleck transformation without explicitly stating that.

The Van Vleck transformation allows each Born-Oppenheimer state to be treated individually while still allowing any number of energetically remote states. Thus using the Van Vleck transformation accounts for interactions between different electronic states as well as interactions between different vibrational states within any given electronic state. It supports the construction of the rotational constant as described above as a function of internuclear separation.

I admit to not having had my coffee yet this morning, but that seems as if it might explain the confusion regarding the Brown and Carrington treatment.ES

Thank you for your reply. As @Twigg said, I have no doubt that B&C and L-B&F are right, I just want to understand the logic behind what they did.

In B&C they do use Van Vleck transformation, and this is related to my original #98 question. To rephrase it, is the Van Vleck transformation equivalent to just doing a perturbation theory expansion of the Hamiltonian? And if not, what is the difference. And if yes, why is my derivation above not working?

 

[EigenState137]

Greetings,

In B&C they do use Van Vleck transformation, and this is related to my original #98 question. To rephrase it, is the Van Vleck transformation equivalent to just doing a perturbation theory expansion of the Hamiltonian? And if not, what is the difference. And if yes, why is my derivation above not working?

There you have it--both Brown and Carrington and Lefebvre-Brion and Field utilize the Van Vleck transformation. That explains both what they did and why they did it. It both simplifies and generalizes the treatment.ES

 

[EigenState137]

Greetings,

An informative introduction to the Van Vleck transformation from Dudley Herschbach at MIT.ES

 

[BillKet]

Greetings,

There you have it--both Brown and Carrington and Lefebvre-Brion and Field utilize the Van Vleck transformation. That explains both what they did and why they did it. It both simplifies and generalizes the treatment.ES

I understand the explanations in their books, I just want to understand why what @Twigg mentioned doesn't seem to work. I am absolutely not trying to argue that what they did in their book is wrong.

 

[BillKet]

Greetings,

An informative introduction to the Van Vleck transformation from Dudley Herschbach at MIT.ES

Thanks a lot, I will look into that.

 

[EigenState137]

I understand the explanations in their books, I just want to understand why what @Twigg mentioned doesn't seem to work. I am absolutely not trying to argue that what they did in their book is wrong.

Greetings,

I am not certain that the approach mentioned by @Twigg cannot be made to work. However, utilizing the Van Vleck transformation provides are more simple and a more general approach to constructing an Hamiltonian that serves to characterize physically significant molecular parameters.

The Van Vleck transformation accounts for interactions between different electronic states of the molecule. It also accounts for interactions between different vibrational states within the same electronic state. Assuming that the energy separation between Born-Oppenheimer states is large relative to the fine structure splittings, the matrix elements of the Hamiltonian can be quite generally treated via second-order nondegenerate perturbation theory. The formalism thus provides definitions of molecular parameters that reflect interactions that appear within the matrix elements of the effective Hamiltonian.

Couplings between different vibrational levels within a given electronic state result from the radial dependencies of the rotational Hamiltonian and the fine structure Hamiltonian yielding the molecular parameters -D_v, A_J, and a_Dv as described by Zare et. al [1].

Couplings between different electronic states are described by the off-diagonal elements of the spin-orbit Hamiltonian and the rotational Hamiltonian. The effects of such interactions are contained within the normal Λ-doubling parameters such as o_v^Π, p_v^Π, and q_v^Π [1, 2].

The molecular parameters mentioned above are the objective of a full and general analysis of the spectrum and serve to characterize the interactions between electronic states or between vibrational states. If the proposed alternative treatment cannot provide the same physical information as simply and as generally, it is not useful.

[1]. R. N. Zare, A. L. Schmeltekopf, W. J. Harrop, and D. L. Albritton, J. Mol. Spectrosc. 46, 37-66 (1973).
[2]. H. Lefebvre-Brion and R. W. Field, "Perturbations in the Spectra of Diatomic Molecules," Academic Press, Orlando, Fl. pp 226-231, 1986.ES

 

[BillKet]

Greetings,

I am not certain that the approach mentioned by @Twigg cannot be made to work. However, utilizing the Van Vleck transformation provides are more simple and a more general approach to constructing an Hamiltonian that serves to characterize physically significant molecular parameters.

The Van Vleck transformation accounts for interactions between different electronic states of the molecule. It also accounts for interactions between different vibrational states within the same electronic state. Assuming that the energy separation between Born-Oppenheimer states is large relative to the fine structure splittings, the matrix elements of the Hamiltonian can be quite generally treated via second-order nondegenerate perturbation theory. The formalism thus provides definitions of molecular parameters that reflect interactions that appear within the matrix elements of the effective Hamiltonian.

