The h Bar

Tobias Fünke

15:00

proving existence of the thermodynamic limit was/is a huge problem in condensed matter, also regarding the stability of matter and so on

sorry, it seems I've misread your first message

Es tut mir leid, Herr Feinmann

Herr Feinmann

@TobiasFünke I should have said "for an ideal gas" (right?)

@TobiasFünke guten morgen

Tobias Fünke

@HerrFeinmann no, it was my mistake. I misread your statement. I wanted to emphasize that you really need the thermodynamic limit (which you've correctly said)

naturallyInconsistent

@ACuriousMind they are poor blackbodies, but they can still emit BBR; just like stars, as you mentioned. As mentioned earlier, the reason for the brightness of soot, is that it is 1) denser 2) has a yellow emission line. The emission line is more important, just as how sodium lamps have bright yellow light even though that is clearly gaseous in density. The effect is that it is concentrating a lot of the energy that would otherwise have had to be spread across the spectrum, to just one colour

Tobias Fünke

your example is fine I'd say (presumably concerning the ideal gas), though not really applicable I think. As far as I understood the question, it really concerns (classical equilibrium) thermodynamics only, no?

Herr Feinmann

15:23

I added it just a side note

Tobias Fünke

yes

Herr Feinmann

@TobiasFünke Herr Feinmann

Nairit Sahoo

15:47

$\delta_{A}B=[A,B]$ so $B$ is $A$-closed if the commutator vanishes and $C:=\delta_A B$ is $A$-exact. Exact implies closed trivially by $d^2=0$. If I want to prove it at the level of commutators we have to use Jacobi's identity.

So I am thinking: is there a relation between Jacobi's identity of commutators and $d^2=0$ which is Bianchi identity?

nvm. Found this. But I hope Jacobi's id has some geometrical significance like Bianchi's which is boundary of a boundary is nothin

Silly Goose

@naturallyInconsistent what is the text that does not use B-O ?

ACuriousMind

@NairitSahoo What are you talking about? $\delta_A^2 \neq 0$ for arbitrary $A$, it is extremely easy to find counterexamples where $[A,[A,B]] \neq 0$. All Jacobi gives you for that is that it is equal to $-[A,[B,A]]$.

Nairit Sahoo

@ACuriousMind Oh yes, you are right. So why does this happen? Is exact=>closed not true in this case but only true in the case of differential forms?

ACuriousMind

@NairitSahoo Why does what happen? Your $\delta_A$ and the exterior derivative $\mathrm{d}$ are completely different operators on completely different spaces, what do you think they have to do with each other?

Nairit Sahoo

@ACuriousMind So $\delta_A$ is not nilpotent here in general. I see

ACuriousMind

15:58

why would it have been nilpotent? You can't just randomly denote some operation as $\delta$ and because that vaguely looks like a $d$ assume it's nilpotent :P

Nairit Sahoo

Hm u r right. my bad. Thx as always

@ACuriousMind But indeed if $A$ was nilpotent does $-[A,[A,B]]$ vanish? I don't see how it vanishes

$\delta_A^2=0$ I meant sorry

Slereah

16:58

finally getting the hang of lattice stuff in categories

The big issue was apparently that the category people tend to not be too clear when they're talking about the category itself versus the category of subobjects

and since the category of subobjects is a poset the terminology is very different

Tobias Fünke

17:32

@imbAF May I give you an advice concerning your questions on the main site? You should really put more work into formatting. This includes using the "double dollar" (instead of the single one) $$ as a Math environment for important equations; but also to use properly the quote function, and to use some highlighting (not too much) with bold and italic fonts. Many of your posts are also rather long, and I think they can be shorten.

I really just wanted to give you that advice, from my experience with the site. Do with that whatever you want.

