The h Bar

2:11 AM

@naturallyInconsistent I can't tell if it's because you were right in Schroeder being a bad book, but I am so over thermo right now. The refrigeration & throttling process section is so boring

Like end of chapter 4

The explanations are so underwhelming for each system, idk what happened. It was going so well lol

I had so much more patience and interest in the chapters preceding this one, it just feels a lot like I'm being talked at instead of engaging with the material

naturallyInconsistent

@Obliv ... refridgeration is the single most important invention in the history of mankind...

But then again, it is just a rehash of Carnot cycle, so that there is no need to keep doing it

Kittel & Kroemer had a good section on refridgeration too

Obliv

Yeah, but I'm not really doing anything special except memorizing some facts about old school refrigeration theory

If I were an engineer or in the 1800s I would probably be way more enthused

I'm also not familiar with chemistry which doesn't help

I do feel better now that I've complained, so I will diligently get back to it :P

Silly Goose

3:25 AM

this is the (unintegrated) transition rate for an electron in a harmonic electromagnetic potential

Is the dipole approximation the following observation:

expanding the exponential, we get terms like $C (\frac{1}{\lambda})^l \langle n \lvert x_j p_k \lvert i\rangle$

so if $\frac{1}{\lambda} << \langle n \lvert x_j p_k \lvert i \rangle$, then we can approximate the exponential with its $0$th order term

naturallyInconsistent

4:29 AM

It definitely is something to do with the Taylor's expansion of the exponential so that you can treat each order of the perturbation separately, obtaining selection rules of opposing parity requirements etc, but I think you have some details wrong. e.g. 0th order.

Silly Goose

5:16 AM

bleb

Obliv

mlem

naturallyInconsistent

meowth

Silly Goose

so in the first two bullet points i write out just mathematically the condition to ignore higher order terms in the expansion of an exponential

but i don't see how this condition relates to the wavelength of the radiation field being large compared to the atomic length scale of the thing being irradiated

the $[-]_{ij}$ just means the matrix entries of $-$ in some basis

naturallyInconsistent

@SillyGoose I think your condition is wrong. $$\frac\omega c\hat{\vec n}\cdot\vec x=\frac{2\pi}\lambda\hat{\vec n}\cdot\vec x\ll1$$ is the condition for truncating the Taylor's expansion, and $\vec x$ is the source of the atomic length scale, $\lambda$ is the wavelength of the radiation field, and the $\frac{\hat{\vec n}\cdot\vec x}\lambda\ll1$ is the statement that the wavelength of the radiation field being large compared to the atomic length scale, as requested.

Silly Goose

5:32 AM

I think I do not understand how $\vec{x}$ related to the atomic length scale?

naturallyInconsistent

@SillyGoose $\vec{\hat x}$ is the position operator for the electron in the atom being considered, and it is being evaluated between two electron bra-ket vectors, both for that particular atom. It thus measures the average position of the electron inside that atom.

Silly Goose

okay i see

wait

but this was what i initially thought but what is really being sandwiched between two bra kets is

$x_jp_k$ for some $j, k = 1,2,3$

not just t he position operator

naturallyInconsistent

I don't think that really changes the correctness of that

Silly Goose

why is that

well hm let's see

naturallyInconsistent

The wavefunction of an atom is appreciably large only near the atom. Both $x$ and $p$ will thus only be appreciably contributing to the integral near the atom. Thus, everything there is bounded by the atomic length scale, in a very handwavy sense.

In particular, $\left|i\right>$ dictates that any contribution must not be too far from the atom, and no matter what, if $\left<f\right|$ is very large, it would then not have a strong overlap integral.

Silly Goose

5:59 AM

where does the sum in 5.329 come from?

I would think the literal integral on the left hand side should equal the right hand side, but without the sum

naturallyInconsistent

@SillyGoose The Dirac delta distribution converts the integral into one single term for each $\omega_{ni}$. There are many of them, thus a sum

Silly Goose

but $\sigma(\omega)$ as a function of $\omega$ has $\omega_{ni}$ fixed doesn't it?

oh i think

sakurai writes 5.328 as equivalent to integrating (or summing over) $\omega_{ni}$

so is sakurai considering discrete transitions $i \rightarrow f$?

i think this is what sakurai means (if we are to not abuse notation)

which seems to be consistent with these notes: farside.ph.utexas.edu/teaching/qm/lectures/node79.html

123

6:38 AM

Hello Everyone...

