I Effective molecular Hamiltonian and Hund cases

[Twigg]

I'm definitely out of my comfort zone here, so take this with a grain of salt. Your result says that the only Stark shift is due to the permanent dipole moment of the molecule, and I don't buy that. I think what's missing is the off-diagonal couplings between degenerate electronic levels, like a $\Lambda$-doublet. There should be some polarizability there that scales inversely with the $\Lambda$-doubling energy splitting, I think? I'm not 100% sure about the scaling, that's just something I think I remember reading in a review paper, but they were talking about $\Omega$-doublets.

My handiness with Wigner algebra is crud, but the angular part looks right. Wolfram says $D^{1}_{00}(\psi,\theta,\phi) = \cos \theta$, so it certainly seems reasonable.

 

[amoforum]

Billket, I'm assuming you went through Section 6.11.6 in B&C? In that case, yes, you'll end up keeping the first term in equation 6.333 (for an intra-electronic transition. They mention the first terms goes to zero for inter-electronic transitions.). And then your matrix element is equation 6.331, where $cos\theta$ is your Wigner matrix element.

Twigg, I believe the $\Lambda$-doubling is contained in the rotational part of the matrix element. It shows up once you pick the basis set, which will include the parity eigenstates. And then when solving for the polarizability, that energy splitting will show up in the denominator, the energy of which is governed by the interaction that splits the parity eigenstates.

 

[BillKet]

Billket, I'm assuming you went through Section 6.11.6 in B&C? In that case, yes, you'll end up keeping the first term in equation 6.333 (for an intra-electronic transition. They mention the first terms goes to zero for inter-electronic transitions.). And then your matrix element is equation 6.331, where $cos\theta$ is your Wigner matrix element.

Twigg, I believe the $\Lambda$-doubling is contained in the rotational part of the matrix element. It shows up once you pick the basis set, which will include the parity eigenstates. And then when solving for the polarizability, that energy splitting will show up in the denominator, the energy of which is governed by the interaction that splits the parity eigenstates.

@Twigg @amoforum thank you for your replies. So I think I did make a mistake for the case of $\Lambda \neq 0$, as there I should first calculate the matrix element for Hund case a and after that combine the Hund case a basis into parity eigenstates. I think I did it the other way around. I will look more closely into that. I also took a look over section 6.11.6, thank you for pointing me towards that. I actually have a quick question about the BO approximation now (unrelated to the EDM calculation). Before equation 6.333 they say: "We now make use of the Born–Oppenheimer approximation which allows us to separate the electronic and vibrational wave functions" and this is the typical statement you see in probably all books on molecular physics. And now I am wondering if I am missing something. Of course BO approximation allows that separation, but after reading the effective hamiltonian chapter it seems like that separation is always true, up to any desired order in PT. BO approximation is basically the zeroth order and that kind of statement implies that the separation is valid only under that very constraining assumption. Isn't that separation always true once we do these PT changes (isn't this the whole point of the effective Hamiltonian)? Along the same lines, I just wanted to make sure I understand how one goes from BO to Hund cases. In BO, one has a wavefunction of the form $|\eta\nu J M>=|\eta>| \nu>Y_{J}^{M}(\theta,\phi)$, where $|\eta>$ is the electronic wavefunction (in the intrinsic frame of the molecule), $|\nu>$ is the vibrational wavefunction and $Y_{J}^{M}(\theta,\phi)$ is the spherical harmonic, showing the rotation of the molecule frame with respect to the lab frame. Then using an identity of the form $Y_{J}^{M}(\theta,\phi)=\sum_{\Omega=-J}^{J}\sqrt{\frac{2J+1}{8\pi^2}}\mathcal{D}_{M\Omega}^{J*}(\theta,\phi)$ (I might be off with that constant) we are able to get the Hund cases, which for case a, for example, based on this equation would become $|\eta>|\nu>|J \Omega M>|S\Sigma>$ where $\mathcal{D}_{M\Omega}^{J*}(\theta,\phi)=|J \Omega M>$ and $|S\Sigma>$ was pulled out by hand for completeness. Is this correct? Thank you!

 

[Twigg]

Thanks for the clarification, @amoforum! And @BillKet, I'd actually be curious to see what you come up with for the stark shift, if you find the time. I tried to spend some time learning this once but my coworkers weren't having it and sent me back to mixing laser dye No pressure, of course!

As far as the BO approximation, when we did spectroscopy we didn't really keep a detailed effective Hamiltonian, we would just re-measure the lifetimes and rotational constants in other vibrational states if there was a need to do so. I think in molecules where the BO violation is weak, you can take this kind of pragmatic approach. Then again, we only thought about molecules with highly diagonal Franck-Condon factors so we never really ventured above $\nu = 2$ or so.

 

[amoforum]

@BillKet Yes, you assume the BO approximation first, then handle the non-adiabatic terms with perturbation theory. i.e. those parameters (or "constants") in the effective Hamiltonian.

As for your second question, that looks right, except I want to clarify: a spherical harmonic is actually a subset of the generalized rotation matrix elements (see Eq. 5.52 in B&C). More generally, you'd start with asymmetric top eigenfunctions (eigenfunctions of Eq. 5.58 in B&C), which for a diatomic would then reduce to symmetric top wavefunctions. B&C Section 5.3.4 might be helpful.