Slereah

ie why the "internal hom" for the negation basically has nothing to do with the actual internal hom in the category

imbAF

17:50

@TobiasFünke Thank you for the suggestion. I will keep it in mind, the next time I post something, to make my thread more appealing to read by others. Put more effort into the visuals so to speak

Tobias Fünke

exactly

DIRAC1930

18:05

What does Schrodinger mean by Pfaff-differential in equation 6.14 on page 48 (page 73 of the pdf) strangebeautiful.com/other-texts/schrodinger-st-struc.pdf

ACuriousMind

@DIRAC1930 It's a bad translation: "Pfaffian form" is a German term for a 1-form, cf. de.wikipedia.org/wiki/Pfaffsche_Form, i.e. he just means that that expression is a 1-form.

DIRAC1930

Thanks

This book is a masterpiece if anyone is interested

Except the last chapter which he introduces his own theory

Roger Penrose learned GR from it

Tobias Fünke

oh, interesting

Herr Feinmann

@ACuriousMind German detected (yeah, I know it was a message specifically about German, but I was taking my revenge for this)

Jan 18, 2023 at 17:41, by ACuriousMind

Italian detected!

DIRAC1930

18:21

I wonder what Schrodingers views on QFT were

Silly Goose

18:42

Does the adjoint representation "commute" with other representations? For example, is it true that $\text{Ad}_{\pi(g)} \pi_(X) = \pi_ (\text{Ad}_{g} X)$ where $\pi:G \to GL(V)$ is a Lie group representation and $\pi_*: \mathfrak{g} \to \mathfrak{gl}(V)$ is the induced Lie algebra representation.

oops ignore

Does the adjoint representation "commute" with other representations? For example, is it true that $\text{Ad}_{\pi(g)} \pi_\star(X) = \pi_\star (\text{Ad}_{g} X)$ where $\pi:G \to GL(V)$ is a Lie group representation and $\pi_\star: \mathfrak{g} \to \mathfrak{gl}(V)$ is the induced Lie algebra representation.

Arjun

19:02

@HerrFeinmann Interesting,maybe an online meeting with the link shared here and anyone who wants to join could : )

Herr Feinmann

19:22

$\text{Ad}_{\pi(g)} \pi(X) = \pi(\text{Ad}_{g} X)$ where
$\pi:G \to GL(V)$ is a Lie group representation and $\pi_*: \mathfrak{g} \to \mathfrak{gl}(V)$ is the induced Lie algebra representation (ftfy @SillyGoose)

Silly Goose

what is ftfy :0

Leaky Nun

@SillyGoose “fixed that for you”

If mirrors “absorb” the photons and then “emit” them after a random interval, wouldn’t the resulting light wave be destructively interfering with itself?

Silly Goose

@HerrFeinmann hm but it should be $\pi_*$, no? $\text{Ad}_g X \in \mathfrak{g}$

Leaky Nun

how can we test whether there is really a delay between absorption and emission?

imbAF

19:46

When it comes to calculating S-matrix elements in QFT, is this way of representing the matrix element, a general case that is valid for whatever process we consider:

$\langle f|\hat{S}|i\rangle$.
$\hat{S}=1+iT$, T is transition matrix.
Then:

$\langle f|T|i\rangle=(2\pi)^4 \delta^{(4)}(P_i - P_f)\mathcal{M}_{fi}$ ?

Herr Feinmann

@SillyGoose I mindlessly fixed the rendering :P

Silly Goose

oh

Herr Feinmann

@SillyGoose "fixed that for you"

You don't reddit enough, my goose

Tobias Fünke

0

Q: Question on the density of states on a $d$-dimensional box $L$ sized

The teacher went through the theory on how to find the density of states of a quantum particule on a $d$-dimensional box of side $L$, spin $J$ and hamiltonian $H=\alpha |\vec{p}|^s$ where $\alpha, s$ are positive constants. But he then used the volume of an hypersphere of radius $R$ to find the n...

does anyone understand what their question is?

I've tried to formulate a reply now several times, but all sound rude, which is not my intent, but I really don't get what the problem is

Herr Feinmann

20:04

@TobiasFünke OP is basically asking: why is the DoS $\frac{d^3pd^3r}{h}$?

I mean, the question is totally obscure and I think OP is mixing his/her personal guesses to facts but I would dare to say that's the question

Tobias Fünke

wait, what is $r$ here?

