The h Bar

Tobias Fünke

09:55

morning

imbAF

10:42

Hi

Can anyone help me with this thread of mine:
https://physics.stackexchange.com/questions/842888/notation-action

@DIRAC1930 I would say is the conservation of a physical quantity as a result of the invariance of Lagrangian under some transformation. i.e invariance of it under rotation is associated with angular momentum conservation

Tobias Fünke

@imbAF didn't the comments help already?

imbAF

What? I was replying to a question DIRAC1930 asked me

But I just saw it now in my notifications

Tobias Fünke

no I meant the comments under your question

Variationsrechnung

Die Variationsrechnung ist ein mathematisches Teilgebiet der Analysis, in welchem kleine Änderungen in Funktionen und Funktionalen studiert werden, um Minima und Maxima von Funktionalen zu bestimmen. Dieses (unendlichdimensionale) Optimierungsproblem mit Anwendungen in der theoretischen und der mathematischen Physik wurde um die Mitte des 18. Jahrhunderts insbesondere von Leonhard Euler und Joseph-Louis Lagrange zu einem Fachgebiet entwickelt. Die Variationsrechnung, ihre verwandten Themen und Anwendungen sind Gegenstand aktueller Lehre, Weiterentwicklung und Forschung. Die Frage „Wie können die…

the german Wiki article is really good I think

imbAF

One of the commets made it seem that is wrong to consider a system of N fields

and talk about the variation of the action

But, in theory, that shouldn't be a mistake

Tobias Fünke

I don't understand. If you understood the variational principle with one degree of freedom in classical mechanics it should be easy easy to generalize it with $N$ degrees of freedom

and the situation is completely analogous to the field case

so: Do you really understand the content of e.g. the Wiki article I've linked?

imbAF

10:50

I mean I understand it

Yes but still I have things that are unclear to me.

And I present my confusion with the questions that I ask

Tobias Fünke

OK. So now do the same steps with two DOF, for example. And then try to re-do in field theory. You will see it is completely analogous.

imbAF

The fact that for L(q,\dot q, \nabla q) I was able to derive an expression for $\delta S$, which is the same as the one in the wiki article for Hamilton principle

Tobias Fünke

Unfortunately I don't really understand your posted question

imbAF

it should highlight that I have an understating

@TobiasFünke for example which one?

@TobiasFünke Well I did that, that is the first part of m thread. Now I want a confirmation from people who know more, and also I want to know whether the notation used by me, is accurate or correct

ACuriousMind

Your question is very hard to understand because you've put a lot of equations into it but fundamentally you simply seem to have misunderstood the notation from the Wiki article (once again, Wiki is not a good resource to learn things you don't know yet from scratch)

Tobias Fünke

10:58

@imbAF I do not understand the post. It is hard for me to follow your thoughts, sorry.

imbAF

Actually, even in my lecture notes, when $\delta S$ was derived, the lecturer considered the case $L(q(t,\vec x),\dot q, \nabla q)$. An over simplified case. I wanted to generalize it to considering vectors

ACuriousMind

There is no such thing as $\frac{\delta S}{\delta \vec q}$ or $\frac{\partial L}{\partial \vec q}$ in careful notation. Equations like $\frac{\delta S}{\delta \vec q} = 0$ in the Wiki article are simply shorthand for the $n$ equations $\frac{\delta S}{\delta q^i} = 0, i = 1,\dots, n$, where the variations/derivatives are completely ordinary w.r.t. a number $q^i$

you don't need to understand "differentiation w.r.t. a vector" or anything

Tobias Fünke

@imbAF but this is not what the wiki article states

it considers $n$ DOF

yes, exactly what ACM said.

ACuriousMind

your question is very long but really the only thing I got from it is that you didn't understand that part

imbAF

Ok, but consider the following case

$S[\vec q + \delta\vec q]$ where $\vec q=(q_1,....q_N)$
Now, in general when you have a function of the form:
$f(x+\delta x)=f(x)+f'(x)\delta x + ...$

Now, if I want to do the same for $S[\vec q + \delta\vec q]=[q]+ \sum_{i=1}^N\frac{\delta S}{\delta q_i}\delta q_i + ...$

would it be accurate?\

ACuriousMind

11:03

If you understood the derivation of the one-variable case, you should be able to derive the n-variable case.

