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8:52 AM
@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
1 hour later…
10:30 AM
I couldn't sleep because I was thinking too much about schur functors and plethysms
but if you can describe your observations, fire away
10:55 AM
Hi Mercio back again!
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
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!)
hmm not sure what you mean with product of the states
<psi0 | psi i> ? I don't remember if those are 0
11:04 AM
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
11:09 AM
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
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
11:11 AM
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
Gamma_psi0 needs not have dimension > 1 ?
11:13 AM
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 ?
11:16 AM
then $\Gamma_q$ might be an $B$
It needed to be dim >1
otherwise no distortion
But not now
Shall we do an example?
when you say Gamma q might be an B do you mean the same B as in Gamma B ?
11:17 AM
so Gamma q tensor Gamma B would have some trivial irrep in it ?
If its any square (e.g. or exactly?)
yeah Gamma B² 's traces are all positive
because square
So we have some elements of second order that are allowed to be non zero
but don't you still need to tensor with Gamma psi0 ?
11:20 AM
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
11:22 AM
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 ?
Total spin=0 is "chemically much more stable"
i have yet to learn about the spin but carry on
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 ... ?
11:25 AM
So that means we now have a set of orbitals, where each of which can take 2 electrons.
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 ?
11:27 AM
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)
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
since you either have squares or nothing
so there will be trivial irreps in the squares but there will also be other things ?
oh wait it's not a square it's an alternate power
11:30 AM
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
11:32 AM
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 ?
11:37 AM
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)
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
11:40 AM
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
11:41 AM
we just know that our wave function can be any of those
any of A1g, B1g , B2g (A2g)
or any linear combination of them ?
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 ?
11:43 AM
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
11:45 AM
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
11:46 AM
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
11:50 AM
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 ?
(we will see that it wants from square to rectangle ...)
(from square to rectangle looks like symmetry breaking)
Because $\Gamma_q = B1g$ solves that.
hmm maybe there are others that solve it better.. though if everyone is 1 dimensional I doubt that
11:54 AM
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
11:56 AM
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
Its actually perturbation theroy
just so you know, I know nothing about that yet
11:57 AM
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
11:59 AM
but lets forget that for the moment
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