The h Bar

01:43

@Mr.Feynman If you look up the papers talking about such things, the Z are defined in the exact sense, IIRC. I mean, I dont remember the full thing about LSZ and Källén–Lehmann spectral representation, but roughly remember that it is defined over the exact stuff. That means that any renormalisation would affect these things too.

IIRC, the wavefunction renormalisation $Z=\left|\left<\psi_0\mid\psi_R\right>\right|^{-2}$, and since the bare and renormalised wavefunctions, over the huge amount of spacetime at which we are integrating the overlap integrals, cancel out to almost zero, this causes this particular renormalisation factor to be infinitely large.

I have never seen someone separate the $Z_m$ and $Z_c$, mass and charge renormalisation parts, out from this particular wavefunction renormalisation, so I am not sure if it is possible for us to get decently sized mass and charge renormalisation; I'd suspect that those are zero (but not so small that they kill the infinity), otherwise some other textbook author would have written it out this way. It must have been unsavoury in some way.

Sigh and now I have covid

@Mr.Feynman This is the part where I really really really really like the Causal Perturbation method. If all you talk about are renormalised quantities, then there are no counterterms, no appearances of Z, no need for all these weird nonsense. Things just work.

Silly Goose

02:17

ther is no unique energy eigenbasis right? e.g. we can always permute the eigenvalues so the hamiltonian remains diagonal?

Silly Goose

02:28

honk begger

3

naturallyInconsistent

@SillyGoose which is said as "unique up to rearrangement"

But really, the Hamiltonian tends to commute with some other operators, and those might have interchangeable parts that do not commute amongst themselves.

e.g. the angular momentum operators

Then it is not unique at all, even if you aligned all the eigenvalues together.

@SillyGoose H O N K

Relativisticcucumber

@SillyGoose at least ur finally off the streets

naturallyInconsistent

@Relativisticcucumber jigglyyyyypuuufff

Relativisticcucumber

04:00

in 1.3 of this pset, we put the real part of the "ansatz expanded schrodinger equation" in the form of the euler equation. if the the Gross-Pittaevskii equation is for a single particle wave function, how can i interpret the meaning of the pressure in this version of the euler equation? the question implies that it's the pressure of an interacting gas, but i dont see how that is engrained in the math here quantum.lassp.cornell.edu/sites/mueller-courses/files/homework/…

Physics Meta

0

Q: What to do with Horrible Early Question

The third question I posted to this site was, in retrospect, awful. I deleted it after an hour of it being posted and learned to pose better questions since. Anyway, I’ve become aware that a deleted question haunts a person’s question score permanently. I’m concerned that there’s no way to fix it...

Ryder Rude

04:59

@SillyGoose hello. whenever the eigenspectrum is degenerate, u r free to choose any basis of each degenerate sub-Hilbert space

as ur eigenbasis

Ryder Rude

05:12

@SillyGoose in the case of non-degenerate eigenspectrum, there is the freedom to multiply each eigenvector of the eigenbasis by a unit complex number

ACuriousMind

06:23

@SillyGoose This is one of these questions where it's relevant why you're asking it. What counts as "unique" here? As nI said, sometimes it's "unique up to rearrangement" (and phase), but in other situations it's things like "for a non-degenerate Hamiltonian in position space, there's a unique basis of real eigenfunction"

Ryder Rude

06:56

sets of are already unordered, so the permutation freedom is not very relevant

the set of eigenvalues is unique. the set of eigenvectors has some freedom (more freedom when degenerate) @SillyGoose

naturallyInconsistent

Actually there is observable consequences for the relative phases of eigenbasis. Of course, the overall full phase is unobservable, i.e. if you stay within just one single eigenfunction, then its phase does not matter, or, say, if you undo the phase change between eigenstates manually, that also does not matter. But if you leave everything else as-is and change the relative phases of the eigenstates, it matters. The alignment of phases of eigenbasis is a really understated part of QM

@Relativisticcucumber you should tell us what you got

user 726941

07:19

@naturallyInconsistent Sorry to hear that, hope you feel better soon.

naturallyInconsistent

@user726941 thanks

Ryder Rude

naturallyinconsistent is once again incorrect. it's the relative phases difference of the components of the state vector that are physically meaningful. one can transform the eigenvectors by any (global or not) and leave predictions invariant (as the observables would also transform under this change of basis to leave predictions invariant)

when you change the relative phases of the state-vector's components, the operators dont transform, which is what leads to different predictions

naturallyInconsistent

@RyderRude If you cannot even bother to read that I have written precisely this thing, then why are you bothering to type anything at all?

