@mercio Hi there!

I don't want to bore people not involved. So I suggest we can also use this chatroom.

Checking possible "simple" examples (molecules) I struggeled over another conjecture, which is maybe less "trivial" than our first two, but vital to the whole thing. My expert colleague confirmed the conjecture, but said he is not aware that ever anyone bothered to "proof" it. So some occasion for a greater contribution from you. Its about the 2nd order JT effect. I'll better start explaining it when you are online. Cheers

I couldn't sleep because I was thinking too much about schur functors and plethysms

but if you can describe your observations, fire away

Hi Mercio back again!

OK

The thing is all "really good" examples are 2nd Order Jahn-Teller effects. But I see that there might be a way that we could reduce these special examples (and maybe any at all) to the 1st order JT effect. I'll start with the 2nd order JT, thats governed by:

$<\psi_0|\frac{d \hat{H}}{d q}|\psi_i>$, with $\psi_0$ ground state and $\psi_i$ virtual=excited states = the rest of the basis functions.

q is the distortion parameter iirc ?

The difference here is that we do not need to compute direct squares of the ground state representation but have to look at prodcuts of pairs of groupd state and some (close by) excited state. $q$ yes

Close by because of the energy denominator (I forogt here).

product with dH^/dq in between

Yes!

I prefer to think first the product of the states and then check if there are $q$

which are "identical"

(btw thanks for the email, its excellent!)

OK?

hmm not sure what you mean with product of the states

<psi0 | psi i> ? I don't remember if those are 0

Well you either can compute the whole direct product and check if it contains the trivial (no thats the scalar pr.) or you do it "stepwise"

Either $\Gamma_{\psi_0} \otimes \Gamma_{\psi_i} \Gamma_{q}$, or

you compute dh/dq |psi i > and check its coordinate on psi0 ?

huh what are those Gammas

the irreps that transform like the $\psi_i$

I'm not sure how this relates with the 2nd order term up there

So either $\Gamma_{\psi_0} \otimes \Gamma_{\psi_i} \otimes \Gamma_{q}$ or do $\Gamma_{q} \subseteq \Gamma_{\psi_0} \otimes \Gamma_{\psi_i} $

that tensor product is really sketchy

there is some kind of product to measure how much an irrep is in a representation

but it's.. oh wait

No its just about if $\Gamma_{trivial}\subseteq\Gamma_{\psi_0}\otimes\Gamma_{\psi_i}\otimes\Gamma_{q} $

thats all

I think it is the number of trivial irreps in $\overline {\Gamma_{\psi_0}} \otimes \Gamma$

are all the representations real ?

Well yes

Because H is hermitian

self adjungated

okay

I still have a nagging doubt but you're probably right

You are maybe right, but we can check that later

There was some detail I miss I think

yeah, so you are looking at tensor products

its about the anti-symmetric / symmetric part again ...

The main point is now

well those don't mean much if you aren't tensoring the same representations

that we now have MUCH more possibilities since we deal with products of pairs

compared to what ?

1st order JT

So now suddenly \psi_0 needs not be degenerate

and suddenly the symmetry of \psi_i plays a role

OK?

Gamma_psi0 needs not have dimension > 1 ?

Yep!

Gamma_psi0 may have dimension 1, and we still see a JT effect ?

Because we have that product of three components

Yes! (A second order one ...)

yeah I'm trying to keep up with all the implicit things you are saying

$\Gamma_A \otimes \Gamma_q \otimes \Gamma_B$

was the 1st order effect only dependent on Gamma_psi0 ?

then $\Gamma_q$ might be an $B$

Yes

mhm

It needed to be dim >1

yes

otherwise no distortion

But not now

OK?

Shall we do an example?

when you say Gamma q might be an B do you mean the same B as in Gamma B ?

Yes

so Gamma q tensor Gamma B would have some trivial irrep in it ?

yes!

If its any square (e.g. or exactly?)

hmmmm

yeah Gamma B² 's traces are all positive

because square

Yes!

So we have some elements of second order that are allowed to be non zero

OK?

but don't you still need to tensor with Gamma psi0 ?

Yes sure

But we assumed thats trivial=A

for this example

ah, that's quite an important assumption lol

then yes

you will get a nonzero term

Well it can be anything in case it equals with the product of the other two

in case it has some irreps in common

So we see we can eventually get nonzero terms even if psi_0 is not dim >1

yes

yes

Now a big new thing:

1. molecules with even numbers of electrons are good and those with odd are not.

(for our examples)

is that a spin thing ?

Yes

Total spin=0 is "chemically much more stable"

i have yet to learn about the spin but carry on

np.

2. How to build many-electron wave functions from orbitals (= one electron wave functions)

?

Answer by product formation.

(very simplistic ...)

I would have said a direct sum ... ?

So that means we now have a set of orbitals, where each of which can take 2 electrons.

mhm

but each orbital corresponds again to an irrep

that sounds like black magic

(What you really do is form an determinant)

not a product but an anti-symmetrized prodcut .. but its not important

does each orbital correspond to a dimension 3 representation ?