 

[BillKet]

@BillKet Yes, you assume the BO approximation first, then handle the non-adiabatic terms with perturbation theory. i.e. those parameters (or "constants") in the effective Hamiltonian.

As for your second question, that looks right, except I want to clarify: a spherical harmonic is actually a subset of the generalized rotation matrix elements (see Eq. 5.52 in B&C). More generally, you'd start with asymmetric top eigenfunctions (eigenfunctions of Eq. 5.58 in B&C), which for a diatomic would then reduce to symmetric top wavefunctions. B&C Section 5.3.4 might be helpful.

@amoforum I looked into more details at the derivation in B&C in section 6.11.6 and I am actually confused a bit. Using spherical tensors for now, the transition between 2 electronic levels would be:

$$E_z\sum_q<\eta'|<\nu'|<r'|\mathcal{D}_{0q}T_q^1(d_{el}+d_{nucl})|r>|\nu>|\eta> = $$
$$E_z\sum_q<\eta'|<\nu'|<r'|\mathcal{D}_{0q}T_q^1(d_{el})|r>|\nu>|\eta> + E_z\sum_q<\eta'|<\nu'|<r'|\mathcal{D}T_q^1(d_{nucl})|r>|\nu>|\eta> = $$
$$E_z\sum_q<\nu'|<\eta'|T_q^1(d_{el})|\eta>|\nu><r'|\mathcal{D}_{0q}|r> + E_z\sum_q<\eta'||\eta><\nu'|<r'|\mathcal{D}T_q^1(d_{nucl})|r>|\nu> $$

The second term is zero because $<\eta'||\eta> = 0$. But the first term is different from the one in B&C equation 6.331. First of all, differently from before (transitions within a given electronic state), $d_{el}$ has component in the intrinsic frame for $q=\pm 1$, not only for $q=0$, so that term is not just $cos(\theta)$ anymore. Why do they ignore the other 2 terms? Also the expectation value $<\eta'|T_q^1(d_{el})|\eta>$ is a function of R (the electronic wavefunctions have a dependence on R), so we can't just take them out of the vibrational integral like B&C do in 6.332. What am I missing?

 

[amoforum]

@amoforum I looked into more details at the derivation in B&C in section 6.11.6 and I am actually confused a bit. Using spherical tensors for now, the transition between 2 electronic levels would be:

$$E_z\sum_q<\eta'|<\nu'|<r'|\mathcal{D}_{0q}T_q^1(d_{el}+d_{nucl})|r>|\nu>|\eta> = $$
$$E_z\sum_q<\eta'|<\nu'|<r'|\mathcal{D}_{0q}T_q^1(d_{el})|r>|\nu>|\eta> + E_z\sum_q<\eta'|<\nu'|<r'|\mathcal{D}T_q^1(d_{nucl})|r>|\nu>|\eta> = $$
$$E_z\sum_q<\nu'|<\eta'|T_q^1(d_{el})|\eta>|\nu><r'|\mathcal{D}_{0q}|r> + E_z\sum_q<\eta'||\eta><\nu'|<r'|\mathcal{D}T_q^1(d_{nucl})|r>|\nu> $$

The second term is zero because $<\eta'||\eta> = 0$. But the first term is different from the one in B&C equation 6.331. First of all, differently from before (transitions within a given electronic state), $d_{el}$ has component in the intrinsic frame for $q=\pm 1$, not only for $q=0$, so that term is not just $cos(\theta)$ anymore. Why do they ignore the other 2 terms? Also the expectation value $<\eta'|T_q^1(d_{el})|\eta>$ is a function of R (the electronic wavefunctions have a dependence on R), so we can't just take them out of the vibrational integral like B&C do in 6.332. What am I missing?

As to your first question:

For an intra-electronic transition, you're coupling two electronic states that have exactly the same electron spatial distribution. Electron population is distributed symmetrically about the molecular axis, so there is no permanent dipole moment perpendicular to the molecular axis for the electric field to interact with. There might be at really short time scales, but then we're not in the Born-Oppenheimer regime anymore.

The same argument applies for inter-electronic transitions, where you're coupling two electronic states that have different electron spatial distributions. Hence, there's usually a dipole moment to interact with (considering symmetry and all that).

As to your second question:

B&C haven't separated the electronic and vibrational integrals in Eq. 6.332 yet. They first apply the Born-Oppenheimer approximation, then separate them. The R-dependence shows up in Eq. 6.333.

 

[BillKet]

As to your first question:

For an intra-electronic transition, you're coupling two electronic states that have exactly the same electron spatial distribution. Electron population is distributed symmetrically about the molecular axis, so there is no permanent dipole moment perpendicular to the molecular axis for the electric field to interact with. There might be at really short time scales, but then we're not in the Born-Oppenheimer regime anymore.

The same argument applies for inter-electronic transitions, where you're coupling two electronic states that have different electron spatial distributions. Hence, there's usually a dipole moment to interact with (considering symmetry and all that).

As to your second question:

B&C haven't separated the electronic and vibrational integrals in Eq. 6.332 yet. They first apply the Born-Oppenheimer approximation, then separate them. The R-dependence shows up in Eq. 6.333.