Ryder Rude

you are talking about "relative phases of the eigenbasis" which is a completely meaningless concept

@naturallyInconsistent it's literally the first sentence

again, each of the eigenvectors of the eigenbasis can be transformed by any phase (global or not). This is just a change of basis and does not lead to different predictions. u should not post ur misunderstandings in the cbat

naturallyInconsistent

@RyderRude just because you are unaware of it does not mean that it does not exist. It appears right away in the off-diagonal parts of the density operator.

Ryder Rude

07:30

i will not explain this again. a change of basis is not physically meaningful

naturallyInconsistent

@RyderRude YES PLEASE DO STAY SILENT FOR ONCE

Ryder Rude

this is not the first time you are posting ur misunderstanding about basic physics (c.f. "special relativity cant handle accelerated frames")

@naturallyInconsistent then dont say wrong things!

naturallyInconsistent

Who is the one who is almost always wrong and never keeps that to himself?

@RyderRude ahahahahahahahahahahahah

Ryder Rude

@naturallyInconsistent im sorry. i read this late. wishing u a good recovery

naturallyInconsistent

@RyderRude I only want that you would forever stop saying anything until you can get a new brain or something. I am way too sick to be dealing with your nonsense.

user 726941

07:43

Upon years of lurking in this chatroom, may I suggest that we develop some mutual respect for one another's discourse? Either that or move to Discord :P

naturallyInconsistent

@user726941 I have spent a lot of effort trying to ignore RyderRude's commenting chains. I have already rarely commented on it. You have just come back after a long hiatus, and missed a lot of the discourse beforehand. It is not the first time that ACM had to do something about RR; it is particularly telling that ACM stated that he is commenting against RR because he did not want the mistakes to be unchallenged.

user 726941

08:14

May I suggest that either or both of you seriuosly consider using the option available in "faq#mute"

naturallyInconsistent

You are far from the first to suggest it. I am not ok with using it. muting comments makes for jarring reading on my end

nickbros123

when we do an inversion of coordinate axes (x'=-x, y'=-y, z'=-z) our system became left handed, if it was initially right-handed. there is this result that says pseudovector (cross product of two euclidian vectors) does not change under inversion, we are taking about a left handed cross product right? or am i missing something..

naturallyInconsistent

@nickbros123 if you want to make sense of these things, the standard physics is just going to continually confuse. The much clearer viewpoint comes from GA, where you can see that the bivector obviously will get both simple vectors negated, and thus the outcome will be invariant under the inversion

nickbros123

what is GA

naturallyInconsistent

and so this is a right handed cross product. The cross product is only ever defined right-handed. And do note that this ties into the fact that physicists and mathematicians actually disagree upon what is the definition of right v.s. left handed. We mean in 3D space we really use our right hands.

@nickbros123 Geometric Algebra

Ryder Rude

08:24

@nickbros123 hello. this post will help : math.stackexchange.com/questions/4659251/…

naturallyInconsistent

whereas mathematicians define right handed coördinates by having the first coördinate cross the 2nd coördinate give the 3rd. This means that if we want to be safe, we have to be careful and work in coördinate systems whereby both physicists and mathematicians would agree that the coördinate system is right-handed

nickbros123

@RyderRude thanks ill look into this. Intuitively, perhaps I may have been wrong all along, I associated handedness with x axis pointing to index finger, thumb to z axis and middle finger for y axis (cos no one likes y axis)

naturallyInconsistent

The standard physics polar coördinates are chosen to be in this subset. But, say, if someone uses the $\chi=\cos\vartheta$ trick, then the coördinate system turns left-handed to mathematicians. The correct choice is really $\chi=-\cos\vartheta$

note that this is also why the Jacobian goes negative in the usual way, whereas the version I chose will keep it positive.

B. Brekke

Let's say $Q$ and $P$ are conjugate variables. How can I tell whether the Poisson bracket $\left\{Q,P\right\} = 1$ or $\left\{P,Q\right\} = 1$? Does it matter? Is it just a convention?

nickbros123

@naturallyInconsistent this is why I just go to griffiths appendix page to check out how he defines the polar coordinates

naturallyInconsistent

08:31

@B.Brekke convention, and still fought over within physics community

@nickbros123 that does not with the trick

nickbros123

hey if im wrong atleast im consistently wrong :]

同等の構成

Okay im still a beginner at physics can someone tell me what happens to the atom in a semiconductor after his hole gets recombined. Does a new hole appear in its place do to thermal energy ?

naturallyInconsistent

@同等の構成 no, what is happening is that the electron-hole pair annihilation converts the energy and momentum into something else, say phonon excitation.

