Mathematics - 2018-08-31 (page 2 of 2)

Ted Shifrin

18:01

@ಠಠ We’re all happy to try to help; you don’t need to pay with upvotes. But thanks :P

@Alessandro: I figured it was something like that. In the US mostly we show up at grad school early and walk up and down the streets looking for housing. At least that’s what I had to do. I guess more things are on line these days, but it’s hard to see how people get along as roommates by just web contact. Glückliche Wünsche.

Alessandro Codenotti

In Italy it's the owner of the flat who looks for people, not the students already living in it

So you have to deal with the owner, as long as you pay the rent whether the flatmates like you or not is irrelevant

Poline Sandra

the $\varepsilon $ is for the convergence of $v_n$

Ted Shifrin

So how do you bound that term? So how are $\|v_n\|$ and $\|v\|$ related?

Poline Sandra

$v_n$ converge to v means that $|| v_n-v||\to 0$

Ted Shifrin

And so?

Poline Sandra

18:16

finally i don't know how to do

Ted Shifrin

Can’t you use the triangle inequality here? Maybe “reverse triangle inequality”?

Poline Sandra

i don't see what i can added

and dellete for the triangle inequality

Ted Shifrin

No, no. You should know a lower bound for $\|x-y\|$. You’ve never seen $\|x\|-\|y\|\le \|x-y\|$?

Poline Sandra

no I don't know it

Ted Shifrin

This is used all the time in analysis. Figure out a proof and then use it!!!

Poline Sandra

18:29

the first inequality is correct or no?

but we need something $\leq$?

math.utah.edu/~pa/math/equations/proof.html

@TedShifrin

Ted Shifrin

OK, so you looked up the proof. Now use it.

So $\|v_n\|\le \|v\| + \|v_n-v\|<\|v\| + \epsilon$.

Poline Sandra

yes

Ted Shifrin

I don’t think your first inequality after we used the triangle inequality was right. Where did the $\epsilon$ come from?

You had $\|v(u_n-u)\|$?

Poline Sandra

the $\varepsilon $ i thinked that I can use the convergence of $v_n$ to $v$ now I don't see how to apply the continuity and the convergence

Ted Shifrin

Sorry. I can’t edit. You had $\|v-n(u(x))-v(u(x))\|$. Doesn’t $\|u\|$ have to come in here?

Poline Sandra

18:45

I don't understand

Ted Shifrin

Go back to the very beginning triangle inequality you wrote down. Explain where that $\epsilon$ came from at the very end.

Poline Sandra

first I do

$$ \|v_n(u_n(x))-v(u(x))\|\le\|v_n(u_n(x))-v_n(u(x))\| + \|v_n(u(x))-v(u(x))\| $$ is this correct?

Ted Shifrin

Yes. Now?

Poline Sandra

I don't know

Ted Shifrin

What did you do with the second term?

Poline Sandra

18:52

I use the convergence of v_n to v , but the convergence say the $||v_n-v|| \leq \varepsilon $ not with $u(x)$

Ted Shifrin

But how do you estimate $\|(v_n-v)(u(x))\|$?

Poline Sandra

I can say by linearity that it is leq ||v_n-v|| |u(x)|

Ted Shifrin

OK, but we don’t want $x$ in there, do we?

The norm on the space of linear maps doesn’t have any $x$ ....

Poline Sandra

19:22

we will pass to the sup on $ ||x||\leq 1$

@TedShifrin

Poline Sandra

20:16

@AlessandroCodenotti please do you know something about norme of linear continuous applications

Alessandro Codenotti

I know something, but what's the question?

Poline Sandra

Let $E,F, G$ be three Banach spaces and let $(v_n)$ be a sequence of continuous linear functions from F to G which converge to $v ,$ and $(u_n)$ be a sequence of linear and continuous functions from $E$ to $F$ which converge to $u.$

> Prove that $v_n\circ u_n$ converges to $v\circ u$ in $\mathcal{L}(E,G)$.

I know that $$\|(v_n \circ u_n)-(v\circ u)\|= \sup_{\|x\|_E\leq 1} \|(v_n\circ u_n)(x)-(v\circ u)(x)\|_G,$$

$$ \|v_n(u_n(x))-v(u(x))\|\le\|v_n(u_n(x))-v_n(u(x))\| + \|v_n(u(x))-v(u(x))\| $$

usukidoll

Just rushed my probability for no reason. The deadline got extended and I just found out today x.x

Oh well that gives me more time to think

Alessandro Codenotti

20:32

Hmm ok, do you know that $||f\circ g||\leq ||f||||g||$?