For the first question:

The dipole moment, as an operator, has 2 components $d_{el}(r)$ and $d_{nucl}(R)$. When the transition is within the same electronic state, what we are left with is $\sum_q <T_q^1(d_{nucl}(R))>$. But for $d_{nucl}(R)$ there is only the q=0 component, so there it is obvious why we drop the $q=\pm 1$ terms. But in the case for transitions between 2 different electronic states, we are left with $\sum_q <\eta|T_q^1(d_{el}(r))|\eta'>$. I am not sure why in this case, for example $<\eta|T_q^1(d_{el}(r))|\eta'>$ would be zero, this is equivalent to $<\eta|x|\eta'>=0$ and $<\eta|y|\eta'>=0$. Is it because of the cylindrical symmetry?

For the second question:

I am a bit confused. Starting from the second integral of 6.332 we have:

$$\int{\int{\psi_\nu\psi_e r \psi_e'\psi_\nu'}}$$

(I dropped some terms, complex conjugates etc. for simplicity). By adding the dependence on different variables we have:

$$\int{\int{\psi_\nu(R)\psi_e(r,R) r \psi_e'(r,R)\psi_\nu'(R)}}$$

which is equal to

$$\int\psi_\nu(R){(\int{\psi_e(r,R) r \psi_e'(r,R))}\psi_\nu'(R)}$$

If we denote $f(R)=\int{\psi_e(r,R) r \psi_e'(r,R) dr}$ the integral above becomes:

$$\int{\psi_\nu(R)f(R)\psi_\nu'(R)}dR$$

but this is not equal to $$(\int{\psi_\nu(R)\psi_\nu'(R)}dR)f(R)$$ we can't just take the $f(R)$ out of that integral, as it depends explicitly on $R$ and I don't see how BO approximation would allow us to do that. BO allowed us to write the function as the product of the electronic and vibrational wavefunctions, but after that doing these integrals is just math.

 

[amoforum]

$$E_z\sum_q<\eta'|<\nu'|<r'|\mathcal{D}_{0q}T_q^1(d_{el}+d_{nucl})|r>|\nu>|\eta> = $$
$$E_z\sum_q<\eta'|<\nu'|<r'|\mathcal{D}_{0q}T_q^1(d_{el})|r>|\nu>|\eta> + E_z\sum_q<\eta'|<\nu'|<r'|\mathcal{D}T_q^1(d_{nucl})|r>|\nu>|\eta> = $$
$$E_z\sum_q<\nu'|<\eta'|T_q^1(d_{el})|\eta>|\nu><r'|\mathcal{D}_{0q}|r> + E_z\sum_q<\eta'||\eta><\nu'|<r'|\mathcal{D}T_q^1(d_{nucl})|r>|\nu> $$

Unfortunately, I can't look at this until later tonight, but I need revisit your derivation above more carefully. Because, I see now that the they way you have it derived, the rotational part forces the electronic part to only $q = 0$ terms, which has to be wrong because inter-electronic transitions exist, and I guess this is the heart of your question. (Maybe the very first line is wrong.)

If you take a look at Eqn 6.331, the sum over all components is clearly still there for the non-rotational components. Read over Section 6.11.4, and revisit how Eqn 6.330 turns into 6.331, and I suspect the discrepancy will show up. i.e. the rotational part got completely separated.

 

[amoforum]

I think I can answer the second question for now. Eqn. 6.333 I believe has some sloppy notation. The second integral should maybe have a different symbol for $R_\alpha$ for the electronic part. It's meant to be at a single internuclear distance, usually the equilibrium distance. So you don't integrate over it. Some other texts might call this the "crude" BO approximation, and Eqn. 6.330 would be the usual BO approximation. Then there's also the Condon approximation which assumes there's no dependence on the nuclear coordinates at all.

 

[BillKet]

I think I can answer the second question for now. Eqn. 6.333 I believe has some sloppy notation. The second integral should maybe have a different symbol for $R_\alpha$ for the electronic part. It's meant to be at a single internuclear distance, usually the equilibrium distance. So you don't integrate over it. Some other texts might call this the "crude" BO approximation, and Eqn. 6.330 would be the usual BO approximation. Then there's also the Condon approximation which assumes there's no dependence on the nuclear coordinates at all.

Thank you for your reply. I will look at the sections you suggested for questions 1. For the second one, I agree that if that $R_\alpha$ is a constant, we can take the electronic integral out of the vibrational integral, but I am not totally sure why can we do this. If we are in the BO approximation, the electronic wavefunction should be a function of $R$, for $R$ not constant, and that electronic integral would be a function of $R$, too. But why would we assume it is constant? I understand the idea behind BO approximation, that the electrons follow the nuclear motion almost instantaneously, but I don't get it here. It is as if the nuclei oscillate so fast that the electrons don't have time to catch up and they just the see the average inter-nuclear distance, which is kinda the opposite of BO approximation. Could you help me a bit understand this assumption that the electronic integral is constant? Thank you!

 

[amoforum]

It is as if the nuclei oscillate so fast that the electrons don't have time to catch up and they just the see the average inter-nuclear distance.

Can you elaborate on how you arrived at this interpretation? Why does it imply that the electrons can't catch up? The "crude" BO approximation gives you a dipole moment result at a specific $R$. If you on average only observe a specific $R_{eq}$ (equilibrium distance), then the electronic integral at $R_{eq}$ will be your observed dipole moment.