同等の構成

@naturallyInconsistent okay thanks

Ryder Rude

@nickbros123 i always do corkscrew (u can still screw the y axis)

@nickbros123 That post is specifically about Griffiths's discussion of this

同等の構成

08:56

Okay so basically if recombination annihilaties electrons and holes in a semiconductor does the covalent bond break and all electrons become free ?

naturallyInconsistent

@同等の構成 you are thinking of the highly energetic annihilation in vacuum. In a semiconductor that is the much less energetic annihilation, and it will not become free

同等の構成

@naturallyInconsistent Got it Ty

Mr. Feynman

09:11

@naturallyInconsistent oh, I'm sorry. Thank you for typing all of that out even if you are ill.

naturallyInconsistent

@Mr.Feynman oh, you are worth it. it is a delight to type this little out

Qmechanic

09:35

@naturallyInconsistent : Get well soon.

2

naturallyInconsistent

thanks, Qmechanic

PandaScientist

13:56

@naturallyInconsistent get well soon (⁠ ⁠◜⁠‿⁠◝⁠ ⁠)

naturallyInconsistent

14:12

@PandaScientist kiss kiss

Slereah

"The naive definition of the category above with a collection of objects and a collection of morphisms results in a wild category"

oh no

naturallyInconsistent

What does this mean, Slereah? I'm ignorant!

Slereah

Category theory be hard because it's supposed to be in some sense the defining block of what mathematics is, including the basis for sets, but categories themselves are defined using sets

it makes everything a big headache

naturallyInconsistent

oh noooo

同等の構成

Guys when voltage hits a pure semiconductor like silicon does it become a positive ion ? I mean when the electrons go in the holes and destroy them do they make the silicon atom a positive ion or does it remain the same since its bonded with other 4 valence electrons from neighbouring atoms ?

naturallyInconsistent

14:21

But then again, I'm so no surprised that it is going to use some set theory

Slereah

Like you have the """collection""" of objects

what is a collection

naturallyInconsistent

@同等の構成 this question does not seem to make enough sense as it is for us to fully explain. An atom is, by definition, not an ion.

@Slereah did they just use set here?

@同等の構成 Voltage does not "hits a pure semiconductor"---what could you possibly mean by this?

Slereah

@naturallyInconsistent Depends on the theory, but typically yeah

But sometimes sets are too small

ie the category of sets would have its collection of objects be al sets

And the set of all sets is problematic

as you well know

naturallyInconsistent

Let us assume that the $1s^22s^22p^6$ electrons of Si are always filled and are too tightly bound to their Si to be moving about. Then we only have the valence 4 electrons to play with. The Si$^{4+}$ ions are arranged in a regular diamond-like array, and the 4 valence electrons make up the bonds between them. The overall structure is neutral.

A hole just means that one of these bonds is missing one electron. An electron is an electron that is in the conduction band, above and around these bonds as a bigger cloud than it should be. When such a conduction band electron drops into a hole, it is just filling up this gap. There is not much weirdness.

@Slereah oops, ouch. Then they really ought to come up with ways to mitigate that and make that the standard.

同等の構成

@naturallyInconsistent Due to room temperature some of the valence electrons leave holes in the valence band of the silicon atoms. When you connect a pure semiconductor to a power supply free electrons from the voltage recombine with the holes left from the thermal energy of the room temperature before you connected it with the power supply. After they recombine the holes and electrons destroy each other and silicon has 1 less electron . Do you guys understand so far ?

naturallyInconsistent

14:32

@同等の構成 this is not what happens. and no, recombination of holes and electrons is a charge conserving operation and you will never get a missing amount of electrons

同等の構成

Yeah i agree about never getting missing amount but does it change the atom(silicon) from being neutral or not thats what im asking

naturallyInconsistent

@同等の構成 No. The hole is somewhat $\text{Si}^+$ and the conduction electron is somewhat $\text{Si}^-$, so their cancellation is a neutralisation reaction.

同等の構成

Guys do you understand when you study what is going on or are you just memorising it 😃

123

15:52

Hello Everyone...

