I Effective molecular Hamiltonian and Hund cases

[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.

 

[DrDu]

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.

Nevertheless, the situation should be clear: at least c(R) is an ordinary function as T^N is diagonal and if it is small everywhere, it can be treated as a perturbation. You first determine the eigenvalues and eigenstates of the unperturbed hamiltonian (Sigma and Pi states times the vibrational states) and then calculate the matrix elements of c.
You can't start and first diagonalize the matrix of the a, b and c, as they aren't numbers, but non-commuting operators. The notation a(R) is missleading in this respect. Better would be $a(P_\mathrm{N}, R)$, where $P_\mathrm{N}=-id/dR$ is the nuclear momentum operator.
The above mentioned van Vleck method tries to diagonalize this operator perturbatively, but I think in the situation at hand this is rather overkill.

 

[EigenState137]

Greetings,

The above mentioned van Vleck method tries to diagonalize this operator perturbatively, but I think in the situation at hand this is rather overkill.

I simply cannot agree with this statement.

From the outset of my participation in this discussion it has been my understanding that @BillKet has endeavored to understand the development of the effective Hamiltonian as presented by Brown and Carrington. The "what" did they do and the "why" did they do it. Those authors utilize the Van Vleck transformation, as do Lefebvre-Brion and Field. To the extent that my understanding of the objectives of this discussion are correct, then the Van Vleck transformation if fundamental to the topic at hand.ES

Edit: Deleted the word sentiment replacing it with statement.

 

[BillKet]

Nevertheless, the situation should be clear: at least c(R) is an ordinary function as T^N is diagonal and if it is small everywhere, it can be treated as a perturbation. You first determine the eigenvalues and eigenstates of the unperturbed hamiltonian (Sigma and Pi states times the vibrational states) and then calculate the matrix elements of c.
You can't start and first diagonalize the matrix of the a, b and c, as they aren't numbers, but non-commuting operators. The notation a(R) is missleading in this respect. Better would be $a(P_\mathrm{N}, R)$, where $P_\mathrm{N}=-id/dR$ is the nuclear momentum operator.
The above mentioned van Vleck method tries to diagonalize this operator perturbatively, but I think in the situation at hand this is rather overkill.

$c(R)$ does contain $T^N$. The way I build this Hamiltonian is by finding the eigenstates of $H_{el}$ and we do have that $<\Sigma|T^N|\Pi>\neq 0$

 

[BillKet]

Greetings,

I simply cannot agree with this sentiment.

From the outset of my participation in this discussion it has been my understanding that @BillKet has endeavored to understand the development of the effective Hamiltonian as presented by Brown and Carrington. The "what" did they do and the "why" did they do it. Those authors utilize the Van Vleck transformation, as do Lefebvre-Brion and Field. To the extent that my understanding of the objectives of this discussion are correct, then the Van Vleck transformation if fundamental to the topic at hand.ES

That's right, my question was in the context of B&C derivation so Van Vleck transformation plays an important role. @Twigg @EigenState137 I am sorry, this thread got so long, but I am not even sure if my question was answered. I still don't see why @Twigg's derivation doesn't agree with the one in B&C, given that the B&C one is for sure right, but @Twigg's one seems correct, too.

 

[DrDu]

$c(R)$ does contain $T^N$. The way I build this Hamiltonian is by finding the eigenstates of $H_{el}$ and we do have that $<\Sigma|T^N|\Pi>\neq 0$

So could you please write down your definitions of a, b and c in the case with many vibrational levels (and how these are defined)?

Greetings,

I simply cannot agree with this sentiment.

From the outset of my participation in this discussion it has been my understanding that @BillKet has endeavored to understand the development of the effective Hamiltonian as presented by Brown and Carrington. The "what" did they do and the "why" did they do it. Those authors utilize the Van Vleck transformation, as do Lefebvre-Brion and Field. To the extent that my understanding of the objectives of this discussion are correct, then the Van Vleck transformation if fundamental to the topic at hand.ES

Why is this a sentiment? I just tried to clarify why using something like diagonalization of a Hamiltonian via the van Vleck transformation might yield the same result as the simpler vibrational averaging proposed by Billket in his concrete example.

 

[EigenState137]

Greetings,

Why is this a sentiment?

Edited to read statement.ES

 

[BillKet]

So could you please write down your definitions of a, b and c in the case with many vibrational levels (and how these are defined)?

Why is this a sentiment? I just tried to clarify why using something like diagonalization of a Hamiltonian via the van Vleck transformation might yield the same result as the simpler vibrational averaging proposed by Billket in his concrete example.

In this paper, Table I, they show something very similar to what I am trying to describe. The only difference is that they use 3 electronic levels, but they still use just only one vibrational level for the same of simplicity. And they do a mix of B&C and @Twigg. They do a Van Vleck transformation on the diagonal, but they average the vibrational levels on the off-diagonal, before fully diagonalizing the Hamiltonian (which they do later), which actually makes me even more believe that @Twigg's derivation and B&C are equivalent.

 

[DrDu]

In this paper, Table I, they show something very similar to what I am trying to describe. The only difference is that they use 3 electronic levels, but they still use just only one vibrational level for the same of simplicity. And they do a mix of B&C and @Twigg. They do a Van Vleck transformation on the diagonal, but they average the vibrational levels on the off-diagonal, before fully diagonalizing the Hamiltonian (which they do later), which actually makes me even more believe that @Twigg's derivation and B&C are equivalent.

I fear I don't have access to this paper.

 

[BillKet]

I fear I don't have access to this paper.

Sorry about that! Here is a screenshot of that table (not sure how to attach the paper here). The table is cut on the right side a bit from the paper (it's quite an old paper).

 

[DrDu]

$c(R)$ does contain $T^N$. The way I build this Hamiltonian is by finding the eigenstates of $H_{el}$ and we do have that $<\Sigma|T^N|\Pi>\neq 0$

Does it? As Sigma and Pi have different symmetry, $<\Sigma(R)|\Pi(R')>=0$ for all R and R', hence also
$<\Sigma(R)|T^N|\Pi(R)>=0$.