Herr Feinmann

(Of course I was sloppy, I meant that $dN$ is equal to that. The density of states is dN/dE)

@TobiasFünke position integration, it gives a volume factor

Tobias Fünke

I see

thanks

Herr Feinmann

20:20

That's my guess, though

Tobias Fünke

yeah, I mean you are right anyway: the question is very obscure

Herr Feinmann

Probably the way OP has seen it is similar to the way one finds the density of states of the EM field

Tobias Fünke

I'll also never understand why so many people have problems to put the relevant information into the question. They either post nothing or put way too much. lol

Relativisticcucumber

todays confusion is about color centers. in ashmerm they introduce the idea of color centers and say that their presence imparts a strong color to an otherwise transparent crystal. i dont understand how this can be. tmu, the reason a crystal is transparent has to do with the resonance of the incident waves with the transitions in the material, so the fact that these defects have a spectrum doesnt explain to me why we see color from these in an otherwise transparent crystal?

Tobias Fünke

I mean how should we know what their teacher is doing

Relativisticcucumber

20:21

@TobiasFünke i think it takes knowledge and experience to frame a question well. sometimes people don't know enough to know what else is out there

Herr Feinmann

@TobiasFünke I think it's a mistake in good faith, though

Tobias Fünke

@Relativisticcucumber yes! just to get this clear: posing a good question is hard

Herr Feinmann

I've seen less effort question

Tobias Fünke

yes, true indeed

Relativisticcucumber

sometimes when i tutor students all they can say is "why" because they are just so confused

Herr Feinmann

20:23

@TobiasFünke I've always thought that posing the right questions is more important than giving answers

In the process of understanding

I hope that that sound cheap :P

Relativisticcucumber

i am an offender in this regard

Tobias Fünke

yes, I agree. one has to think clearly before one is able to asking a good and well-posed question

Relativisticcucumber

there's some interesting philosophy here too bc some people think in words, but i think in images, so it can be very hard to translate this into a question

Herr Feinmann

What I meant is that a good asker is better for science than a good answerer

If someone asks the question be sure that one day someone will be there to answer

Tobias Fünke

but what I mean is that I see soooo many posts about what happens in a lecture, for example, and then no relevant information is given (which course, notation explained, derivation steps etc.) it is often just "I did not understand why my teacher did X"...

@HerrFeinmann to some degree yes. To be successful in academia (which is not necessarily the same as "doing good science"), one also has to identify knowledge/research gaps and asking relevant questions

Herr Feinmann

20:26

I have made similar posts in the past but the result was that I probably gave too much material and no one wanted to answer :P

Tobias Fünke

but to find suitable ansatzes for answers is important too

Relativisticcucumber

@TobiasFünke the other week one of my students requested that we discuss volume

Tobias Fünke

@Relativisticcucumber what do you mean with "volume"? :d

Silly Goose

is the delta potential $\delta(x)$ as an abstract linear operator $\lvert x = 0 \rangle \langle x = 0 \lvert$?

Tobias Fünke

@Relativisticcucumber yes, I see that

Relativisticcucumber

20:27

@TobiasFünke thats exactly what i said

or "can we discuss integrals"

love these things xD

Tobias Fünke

@SillyGoose can you expand? What is the level of rigor you are discussing objects like $|x\rangle\langle x|$?

Silly Goose

I think maybe I have something slightly off, because in the position basis $\langle x \lvert 0 \rangle \langle 0 \lvert x' \rangle = \delta(x) \delta(x')$

Relativisticcucumber

another common offender is "i understand how you did it, but can you tell me why my teacher did it this way"

i despise this request

Silly Goose

@TobiasFünke I am (rare is the occasion!) looking to avoid all technicalities and use the bare minimum machinery to obtain something that works out.

Tobias Fünke

@Relativisticcucumber hehe yeah

Relativisticcucumber

20:29

@SillyGoose gasps

Tobias Fünke

@SillyGoose well, the RHS applied to some vector $\psi$ simply gives $\psi(x) |x\rangle$.