Tobias Fünke

@ACuriousMind that's what I pointed out already ^^

imbAF

That is what I think I am doing

above

ACuriousMind

No, you're asking us if it's "accurate"

you're not trying to derive the equation

you're just asking if it'S true

imbAF

lmao

Yes, I ask for a confirmation after what I think I did was the derivation

ACuriousMind

but if you understood why these Taylor-series-like equations hold in the simpler cases, you should be able to understand why or why not they hold in this more complex case

Tobias Fünke

11:05

@imbAF what does the q in the brackets mean?

imbAF

I already gave that $\vec q=(q_1,....q_N)$

How exactly isn't this clear?

Tobias Fünke

[𝑞]

imbAF

I am confused. What am I saying that is so unclear

Tobias Fünke

You wrote [𝑞]

Could you please not be so hostile? I am trying to understand what you are doing here

imbAF

Ahhh, I see

It's a typo. It is meant to be $S[q]$

Tobias Fünke

11:07

Ok

Signor Feynman

Just a minor thing to point out. I wouldn't call $\partial/\partial\vec{q}$ an uncareful notation . It's just an alternate notation for $\nabla_{\vec{q}}$

imbAF

But I cannot edit it now

ACuriousMind

@SignorFeynman but that would be the directional derivative in the direction of the vector field $\vec q$, that's not what is meant here

Signor Feynman

@ACuriousMind some use that notation for the gradient :P

ACuriousMind

well those some are bad at notation :P

Signor Feynman

11:09

The subscript is just there because you only take a "partial" gradient (i.e. not respect to all variables)

Tobias Fünke

I would call it bad notation because it is unclear lol --of course just IMHO

ACuriousMind

in any case, for what I said it doesn't really matter if you agree with my opinion about the notation :P

Signor Feynman

@ACuriousMind Indeed, but I learned about it way before I used the connection symbol

$\nabla$

It could be nastier looking, like $\mathrm{grad}_{\vec{q}}$

ACuriousMind

@SignorFeynman unaesthetic as it is, at least it's unambiguous :P

Signor Feynman

If Weinberg were among us, we could ask him advice about the worst possible notation

imbAF

11:11

I have done, taylor expansion for the multipole case, and of course the higher the order the more "under sum" terms you have. It's just different with functional

But ok

ACuriousMind

@SignorFeynman Weinberg has an extremely consistent approach to notation (which happens to produce abominations): You just write out every index and every dependency on every variable explicitly.

imbAF

That's good actually. Then you can introduce a simple symbol to replace one with many symbols, and the reader has an idea of what that simple symbol contains or implies

That is actually very good

ACuriousMind

no, it's atrocious because you can't read equations with 6 indices on every symbol, the human mind is not made for it :P

Hm, I guess I'm calling Weinberg an alien with that

Perhaps the truth is more nuanced? No, it's the other physicists who are aliens!

Signor Feynman

@ACuriousMind AHAHAHAAH thinking about it, it's so true

So simple and yet so excruciating

My biggest gripe with that is when you have functional derivatives and write the argument of the field

ACuriousMind

@SignorFeynman I'm afraid to say that makes a certain amount of sense to me. Like, the basic property physicists want to use is $\frac{\delta \phi(x)}{\delta \phi(x')} = \delta(x-x')$. Ignoring the already horrible use of $\delta$ in two meanings here, how do you write that without the field argument?

like, I wouldn't use that notation if I'm dealing with the theory of variational calculus rigorously, but for the way physicists do it, I think it's clear why the notation is like that

User198

11:30

When I have an operator in Schrodinger picture $\hat A_s$ and I want it in Heisenberg picture:
why does $\hat A(t)_H= U^\dagger \hat A_S \hat U$ change from time indepentant to time dependant operator?

ACuriousMind

@User198 It's not clear what you're asking. If you write $A(t)_H$ on the l.h.s., you should also write $U^\dagger(t)$ and $U(t)$ on the r.h.s. What's the question, exactly?

User198

Why does acting on the $\hat A_S$ with unitary operators in such a fashion turn it into a $A(t)$ ?

Should I just regard it as a definition?