Poline Sandra

yes when f and g are linear

then it is $\leq ||v_n-v|| ||u_n(x)||+ ||v|| ||(u_n-u)(x)||$

@AlessandroCodenotti

Alessandro Codenotti

$||v_n-v||$ and $||u_n-u||$ are going to $0$ as $n\to\infty$

Poline Sandra

and $u_n$ and v are bounded right?

$||u_n||\leq ||u_n-u||+||u||$

@AlessandroCodenotti

Alessandro Codenotti

@PolineSandra yep

Poline Sandra

thank you very much

Bob

20:44

I posted my incorrect solution to a question about about sample variance and I got a correct answer. That is, I believe I know how to do the problem. I then did a very similar question and I got the wrong answer. I then posted the second problem and my answer to math stack exchange. Is that considered a duplicate question?

Manolis Lyviakis

math.stackexchange.com/questions/2901105/… if you got the time

Bob

what does the term pre imagine mean?

Rudi_Birnbaum

@mercio are you 'round?

Bob

what does the term pre image mean?

Alessandro Codenotti

21:01

If $f:X\to Y$ is a function and $y\in Y$ the preimage of $y$ through $f$, denoted with $f^{-1}(y)$ is the set $\{x\in X\mid f(x)=y\}$, informally the stuff that gets mapped to $y$

Rudi_Birnbaum

@Bob Its the same as "fiber"

Bob

so if $f(x,y,z) = x + 2$ then what is the preimage in this case?

is it $y = x + 2$?

Abdullah UYU

Afai understand you must provide a $y\in Y$ to be able to talk about preimage, am I right @AlessandroCodenotti?

Because it is expressed with some $y$, i.e., "...the preimage of $y$...".

Jasper Loy

@Bob If a function f maps X to Y, and E is a subset of Y, then the preimage of E under f is the set of elements in X that are mapped under f to some element in E.

Bob

Do I have my example right?

Jasper Loy

21:12

I don't understand what your example means because you have not specified the domain and codomain of the function.

Abdullah UYU

Why we define $E\subset Y$ instead of taking directly some element $\in Y$ @JasperLoy? Is there a motivation for that?

Jasper Loy

@AbdullahUYU Just to make things more general if needed. Of course you can also restrict that E to one element only, similar thing.

Bob

do I have the preimage correct then?

Abdullah UYU

Ah, ok @Jasper

mercio

@Rudi_Birnbaum I'm back

Abdullah UYU

21:14

@Bob You didn't specify a subset or an element of $Y$.

Manolis Lyviakis

what do you call a smooth function that is 1-1 onto and has a smooth inverse?

Jasper Loy

@Bob What is the domain and codomain of your function?

Abdullah UYU

And yes, yet another lack of info ^

mercio

a smootheomorphism

Jasper Loy

Before you attempt to do the exercises you need to understand the concept, so you need to read your book or notes.

There is no point doing questions if you do not know the concept first.

Rudi_Birnbaum

21:15

Hi our JT-colleague mailed back, is interested and I have a first choice for some sample molecules. So that's on its way. But I got one more question.

Manolis Lyviakis

diffeomorphism **

mercio

@ManolisLyviakis apparently it's a diffeomorphism

Rudi_Birnbaum

takes a deep breath

Manolis Lyviakis

yes thank you

Bob

for my function, the domain is $R^3$ and the codomain is $R$

mercio

21:16

I thought that was C1 both ways but apparently it means smooth both ways

are you going to do real experiments with real molecules ?

Jasper Loy

@Bob OK, and it maps the triple (x,y,z) to x+2. Now before you can ask for the preimage, what is the set whose preimage are you talking about?

Rudi_Birnbaum

@mercio: The Jahn-Teller effect I described last time.

Bob

R^3

Rudi_Birnbaum

"are you going to do real experiments with real molecules ?" thats not indicated, for reasons I can explain later.

The Jahn-Teller effect I described last time is only one instance of a JT-effect.

mercio

the Jahn Teller effect about the molecule distorting to get a less symmetric hamiltonian and distinct energy levels for the ground state and the virtual state ?