Important is only that the symmetry of the total wavefunction is the tensor product of the orbitals in the power of the electrons they have (0,1, or 2)

yes

(again)

That means in essence that (most by far) proper molecules have ground state wave functions that are trivial irreps

mhm so you take the (0,1,or 2)-th alternate power of the representation

yes

since you either have squares or nothing

so there will be trivial irreps in the squares but there will also be other things ?

right.

oh wait it's not a square it's an alternate power

I think only trivials unless one ore more dim>1

I think the trivial irrep is more likely to be in the symmetric part

and those are very particular cases

well if it has dim 1 then its alternate square has dimension 0

not sure what that means physically

if it has dim 2 then the alternate square is the determinant

That closed shell molecules have trivially symmetric irrep wavfunction reps

could be trivial, could be nontrivial

I am here on shaky grounds as well, but the main point, and thats for me highly non trivial is

that when you have 2 electrons in a 2 dim rep, then you have a set of possible states,

eg take $E_u$ in $D4h$.

if I read it right its alternate square is A2g ?

No either A1g, B1g or B2g (A2g not because of spin reasons)

The whole wave function MUST be antisymmetric with exchange of electrions

and we now look only at the spatial component

but there is the spin component as well

:(

and in case both electrons have "the same spin" the spin function itself is anti-symmetric

do you mean symmetric ?

and finally all the shite has to be multiplied with the spin function as well

that reminds me of supersymmetry

?

cuz the other day i was reading some stuff talking about Lie superalgebras

but nevermind

so in case both electrons have the same spin, we look into the alternate square, and in case they have opposite spin we look into the symmetric square ?

OK so for simplicity the spatial part must be any of the decomposition of the square (in case of two electrons)

Yes

And for other reasons we do not look at the same spin case now.

Take is as arbirtrary.

(Truth is they are mostly much higher in energy)

and thus are no candidates for the ground state

maybe I misread something earlier cuz of no sleep lol

hello new person

bye

lol

can we continue or do you have questions?

(sure you have ...

alright, so in the symmetric square we will have some trivial irrep piece

no wait

the A1g in your example ?

thats not speciaö

aw

we just know that our wave function can be any of those

any of A1g, B1g , B2g (A2g)

or any linear combination of them ?

OK?

well not A2g if they have different spin

Because H commutes with the group

the eigenfunction of H can only be in one certain irrep.

of course we can prepare a state thats any linear combination

but thats not gonna be an eigenfunction with defined energy

does it have to be irreducible ?

yes, like the orbitals

I will take it on faith then

in case its an eigenfunctio

but we only consider stuff we measure

and after measurement its surely in an eigenstate

so it must be

but how do i know what is a measurement and what isn't

it frustrates me a bit that you need faith here ;-)

All we observe is measured

if we observe the same experiment, will we observe the same thing

all what has an self adjungated operator can be measured

I have been lazy in my learning of QM, I'm sorry

on that hilbert space

:-)

OK so we have these E orbitals (its two orbitals, we say)

so how do we know which one it will be ?

You don't

unless you do the numerics

mhm

symmetry alone doesn't help you here

it can just help a bit, the higher the more it can

OK now as a matter of fact B1g is the ground state

and the NEXT excited state (as a matter of fact) is A1g

((BUT we will see that we could now just try any of the possible combinations ...))

((for ground and excited states, I mean))

That means that we seek q such that

if we compute the wavefunctions and compare their energy, the B1g one has lowest energy ?

$A1g\otimes\Gamma_{q}\otimes{B1g}$ contains thr trivial

@mercio yes!

And the next one is A1g.

doesn't that depend on the whole molecule and not just the point group ?

just to be sure

Yes you need the FULL numerics

okay

the whole molecule

So we search $q$ s.t. $A1g \subseteq A1g\otimes\Gamma_{q}\otimes{B1g}$

so the distortion will have the molecule pick the excited state that is compatible with its symmetry ?

Yep!

(we will see that it wants from square to rectangle ...)

(from square to rectangle looks like symmetry breaking)

Because $\Gamma_q = B1g$ solves that.

OK?

hmm maybe there are others that solve it better.. though if everyone is 1 dimensional I doubt that

In general yes, in our case here, just check $D_{4h}$

I'm not sure what dimension Gamma q is

here 1

or rather, the symmetry of the ground state will push the molecule to distort in a way that has the compatible symmetry ?

I'm not sure if I had my cause effect right the first time

Yes, thats exactly 2nd order Jahn-Teller

cuz I said 2 opposite things and you said yes to both

Well I don't worry about cause and effect in JT

o..o ah

Because the undistorted actually exists only in a transient fashion

Its more like a starting point for the consideration

like a virtual thing in Feynmann diagrams ?

like a proof with contradiction, I guess

Yes!

Its actually perturbation theroy

just so you know, I know nothing about that yet

NP. Me not much either ...

(all of the terms in the perturbaition series correspond to a picture)

(1st order JT effect would be a little circle, I guess ...

2nd order is then some crossing or something

:|

You can use that to see how the different order terms looks like

well if I had any experience with those pictures I might do that

there is a 1:1 from the pictures to the terms

interesting

but lets forget that for the moment