Physics Meta

16:13

0

Q: Why is my question off-topic?

The question doesn't involve the application of any physical principles to the design or manufacture of any real-world product, nor does it involve any consideration for practical utility*, costs, material availability, or existing technologies. Additionally, the creation of nano-scale hook and l...

discussion scope specific-question closed-questions close-reasons

Physics Meta

17:03

1

Q: apparently incomplete support of mhchem in physics.se

$\require{mhchem}$ Inspired by a question on chemistry.se (now closed, because it was a cross post from physics.se here), I observe the mhchem extension can be used here provided the question/answer contains once $\require{mhchem}$* to render e.g., \ce{C6H12O6} (sum formula of e.g. glucose) as $...

support bug

Physics Meta

17:53

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Q: Help finding a question about charged metal block expanding in volume

Either myself or someone else had asked a question here about how the volume of a metal block changes as it becomes positively charged (extracting electrons). Going through my question history I couldn’t find it so I assume someone else asked it. Using the search bar for “charged block volume”, I...

discussion support

nickbros123

18:03

I find myself in this loop of revising the whole damn book every now and then. I simply cannot trust myself to retain information, and I have this compulsion to have the entire chapter + all the chapters before that on my finger tips before I continue solving problems of that chapter. I don't know how to rid myself of this mentality

Physics Meta

18:18

0

Q: How to resolve multiple open duplicates of essentially the same question

There are 4 variations of essentially the same question about the effect of pressure on dew point. From oldest to newest they are: How does athmospheric pressure influence dewpoint? To calculate the dewpoint ºC increase in a compressed air line How does atmospheric pressure affect dew point? Wha...

discussion duplicates vote-to-close

ACuriousMind

man meta's going wild today

user 726941

indeed

user 726941

18:39

@nickbros123 Eventually you'll develop confidence the longer you keep working at it.

Relativisticcucumber

18:52

@naturallyInconsistent ill write up a post on the main site including that since it seems that might be a better place for a more detailed inquiry :) + hope you recover soon -- covid is the worst!!! :((

imbAF

19:02

Is it correct to say that in Quantum mechanics the probability density is replaced with the density matrix?

fqq

@imbAF it's not a precise enough statement for this kind of question

imbAF

sorry

phase space*

in classical mechanics

and density operator in quantum mechanics

ACuriousMind

can you formulate that as a complete sentence?

imbAF

Is the quantum mechanical analogue of the probability density, the density matrix?

ACuriousMind

you mean of the classical idea of a probability density on phase space? Yes.

imbAF

19:16

yes

ACuriousMind

note that the square of the wavefunction is also a probability density, so you're really not being precise enough

imbAF

I mean

clasically

the pdf is the square amplitude of the wavefunction

ain't those two synonymous, classically ?

ACuriousMind

there is no wavefunction, classically

imbAF

But there is the probability density

?

ACuriousMind

a classical probability density is just a function $\rho(x,p)$ on phase space, this is not a square of a wavefunction

imbAF

19:18

I see

ACuriousMind

note that wavefunctions are not functions of $x$ and $p$, but of $x$ or $p$; their square has nothing to do with any classical $\rho(x,p)$

there is a "$\rho(x,p)$" associated with a wavefunction - the Wigner quasi-probability distribution - but it's not an actual probability density.

imbAF

Why is important to mention

what you just said?

ACuriousMind

I'm just pointing out various related but not identical objects that one might call "probability density" to make it clear that your initial phrasing was too ambiguous

imbAF

I see

ACM I have a question which is a bit tricky, so I'll try to explain it with a simple example. It has been bothering me for some days. Mostly how it's phrased

Initially, would this statement be accurate:
For an eq. to be form invariant under LTs or said otherwise Lorentz covariant, it's components/physical quantities (whatever the correct terminology here) need to be lorentz covariant quantities

ACuriousMind

19:36

I'm not really sure what that's supposed to mean

imbAF

what does it mean for an equation to be form invariant under LTs?