Jasper Loy

21:18

@Bob So you are asking for the preimage of R^3, are you sure?

Rudi_Birnbaum

Its the archetypical but less common one, its called 1st order Jahn-Teller effect.

Bob

yes

Rudi_Birnbaum

@mercio yes!

Abdullah UYU

That is the entire domain. @Bob

But anyway, it is valid.

Oh, pardon. It isn't.

Manolis Lyviakis

a diffeomorphism exist only between manifolds does $R^n$ count?

mercio

21:19

(and I totally missed Manolis sniping my answer)

Bob

I thank you for that math lesson

I am going to have dinner

Rudi_Birnbaum

The derivation of the JT effect goes via expanding the energy in a series.

Bob

bye

mercio

yes, 1st order perturbation thing

Rudi_Birnbaum

depending on the distortion q

Manolis Lyviakis

21:20

:D

mercio

:D

Rudi_Birnbaum

@mercio yes, again

mercio

with the first term being the angular momentum in the case of a magnetic field

Rudi_Birnbaum

I'll quickly get the equation one second

@mercio nonono

mercio

what ?

Rudi_Birnbaum

21:21

now you mix up the magnetic response

with the geometric distortion

mercio

uuuuh

Abdullah UYU

$\mathbf{R}^3\not\subset \mathbf{R}$. So it is not a valid chose of $E$. @Bob

Rudi_Birnbaum

you do both via series expansion

they are VERY famous in quantum chemistry ...

mercio

@ManolisLyviakis I think R^n counts as a manifold yes

oh

there is perturbation of the hamiltonian

Manolis Lyviakis

oh ok

mercio

21:22

and perturbation of the wavefunction

I might have mixed them up

Rudi_Birnbaum

one second

mercio

is that right ?

Rudi_Birnbaum

\begin{align*}E_i(q)&=E^0_i+q\{\Psi_i|\partial H/\partial q|\Psi_i\}+\frac{q^2}{2}\{\Psi_i|\partial^2 H/\partial q^2|\Psi_i\}\\
&+q^2\sum_{j\ne i}\frac{\{\Psi_i|\partial H/\partial q|\Psi_j\}}{E^0_i - E^0_j}\end{align*}

mercio

why did I ever decide to read up on quantum mechanics

Rudi_Birnbaum

crap

mercio

21:24

I recognize that last term but

the indexing has me confused

ooh

Alessandro Codenotti

@ManolisLyviakis Sure, $\Bbb R^n$ is a very nice manifold!

Rudi_Birnbaum

I had here Dirac brakets instead of curly braces, but do you know what that is here

?

Dirac brackets are the scalar products

mercio

the scalar product of psi i with dH/dq of psi i and so on ?

Rudi_Birnbaum

I'll fix it

Abdullah UYU

@mercio Are you French?

Rudi_Birnbaum

21:25

@mercio yes

mercio

is dH/dq auto adjoint ?

@AbdullahUYU maybe but don't tell leaky nun

Abdullah UYU

:)

Rudi_Birnbaum

\begin{align*}E_i(q)&=E^0_i+q<\Psi_i|\partial H/\partial q|\Psi_i>+\frac{q^2}{2}<\Psi_i|\partial^2 H/\partial q^2|\Psi_i>\\
&+q^2\sum_{j\ne i}\frac{<Psi_i|\partial H/\partial q|\Psi_j>}{E^0_i - E^0_j}\end{align*}

Abdullah UYU

"...has me confused..."

Rudi_Birnbaum

@mercio yes

mercio

21:27

is $q$ the position ?

Rudi_Birnbaum

@mercio yes

finally.

mercio

where is the magnetic field in there ?

Rudi_Birnbaum

no magnetic field, different story

we just talk about distortion dependence of energy here

mercio

I'm not sure I'm getting it

Rudi_Birnbaum

No one except me and you knows that there is a connection with magnetism yet...

mercio

21:29

I certainly don't know that

dH/dq ?

what is Ei0 ?

Rudi_Birnbaum

The energy is of state $i$ is $<\Psi_i| H|\Psi_i>$

mercio

is psi i an eigenfunction for H ?