123

6:49 AM

@lucabtz Thanks for the reply. Now i understand the solution after your comment.

lucabtz

9:05 AM

@123 great

lucabtz

9:17 AM

@Relativisticcucumber the Hall book you're reading is Lie groups, Lie algebras and Representations right? Might do some revision of Lie theory and the book you're asking questions about seems good

first time i see these tables

naturallyInconsistent

@lucabtz this is so useless... And the difficulty is in doing the subtraction of multiples of $2\pi$ accurately and precisely...

lucabtz

its useless now, it was useful before calculators were a thing

naturallyInconsistent

Today's experiments had the Geiger counter go crazy. >80k counts per minute and the Geiger counter itself saturated

lucabtz

its an old book so at the time this was their calculator basically

Relativisticcucumber

9:44 AM

@lucabtz the more ppl in this chat who read hall, the better for me

and yes, it is

lucabtz

great might give it a read soon

Sir Crackpot

10:06 AM

@Relativisticcucumber are you creating an army

lucabtz

Lie army

Sir Crackpot

@lucabtz you only need a tiny soldier to build the army

lucabtz

tiny soldiers near the identity

Sir Crackpot

Maybe a bunch of soldiers yeah

lucabtz

non-commutative soldiers

unless you want an abelian group

Slereah

10:22 AM

He was this close to making an analogy but then gave up midway

lucabtz

the explanation is left as an exercise to the reader

Slereah

""One-dimensional",like "connected", is actually a philosophical concept, related to the minimal Hegelian level of figures which must be considered within an arbitrary space in order to determine that space's connectedness."

Ryder Rude

10:47 AM

it seems like Hegel's philosophy has contributed a lot to math

Slereah

Lawvere would certainly say so

Ryder Rude

why do philosophers overanalyse ideas that are understandable in an easy way? do they make any progress that way

for instance, i take a "set" for granted, and the idea can be understood easily

imo it is impossible to define a set in a non circular way

Slereah

that is the job

Ryder Rude

i just dont think theres any progress to be made in analysing these ideas. u r never going to be able to define a set in a non circular way

philosophers may over-analyse the word "existence"

or "being"

Slereah

Who do you think invented sets in the first place

Ryder Rude

10:54 AM

thats a good point, if Cantor was a philosopher

probably was

Slereah

I've read the original paper and he certainly didn't shy from overanalysing

Ryder Rude

oh

today, students are taught his theory without any overanalysis

Slereah

Plus sets are nothing but the modern version of universals and particulars :p

Ryder Rude

oh

some analysis may be necessary. e.g. Cantor's original theory had Russel's paradox

hard to say when the analysis is overanalysis

Slereah

His original theory had other issues

Silly Goose

10:58 AM

Do u guys think logic is not an innate human framework

Slereah

No idea, I've never been able to desincarnate myself from my human form

Silly Goose

It seems that we are hard wired to see things in what we call 3D

but we are free to construct whatever logic we’d like

Ryder Rude

logic is innate to the universe in some philosophies like the mathematical universe hypothesis

Slereah

Some people can't see properly

There are cases of people who can't really interpret their vision as a space

ie en.wikipedia.org/wiki/Visual_agnosia

3D is sort of "learned"

As far as the brain is concerned, there isn't any dimensionality from what the eyes send

It's just a bunch of points of light

Silly Goose

From what do we learn 3D from though 🤔

If we didn’t have clocks would we still think of time 0.o

Slereah

11:09 AM

The structure of space is mostly learn from kinesthesia

The sensation of tension of your limbs

And this is then extrapolated to vision

Ryder Rude

can a baby not comprehend space on their first day?