 

[amoforum]

@amoforum I looked into more details at the derivation in B&C in section 6.11.6 and I am actually confused a bit. Using spherical tensors for now, the transition between 2 electronic levels would be:

$$E_z\sum_q<\eta'|<\nu'|<r'|\mathcal{D}_{0q}T_q^1(d_{el}+d_{nucl})|r>|\nu>|\eta> = $$
$$E_z\sum_q<\eta'|<\nu'|<r'|\mathcal{D}_{0q}T_q^1(d_{el})|r>|\nu>|\eta> + E_z\sum_q<\eta'|<\nu'|<r'|\mathcal{D}T_q^1(d_{nucl})|r>|\nu>|\eta> = $$
$$E_z\sum_q<\nu'|<\eta'|T_q^1(d_{el})|\eta>|\nu><r'|\mathcal{D}_{0q}|r> + E_z\sum_q<\eta'||\eta><\nu'|<r'|\mathcal{D}T_q^1(d_{nucl})|r>|\nu> $$

Okay, here's my stab at the first question:

The derivation is completely written out in Lefebvre-Brion/Field in Section 6.1.2.1, and it looks like yours is consistent.

Now as to why there's only cos$\theta$ in B&C's version. I suspect this is all because of Eqn. 6.330 in B&C. Notice that the rotational wavefunctions are spherical harmonics for both the initial and final states. The symmetric top wavefunctions reduce to spherical harmonics if $\Omega = 0$. (See the text above Eqn 5.52 and reconcile that with Eqn. 5.145). This is a very constraining assumption, because that means both states must be $\Omega = 0$, like $^1\Sigma$ states. (I guess we can constrain ourselves to $M = 0$ states too?) And if that's the case, then the 3j-symbol has $\Omega = 0$ and $\Omega' = 0$ in its bottom row, meaning $q$ must equal zero for it to not vanish.

So then the only thing I can't reconcile is the sentence after Eqn. 6.331 that says $\Delta J = 0$ is allowed. To me that's only true if you have symmetric top wavefunctions, because then you can have a change in both $\Omega$ and $J$ that adds to zero.

I wouldn't be surprised that this detail was glossed over, considering that the main point they wanted to get across in that section was the electronic-vibrational stuff like Franck-Condon factors and allowed electronic transitions in homonuclears.

 

[BillKet]

Can you elaborate on how you arrived at this interpretation? Why does it imply that the electrons can't catch up? The "crude" BO approximation gives you a dipole moment result at a specific $R$. If you on average only observe a specific $R_{eq}$ (equilibrium distance), then the electronic integral at $R_{eq}$ will be your observed dipole moment.

I guess I don't understand what is the mathematical approximation used that allows you to assume that the electronic integral, which is a function of $R$ can be approximated to be constant. In the BO approximation you would use the adiabatic approximation, but I am not sure here, formally, what allows you to do that. Intuitively, if you have, say, the function $cos^2$, but your response time to this oscillation is too slow, what you see is the average over many periods which is $1/2$. Given that that electronic integral sees just the average internuclear distance, I assumed it is something similar to this i.e. the electrons see just an average of the internuclear distance.

 

[BillKet]

Okay, here's my stab at the first question:

The derivation is completely written out in Lefebvre-Brion/Field in Section 6.1.2.1, and it looks like yours is consistent.

Now as to why there's only cos$\theta$ in B&C's version. I suspect this is all because of Eqn. 6.330 in B&C. Notice that the rotational wavefunctions are spherical harmonics for both the initial and final states. The symmetric top wavefunctions reduce to spherical harmonics if $\Omega = 0$. (See the text above Eqn 5.52 and reconcile that with Eqn. 5.145). This is a very constraining assumption, because that means both states must be $\Omega = 0$, like $^1\Sigma$ states. (I guess we can constrain ourselves to $M = 0$ states too?) And if that's the case, then the 3j-symbol has $\Omega = 0$ and $\Omega' = 0$ in its bottom row, meaning $q$ must equal zero for it to not vanish.

So then the only thing I can't reconcile is the sentence after Eqn. 6.331 that says $\Delta J = 0$ is allowed. To me that's only true if you have symmetric top wavefunctions, because then you can have a change in both $\Omega$ and $J$ that adds to zero.

I wouldn't be surprised that this detail was glossed over, considering that the main point they wanted to get across in that section was the electronic-vibrational stuff like Franck-Condon factors and allowed electronic transitions in homonuclears.

Oh I see, that makes sense. Thanks a lot! I still have a quick question about the electronic integral. In order to have transitions between different electronic states, as you mentioned, terms of the form $$<\eta'|T_{\pm 1}^1(r)|\eta>$$ should not be zero. But this is equivalent to $$<\eta'|x|\eta>$$ not being zero (and same for $y$). However, the electronic wavefunctions have cylindrical symmetry, so they should be even functions of x and y (here all the coordinates are in the intrinsic molecular (rotating) frame). Wouldn't in this case $$<\eta'|x|\eta>$$ be zero?

 

[amoforum]

I guess I don't understand what is the mathematical approximation used that allows you to assume that the electronic integral, which is a function of $R$ can be approximated to be constant. In the BO approximation you would use the adiabatic approximation, but I am not sure here, formally, what allows you to do that. Intuitively, if you have, say, the function $cos^2$, but your response time to this oscillation is too slow, what you see is the average over many periods which is $1/2$. Given that that electronic integral sees just the average internuclear distance, I assumed it is something similar to this i.e. the electrons see just an average of the internuclear distance.

