OK. Do you see why we need B1g now?

for q

?

because the ground state is B1g and Gammaq wants to annihilate it so it can get a nonzero term

yes. and because its the only irrep in the group the gives the trivial when puliplied be B1g

multiplied

is it ?

Yes gernot-katzers-spice-pages.com/character_tables/D4h.html

yes

:-)

In other cases we would have had more alternatives

such that potential square molecules can be rhomboids as well ...

so are you oging to ask me why A1g only appears on the boxes on the diagonal ?

Eg

No

aaw

that would be too simplisitc

We proceed. D4h distorts along B1g to give D2h, in this case

Thereby A1g gets Ag, B1g gets Ag and B2g gets B1g

in the new D2h (in that orientation).

wowow, what ?

ah

:-), yes?

no I'm confused

:-(

the things in the columns are representations of the subgroup, right ?

which columns?

in the pink table

yes, right!

so B1g now refers to an irrep of D4h AND an irrep of D2h

unfortunately yes

how do I know it distorts on D2h ?

and also, I see there are two embeddings of D2h in D4h ?

One can check which elements are trivial under B1g

stay unchanged

we pick the subgroup ... of elements that are trivial under B1g ?

Yes!

... so the kernel of that irrep ?

Could be.

Yeah

cool!

hmm okay

then B1g descents into a trivial irrep

yes!

while B2g

A1g was already trivial

and B2g does something else

into B1g in D2h

okay ?

So that was an example of a second order JT effect. I claim now that there is a 1:1 correspondence to a 1st order JT effect hidden.

what do you mean by that ?

Because the new set of wave functions we have: Ag,Ag,B1g is acually again only based on orbitals

?

and it happens that in our case the "two" $E$ orbital where distorted in 1st order JT fashion to B2g and B3g in D2h from which we can build our 2Ag and B1g by squaring

But maybe we do it slower,

Just assume we did not start with two electrons but just with one, OK?

So we have one electron in E.

under D4h.

Thats a 1st order JT case

in the pink table, Eu descents into B2u+B3u and not into B2g+B3g

Oh sorry, my fault!

you are tright

squaring and descent commute

wait!

we can either have $E_u$ or $E_g$, no $E$ to begin with.

There is no $E$ without index in D4h

I am not sure what you are trying to say with that

When we start with $E_g$ it descends to $B2g$ and $B3g$ in D2h, OK?

Eg -- descent -> B2g + B3g -- alternate square --> ?

Eg -- alternate square -> A2g -- descent --> B1g

Eg -- descent -> B2g + B3g . STOPP

now suppose we want 2 electrons in the D2h orbitals

tensor product of the two

Yes!

which is also the alternate square of the direct sum

(maybe, but not relevant now) Ag, Ag,

Ag Ag ?

B2g^2 = Ag

B3g^2 = Ag

yeah but idk why that is relevant

And those Ag's are now so to say "the same" as the "Ag" we got in the second order JT for the 2 electron case.

!!

those Ags are the trivial irrep

Well unfortunately

yes

But my point is that there is a general principle

conjecture ..

when did we finish the second order JT for the 2 electron case ?

Yes, we know all we can.

We know B1g, we know it gets D2h

we know what the irreps become

thats all.

how does that tell us Ag

A1g became Ag

B1g will always descent into the trivial irrep if you pick the subgroup that is the kernel of B1g

but A1g was trivial to begin with

Yes, thats the sad thing here.

and B2g became B1g

which is not Ag

yes

if you square 1-dimensional representations you will always get the trivial irrep

and if you descent from the trivial irrep you will always get the trivial irrep

and if you descent from an irrep by choosing its kernel as the subgroup you will always get the trivial irrep

What I mean to say is if I know a 1electron 1st order JT system

I know automatically a 2-electron 2nd order JT system.

:(

Think about the case: 1 electron in $E_g$ in D4h.

It gives you the 1st order JT. OK?

maybe it will distort along D2

no I don't remember what we do with the Eg

for one electron Eg gives an Eg wave function

for two electrons it gives anyone of Eg x Eg

(the decomp.)

:|

then you pick one of the pieces of that product at random and descent along its kernel

yes

then some of them become trivial and some others don't

yes

then I am quite lost

That was the two electron case, OK?

:|

yes ... ?

We have the orbital Eg. Having that we can contruct an one-electron wavefunction, or a two electron wave functiuon, OK?

The orbital is just an empty box.

a basis

function

is a two electron wavefunction a kind of probability measure on R^6 ?

Once having one-electron orbitals, no matter wherefrom, we can built many electron wave functions. Eg an 1-electron wave function, thats already the orbital, or

yes

exactly

\psi_1(x1,y1,z1) \psi_2(x2,y2,z2)

yes

would be a (too) simple two electron wave function

too simple because it disrepects anti-commutativity

$\Psi(1,2)=-\Psi(2,1)$

yes so you remove another term ?

no

substract*

yes!

you from the derminant

$\psi_1(x_1,y_1,z_1) \psi_2(x_2,y_2,z_2) - \psi_2(x_1,y_1,z_1) \psi_1(x_2,y_2,z_2)$

mhm ?

And that is one of $\Gamma_{\psi_1}\otimes\Gamma_{\psi_2}$

unnnghhr

(I think the anti-symmetrization should not matter here, no?)

It should also work with the simple product

is Gamma_psi1 a representation on R^3 ?

yes

shouldn't that tensor product be a representation on ... R^9 via bilinear functions

an orbital $\Bbb R^3 \to [-1,1]$

i was expecting complex numbers there

but if it's real

still, dunno why it would be bounded

in the case of hermitian operators they always can be made purely real

bounded, because of convention

the are normalized

its L² norm is 1

that works since they are integrable in some sense ...

dunno which integral keep mixing up those words ...