Silly Goose

Is there a reason to think of time as some fundamental thing—it seems more like a notion invented for convenience

Slereah

I don't think we ever asked a baby

Ryder Rude

i thought the experience on a baby's first day is not radically different from human experience

only neuroscientists can say for sure

@SillyGoose in some philosophies of relativity, all of our life exists simultaneously and the flow of time is just a biological illusion

by "simultaneously", I mean all of spacetime exists

not that "all of our life exists at the same moment of time"

Silly Goose

Like arrival

Ryder Rude

11:16 AM

i think this philosophy may be flawed becuz we dont understand QM indeterminism yet completely

Silly Goose

Blebs

Do you guys know if the following model exists: suppose you have a differential equation for which $\psi$ is a solution. 1. Supply an initial value, determining a $\psi$. 2. After evolving for time interval $\Delta t$, map $\psi(\Delta t)$ to a random point. 3. Repeat step 2.

Well it probably exists, but perhaps it has a name

Ryder Rude

what do u mean by "map $\psi(dt)$ to a random point'?

there r some theories which add stochaisticness to field diff eqns. Oppenheim's stochaistic gravity is a recent one

Silly Goose

Oh i should say map it to a random point and use this point as a new initial condition to determine a new solution for the next time step

Oh perhaps stochastic differential equations is what i am looking for

Ryder Rude

11:31 AM

this is another such theory en.wikipedia.org/wiki/…

it's collapsing the QM wavefunction spontaneously in small time intervals

in a random way

Silly Goose

Oh okay cool

Ryder Rude

11:45 AM

crazy that the brain does not make the first day a permanent memory

"new" things usually make a lasting impact

but the newest of new things is forgotten

Slereah

12:24 PM

What would it remember

It does not at that point even have the abstract notions necessary for it

Ryder Rude

12:44 PM

oh

i just didnt think those days would be so different

Slereah

The brain does a lot of preprocessing that people are typically not aware of unless they're on drugs or have brain damage

The world becomes a lot harder to understand and remember without it

Ryder Rude

scary

Slereah

You don't even really have the notion of what an object is, by default

Ryder Rude

ive heard of lack of object permanence

Slereah

Some abstraction of some set of sensations that persists

Why would you assume that two entirely different sets of visuals represent the same object

It is very difficult to teach neural networks that certainly

Ryder Rude

12:47 PM

it says perhaps babies cant feel pain either

Slereah

No that is an old theory that did not pan out

Ryder Rude

@Slereah adults take this idea for granted

Slereah

with fairly unfortunate consequences

Ryder Rude

oh

Slereah

that is what happens with visual agnosia

People are unable to assemble visual clues into understanding that it forms an object

Ryder Rude

12:48 PM

oh

evolutionary mechanisms keep us away from so many unnecessary details

Slereah

Not much to do with evolutionary mechanisms, you just can't be aware of everything going on in your brain

Otherwise you'd have to be aware of what's going on in those mechanisms themselves and so on

Plus it is a lot of sugar to keep a bunch of neurons going

Ryder Rude

now i think sets and objects may not be fundamental to the universe's ontology and are just abstractions that are formed in subjective experience

but the computationalist theory says the ontology itself is a computational object

Computationalists uses this "hidden brain processes" stuff to support the hypothesis

the idea is that it's extremely complex processes happening and it's nothing like what we think it is

but i think the correct understanding of the situation is that "sets, objects, neurons, connections" themselves are "what we think it is". these are artifacts of the experience

so i think what's out there is some physical process that's understood by us as a computation

according to Russel, the "subjective experience" is our only direct contact with the nature of the blood and bones of the universe

Slereah

1:22 PM

> The apodictic certainty of all geometrical principles and the possibility of their a priori construction are grounded in this a priori necessity.
For if this representation of space were a concept acquired a posteriori, which was drawn out of general outer experience, the first principles of mathematical determination would be nothing but perceptions. They would therefore have all the contingency of perception, and it would not even be necessary that only one straight line lie between two points, but experience would merely always teach that. What is borrowed from experience always has o…

Oh Kant you fool

You don't even know what you're starting

Ryder Rude

1:42 PM

what is he saying

Slereah

that space is a priori and not a posteriori

He thinks that this is not something you can learn from experience

Ryder Rude

oh

Silly Goose

for the quantum mechanical treatment of the photoelectric effect, why do we consider the "out states" to be similar energy and similar direction of momentum? is it just a general principle that (from the system POV) non conservative transitions are less favorable? I.e., not conserving direction of momentum is not as favored as conserving direction of momentum

zxen

2:18 PM

I would like to request some clarification regarding areas in surface tension calculations.