Couplings between different vibrational levels within a given electronic state result from the radial dependencies of the rotational Hamiltonian and the fine structure Hamiltonian yielding the molecular parameters -D_v, A_J, and a_Dv as described by Zare et. al [1].

Couplings between different electronic states are described by the off-diagonal elements of the spin-orbit Hamiltonian and the rotational Hamiltonian. The effects of such interactions are contained within the normal Λ-doubling parameters such as o_v^Π, p_v^Π, and q_v^Π [1, 2].

The molecular parameters mentioned above are the objective of a full and general analysis of the spectrum and serve to characterize the interactions between electronic states or between vibrational states. If the proposed alternative treatment cannot provide the same physical information as simply and as generally, it is not useful.

[1]. R. N. Zare, A. L. Schmeltekopf, W. J. Harrop, and D. L. Albritton, J. Mol. Spectrosc. 46, 37-66 (1973).
[2]. H. Lefebvre-Brion and R. W. Field, "Perturbations in the Spectra of Diatomic Molecules," Academic Press, Orlando, Fl. pp 226-231, 1986.ES

Thanks a lot for this. I was wondering, if I have 2 levels that are very close by and strongly interacting (for example a $\Sigma$ and $\Pi_{1/2}$) such that I can't use PT i.e. electronic off diagonal matrix element are large (assume that the other electronic states are far away from these two). How should I proceed? It is here where I thought that @Twigg derivation makes sense, as in this case I would actually diagonalize the vibrational levels that I need in the 2 electronic states without using PT at all. But given the Van Vleck transformation, I am not sure how to proceed in this situation i.e. when PT can't be used.

 

[DrDu]

Hello again. So I read more molecular papers meanwhile, including cases where perturbation theory wouldn't work and I want to clarify a few things. I would really appreciate your input @Twigg @amoforum. For simplicity assume we have only 2 electronic states, $\Sigma$ and $\Pi$ and each of them has only 1 vibrational level (this is just to be able to write down full equations). The Hamiltonian (full, not effective) in the electronic space is:

$$
\begin{pmatrix}
a(R) & c(R) \\
c(R) & b(R)
\end{pmatrix}
$$

where, for example $a(R) = <\Sigma |a(R)|\Sigma >$ and it contains stuff like $V_{\Sigma}(R)$, while the off diagonal contains stuff like $<\Sigma |L_-|\Pi >$. If we diagonalize this explicitly, we get, say, for the $\Sigma$ state eigenvalue:

$$\frac{1}{2}[a+b+\sqrt{(a-b)^2+4c^2}]$$

Assuming that $c<<a,b$ we can do a first order Taylor expansion and we get:

$$\frac{1}{2}[a+b+(a-b)\sqrt{1+\frac{4c^2}{(a-b)^2}}] = $$

$$\frac{1}{2}[a+b+(a-b)(1+\frac{2c^2}{(a-b)^2})] = $$

$$\frac{1}{2}[2a+\frac{2c^2}{(a-b)})] = $$

$$a+\frac{c^2}{(a-b)} $$

Here by $c^2$ I actually mean the product of the 2 off diagonal terms i.e. $<\Sigma|c(R)|\Pi><\Pi|c(R)|\Sigma>$This is basically the second order PT correction presented in B&C. So I have a few questions:

1. Is this effective Hamiltonian in practice a diagonalization + Taylor expansion in the electronic space, or does this happened to be true just in the 2x2 case above?

2. I am a bit confused how to proceed in a derivation similar to the one above, if I account for the vibrational states, too. If I continue from the result above, and average over the vibrationally states, I would get, for the $\Sigma$ state:

$$<0_\Sigma|(a(R)+\frac{c(R)^2}{(a(R)-b(R))})|0_\Sigma> = $$

$$<0_\Sigma|a(R)|0_\Sigma>+<0_\Sigma|\frac{c(R)^2}{(a(R)-b(R))}|0_\Sigma> $$

where $|0_\Sigma>$ is the vibrational level of the $\Sigma$ state (again I assume just one vibrational level per electronic state). This would be similar to the situation in B&C for the rotational constant in equation 7.87. However, if I include the vibration averaging before diagonalizing I would have this Hamiltonian:

$$
\begin{pmatrix}
<0_\Sigma|a(R)|0_\Sigma> & <0_\Sigma|c(R)|0_\Pi> \\
<0_\Pi|c(R)|0_\Sigma> & <0_\Pi|b(R)|0_\Pi>
\end{pmatrix}
$$