ACuriousMind

That's why I said you should also write $U(t)$

of course, if you let time-dependent operators $U(t)$ act on the time-independent $A_S$, the result is itself time-dependent in general

it's not clear what more of a "why" you're looking for

User198

I am looking for some motivation on why do we do $U(t)^\dagger$ on the left of it and $U(t)$ on the right of $\hat A_S$, it seems out of the blue.

So just because $U(t) is time dependant it turns $\hat A_S$ into a time dependant quantity?

ACuriousMind

You're not really asking why it's "time-dependent"

Signor Feynman

11:36

@ACuriousMind it does, in the sense that it works with distributions and makes things easier, without having to differentiate integral functionals only. That being said, it's the same degree of wrong of conflating $f$ and $f(x)$, which I think that can lead to some confusions with notation, especially if you learn directly that way

ACuriousMind

you want to know why this specific $A_H(t)$ is something we should consider

then the question becomes: How do you even know about this equation of $A_H(t)$ without having derived it, apparently?

Signor Feynman

If you learn directly to functionally differentiate like that, you will get stuff like $\nabla\delta(x-y)$, which makes me uncomfortable without an integral

Much more than a normal delta :P

ACuriousMind

@SignorFeynman Why? Just like the $\delta$ gives the value at a point when integrated against a function, its derivatives give the values of derivaties at a point. There's nothing wrong with it

The derivatives of the $\delta$ are perfectly well-defined even in the rigorous theory of distributions

User198

Hm. My question is than this:

When considering a unitary transformation of a state $\psi$ if we want $\psi '$ we get it by $\psi '= \hat U(t) \psi$. Ok.

For operators we do:

$\hat A'=\hat U \hat A U^\dagger$

Why do we do it in that way for transforming operators?

ACuriousMind

@User198 That means you have not seen any derivation of the Heisenberg equation of motion?

User198

11:45

It reminds me of kind of when you act on a vector with a rotation matrix from one side and its transponse on the left to make a rotation.

ACuriousMind

Texts usually don't drop the formula for $A_H$ in terms of $U$ and $A_S$ out of the blue, but they derive it, so I'm puzzled where this question comes from

Signor Feynman

@ACuriousMind well, actually there should be a minus due to integration by parts

ACuriousMind

@SignorFeynman as usual I am talking modulo signs :P

Tobias Fünke

@SignorFeynman and actually, the "integration by parts" is just a mnemonic :p. the derivative of distribution is defined to be as it is

Signor Feynman

For the case of the delta, it appears or not depending on whether you differentiate wrt $x$ or $y$

@TobiasFünke I forgot the " "

Tobias Fünke

11:51

yeah yeah, I just wanted to join the nitpicking :d hehe

ACuriousMind

for instance, in terms of expectation values, it's easy to see that this is the only possibility: For a Schrödinger picture density matrix $\rho(t) = U^\dagger(t) \rho U(t)$ and a Schrödinger picture operator $A$ we have $\langle A\rangle (t) = \mathrm{Tr}(U^\dagger(t) \rho U(t) A)$ as the expectation value of $A$ at time $t$. This expression is not picture-dependent.

In the Schrödinger picture you package the first three terms in the trace into the $\rho(t)$, in the Heisenberg picture you use cyclicity of the trace and then package the last three terms in $\mathrm{Tr}(\rho U(t)AU^\dagger(t))$ into the Heisenberg operator $A_H(t)$.

Signor Feynman

That minus would be the one of $\nabla_x\delta^3(\vec{x}-\vec{y})=-\nabla_y\delta^3(\vec{x}-\vec{y})$

And since I'm obsessed with minus signs, I hate that :P

skullpatrol

some would even call them lowered minus sign to added a lord-like tone to the obsession :P

ACuriousMind

what

Signor Feynman

I didn't understand, but I read that with in a upper-class British accent

With a disgusted tone for "lowered"

skullpatrol

11:59

you got it, then

ACuriousMind

but...why is the minus sign "lowered"? It's just a minus.

skullpatrol

because you haven't seen the raised minus and plus signs before numbers

they're found in old text

Tobias Fünke

quora.com/…

do you mean that?

holy

skullpatrol

2

Q: About the raised negative sign in some basic textbooks

In a math document recently (a UK A level test paper from the EdExcel board), I noticed that the negative/minus sign was raised and aligned to the top of the number. I'm interested to know whether this is a local or regional habit, how common or widespread it is, and whether it's considered a goo...

pretty much faded away by now

User198

12:25

No until now. I think I understand it now. We use unitary transformations because they preserve the inner product, so measurements will always be correct.