In a capillary tube (liquid is water) the meniscus has a curved area (either concave or convex to water, doesn't matter). So when we calculate $F=\Delta PA$, which area is this? In my notes, this has been considered the projected area of the air-liquid interface.

When we calculate $dW=\gamma dA$, where gamma is the surface tension, then which area is this? I think this is the area of the glass-water interface.

Slereah

2:39 PM

> A close friend of Einstein’s has told me that many of the physicist’s greatest ideas came to him so suddenly while he was shaving that he had to move the blade of the straight razor very carefully each morning, lest he cut himself with surprise. And a well-known physicist in Britain once told Wolfgang Köhler, “We often talk about the three B’s, the Bus, the Bath, and the Bed. That is where the great discoveries are made in our science.”

Obliv

3:14 PM

so before buses it was just the bed and the bath?

archimedes was in a bath when he discovered buoyancy I think, so that checks out

Slereah

You had to take the stagecoach

Obliv

what about before stagecoaches were invented? and baths

and beds

Slereah

Just go for a walk

That was the big thing of the early philosophers

Obliv

I'm gonna spend all my time in the 3 b's to maximize my chances

Slereah

that's why they were called the Peripatetic school

the school of walking around

Obliv

3:17 PM

i wonder what the equivalent procedures are for other disciplines

lucabtz

webinar on quantum sensing just finished

I just discover people quantize the LC oscillator using the correspondence principle

i mean it makes sense, but never thought about it or saw this before

Silly Goose

what is the meaning of a cross section $\sigma$ in quantum mechanics? I am having trouble understanding the meaning of the units being $(\text{length})^2$

Slereah

@SillyGoose Number of reactions per volume over a certain length

Silly Goose

I have seen that $\sigma$ should be proportional to the probability of an interaction (or whatever you are modeling) happening, so presumably we should divide this $\sigma$ by some quantity with units $(\text{length})^{-2}$ to get a probability

Slereah

The typical example is particles going through a wall

Silly Goose

3:23 PM

so what if I wanted to consider just a single electron subject to electromagnetic radiation? Do I just divide $\sigma$ by the classical cross-sectional area of an electron or something?

Slereah

Typically you'll consider something like the surface area of the beam going in

If you mean as it relates to a single particle I forget the motivation

I remember Peskin had some section on the topic

Silly Goose

@Slereah is this a beam of "target" particles?

hm well maybe sakurai talks about it more in chapter 6. he just starts using cross sections in chapter 5 without talking about what they really mean :P

Slereah

The analogy in the case of a beam is that a beam of a certain cross section will sweep a certain volume given a certain length of travelling inside the target

And therefore will encounter some statistical average number of particles in the target in that volume

or pass by it within some distance, anyway

How it relates exactly to the S-matrix I forget

For a fun analogy of quantities of surface dimension due to this you can consider the case of fuel consumption btw :

what-if.xkcd.com/11

naturallyInconsistent

3:39 PM

@SillyGoose It is a horrendous fact of life that there is no concept of area of an electronic orbital that makes any kind of sense. Like, if you take any expected orbital area and divide that out, then you expect that the probability densities will be substantially less than one and integrate to one. But in fact, the probability density depends strongly on energy of interaction and can exceed one. We are forced to accept that probability and area melded into one unholy mess.

naturallyInconsistent

3:54 PM

@zxen in the surface tension case, that is definitely the curved glass-water interface.

@SillyGoose did you actually compute the correct interaction, or just an approximation using semi-classical stuff? Because if you actually do the solid state calculations, you would have to consider umklapp scattering separately from normal scattering, etc. It will be quite messy.

