Effective molecular Hamiltonian and Hund cases

[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
$<\Sigma(R)|T^N|\Pi(R)>=0$.

 

[DrDu]

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).

View attachment 286277

What is $\xi$ and $\eta$?

 

[BillKet]

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

Actually, for the case of $\Sigma$ and $\Pi$ won't the off diagonal be zero whether I use adiabatic or diabatic? What I mean is, if I use adiabatic, I get the eigenfunctions of $H_{el}$, but the off-diagonal in $T^N$ is zero, as you mentioned. If I use diabatic, I get the eigenfunctions of $T^N$, but the off-diagonal in $H_{el}$ would also be zero, as they have different symmetries again.

 

[BillKet]

What is $\xi$ and $\eta$?

Sorry! $\xi$ is the SO hamiltonian term and $\eta$ the rotational one. So for example $\xi = <\nu_\Sigma|<\Sigma|H_{SO}|\Pi>|\nu_\Pi>$ and $\eta = <\nu_\Sigma|<\Sigma|H_{rot}|\Pi>|\nu_\Pi>$

 

[DrDu]

Actually, for the case of $\Sigma$ and $\Pi$ won't the off diagonal be zero whether I use adiabatic or diabatic? What I mean is, if I use adiabatic, I get the eigenfunctions of $H_{el}$, but the off-diagonal in $T^N$ is zero, as you mentioned. If I use diabatic, I get the eigenfunctions of $T^N$, but the off-diagonal in $H_{el}$ would also be zero, as they have different symmetries again.

As we discussed already in a previous thread
https://www.physicsforums.com/threa...oppenheimer-approximation-doesnt-work.999950/
this is not the case in the adiabatic representation. As the sigma and pi levels intersect, the (non-diagonal) non-adiabatic coupling becomes even singular at the intersection point.
In the diabatic approximation, only small non-diagonal terms, like the SO-coupling remain.

 

[DrDu]

Sorry! $\xi$ is the SO hamiltonian term and $\eta$ the rotational one. So for example $\xi = <\nu_\Sigma|<\Sigma|H_{SO}|\Pi>|\nu_\Pi>$ and $\eta = <\nu_\Sigma|<\Sigma|H_{rot}|\Pi>|\nu_\Pi>$

Ok, so they actually use the vibrational averaging approach in the diabatic representation of the Pi and Sigma states.

 

[BillKet]

Ok, so they actually use the vibrational averaging approach in the diabatic representation of the Pi and Sigma states.

I am pretty sure they are adiabatic. Here is a paper giving a theoretical description of the experimental results in the paper I previously mentioned and they explicitly say that the electronic curves are adiabatic.

But whether they are adiabatic or diabatic, I am not sure how that answers by question about the difference between the 2 approaches.

 

[DrDu]

Concerning the difference between first diagonalizing the hamiltonian and then do vibrational averaging and doing first vibrational averaging: The first possibility is quite hard and can also only be performed perturbationally. Analytical solutions are only available in favourable situations.
For example assuming that the vibrational Pi and Sigma states are all harmonic and only shifted relative to each other and that also the SO coupling depends at most linearly on R.
So if $H=H_0 + \lambda H'$, where $H'$ results from the Spin-Orbit coupling,

$H_0 =\omega a^+ a \sigma_0 + (\alpha+\beta (a+a^+))\sigma_z,$$H' =(\gamma+ \delta(a+a^+))\sigma_x.$

Where $a$ and $a^+$ are the usual HO anihilation and creation operators which can be expressed in terms of R and d/dR. $\sigma_0$ is the 2x2 unit matrix while the other sigmas are the usual Pauli matrices.

We now try to diagonalize $H$ via a unitary transformation
$\exp(-iS)H\exp(iS)$ with

$S=\lambda S_1+ \lambda^2 S_2 \ldots$.
In first order

$i[H_0,S_1]=H'.$

This equation is trivial to solve in an eigenbasis of $H_0$, i.e. using vibrational averaging.
To solve it in terms of R and d/dR, or equivalently in terms of a and $a^+$,
we make for $S_1$ the ansatz

$S_1=(p+qa+ra^+) \sigma_x +(s+ta+ua^+) \sigma_y$

Evidently, the six coefficients p,q,r,s,t and u can be determined comparing terms on the left and right hand side of

$i[H_0,S_1]=H'.$

Now if $\psi_0$ is a zeroth order eigenstate, then

$E= \frac{ <\psi_0| \exp(-i\lambda S_1)H\exp(i\lambda S_1)| \psi_0>}{<\psi_0| \psi_0>}$

is the correct energy eigenvalue up to and including order $\lambda^3$.

What I really want to say is that the non-commutability of the a and $a^+$ or of R and d/dR complicates the diagonalization of the hamiltonian considerably as compared to the case, were the terms depend only on R alone.