If I do the diagonalization and Taylor expansion as before, I end up with this:

$$<0_\Sigma|a(R)|0_\Sigma>+\frac{<0_\Sigma|c(R)|0_\Pi><0_\Pi|c(R)|0_\Sigma>}{(<0_\Sigma|a(R)|0_\Sigma>-<0_\Pi|b(R)|0_\Pi>)} $$

But this is not the same as above. For the term $<0_\Sigma|c(R)|0_\Pi><0_\Pi|c(R)|0_\Sigma>$, I can assume that $|0_\Pi><0_\Pi|$ is identity (for many vibrational states that would be a sum over them that would span the whole vibrational manifold of the $\Pi$ state), so I get $<0_\Sigma|c(R)^2|0_\Sigma>$, but in order for the 2 expression to be equal I would need:

$$\frac{<0_\Sigma|c(R)^2|0_\Sigma>}{(<0_\Sigma|a(R)|0_\Sigma>-<0_\Pi|b(R)|0_\Pi>)} =
<0_\Sigma|\frac{c(R)^2}{(a(R)-b(R))}|0_\Sigma>
$$

Which doesn't seem to be true in general (the second one has vibrational states of the $\Pi$ states involved, while the first one doesn't). Again, just to be clear by, for example, $<0_\Sigma|a(R)|0_\Sigma>$
I mean $<0_\Sigma|<\Sigma|a(R)|\Sigma>|0_\Sigma>$ i.e. electronically + vibrational averaging.

What am I doing wrong? Shouldn't the 2 approaches i.e. vibrational averaging before or after the diagonalization + Taylor expansion give exactly the same results?

I read again what you are trying to do here. Your first expression for the Hamiltonian ("full, in electronic space") is actually the potential energy matrix for the nuclei in a diabatic representation. This makes sense, as the Sigma and Pi states are of different symmetry as long as Lambda coupling etc are negligible. Hence c(R) is also very small and only non-negligible at $R_\mathrm{is}$ where the two curves a and b intersect $a(R_\mathrm{is})= b(R_\mathrm{is})$. Especially, c(R) may be set to the constant value $c(R_\mathrm{is})$.
Nevertheless the diagonalization of the potential energy matrix, which yields the adiabatic potential, has dramatic effects on both the electronic and the nuclear wavefunctions, as it interchanges the states upon crossing the point $R_\mathrm{is}$. Let the potential energy matrix be $V(R)$, then this matrix is diagonalized by a unitary transformation $U^\dagger(R)V(R)U(R)=W(R)$, where $W(R)$ is the diagonal matrix of the adiabatic potential energy surfaces. However, this transformation does not diagonalize the nuclear hamiltonian $T_\mathrm{nuc}+V(R)$, as $U(R)^\dagger T_\mathrm{nuc}U(R)-T_\mathrm{nuc}$ is not zero and becomes very large at $R_\mathrm{is}$.
So the correct recipe is to use the vibrationally averaged (averaged before diagonalization) Hamiltonian setting $c=c(R_\mathrm{is})$, so that the matrix elements of c become proportional to Frank-Condon factors:

$<\nu_\Sigma|T_\mathrm{nuc}+a(R)|\nu_\Sigma>+c^2\sum_{\nu_\Pi}{(R_\mathrm{is})\frac{|<\nu_\Sigma|\nu_\Pi>| ^2}{(<\nu_\Sigma|T_\mathrm{nuc}+a(R)|\nu_\Sigma>-<\nu_\Pi|T_\mathrm{nuc}+b(R)|\nu_\Pi>)}$

To obtain the same result in the adiabatic representation (i.e. first diagonalizing the potential energy matrix, then averaging over vibrational states), one has to evaluate the non-adiabatic couplings using the Hellmann-Feynman theorem .

 

[BillKet]