I'd say it's more of a physical approximation than a mathematical one. For low vibrational states (shorter internuclear distances), the region of the dipole moment function is relatively flat. So just picking the equilibrium distance actually approximates it pretty well. At high vibrational states, where you'd sample large internuclear distances, the curve starts to get wobbly on the outskirts and the approximation breaks down. This makes sense because you'd expect BO breakdown at higher vibrational energies.

 

[amoforum]

Oh I see, that makes sense. Thanks a lot! I still have a quick question about the electronic integral. In order to have transitions between different electronic states, as you mentioned, terms of the form $$<\eta'|T_{\pm 1}^1(r)|\eta>$$ should not be zero. But this is equivalent to $$<\eta'|x|\eta>$$ not being zero (and same for $y$). However, the electronic wavefunctions have cylindrical symmetry, so they should be even functions of x and y (here all the coordinates are in the intrinsic molecular (rotating) frame). Wouldn't in this case $$<\eta'|x|\eta>$$ be zero?

Time to look at some molecular orbitals! Only $\Sigma$ states have cylindrical symmetry, which as you've pointed out, means $\Sigma$ to $\Sigma$ transitions are not allowed, unless you go from $\Sigma^+$ to $\Sigma^-$, the latter of which is not symmetric along the internuclear axis, but still only $q = 0$ transitions allowed.

Take a look at some $\Pi$ or $\Delta$ orbitals. They are absolutely not cylindrically symmetric, because they look like linear combinations of $p$ and $d$ orbitals.

 

[BillKet]

Time to look at some molecular orbitals! Only $\Sigma$ states have cylindrical symmetry, which as you've pointed out, means $\Sigma$ to $\Sigma$ transitions are not allowed, unless you go from $\Sigma^+$ to $\Sigma^-$, the latter of which is not symmetric along the internuclear axis, but still only $q = 0$ transitions allowed.

Take a look at some $\Pi$ or $\Delta$ orbitals. They are absolutely not cylindrically symmetric, because they look like linear combinations of $p$ and $d$ orbitals.

Thanks for the vibrational explanation! I understand what you mean now.

I should check molecular orbitals indeed, I kinda looked at the rotational part only. But if that is the case, then it makes sense. Thank you for that, too!

 

[Twigg]

I have a feeling that mathematically the approximation of taking $\frac{\partial \mu_e}{\partial R} \approx 0$ could be obtained from the BO approximation with the adiabatic theorem, taking the dynamical and Berry's phases evolved to be negligibly small since the nuclei barely move over a transition lifetime. I could just be crazy though. I never put a lot of thought into it before.

 

[BillKet]

Time to look at some molecular orbitals! Only $\Sigma$ states have cylindrical symmetry, which as you've pointed out, means $\Sigma$ to $\Sigma$ transitions are not allowed, unless you go from $\Sigma^+$ to $\Sigma^-$, the latter of which is not symmetric along the internuclear axis, but still only $q = 0$ transitions allowed.

Take a look at some $\Pi$ or $\Delta$ orbitals. They are absolutely not cylindrically symmetric, because they look like linear combinations of $p$ and $d$ orbitals.

I came across this reading, which I found very useful in understanding the actual form of the Hund cases (not sure if this is derived in B&C, too), mainly equations 6.7 and 6.12. I was wondering how this would be expanded to the case of nuclear spin (call it $I$). Given that in most cases the hyperfine interaction is very weak, we can assume that the basis we build including $I$ would be something similar to the Hund case b) coupling of $N$ and $S$ in 6.7 i.e. we would need to use Clebsch–Gordan coefficients.

So in a Hund case a, the total basis wavefunction after adding the nuclear spin, with $F$ the total angular momentum we would have:

$$|\Sigma, \Lambda, \Omega, S, J, I, F, M_F>=\sum_{M_J=-J}^{J}\sum_{M_I=-I}^I <J,M_J;I,M_I|F,M_F> |\Sigma, \Lambda, \Omega, S, J, M_J>|I, M_I>$$

where $|\Sigma, \Lambda, \Omega, S, J, M_J>$ is a Hund case a function in the absence in nuclear spin. For Hund case b we would have something similar, but with different quantum numbers:

$$|\Lambda, N, S, J, I, F, M_F>=\sum_{M_J=-J}^{J}\sum_{M_I=-I}^I <J,M_J;I,M_I|F,M_F> |\Lambda, N, S, J, M_J>|I, M_I>$$

with $|\Lambda, N, S, J, M_J>$ being a Hund case b basis function in the absence of nuclear spin. Is this right? Thank you!

 

[Twigg]

Yep, that's right. Nuclear angular momentum is just tacked on at the end of the hierarchy (though it need not be the smallest spectral splitting) with another addition of angular momenta.

 

[amoforum]

I came across this reading, which I found very useful in understanding the actual form of the Hund cases (not sure if this is derived in B&C, too), mainly equations 6.7 and 6.12. I was wondering how this would be expanded to the case of nuclear spin (call it $I$). Given that in most cases the hyperfine interaction is very weak, we can assume that the basis we build including $I$ would be something similar to the Hund case b) coupling of $N$ and $S$ in 6.7 i.e. we would need to use Clebsch–Gordan coefficients.