And it is U dagger on the left because that is how a bra transforms under a unitary transformation and U on the right because that is how a ket transforms under a UT.

2.) Do some people do unitary transformations on the Hamiltonian, and calculate the evolution on a system that way, instead of solve the Schrodinger equation?

Tobias Fünke

@User198 what do you mean?

Why should $UHU^*$ give the same as the SE?

User198

I read this on the wiki en.wikipedia.org/wiki/…

Often, however, the Schrödinger equation is difficult to solve (even with a computer). "Therefore, physicists have developed mathematical techniques to simplify these problems and clarify what is happening physically. One such technique is to apply a unitary transformation to the Hamiltonian. Doing so can result in a simplified version of the Schrödinger equation which nonetheless has the same solution as the original."

Tobias Fünke

yes you can do that (what is described in the Wiki article)

but what you asked seems not to be what Wiki says?!

ACuriousMind

why is everyone here trying to learn physics from Wiki intros :P

User198

xD

Tobias Fünke

12:29

I don't get it either

ACuriousMind

Wikipedia is not a textbook, it will not teach you an actual, coherent understanding physics at the same level as a lecture or textbook

User198

Yes yes I understand that. But it is a super fast resource

But I agree with you 100%

ACuriousMind

picking random articles and acting as if the colloquial sentences in the intros (Wiki intros are intentionally not written to be technical) can be picked apart or interpreted in a direct technical sense is not useful

Tobias Fünke

+++

@User198 super fast but as it seems super fast to misunderstand (?)

User198

True true

ACuriousMind

12:33

I find engaging with the questions this generates very difficult. The questions use all the technical terms you learned from Wikipedia, but there is none of the technical understanding behind using that term I would expect a student of physics to have. So I answer as I would answer a student, and what ensues is a lengthy conversation full of confusion

Tobias Fünke

yes

Anyway, what the Wiki article is about that you can change the problem of determining the evolution of a state by transforming both the Hamiltonian and the state

and then it sometimes is easier to solve this transformed problem, and then simply use the inverse transformation to get the evolution you are actually interested in

Rabi Oscillations is the typical intro example

User198

@ACuriousMind Understandable. I will restrain for asking such questions furthermore. Although some of the answers I got in this chat were helpful, so I am thankful for that.

@TobiasFünke Ah ok thanks.

imbAF

13:11

On the topic of functionals, would it be right to say that action variation, is the variation of a functional that takes a vector valued function as an argument?

Tobias Fünke

$S$ takes as input some path /function (which indeed can be vector-valued)

the first variation of a functional is a functional again

imbAF

the first variation of a functional?

Tobias Fünke

...

Ok

How do you define variation?

imbAF

You mean $\delta S$?

Tobias Fünke

what is action variation?

imbAF

13:17

$\delta S$

Tobias Fünke

what does $\delta S$ mean, mathematically?

imbAF

A change/variation of the value of the functional, for a change in value of the function, that the functional takes as an argument

Tobias Fünke

I can guess what you mean, but that is not mathematically precise

do you recognize what you say in the equation/definition here? :

First variation

In applied mathematics and the calculus of variations, the first variation of a functional J(y) is defined as the linear functional δ J ( y ) {\displaystyle \delta J(y)} mapping the function h to δ J ( y , h ) = lim ε → 0 J ( y + ε h...

imbAF

It's a bit confusing because it mentions the linear functional $\delta J(y)$ but afterwards it considers $\deltaJ(y,h)$

But other than that, yes, I am familiar with the expression

Tobias Fünke

good

$\delta J(y): h\mapsto \deltaJ(y,h):=\ldots$ where $\ldots$ means the definition given in the article.

although I don't think it has to be linear (but we should anyway ignore this for now)

Re-reading your question, I guess one can safely ignore the first part of the sentence, and the question thus reads: would it be right to say that action functional takes a vector valued function as an argument --- ? is that your question? I don't see why you would bring the variation here into play

imbAF

13:31

Just to add something, in case is of importance. In the lecture, we considered a slightly different way of expressing (first) variation:
For $F(f)$
$\frac{\delta F}{\delta f(x)}=\lim_{\epsilon\rightarrow 0}\frac{F[f(x')+\epsilon\delta(x-x')]}{\epsilon}$

