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08:00 - 12:0012:00 - 16:00

12:00 PM
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
is it ?
12:02 PM
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 ?
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 ?
12:05 PM
:-), yes?
no I'm confused
the things in the columns are representations of the subgroup, right ?
which columns?
in the pink table
12:07 PM
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 ?
12:09 PM
... so the kernel of that irrep ?
Could be.
hmm okay
then B1g descents into a trivial irrep
while B2g
A1g was already trivial
and B2g does something else
12:11 PM
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
12:16 PM
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
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
12:20 PM
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
which is also the alternate square of the direct sum
12:22 PM
(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
12:24 PM
Well unfortunately
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
12:27 PM
Yes, thats the sad thing here.
and B2g became B1g
which is not Ag
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.
12:31 PM
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
then some of them become trivial and some others don't
12:34 PM
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
is a two electron wavefunction a kind of probability measure on R^6 ?
12:38 PM
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
\psi_1(x1,y1,z1) \psi_2(x2,y2,z2)
would be a (too) simple two electron wave function
too simple because it disrepects anti-commutativity
yes so you remove another term ?
12:40 PM
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 $\Gamma_{\psi_1}\otimes\Gamma_{\psi_2}$
(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 ?
12:45 PM
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
12:48 PM
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"?
So again we have ONE empty orbital
empty ?
from that we can build many-electron wave functions
12:52 PM
when we build the one-electron wave function, than its "the same" as the orbital.
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 ....
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
12:55 PM
because of a different law of QM, 2 electrons may never agree in ALL quantum numbers
spin and energy/momentum ?
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
12:57 PM
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
1:02 PM
And those have to be multiplied to the orbitals to give spin orbitals and
$\alpha(\omega)(\alpha) = 1/2$ ?
history, physics
I had a 50% chance
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
1:07 PM
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
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.
1:13 PM
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).
Puh I am getting tired, I suggest I go home have a little break and if you like we continue in the evening?
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 ?
1:27 PM
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
what was the first order term again ?
I just meant with respect to wave function symmetry
I don't even know what to do with it
1:31 PM
check if it must be zero or not by checking
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!
2 hours later…
3:41 PM
@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 ${\Gamma_{\psi_0}}^2$).
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