Silly Goose

I only computed what sakurai computed above @naturallyInconsistent

@Slereah omg i think ive seen this before

which if by semi-classical stuff you mean using density of states and also using classical electromagnetic potential, yes

naturallyInconsistent

@SillyGoose yeah. if you did it exactly youd have crystal momentum and other headaches

Silly Goose

bleb

so that would be in a solid state textbook you say?

i wonder what all these sections in sakurai are even for

Arjun

4:16 PM

Is the born's probablity rule in momentum space $P(k) = |ψ˜(k)|^2$,where ψ˜(k) is the fourier transform of ψ(x),an independent rule from $P(x) = |ψ(x)|^2$? (here P(a) means pdf of the observable a)

i.e it can not be somehow derived from $P(x)=|ψ(x)|^2$ right?

Ryder Rude

@Arjun yes. it cannot be derived. in general, one postulates the Born rule for an arbitrary observable $A$. Postulating it for one observable does not imply the rule for others.

@Arjun see this for the general postulate en.m.wikipedia.org/wiki/Born_rule

Silly Goose

4:32 PM

Im confused shouldn’t there be zero probability for transitions up an energy level if i turn on a constant potential that is less than all of the energy spacings of my system

but this doesn’t seem to be what time dependent p theory predicts

And i mean not magnitudes less than the smallest energy spacing, i mean literally less than the smallest energy spacing

Ankit

@everyone I had some confusion with dirac notations in QM

Say there is an operator O^ such that <m|O^ = m<m|.. then can i say that O^|m> = m|m> ???

I don't really think its true as I couldn't get it when i was trying to see it using matrices but my instructor gave us something similar to this to be true

Someone please clarify

Silly Goose

Do you know what the $\dagger$ operation is

And how it acts on things

Sanjana

4:48 PM

@RyderRude But I dont think Arjun is asking whether Born rule can be derived from a more fundamental principle (which btw in some sense one can do using Gleason's theorem+non-contextuality+some other subtle assumptions which nobody states :) but its not mainstream i must say)...

@Arjun I think it can be done using Parseval's identity if I am not missing something

Ankit

@SillyGoose yeah to some extent

Arjun

@Sanjana Are you saying born's rule for one observable can be derived from born's rule for another observable using parseval's identity if i'm not mistaken?

@RyderRude thanks..even i thought the same ..but there is this one problem in one of mit's problem sets which claims that "The relation P(k) = |ψ˜(k)|^2, as discussed in lecture, thus follows from the Born relation,P(x) = |ψ(x)^|2 ." see problem 4

ocw.mit.edu/courses/8-04-quantum-physics-i-spring-2013/…

Silly Goose

5:04 PM

@Ankit is $O$ a hermitian operator?

You can consider what $(\langle m \lvert O)^\dagger = (m\langle m \lvert)^\dagger$ is

Sanjana

5:18 PM

@Arjun Not just any observable, but if the state projections are related by a Fourier transform. This should be given in most intro books to QM. E.g. See pg. 125 of Zettili eqn (2.309). Also I have assumed you are talking of not just probability densities but probability itself (density integrated) since Born rule talks of them.

Ankit

@SillyGoose yeah if i take dagger both sides and then use the fact that the eigenvalue of adjoint operator is conjugate of the eigenvalue of the given operator , I will get the second equivalence which I was asking for

But I couldn't really see this equivalence from matrix representation

Arjun

@Sanjana thanks for the reference..but i was actually concerned with probability densities themselves : ) nonetheless i'll look into zettili ..

Sanjana

@Arjun Hmm... But the MIT OCW link you sent also is concerned with probability not probability density.

Arjun

@Sanjana See the claim at the end of problem 4

"The relation $P(k) = |ψ˜(k)|^2$, as discussed in lecture, thus follows from the Born relation,
$P(x) = |ψ(x)|^2$"

@Sanjana would you recommend zettili as the first read for QM?