Herr Feinmann

It's time for me to watch the weekly episode of Daima

Tobias Fünke

@HerrFeinmann yeah as I've said. It is hard to give the right amount of information, and it is especially hard if one understands little of the topic one asks about. I agree

Herr Feinmann

I think I managed to get a decent understanding of magnetic work for superconductivity. Physics makes sense again, no need to quit :P

Tobias Fünke

@SillyGoose The question is what exactly you mean here with linear operator at all. What should be the action on vectors/wave function?

@HerrFeinmann hehe good job. great. enjoy the pleasure of success!

If you want to define an operator (no rigor here at all on this level) as e.g. $\delta: \psi\mapsto \delta_\psi$, where $\delta_\psi(x):=(\delta \psi)(x)= \delta(x) \psi(0)$, then one could write $\delta=|x\rangle\langle 0|$ or so. Is that what you want?

Herr Feinmann

20:33

T$\mathfrak{H}anks$

Tobias Fünke

haha

don't make fun of fraktur

$\mathfrak h$ is my favorite QM symbol lol

I use it for "single-particle Hilbert space"

Herr Feinmann

It's an upper case H! It's used for magnetic field with no magnetization

Tobias Fünke

yes I know (that it is an uppercase H) $\mathfrak H$

Herr Feinmann

@TobiasFünke for me it will always be my second choice for a generic Lie algebra :P

Tobias Fünke

aha

why not something related to L?

$\mathfrak L$

Silly Goose

20:35

@TobiasFünke i just want a potential operator that obstructs the wave function from having nonzero value in a certain subset of space

Herr Feinmann

The standard notation uses g first (lazy to type mathjax), then h, then k

Just like you use x first, then y and then z

Tobias Fünke

I see

Silly Goose

i think the delta potential is used in textbook QM, and i thought this would furnish an example. but i am not sure if it is a good example because im not sure what the delta potential is in abstract formalism (not in position basis)

Tobias Fünke

@SillyGoose Well, for finite intervals $\Delta$, one certainly has $\int_\Delta \mathrm dx |x\rangle\langle x|$.

Herr Feinmann

Back when I hadn't realized what it was (and when I didn't understand the difference between $H$ and $\mathfrak{H}$), I didn't know how to read it. Sometimes I would joke about that with my friends and call it "horse"

Tobias Fünke

20:37

the latter projects the wave function to the subset $\Delta \subset \mathbb R$

Herr Feinmann

(My professor's handwriting made it worse)

Tobias Fünke

haha

Herr Feinmann

I once asked a question during a lecture calling it "that crooked H"

Silly Goose

okay so i guess the delta potential might look like $(\int_{\mathbb{R}} dx \lvert x \rangle \langle x \lvert) - \lvert 0 \rangle \langle 0 \lvert$ then?

Which is heuristically $\mathbb{I} - \lvert 0 \rangle \langle 0 \lvert$

Tobias Fünke

Silly Goose

20:41

But I mean I can WLOG subtract off $\mathbb{I}$ leaving the Hamiltonian as $H := -\frac{\hat{p}^2}{2m} - \lvert 0 \rangle \langle 0 \lvert$

Tobias Fünke

I still don't understand what you are aiming for :d but right now it seems that the "operator" you propose is indeed a good choice, since it corresponds to multiplication with the Dirac delta in position space, as far as I can see

Silly Goose

i am trying to see if one can describe a potential-induced obstruction of the wave function as instead a space-induced obstruction of the wave function

i.e. if i can interconvert between a potential term and an effective space on which the theory is taking place

Tobias Fünke

oO I still don't understand, but no problem.

@SillyGoose where is the problem now?

I think problems like this will always arise in this non rigorous fashion, no?

Silly Goose

i guess it might be simpler to pick out all of the wave function except say a gaussian plane wave centered at the origin

Tobias Fünke

yes

probably

intuitively, if $|\phi\rangle$ is a very narrow Gaussian centered at $0$ in position space, then approximately $|\phi\rangle\langle \phi|\psi\rangle \approx \psi(0) |\phi\rangle$. Does that make sense?

Silly Goose

20:51

yesh

Well working with some gaussian is probably more realistic anyway.