ACuriousMind

ugh, I hate that terminology, but what people usually mean is that it's an equation like $A^{\mu\nu\dots\sigma} = B^{\mu\nu\dots\sigma}$ that looks like that in every Lorentz-inertial frame

for that both $A$ and $B$ have to transform in the same way under Lorentz transformations (but them having the same index positions already guarantees that unless you're using index notation incorrectly/confusingly, which is why I find the terminology kinda pointless)

imbAF

I see

I derivated the D.E in two frames, where one is the moving frame relative to the rest frame of some aribtrary event

and the equation looks the same

only after you make some substitutions, appropriate ones

so that

$S(\Lambda)\gamma^\mu S^{-1}(\Lambda)=\gamma^\nu$

If you start wit the D.E in the rest frame O, and try to transform it so that you get the D.E in the moving frame O', this term $S(\Lambda)\gamma^\mu S^{-1}(\Lambda)$ is present

But than I realized that by finding solutions to $S(\Lambda)$, you essentially proved the form invariance of the D.E

with the only difference having primed and unprimed symbols

representing lorentz covariant physical quantities

One more thing though, is it possible, similarly to how it's possible to derivate the S.E from the K.G equation, to derive the S.E from the D.E ?

Silly Goose

I am a bit confused as to how the generic position vector $\vec{r}$ is equal to the polar coordinate $\rho \hat{\rho}$. In this problem I am solving there seems to be a $\hat{\phi}$ component to the position vector

i feel i have a big misunderstanding about position in polar coordins :P

ACuriousMind

19:53

@SillyGoose not sure how you expect anyone to help you since you didn't include figure 1.11(b) :P

Silly Goose

heh here it is

thisi s also the specific problem (I dont want a solution, just to resolve my confusion about modeling motion in polar coords)

in particular, in part b) I get a position vector $\vec{r} = \rho \hat{\rho} - \omega t \hat{\phi}$ where $\rho$ is not important and $\omega$ is just given; i.e. I get a radial and angular component to the position

ACuriousMind

okay, so, the position vector in polar coordinates isn't actually a vector you can usefully express in the basis vectors

pretty much by definition the position vector of a point at distance $r$ is just $r\hat{r}(r,\phi)$

Silly Goose

i see...

ACuriousMind

see math.stackexchange.com/a/1023552/143136

Silly Goose

hm so i guess they are asking to write down the coordinates of position and not the position vector

and so this position vector business is actually not all too relevant here; though, still good to resolve my confusion

Mr. Feynman

20:41

@naturallyInconsistent With qmechanic's bless you've gained a 2x healing factor

@ACuriousMind when I chuckle when I see messages of yours in the starboard like the one currently showing. I read and wonder "what could have possibly caused that reaction?" and if I may say so I'm a good guesser :P

we're getting a lot of those mildly irritated ACM's starred messages lately

Relativisticcucumber

21:47

i know that if we perturb a system, this leads to energy splitting via adding terms to the hamiltonian, impacting the energy levels. however, i dont understand why tunneling should lead to energy splitting if there is nothing changing about the system when it tunnels -- tunneling just describes a possible solution when there is a set potential barrier, right?

ACuriousMind

uh, yes?

who claims "tunneling should lead to energy splitting"?

Relativisticcucumber

the context is this but i might be misunderstanding

"Ammonia exhibits a quantum tunnelling due to a narrow tunneling barrier,[12] and not due to thermal excitation. Superposition of two states leads to energy level splitting, which is used in ammonia masers."

from wiki on nitrogen inversion

ACuriousMind

Those are two separate sentences and there is no "because" or something like that between them

It's just saying ammonia has a) tunneling b) energy splitting

Relativisticcucumber

hm im not seeing why the superposition of states leads to energy level splitting though?

ACuriousMind

I mean those two sentences in the Wiki article are not terribly good?

it's just not a well-written article

Relativisticcucumber

21:54

well i was reading another source and then tried to clarify on wiki

and this other source seems to show the states and mentions the splitting but

hmmm

i need to trace my confusion to its origin

i will be back

ACuriousMind

@Relativisticcucumber if you want to figure out what the Wiki article is really trying to say, just read the source it links

the idea is that you have two states which "should" be degenerate energy eigenstates but due to a tunneling amplitude between them you get two split energy eigenstates that are superpositions of the "naive" states instead

Chemistry

23:32

Hi everyone. I was wondering if you could help. I prefer the chat option as I need a constant back-to-back messaging to understand this.

I'm trying to understand why in the Lagrangian, it's said that velocity and position are independent variables. It' said that they're independent in the Lagrangian, but they get linked later on.

imagine we have L = 1/2 m x'^2 - mgx

Are we saying that at this very step, when we write the Lagrangian, before moving forward, x' is not a derivative of x ? Is this what we mean when we say they're independent variables ?

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