So in a Hund case a, the total basis wavefunction after adding the nuclear spin, with $F$ the total angular momentum we would have:

$$|\Sigma, \Lambda, \Omega, S, J, I, F, M_F>=\sum_{M_J=-J}^{J}\sum_{M_I=-I}^I <J,M_J;I,M_I|F,M_F> |\Sigma, \Lambda, \Omega, S, J, M_J>|I, M_I>$$

where $|\Sigma, \Lambda, \Omega, S, J, M_J>$ is a Hund case a function in the absence in nuclear spin. For Hund case b we would have something similar, but with different quantum numbers:

$$|\Lambda, N, S, J, I, F, M_F>=\sum_{M_J=-J}^{J}\sum_{M_I=-I}^I <J,M_J;I,M_I|F,M_F> |\Lambda, N, S, J, M_J>|I, M_I>$$

with $|\Lambda, N, S, J, M_J>$ being a Hund case b basis function in the absence of nuclear spin. Is this right? Thank you!

There's also a nice discussion in B&C Section 6.7.8 about the different ways $I$ couples in Hund's cases (a) and (b). For example, if $I$ couples to $J$ in Hund's case (b), that's actually called Hund's case (b$_{\beta J}$), which is one of the different ways it can couple in.

 

[BillKet]

Thanks for the clarification, @amoforum! And @BillKet, I'd actually be curious to see what you come up with for the stark shift, if you find the time. I tried to spend some time learning this once but my coworkers weren't having it and sent me back to mixing laser dye No pressure, of course!

As far as the BO approximation, when we did spectroscopy we didn't really keep a detailed effective Hamiltonian, we would just re-measure the lifetimes and rotational constants in other vibrational states if there was a need to do so. I think in molecules where the BO violation is weak, you can take this kind of pragmatic approach. Then again, we only thought about molecules with highly diagonal Franck-Condon factors so we never really ventured above $\nu = 2$ or so.

@Twigg, here is my take at deriving the Stark shift for a Hund case a. In principle it is in the case of 2 very close $\Lambda$-doubled levels (e.g. in a $\Delta$ state as in the ACME experiment) in a field pointing in the z-direction, $E_z$. Please let me know if there is something wrong with my derivation.

$$H_{eff} = <n\nu J M \Omega \Sigma \Lambda S|-dE|n\nu J' M' \Omega' \Sigma' \Lambda' S'>=$$
$$<n\nu J M \Omega \Sigma \Lambda S|-E_z\Sigma_q\mathcal{D}_{0q}^1T_q^1(d)|n\nu J' M' \Omega' \Sigma' \Lambda' S'>=$$
$$-E_z\Sigma_q<n\nu J M \Omega \Sigma \Lambda S|\mathcal{D}_{0q}^1T_q^1(d)|n\nu J' M' \Omega' \Sigma' \Lambda' S'>=$$
$$-E_z\Sigma_q<n\nu|T_q^1(d)|n\nu><J M \Omega \Sigma \Lambda S|\mathcal{D}_{0q}^1| J' M' \Omega' \Sigma' \Lambda' S'>$$

For the $<n\nu|T_q^1(d)|n\nu>$, given that we are in a given electronic state, the difference between $\Lambda$ and $-\Lambda$ can only be 0, 2, 4, 6..., (for a $\Delta$ state it would be 4) so the terms with $q=\pm 1$ will give zero. So we are left with

$$-E_z<n\nu|T_0^1(d)|n\nu><J M \Omega \Sigma \Lambda S|\mathcal{D}_{00}^1| J' M' \Omega' \Sigma' \Lambda' S'>$$

If we use the variable $D$ for $<n\nu|T_0^1(d)|n\nu>$, which is usually measured experimentally as the intrinsic electric dipole moment of the molecule (I might have missed a complex conjugate in the Wigner matrix, as it is easier to type without it :D) we have:

$$-E_zD<\Sigma S||\Sigma' S'><\Lambda||\Lambda'><J M \Omega |\mathcal{D}_{00}^1| J' M' \Omega' >$$

From here we get that $S=S'$, $\Sigma=\Sigma'$ and $\Lambda = \Lambda'$, which also implies that $\Omega = \Omega'$. By calculating that Wigner matrix expectation value we get:

$$-E_zD(-1)^{M-\Omega}
\begin{pmatrix}
J & 1 & J' \\
\Omega & 0 & \Omega' \\
\end{pmatrix}
\begin{pmatrix}
J & 1 & J' \\
M & 0 & M' \\
\end{pmatrix}
$$

This gives us that $M=M'$ and $\Delta J = 0, \pm 1$. If we are in the $\Delta J = \pm 1$ case, we connect different rotational levels, which are much further away from each other relative to $\Lambda$-doubling levels, so I assume $\Delta J = 0$. The expression above becomes:

$$-E_zD(-1)^{J-\Omega}\frac{M\Omega}{J(J+1)}$$

Now, the parity eigenstates are linear combinations of hund a cases:

$$|\pm>\frac{|J M S \Sigma \Lambda \Omega>\pm|J M S -\Sigma -\Lambda -\Omega>}{\sqrt{2}}$$

If we build the 2x2 Hamiltonian in the space spanned by $|\pm>$ with the Stark shift included it will then look like this (I will assume the ACME case, with $J=1$ and $\Omega = 1$):

$$
\begin{pmatrix}
\Delta & -E_zD M\\
-E_zD M & -\Delta \\
\end{pmatrix}
$$

Assuming the 2 levels are very close we have $\Delta << E_zD$ and by diagonalizing the matrix we get for the energies and eigenstates (with a very good approximation): $E_{\pm} = \pm E_zD M$ and $\frac{|+>\pm|->}{\sqrt{2}}$. Hence the different parities are fully mixed so the system is fully polarized.