Rudi_Birnbaum

Sure

the full set of eigenfunctions forms a complete basis

an orthonormal Hilbert space basis

mercio

that might have been a dumb question lol

Rudi_Birnbaum

For that I can use the \psi_i s to expand solutions of perturbed Hamiltonians, OK?

mercio

21:32

no

I don't know how the hamiltonian is perturbed

E is the energy ?

Rudi_Birnbaum

forget that I explain that wrongly

no perturbation (yet)

mercio

Ei0 is the energy of state i ?

Rudi_Birnbaum

E=<psi_i|H|psi_i> at q=0

now we develop the thing in a Taylor series around q=0

mercio

okay

but

Rudi_Birnbaum

Starting from $E=<\Psi_i| H|\Psi_i>|q=0$

Manolis Lyviakis

21:34

Smooth surface= there exist a $φ: U \subset R^2 \rightarrow M$ paramtrization of the M surface that is smooth . ??

mercio

how do things depend on q

Rudi_Birnbaum

The Hamiltonian depends on q, but how does not matter.

We just form the derivative anyway ...

mercio

while keeping the wavefunctions constant ?

Rudi_Birnbaum

Yep

mercio

ungh

Rudi_Birnbaum

21:36

Well no

we say \psi_i

Manolis Lyviakis

(quick answer to my question please?)(im having trouble with the theory)

Rudi_Birnbaum

We develop psi' (the perturbed psi) into a linear combination of \psi_i s

one second I get a book

mercio

I'm having a really hard time getting how this is going to be helpful to describe the perturbed eigenfunctions of the perturbed hamiltonian

Rudi_Birnbaum

ibb.co/eq2bFK

mercio

I don't think I can check that without having the taylor development of psi i ?

Rudi_Birnbaum

21:45

yes right. see here en.wikipedia.org/wiki/… ; here $|n>:=\psi_n$

$\lambda$ is there the "explicit" parameter for the perturbation. And the idea for the perturbed psi is to use a expansion in terms of the \psi_i basis.

Manolis Lyviakis

how can i code commutative diagrams in chat?

mercio

does d/dλ commute with uuh everything else ?

well the scalar product is linear

so it might be okay

ungh

no

Rudi_Birnbaum

@mercio yes,

mercio

I don't know why the first terms on both sides cancel each other

Rudi_Birnbaum

@mercio multiply from the left with $<n^{(0)}|$

mercio

21:55

yeah but I'm really doubtful at <n0|H|n1> = d/dλ(<n0|H|n0>) and same for the other one

but it IS linear

so

the page says to multiply with <n0|

Manolis Lyviakis

$ \begin{array} 00&\stackrel{f_1}{\longrightarrow}&A\\
\downarrow{g_1}&&\downarrow{g_2\\
0&\stackrel{h_1}{\longrightarrow}&0
\end{array} $

mercio

<n0|H|n1> = d/dλ(<n0|H|n0>) can't be true because H depends on λ too

Rudi_Birnbaum

only H

n0 is the unperturbed ground state

mercio

<n0|H|n1> = d/dλ(<n0|H|n>) I mean, maybe

and that can't be true

it would be <n0|V|n0> + <n0|H|n1> wtf am i writing

Rudi_Birnbaum

one moment where do you get <n0|H|n1> = d/dλ(<n0|H|n0>) from?

mercio

21:59

however, <n|H|n> = E

if i differentiate that

Rudi_Birnbaum

no thats not how that works

mercio

i get <n0|H|n1> + <n0|V|n0> + <n1|H|n0> = E1

where did E0 go

Rudi_Birnbaum

<n0|H|n1> = (<n1|H|n0>)* =

=(<n1|E0|n0>)*= E<n0|n1>*=E 0 = 0

because H|n0>=E|n0> and

<n0|n1>=0

mercio

yes

Rudi_Birnbaum

<n_i|n_k>=delta_i,k

Manolis Lyviakis

22:03

Ok here is the question

mercio

<n0|n1> = 0 from the normalization condition

Rudi_Birnbaum

yes

its a result

mercio

so i get <n0|V|n0> = E1

Rudi_Birnbaum

from being an eigenfunction to H

right

mercio

okay

whatever hocus pocus they did gave a good result it seems >:|

Rudi_Birnbaum

22:04

similarly you get E2

thats gonna contain the summation over all $i$

now get back to the JT page from the book

Manolis Lyviakis

Let $f:R^3 \rightarrow R^3$ difeomorphism . Prove that for every smooth surface M the set f(M) is also a smooth surface. So to prove that f(M) is a smooth surface a i need to find a smooth parametrization $ψ:W\subset R^2 \rightarrow f(M)$ for f(M) to be surface.that is smooth with continuous inverse 1-1 and onto. so i need this $\begin{array} MM&\stackrel{f}{\longrightarrow}&f(M)\\