Tobias Fünke

Well, working with the dirac delta should also be fine, to some extent, when keeping in mind the subtle issues

Relativisticcucumber

german word of the day: farbzentrum

Tobias Fünke

hehe yes

Silly Goose

21:06

okay I think this is a sketch of a proof that this Hamiltonian obstructs the system having any nonzero component of some arbitrary state $\lvert g \rangle$. In particular, this can be (in the position basis) a gaussian centered at $x = 0$.

so long as $\lvert g \rangle$ is not an eigenstate of $\hat{p}^2$

Tobias Fünke

for anyone wondering how Max Planck spoke:

Max Planck - Selbstdarstellung im Filmportrait (1942)

►

@SillyGoose Your Hamiltonian is a special case of a so-called rank-one update

a certain perturbation which was and is studied a lot. It appears in my contexts, also in physics

@SillyGoose Are you sure that your derivation is correct?

qwerty

21:25

I only just started writing answers so I'm a bit new to it. is there any way to get feedback why people didn't like something or presumably thought it was wrong?

Tobias Fünke

Take e.g. $\psi$ eigenstate of $H$ and $g=\psi$. Then your claim is wrong.

or do I miss something?

@qwerty Hi qwerty

Well, not really. If you received a downvote, you could write a comment under your own answer asking for clarification

qwerty

@TobiasFünke morning

Tobias Fünke

but in my experience this does not help in most situations. anyone caring for constructive feedback (if it is worth to spend time writing a comment) will leave a comment with suggestions/critique etc.

or do I misunderstand your question?

qwerty

@TobiasFünke no, you understood. thanks

Allie

yay

i got a 4.0

Tobias Fünke

21:28

nice, congratz!

qwerty

@TobiasFünke I wonder if it's bad form to ask here

Tobias Fünke

@SillyGoose ah, sorry, I missed the point indeed ^^ but still the result seems weird. especially if you go to finite dimensions

Arjun

@Allie That's impressive,congrats!

Tobias Fünke

@qwerty I don't know any rule. Why should it be?

Allie

:3 thank you besties

time to upload this to all my grad apps

qwerty

21:29

@TobiasFünke dunno just I'm always worried about doing smthg wrong

Tobias Fünke

haha

I mean now we could simply check all your questions/answers to find out what you mean...

so you can just post it; at least from a practical POV... again, I don't know the rules. But in worst case a mod (presumably ACM) will just delete the message, no?

qwerty

-1

A: In general relativity, how do we know when the coordinates we compute are physical observables?

the coordinates we mathematically use are just "labels", that can change and live on a curved surface. There is something that other answers, I feel, have not stressed: You, the observer, are always locally in a little tiny patch of Minkowski spacetime. So you are free to pick, for your own lab...

Tobias Fünke

@SillyGoose yeah, I think there is a mistake in $(1a)$. The operator on the RHS will generally take you out of the $g$ subspace

Silly Goose

@TobiasFünke but isn't this what is wanted?

Tobias Fünke

I don't understand.

If you want to take the $g$ and orthogonal $g$ components of eq. 1, then no, this is not what you are doing

you take the g component of the LHS, but still equate it fully to the RHS

Silly Goose

21:34

oh yes i made a mistake with the denominator of LHS i think

at the least

more like this i think as you said

oops forgot some constants

but anyway the form i think is more correct now

Tobias Fünke

Ok. As far as I can see, you neither made use of the infinite-dimensionality of the space under consideration nor of the specific form of the Hamiltonian, correct?

Silly Goose

I think not the infinite dimensionality of the space .But I did make use of the specific form of the hamiltonian i think. unless i misunderstand you

Tobias Fünke

where?

Silly Goose

Tobias Fünke

how does your conclusion change if you replace $H$ by $H=H_0+|g\rangle\langle g|$?

Silly Goose

21:45

also i think this is the correct version

Tobias Fünke

(I mean a generic hermitian rank one update)

Silly Goose

@TobiasFünke sorry what is $H_0$?