 

[Twigg]

Thank you! I really appreciate it! Your derivation helped put a lot of puzzle pieces together for me.

I was able to get the polarizability out of your 2x2 Hamiltonian. It has eigenvalues $$E_{\Lambda,M} = \frac{\Lambda}{|\Lambda|} \sqrt{\Delta^2 + (E_z DM)^2} \approx \frac{\Lambda}{|\Lambda|} (\Delta + \frac{1}{2} \frac{E_z ^2 D^2 M^2}{\Delta} +O((\frac{E_z DM}{\Delta})^2))$$
From this, polarizability is $$\alpha = \frac{\Lambda}{|\Lambda|}\frac{D^2 M^2}{2\Delta}$$, since the polarizability is associated with the energy shift that is quadratic in electric field. This seems to be in full agreement with what that review paper was saying (one of these days, I'll find that paper again).

I might have missed a complex conjugate in the Wigner matrix, as it is easier to type without it :D

I can never reproduce something I derived using Wigner matrices because of all the little mistakes here and there. They're just cursed. I'd sell my soul for a simpler formalism

By the way, I found a thesis from the HfF+ eEDM group that derives the Stark shift, and it exactly agrees with your expression for no hyperfine coupling ($F=J$ and $I=0$). Nice work!

 

[BillKet]

Thank you! I really appreciate it! Your derivation helped put a lot of puzzle pieces together for me.

I was able to get the polarizability out of your 2x2 Hamiltonian. It has eigenvalues $$E_{\Lambda,M} = \frac{\Lambda}{|\Lambda|} \sqrt{\Delta^2 + (E_z DM)^2} \approx \frac{\Lambda}{|\Lambda|} (\Delta + \frac{1}{2} \frac{E_z ^2 D^2 M^2}{\Delta} +O((\frac{E_z DM}{\Delta})^2))$$
From this, polarizability is $$\alpha = \frac{\Lambda}{|\Lambda|}\frac{D^2 M^2}{2\Delta}$$, since the polarizability is associated with the energy shift that is quadratic in electric field. This seems to be in full agreement with what that review paper was saying (one of these days, I'll find that paper again).
I can never reproduce something I derived using Wigner matrices because of all the little mistakes here and there. They're just cursed. I'd sell my soul for a simpler formalism

By the way, I found a thesis from the HfF+ eEDM group that derives the Stark shift, and it exactly agrees with your expression for no hyperfine coupling ($F=J$ and $I=0$). Nice work!

I am glad it's right! :D Please send me the link to that paper when you have some time. About the polarization, I am a bit confused. Based on that expression it looks like it can go to infinity, shouldn't it be between 0 and 1 (I assumed that if you bring the 2 levels to degeneracy you would get a polarization of 1)?

Side note, unrelated to EDM calculations: I am trying to derive different expressions in my free time just to make sure I understood well all the details of diatomic molecules formalism. It's this term for the Hamiltonian due to parity violation. For example in this paper equation 1 (I just chose this one because I read it recently, but it is the same formula in basically all papers about parity violation) gets turned into equation 3 after doing the effective H formalism. I didn't get a chance to look closely into it, but if you have any suggestions about going from 1 to 3 or any paper that derives it (their references don't help much) please let me know. I guess that cross product comes from the Dirac spinors somehow but it doesn't look obvious to me.

 

[Twigg]

Here's that thesis. I was looking at equation 6.11 on page 103. Also, I used $\Lambda$ instead of $\Omega$ in my last post, just a careless error.

I don't have APS access right now, so I can't see the Victor Flambaum paper that is cited for that Hamiltonian. Just looking at the form of that Hamiltonian, the derivation might have little to do with the content of Brown and Carrington because it's talking about spin perpendicular to the molecular axis.

If you're reading papers on parity violation, this one on Schiff's theorem is excellent if you can get access. I used to have a copy but lost it. Also, talk about a crazy author list What is this, a crossover episode?

 

[Twigg]

Just noticed I missed your question about polarizability. I'm not sure why it would be limited between 0 and 1. Are you thinking of spin polarization? What I mean here is electrostatic polarizability $\vec{d}_{induced} = \alpha \vec{E}$. It only appears to go to infinity as $\Delta \rightarrow 0$ because the series expansion I did assumed $\Delta \gg E_z DM$. The reason for this inequality is that polarizability is usually quoted for $E_z \rightarrow 0$ by convention.

 

[amoforum]

For more Stark shift/polarization stuff, also take a look at the Gurevich thesis if you haven't already: http://www.doylegroup.harvard.edu/files/bufferpubs/theses/yvg_thesis.pdf

Equation 3.22 shows the bit about polarization going from 0 to 1.