\uparrow{φ}&&\uparrow{ψ}\\
U&\stackrel{??}{\longrightarrow}&W
\end{array}$

mercio

was psi i the ith differential of psi ?

or the ith element of an orthnormal basis ?

cuz the book also has a sum

Rudi_Birnbaum

its always the i th of the basis

mercio

but it seemed to be it summed over the basis

Rudi_Birnbaum

for psi

yes

Manolis Lyviakis

22:07

since M is a surface already i know there exist φ parametrization smooth.

mercio

I haven't finished reading why the other basis elements have any influence on our distortion of psi i

Rudi_Birnbaum

the pert. wave function you develop in the basis

while the operator you do derivatives

Abcd

Is code/pseudo-code allowed as a Math.SE answer?

mercio

it perturbs along its orthgonal in the eigenspace ???

Rudi_Birnbaum

@mercio that you'll see from the dervation of the second order property.

mercio

22:08

that's an assumption that's coming up out of nowhere

hmmmm

Manolis Lyviakis

so i need to find ψ

Rudi_Birnbaum

because the perturbed hamiltonian has new eigenfunctions which need to be described as series over the whole basis (eigenfunctions of the unperturbed)

doing a bad job in explaining here, sry

mercio

resolution of the identity only involves the eigenspace ?

Rudi_Birnbaum

yes

exactly

mercio

that is assuming that V|n0> is in the eigenspace

why would that be true

does H commute with V ?

Rudi_Birnbaum

22:13

you add V to H, cannot multiply, I guess

\lambda V is small

mercio

irrelevant

Rudi_Birnbaum

very very small ? ;-)

mercio

>:|

Rudi_Birnbaum

no idea that problem never occured to me

Why is that assuming that V|n0> is in the eigenspace

??

mercio

cuz they are writing V|n0> = sum of (ith coordinate of V|n0> in the basis) * ith basis element

you chemists and your fancy names

Rudi_Birnbaum

22:18

I mean normally H = H_0 + dH/dq * dq = H_0 + H_1 (=V)

mercio

I guess that V|n0> being in the eigenspace is a thing we can assume for now

maybe there is a physical reason behind it

but I'm not finding a justificatoin in the page so far

Rudi_Birnbaum

You know we use the hydrogen eigenfunctions to describe everything at all ...

Abdullah UYU

I changed my profile picture, why it didn't change in chat?

mercio

aaah

Rudi_Birnbaum

OK the main point of that all would be give some justification of 7.8 here ibb.co/eq2bFK

mercio

22:22

okay they assume that the psi i is an orthnormal basis of the hilbert space

bold if you ask me

then okay 7.8 might make sense

Rudi_Birnbaum

@mercio I don't recall ever claiming the opposite ;-)

mercio

well I'm irked with the fact that they write finite sums

which implies a finite dimensional hilbert space

Rudi_Birnbaum

no thats infinite

mercio

which is irky when you are introduced with L²(R)

Rudi_Birnbaum

if it doesn't give limits it means all there is

Manolis Lyviakis

22:24

i think $φ\circ f$ will work on my problem

Rudi_Birnbaum

thank god I'm not ...

mercio

ah maybe it's a discrete summation then

instead of an integral

there are different kind of infinite sums after all

okay

Rudi_Birnbaum

Sure discrete, thats the point of sigma instead of th german-s

mercio

I wrongly assumed the sums were finite

Rudi_Birnbaum

, isn't it?

mercio

22:25

okay then I believe you in 7.8

Rudi_Birnbaum

huray!

Now first order JT is when the term linear in q dominates

exactly that

mercio

<psi i | V |psi i > ?