Sanjana

5:45 PM

@Arjun Oh. Well then I am not sure what the comment at the end of the exercise trying to imply. Depending on how you defined $\mathbb{P}(k)$ in the lectures, the exercise seems to imply somehow by putting $f(\hat{p})=\hat{\mathbb{1}}$ in the last relation, that $\mathbb{P}(k)=|\tilde{\psi}(k)|^2$ which, I thought, the definition of $\mathbb{P}(k)$ is in the first place.

@Arjun Depends on your interests. But if it is along the lines of "shut up and calculate" and you want the calculations done for you in detail then, yes. But it is not absolutely "modern".

Idk if you already know this but there's a fat book by Zwiebach on QM too if you are following his course at MIT.

Arjun

@Sanjana In the lectures they defined P(k) as the density

Sanjana

@Arjun I mean what is this in terms of equations? I thought $\mathbb{P}(k)=|\tilde{\psi}(k)|^2$ follows by definition.

Arjun

@Sanjana Could you rephrase your question.. i don't get the "what is this in terms of equations? " part..and $P(k)=|ψ~(k)|^2$ is a postulate.. true..my original queastion was if this could somehow be derived from the corresponding relation for x ,i.e P(x)=|ψ(x)|^2

Sanjana

@Arjun I was just asking what you mean when you say $\mathbb{P}(k)$ is the density in terms of equations. As I said in the previous text $\mathbb{P}(k)=|\tilde{\psi}(k)|^2$ should be the definition and you agree to that too. But if that is your definition, there's nothing to derive; and if you want to derive $\mathbb{P}(k)=|\tilde{\psi}(k)|^2$ from $\mathbb{P}(x)=|\psi(x)|^2$, what is your definition for $|\mathbb{P}(k)|^2$ then?

Arjun

6:04 PM

In the lectures they claim that the corresponding probability density in the k space is |ψ˜(k)|^2 ..and i was wondering if it somehow could be derived from the probability density in the x space, this thought was provoked by that claim given in the problem set,which made me think if probability density in x space was more fundamental than the other one and if we could somehow derive it

@Sanjana How does putting f(p)=1 give P(k)=|ψ~(k)|^2?

Sanjana

@Arjun If you define $\mathbb{P}(k)$ as whatever weight one must include when finding expectation values in momentum space, then appearance of $|\tilde{\psi}(k)|^2$ in the last equation in that exercise "proves" that $\mathbb{P}(k)=|\tilde{\psi}(k)|^2$

Arjun

But taking the weight as $\mathbb{P}(k)=|\tilde{\psi}(k)|^2$ giving me the correct expectation value for momentum doesen't necessarily mean that it is the probability density,because probability density function apart from giving me the correct expectation value must also give me the correct probabilities when integrated right?

Sanjana

6:20 PM

@Arjun And that is guaranteed by Parseval's theorem

@Arjun :65331028 The way I see it is this: You ask "What is the probability of finding a state having momenta in the range $k$ to $k+dk$, given that the state is described in position basis with $\psi(x)$?" The answer is indeed $\mathbb{P}(k)=|\tilde{psi}(k)|^2$ and that indeed follows from $\mathbb{P}(x)=|\psi(x)|^2$ and the definition of $\tilde{\psi}(k)$.

@Arjun Probability density in any space is not more fundamental than any other one because the change of "spaces" is actually a change of bases and you are free to choose any basis you like

Arjun

@Sanjana I'm still not sure ..could you send me a link to an explicit proof of the above claims

123

6:41 PM

Pls see above example from knk , does masses of bola interact with each other by some force equal and opposite?

Sanjana

@Arjun Section 3.4 of Griffiths might help. The problem (?) is that most intro books/lectures make it look like $\psi(x)$ as fundamental where it is merely a projection $\psi(x)=\langle x | \psi \rangle$. When we start to see that, then the Born rule is stated in terms of $| \psi \rangle$ and $\langle \psi |$instead and then you get generalized statistical interpretation without making "position" look special or something...

@Arjun Shankar is also a good choice if you want to do stuff starting with linear algebra

bolbteppa

9:56 PM

@RyderRude more of this crankish propaganda

DIRAC1930

11:30 PM

GSW 1 reads like a normal physics book

I'm genuinely surprised how normal it seems compared to more modern stuff

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