 

[BillKet]

So I tried to derive the Zeeman effect for a Hund case b, with the nuclear spin included in the wavefunction. The final result seems a bit too simple, tho. I will look only at the $S\cdot B$ term and ignore the $g\mu_B$ prefactor. For a Hund case b, the wavefunction with nuclear spin is:

$$|NS\Lambda J I F M_F> = \Sigma_{M_J}\Sigma_{M_I}<JM_JIM_I|FM_F>|NS\Lambda J M_J>|IM_I>$$

And we also have:

$$|NS\Lambda J M_J> = \Sigma_{M_N}\Sigma_{M_S}<NM_NSM_S|JM_J>|NM_N\Lambda>|SM_S>$$

where $<JM_JIM_I|FM_F>$ and $<JM_SIM_N|JM_J>$ are Clebsch-Gordan coefficients. Now, calculating the matrix element we have:

$<NS\Lambda J I F M_F|S\cdot B|N'S'\Lambda' J' I' F' M_F'>$

I will assume that the magnetic field is in the z direction. Also, given that we are in Hund case b we can look at the spin quantized in the lab frame, so we don't need Wigner rotation matrices, so we get $S\cdot B = T_{p=0}^1(S)T_{p=0}^1(B) = B_zS_z$, where both $B_z$ and $S_z$ are defined in the lab frame, with $S_z$ being an operator, such that $S_z|SM_S> = M_S|SM_S>$. So we have:

$$<NS\Lambda J I F M_F|B_zS_z|N'S'\Lambda' J' I' F' M_F'>=$$

$$B_z (\Sigma_{M_J}\Sigma_{M_I}<JM_JIM_I|FM_F><NS\Lambda J M_J|<IM_I|)S_z(\Sigma_{M_J'}\Sigma_{M_I'}<J'M_J'I'M_I'|F'M_F'>|N'S'\Lambda' J' M_J'>|I'M_I'>)$$

As $S_z$ doesn't act on the nuclear spin we get:

$$B_z \Sigma_{M_J}\Sigma_{M_I}\Sigma_{M_J'}\Sigma_{M_I'}<JM_JIM_I|FM_F><J'M_J'I'M_I'|F'M_F'><IM_I||I'M_I'><NS\Lambda J M_J| S_z |N'S'\Lambda' J' M_J'> = $$

$$B_z \Sigma_{M_J}\Sigma_{M_J'}\Sigma_{M_I}<JM_JIM_I|FM_F><J'M_J'IM_I|F'M_F'><NS\Lambda J M_J| S_z |N'S'\Lambda' J' M_J'> = $$

(basically we got $I=I'$ and $M_I = M_I'$). For the term $<NS\Lambda J M_J| S_z |N'S'\Lambda' J' M_J'>$ we get:

$$(\Sigma_{M_N}\Sigma_{M_S}<NM_NSM_S|JM_J><NM_N\Lambda|<SM_S|)S_z(\Sigma_{M_N'}\Sigma_{M_S'}<N'M_N'S'M_S'|J'M_J'>|N'M_N'\Lambda'>|S'M_S'>)$$

As $S_z$ doesn't act on the $|NM_N\Lambda>$ part we have:

$$\Sigma_{M_N}\Sigma_{M_S}\Sigma_{M_N'}\Sigma_{M_S'}<NM_NSM_S|JM_J><N'M_N'S'M_S'|J'M_J'><NM_N\Lambda||N'M_N'\Lambda'><SM_S|S_z|S'M_S'>$$

From which we get $N=N'$, $M_N=M_N'$ and $\Lambda=\Lambda'$. So we have:

$$\Sigma_{M_N}\Sigma_{M_S}\Sigma_{M_S'}<NM_NSM_S|JM_J><NM_NS'M_S'|J'M_J'><SM_S|S_z|S'M_S'> = $$

$$\Sigma_{M_N}\Sigma_{M_S}\Sigma_{M_S'}<NM_NSM_S|JM_J><NM_NS'M_S'|J'M_J'>M_S'<SM_S||S'M_S'> = $$

And now we get that $S=S'$ and $M_S=M_S'$ so we have:

$$\Sigma_{M_N}\Sigma_{M_S}<NM_NSM_S|JM_J><NM_NSM_S|J'M_J'>M_S = $$

$$\delta_{JJ'}\delta_{M_JM_J'}M_S$$

So we also have $J=J'$ and $M_J=M_J'$. Plugging in in the original equation, which was left at:

$$B_z \Sigma_{M_J}\Sigma_{M_J'}\Sigma_{M_I}<JM_JIM_I|FM_F><J'M_J'IM_I|F'M_F'><NS\Lambda J M_J| S_z |N'S'\Lambda' J' M_J'> = $$

$$B_z \Sigma_{M_J}\Sigma_{M_I}<JM_JIM_I|FM_F><JM_JIM_I|F'M_F'>M_S = $$

$$B_z M_S \delta_{FF'}\delta_{M_FM_F'}$$

So in the end we get $F=F'$ and $M_F=M_F'$, so basically all quantum numbers need to be equal and the matrix element is $B_zM_S$. It looks a bit too simple and too intuitive. I've seen mentioned in B&C and many other readings that hund case b calculations are more complicated than Hund case a. This was indeed quite tedious, but the result looks like what I would expect without doing these calculations (for example for the EDM calculations before I wouldn't see that $\frac{1}{J(J+1)}$ scaling as obvious). Also, is there a way to get to this result easier than what I did i.e. figure out that $B_zM_S$ should be the answer without doing all the math? Thank you!

 

[amoforum]

I haven't gone through your derivation yet, but yes, there's a way easier method, which is how B&C derive all their matrix elements.

Look at equation 11.3. Its derivation is literally three steps, by invoking only two equations (5.123 first and 5.136 twice, once for $F$ and once for $J$). The whole point of using Wigner symbols is to avoid the Clebsch-Gordan coefficient suffering.

By the way, almost every known case is in the B&C later chapters for you to look up. Every once in a while it's not. It happened to me actually, but I was able to derive what I needed using the process above.