Tobias Fünke

any hermitian matrix

Silly Goose

also thanks for the class of Hamiltonian i will look into these more

Tobias Fünke

yeah, I used these sooo many times already

very useful to know+

Silly Goose

21:47

For my remark to be unchanged, I would require $H_0 \lvert g \rangle \langle g \lvert = 0$ I think

Tobias Fünke

Here is a small $2\times 2$ counter example. Let $A:=\mathrm{diag}(1,-1)$, and consider $v:=(1,1)$ to construct the hermitian rank-one update $R:=vv^T$ (a matrix with every entry equal to one). For the vector $x=(1+\sqrt 2,1)$ it holds that $Rx=v \langle v,x\rangle\neq 0$, but $x$ is an eigenvector of $A+R$.

.. and $v$ is not an eigenvector of $A$: $Av=(1,-1)$

All I want to point out is that your very first result seemed way too strong. (ofc I might be wrong, so feel free to correct me)

Tobias Fünke

22:08

also your second or new result (?) does not hold: We have $\langle v,Av\rangle=0$, but $1+E\neq 0$, where $E=1+\sqrt 2$ is the eigenvalue associated to $x$. --did I misunderstand anything here?

Silly Goose

Yes i think you are right

I will check to see if your example satisfies the new condition i came to

Tobias Fünke

OK

A simple (but perhaps relatively useless) condition is this: Write $H=H_0-R$, with $H_0$ hermitian and $R=|v\rangle\langle v|$. Then if $Hx=Ex$ for non-zero $x$, it necessarily holds that $(H_0-E)x=Rx$, and if $E$ is not an eigenvalue of $H_0$ then $x=(E-H_0)^{-1}Rx = \langle v,x\rangle (E-H_0)^{-1}|v\rangle$.

From that, you can e.g. conclude: a) it must hold that $\langle v,x\rangle\neq 0$ and b) that $1=\langle v|(E-H_0)^{-1}|v\rangle$.

In general, if $\langle v,x\rangle=0$, then ofc. $(E,x)$ is an eigenpair of $H$ iff it is an eigenpair of $H_0$.

I hope I did not do any mistakes, I leave it you to double check if you think it seems interesting :p

lol I messed up with bra-ket notation haha; I hope it is still clear

and with some minus signs... but ok, you get the point

BTW, the (minus sign corrected) result should follow from the matrix determinant lemma. As I said: It is a well-studied --yet extremely useful-- topic.

indeed, you made a mistake (?) in drawing the correct conclusion from (1a) (assuming this equation is correct); from 1a it simply follows that $g^T (E+p^2/2m)^{-1} g=-1/\lambda$. It is exactly what I've derived above (modulo my minus sign typo)

Silly Goose

22:31

right

Tobias Fünke

good

Silly Goose

except yours is slightly more general with $H_0$ as opposed to the kinetic term

Tobias Fünke

sure

but mine assumes finite-dimensions. There are also results for infinite-dimensional rank one updates, but I've never used those before

(but your calculations are not rigorous a priori anyway, so I guess the finite-dimensional constraint thus far is not too restrictive)

sorry to ask that again, but what exactly was/is your goal in proving? what do you mean with "obstruct"?

Relativisticcucumber

23:28

i found an interesting new (to me) book - Composite Fermions - does anyone have anything to say about this book?

Tobias Fünke

never heard of it before, sorry

Silly Goose

@TobiasFünke i wanted to answer the question: Given a Hamiltonian $H$ on Hilbert space $\mathcal{H}$ that obstructs the system from occupying a fixed state $\lvert g \rangle \in \mathcal{H}$ as in $\langle g \lvert \psi \rangle = 0$ for all physical $\lvert \psi \rangle$, is there a dual description in terms of restricting the corresponding classical configuration space, e.g., from $\mathbb{R}^3$ to $\mathbb{R}^3 \backslash\{p_1, p_2, ...\}$

Tobias Fünke

hmmh ok

Silly Goose

i guess this is necessarily working with infinite-dimensional Hilbert spaces

or perhaps there is some sort of lattice analogue

Tobias Fünke

and with physical states you mean eigenstates of the Hamiltonian?

@SillyGoose why?