Tobias Fünke

yes, that is the "physicist way"

Functional derivative

In the calculus of variations, a field of mathematical analysis, the functional derivative (or variational derivative) relates a change in a functional (a functional in this sense is a function that acts on functions) to a change in a function on which the functional depends. In the calculus of variations, functionals are usually expressed in terms of an integral of functions, their arguments, and their derivatives. In an integrand L of a functional, if a function f is varied by adding to it another function δf that is arbitrarily small, and the resulting integrand is expanded in powers of δf,...

imbAF

@TobiasFünke yes, that is my question.

Tobias Fünke

@imbAF ... but then why not formulate it so simple?

imbAF

Because we are in the topic of variation

Tobias Fünke

and you for sure know the answer already. You yourself this morning wrote $S(q)$ where $q$ was a vector valued function

imbAF

13:33

Correct

Tobias Fünke

Good

So can you ask your question again, in a precise way?

imbAF

My question has to do with the variation, in the end but I would say

For a functional, that has a vector valued function as an argument, how to get the expression for the first variation of it?

Tobias Fünke

why do you think it would look different?

imbAF

Nothing apart from the notation, to signalize that, the argument is a vector value function

And that is what interests me. The notation

Tobias Fünke

13:49

I cannot follow what you are trying to ask here

Why don't you ask this then directly?

And what do you mean with you are interested in the notation. You can use whatever you want. There is no fixed notation

imbAF

Ok, I will ask directly

This expression doesn't satisfy me. I want to know how the expression looks like, when we EXPLICITLY write down the components of the vector valued function, that is the argument of the functional S.

Tobias Fünke

Why not taking $N=2$, and start from the definition

to see

ACuriousMind

@imbAF The $\cdot$ in the last expression are scalar products. What problem do you have converting that into an explicit expression in terms of components?

Tobias Fünke

I am afraid that the notation I'd use is not what you'd like

imbAF

@ACuriousMind I think I am running on loops here. If I do what you say, and dare to ask, if what I wrote down is correct or not, I will be told, aren't you sure, why are you asking, you should be 100% certain.

Tobias Fünke

13:57

Ok. I suggest this (probably for the third time): Forget the Wiki article

Take an action depending on $q_1,q_2$

derive completely analogous what you have done already for Lagrangians/actions which depend only on $q_1$ (or $q$ for that matter)

Then you can see what notation suits you the most and which notations have certain advantages or disadvantages

imbAF

I did do that, and I got a result, which should be 100% correct

Tobias Fünke

good

have you checked for e.g. two non-interacting particles?

as a sanity check?

ACuriousMind

@imbAF What I am trying to do is to get you to ask the question in a way that reveals what the actual problem in the understanding is. You know the definition of a scalar product in terms of components. You can apply that definition to the last expression in your screenshot. You get a result. Which step of that are you uncertain about, i.e. from where does the doubt arise because of which you are trying to ask this here?

Tobias Fünke

you should get uncoupled Newton's equations

imbAF

@TobiasFünke I don't know how to do that, the two non-interacting particles. Or maybe I do, and just don't know what you mean

Tobias Fünke

14:02

take two free particles of masses $m_1$ and $m_2$. write down the Lagrangian

apply your result to see what EOM you get out

of course you can even test more general cases

imbAF

The problem, which I will try to highlight is that:

I am trying to understand something. My starting point is A and i need to land to point Z. In between there are multiple instances (points) where question marks pop. So when I ask a question, that doesn't mean that suddenly all the rest of the instances where I have unclarity are suddenly clear. I have no issue in writing $\delta S$ when S has a vector valued function as an argument. Just introduce a sum notation on the RHS , index the components you derivate the function with and the infinitsimal changes you multiply them with and is don…

Tobias Fünke

14:22

?

ACuriousMind

yeah, I don't understand that response either

Tobias Fünke

sometimes I feel "mean", because most of my replies are "I don't understand"--but I honestly do not get what they want to say

Signor Feynman

14:57

@TobiasFünke nah, it's not mean if you genuinely don't understand

I like to say "I don't understand" in real life until someone says a sentence in a way I like. That's mean :P

Tobias Fünke

lol

skullpatrol