I read again what you are trying to do here. Your first expression for the Hamiltonian ("full, in electronic space") is actually the potential energy matrix for the nuclei in a diabatic representation. This makes sense, as the Sigma and Pi states are of different symmetry as long as Lambda coupling etc are negligible. Hence c(R) is also very small and only non-negligible at $R_\mathrm{is}$ where the two curves a and b intersect $a(R_\mathrm{is})= b(R_\mathrm{is})$. Especially, c(R) may be set to the constant value $c(R_\mathrm{is})$.
Nevertheless the diagonalization of the potential energy matrix, which yields the adiabatic potential, has dramatic effects on both the electronic and the nuclear wavefunctions, as it interchanges the states upon crossing the point $R_\mathrm{is}$. Let the potential energy matrix be $V(R)$, then this matrix is diagonalized by a unitary transformation $U^\dagger(R)V(R)U(R)=W(R)$, where $W(R)$ is the diagonal matrix of the adiabatic potential energy surfaces. However, this transformation does not diagonalize the nuclear hamiltonian $T_\mathrm{nuc}+V(R)$, as $U(R)^\dagger T_\mathrm{nuc}U(R)-T_\mathrm{nuc}$ is not zero and becomes very large at $R_\mathrm{is}$.
So the correct recipe is to use the vibrationally averaged (averaged before diagonalization) Hamiltonian setting $c=c(R_\mathrm{is})$, so that the matrix elements of c become proportional to Frank-Condon factors:

$<\nu_\Sigma|T_\mathrm{nuc}+a(R)|\nu_\Sigma>+c^2\sum_{\nu_\Pi}{(R_\mathrm{is})\frac{|<\nu_\Sigma|\nu_\Pi>| ^2}{(<\nu_\Sigma|T_\mathrm{nuc}+a(R)|\nu_\Sigma>-<\nu_\Pi|T_\mathrm{nuc}+b(R)|\nu_\Pi>)}$

To obtain the same result in the adiabatic representation (i.e. first diagonalizing the potential energy matrix, then averaging over vibrational states), one has to evaluate the non-adiabatic couplings using the Hellmann-Feynman theorem .

Thank you. I am no sure why you are saying that the representation is diabatic. I actually followed the B&C logic and they use only adiabatic states. So in the Hamiltonian above, for example, $a(R)$ would include $V_{\Sigma}(R)$ while $c(R)$ would include $T^N$ and $A(R),B(R)$. Actually in general you can write a matrix like that both in the diabatic and adiabatic representation, you will always have off-diagonal terms (e.g. from SO coupling, rotation). Why would that matrix be in the diabatic representation?

 

[DrDu]

Thank you. I am no sure why you are saying that the representation is diabatic. I actually followed the B&C logic and they use only adiabatic states. So in the Hamiltonian above, for example, $a(R)$ would include $V_{\Sigma}(R)$ while $c(R)$ would include $T^N$ and $A(R),B(R)$. Actually in general you can write a matrix like that both in the diabatic and adiabatic representation, you will always have off-diagonal terms (e.g. from SO coupling, rotation). Why would that matrix be in the diabatic representation?

I don't know the B&C book. In general, in the adiabatic representation the potential energy seen by the nuclei is diagonal while there are non-diagonal non-adiabatic coupling terms which contain the nuclear momentum operator -id/dR. If the latter coupling terms are removed by a unitary transformation between electronic states of interest, non-diagonal potential terms are introduced. This is the diabatic representation. I supposed that a(R), b(r) and c(R) are functions of R, not operators containing also e.g. $T^N$.
Anyhow you assume that $a$ and $b$ are linked to Sigma and Pi states which is incompatible with adiabatic states if the order of the electronic Sigma and Pi states changes as a function of R.

 

[BillKet]

I don't know the B&C book. In general, in the adiabatic representation the potential energy seen by the nuclei is diagonal while there are non-diagonal non-adiabatic coupling terms which contain the nuclear momentum operator -id/dR. If the latter coupling terms are removed by a unitary transformation between electronic states of interest, non-diagonal potential terms are introduced. This is the diabatic representation. I supposed that a(R), b(r) and c(R) are functions of R, not operators containing also e.g. $T^N$.
Anyhow you assume that $a$ and $b$ are linked to Sigma and Pi states which is incompatible with adiabatic states if the order of the electronic Sigma and Pi states changes as a function of R.

I am not sure I understand. What I mean by that Hamiltonian is that you solve the Schrodinger equation for the electrostatic Hamiltonian and you get 2 eigenvalues as function of R: $V_{\Sigma}(R)$ and $V_{\Pi}(R)$, then you compute the matrix elements of other terms in the molecular Hamiltonian in this basis. For example from the SO Hamiltonian you will have both diagonal terms i.e. $<\Sigma|H_{SO}(R)|\Sigma>$ and $<\Pi|H_{SO}(R)|\Pi>$ but also off-diagonal terms $<\Sigma|H_{SO}(R)|\Pi>$, so you end up with the general form the of the Hamiltonian I mentioned, while using adiabatic states i.e. eigenstates of the electrostatic Hamiltonian.

Also $a(R), b(R)$ and $c(R)$ do contain $T^N$. This is what gives the kinetic energy of the vibrational states after the vibrational averaging.