(i should never have asked if q was the position)

(i see it's the perturbation factor now)

Rudi_Birnbaum

7.8 is in the book page here ibb.co/eq2bFK

mercio

yes

dH/dq

Rudi_Birnbaum

q is some combination of nuclear positions, eg. any polynomials ...

mercio

22:28

:c

do you separate the distortion perturbation and the magnetic field perturbation ?

Rudi_Birnbaum

like make all bonds longer by q %.

mercio

are we only looking at the distortion perturbation for now ?

Rudi_Birnbaum

forget the magnetic field, nothing I talked about today is concernced with it

mercio

okay okay

Rudi_Birnbaum

yep

So q is some distortion.

For example you can form a symmetry adapted set of structure coordinates.

mercio

22:30

so if you have a large eigenspace

that third term will explode ?

because there is another basis element with the same energy ?

in the unperturbed state

and it will take over the first order term ?

Rudi_Birnbaum

Well when the second gets important thats called 2nd order JT effect

that is what I wanted to tell you today

mercio

if all your eigenspaces have dimension 1, this can't really happen, right ?

Rudi_Birnbaum

@mercio actually thats a really good point.

mercio

since we are talking about symmetrybreaking here

Rudi_Birnbaum

"all your eigenspaces " what do you mean, the psi_i s?

mercio

22:33

if the psi i all have distinct energy

Rudi_Birnbaum

Yes thats Chemistry

mercio

it will be "okay"

but if there is a big eigenspace

like in the hydrogen atom for example ?

Rudi_Birnbaum

actually thats small I guess

mercio

iirc the nth energy level has dimension 2n-1 or something like that ?

Rudi_Birnbaum

for real molecules the spaces are much much larger, though never contiuous,

at least we do not care about that

@mercio Well of course

there is degeneracy for certain levels

e.g. when we have E type irreps

or T,H or G. Then some levels have higher (sub)dimension than 1

mercio

22:36

what I am getting here is that the wavefunction will perturb inside the eigenspace of its energy level very very easily

Rudi_Birnbaum

in fact without effort

thats paramagneticity btw.

mercio

and it will use that to counteract whatever distortion the hamiltonian gets

Rudi_Birnbaum

some aspect of it ...

Hmm yes thats a neat way to see that

nice :-)

mercio

the most significant match will be using that and the coordinates among the other energy levels will probably be orders of magnitude smaller

Rudi_Birnbaum

WHat I wanted to say, it turned out in the decades of JT research, that in deed as you suggested even very high orders sometimes are relevant.

But not today for us.

mercio

22:38

you should pay me tbh :*)

Rudi_Birnbaum

Well I am actually searching for a post doc ;-)

mercio

I guess we will both go hungry then

Rudi_Birnbaum

lol

OK when the first order term is 0 by symmetry

mercio

so initially the JT effect was not particularly about large eigenspaces, right ?

when it was first studied

I think Im going to sleep soon

Rudi_Birnbaum

I still don't get that point exactly, what is large?

mercio

22:41

hydrogen atom's eigenspaces that have dim > 1

Rudi_Birnbaum

you mean degenerate

?

mercio

I'm not sure but maybe I mean that

Rudi_Birnbaum

Degenerate energy levels

In quantum mechanics, an energy level is degenerate if it corresponds to two or more different measurable states of a quantum system. Conversely, two or more different states of a quantum mechanical system are said to be degenerate if they give the same value of energy upon measurement. The number of different states corresponding to a particular energy level is known as the degree of degeneracy of the level. It is represented mathematically by the Hamiltonian for the system having more than one linearly independent eigenstate with the same energy eigenvalue. In classical mechanics, this can be...

mercio

yes

Rudi_Birnbaum

No that was the first idea

mercio

22:42

ah

well anyway that will be all for today from me

Rudi_Birnbaum

The idea is symmetry breaking due to mismatch of nuclear position symmtrey and electronic symmetry

OK, thanks again for tuning in!!

why is 2 large when 1 is not?

Rudi_Birnbaum

22:58

@mercio Oh I am so stupid! I know now what you mean!! If you have degeneracy in the basis space, this derivation is wrong and you need do it more carefully. This is called "degenerate perturbation theory". There you see that you never get 1/(e_i-e_i). E.g. here en.wikipedia.org/wiki/… for details.

you then have to diagonalize the pert. hamiltonian in the degenerate subspace.

Li357

23:31

Can the noun "proof" refer to some statements that disprove as well as prove?

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