Silly Goose

23:39

i am not sure. i think the statement is necessarily false if talking about eigenstates of Hamiltonian. because the eigenstates of the Hamiltonian (if hermitian) should span $\mathcal{H}$, so it would not seem possible for a state $\lvert g \rangle \in \mathcal{H}$ to be excluded from being occupied; i.e., generically if $\lvert g \rangle \in \mathcal{H}$ there exists a linear combination of energy eigenstates that has non-zero $\lvert g \rangle$ component.

Tobias Fünke

exactly this is what I'd have pointed out now. $\langle g,\psi_n\rangle=0$ for all $n$ implies $g=0$, this holds for any Hilbert space and orthonormal basis (independent of the dimension of the (separable) complex Hilbert space)

Silly Goose

@TobiasFünke since i am wanting to connect to real space (classical configuration space)

Tobias Fünke

@SillyGoose ah, sure, makes sense.

Silly Goose

although, i am not really familiar with infinite-dimensional quantum mechanics, so I could just be making a lot of false conclusions.

Tobias Fünke

well before you proceed, you first have to define what you mean with "physical states", i.e. define a suitable subset of states and so on... my two cents

Silly Goose

23:43

@TobiasFünke Also, I am thinking that the conclusion i came to that you wrote down here is a bit strange. it seems to place major constraints on $m, \lambda$, and $E$ since the ExpVal of $\hat{p}^2$ should be positive. I am wondering if I did something wrong then.

Tobias Fünke

@SillyGoose nono, your conclusion above (about $g$) is totally correct. If you have an orthonormal system $(e_n)_n$ in a complex (separable) Hilbert space, then it is a basis if and only if $\langle e_n,x\rangle=0$ for all $n$ implies $x=0$.

@SillyGoose No. I am quite sure that the result I posted is correct (except for the minus sign). Yes, of course you should expect non-trivial constraints for generic Hamiltonians/vectors $g$

Silly Goose

hm okay i guess it makes sense. the constraint i get just puts an upper bound on the energy, which seems sensible for a potential to impose

Tobias Fünke

well, yes. in general, you have Weyl's inequalities

and for rank-one updates I think you can conclude even more

is $\lambda>0$ in your case?

Silly Goose

i am letting it be free (but real i suppose)

the constraint i get is $-\frac{1}{\lambda} \geq E$

Tobias Fünke

can you post the derivation?

do you assume that $g$ is normalized?

Silly Goose

23:49

err wait i found another error bleb

Leaky Nun

a bit of a broad question, but, why is Schrödinger's equation linear?

Tobias Fünke

@LeakyNun I don't understand the question. Can you be a bit more specific? First of all, do you mean the time-dependent or time-independent SE?

Leaky Nun

both are linear right

Tobias Fünke

Second, "why" is the wrong word, no? By definition/construction it is so. -- We use SE because it correctly predicts what we observe (to some extent)

Silly Goose

Leaky Nun

23:52

i guess to expand on what i mean by "why", it's more like, "why" should we expect quantum particles to behave according to linear algebra?

Tobias Fünke

aha

good question

do you know the paper "The unreasonable effectiveness of mathematics in the natural sciences" by E. Wigner?

because you could ask this question in a much broader context

Leaky Nun

i don't know that paper, but sounds interesting

and what was his answer?

Tobias Fünke

you should read it

The Unreasonable Effectiveness of Mathematics in the Natural Sciences

"The Unreasonable Effectiveness of Mathematics in the Natural Sciences" is a 1960 article written by the physicist Eugene Wigner, published in Communication in Pure and Applied Mathematics. In it, Wigner observes that a theoretical physics's mathematical structure often points the way to further advances in that theory and to empirical predictions. Mathematical theories often have predictive power in describing nature. == Observations and arguments == Wigner argues that mathematical concepts have applicability far beyond the context in which they were originally developed. He writes: "It ...

there is also a Wiki page, apparently

@LeakyNun I mean, one could, perhaps, argue that this or that experiment suggests so (that linear algebra might be a suitable framework to formulate a theory). But I think historically this is not how QM grew. For example, Heisenberg did not even know matrices, but still developed what is nowadays known as "matrix mechanics".

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