I guess in full generality there are always subspaces which are

and those matter in Chemistry

remember the "continuum"?

...

okay

So again we have ONE empty orbital

empty ?

from that we can build many-electron wave functions

okay

when we build the one-electron wave function, than its "the same" as the orbital.

yes

But when we want to put in more electrons we have to form the tensor products of the irreps

1-d orbitals can take 2 electrons at most

2d can take 4 at most ....

no

but that does not matter

We can put even 4 electrons into any "E-type" orbital

because its an empty box double the dimension of an 1-D irrep

Never mind we do not need it here

why can't I cube the 1-dimensional irrep ?

and put 3 electrons in it

because of a different law of QM, 2 electrons may never agree in ALL quantum numbers

spin and energy/momentum ?

yes!

well

spin (+ spin-orientation), angular momentum, orientation of angular momentum and energy

so 2 opposite spin electrons in a 1d orbital corresponds to the square

2 same spin electrons in a 2d orbital... the alternate square ? that's also a 1d representation

Then each has spin up or spin down

2 opposite spin electrons in a 2d orbital -> symmetric square

3 electrons -> idk

but remember the WHOLE thing, and now the full truth: inculding the spin part has to be anti-symmetrioc

So you actually have $\psi(x,y,z,\omega)$ for one electron

4 electrons -> 22 schur functor

with $\omega$ being the spin variable which has two possible values: $\alpha$ or $\beta$

and the spin eigenfunctions can map those to the set $\{-1/2,+1/2\}$.

S.t. that there are two possible spin eigenfunctions which are called "funniyl" also $\alpha(\omega)$ and $\beta(\omega)$

that's some terrible names

And those have to be multiplied to the orbitals to give spin orbitals and

$\alpha(\omega)(\alpha) = 1/2$ ?

history, physics

YES!

I had a 50% chance

wait

no

one second

iirc the 22 schur functor applied to a 2dim vector space has dimension 1

not sure because I only read that this morning

but my wild guess would be that 4 electrons in your 2d orbital give a 1d representation

and as for the 21 schur functor

You may look here en.wikipedia.org/wiki/Spin-%C2%BD#Mathematical_description meahwhile I have to go for little boys ...

would give a 2d representation

I'm going to stare at that gif

for weeks

The point is that

the dimension of the spin-function space is 2.

and the spin of each particle can be in a superposition of the two

ah

and when you measure its either up or down

the spinfunctions are orthonormal

one has eigenvalue -1/2

the other +1/2 and thats it.

yes

now each electron requires one more "variable" to be described.

the "spin"-coordinate, over which you in addition "integrate" when you do csalar prodcuts

And now the anti-symmetricty of the whole wave function is required for the whole function space including the spin-coordinate.

In case of doubly occupied non-degenerate orbitals everything simplifies so strongly that you don't have to worry about spin, but thats involved to see (at least for me).

x..x

Puh I am getting tired, I suggest I go home have a little break and if you like we continue in the evening?

maybe

Important would be to see how to build the many electron wave function (symmetry) from the ("empty") orbitals.

And thats how I said simply by product formation.

And now you do the D4h Eg JT example for 1 and then for 2 electrons

then one should see that for each 2nd order example there should be a 1 to 1 map to a 1st order case

that should be non trivial and is in some way very important for us. since the good real world expamples are 2-e cases ...

and our ideas where derived for the one-electron world

Accodring to my JT specialist that is something also others have observed but nothing is worked out (and there is also the sad thing the in the end you can step down in all cases until C1 and A ....; but that must not worry us since it applies for the whole field so to say, and thats not dead, far from it actually)

for 1 electron you said we do nothing to it then it was just Eg ?

I mean we could go through the whole thing how to really build the many electron wave functions but thats lengthy and in the end we need only to know that we have to form the tensor product

Well and then the 1st Order JT.

but I don't know to what subgroup it distorts

And for the two electron case you get a set of possible wave function

@mercio Nono for that you go through the 1. order JT

proeceedure

what was the first order term again ?

I just meant with respect to wave function symmetry

<psi_0|dH/dq|psi_0>

I don't even know what to do with it

check if it must be zero or not by checking

$\Gamma_0\otimes\Gamma_0\otimes\Gamma_q$

so Gammaq wants to be Gamma0²

to get the highest number

In that case it can be non-zero

I don't know if its exact

If it exactly in that case can be non-zero.

the "can" is still relevant.

since it can be always zero by "incidence"

And only in case of non-1d irreps for psi_0 can be other then trivial.

I am just too tired

trivial doesn't count since we project that out of q.

mee too

cu later!

@mercio I think I start to understand it.

1e case: $<\psi_0|q|\psi_0>$

2e case: $<\Psi_0|q|\Psi_i>$

where $\Gamma_{\Psi_0},\Gamma_{\Psi_i}\in{\Gamma_{\psi_0}}^2$.

Then our unique 2e counterpart for any 1e JT case is $\Gamma_{\Psi_i}=1$

I mean the trivial irrep, which always is contained in ${\Gamma_{\psi_0}}^2$.

That is 1e case $\Rightarrow$ 2e case. The other direction then will not work. Thats OK as well. In that case these special 2e cases deserve a name I'd say.

(Mind that $q=q$ and is determined by the 1e case up to choice out of the decomposition of ${\Gamma_{\psi_0}}^2$).