The h Bar

Alex

17:00

Hi @ACuriousMind, not sure if you remember but we discussed how $\langle x| s \rangle $ doesn't make sense in QM because spin states don't exist in the same space as position states. Would I be right in stating then that it doesn't make sense to ask if they are commutable since he commutator doesn't have a set of common functions to evaluate?

John Rennie

If you hadn't already guessed I'm still trying to understand what vacuum fluctuations actually are ...

Slereah

Basically the vacuum is like $\approx e^{itH_{int}} |0\rangle$

With $t$ sent to infinity

or something, I forget

John Rennie

So you end up with something that can be written as a sum of Fock states?

Slereah

Yes

ACuriousMind

@Alex Well, "commuting" you have to ask for operators, not states. But you can ask whether position and spin commute - when the physicist writes $x$ and $s$ as operators on a state in the combined spin-position space, then they often mean the operators $x\otimes 1$ and $1\otimes s$ (which commute and therefore allow us to use basis kets $\lvert x,s\rangle$).

Slereah

17:01

There is a little physicist magic involved

At some point Peskin says "Since the perturbation is small we can assume that the Fock states are close to the interacting states"

That is where the magic trick happens

(The trick is that you can't assume)

Or you can I guess, since it works after all

John Rennie

I think at this point i'll see how much of Peskin and Schroeder chap 4 I can understand ...

Slereah

It's not too hard to understand

There's a lot of formulas to work out though

You have to expand field operators a lot

John Rennie

So the interacting vacuum doesn't have a number of particles because when you act with the particle number operator you're effectively acting on a superposition of Fock states?

DanielSank

@JohnRennie Oh pick me pick me!

John Rennie

@DanielSank yes?

ACuriousMind

17:04

If we ignore all the rigorous problems, one can express the interacting vacuum in terms of the free vacuum as $\lvert \Omega\rangle = \lim_{t\to\infty(1-\mathrm{i}\epsilon)} \left(\exp(-\mathrm{i}e_\Omega t)\langle\Omega\vert 0\rangle\right)^{-1} \exp(-\mathrm{i}HT)\lvert 0\rangle$.

Slereah

Well that is true even in free theories

A lot of states are superpositions

And "particle number" isn't really an observable

DanielSank

@JohnRennie Are you trying to understand spontaneously created particles, or vacuum fluctuations in a more pedestiran sense of e.g. why an LC circuit in its ground state has nonzero mean square voltage?

Slereah

Since particle states aren't physical

ACuriousMind

Which, apart from the ugly normalization factor, is the $\exp(-\mathrm{i}HT)\lvert 0\rangle$ with sending $t$ to infinity that @Slereah mentioned.

John Rennie

and the operator $\exp(-\mathrm{i}HT)$ acting on $\lvert 0\rangle$ produces some sum of Fock states?

Slereah

17:06

yes

Alex

@ACuriousMind Okay thanks.

ACuriousMind

@JohnRennie Well...yes. The problem is that rigorously the $\lvert \Omega\rangle$ and the $\lvert 0\rangle$ can't live in the same space, but if we assume they do, then yes.

John Rennie

@ACuriousMind So when we draw those vacuum diagrams we are actually calculating the expression you wrote down?

Slereah

The vacuum diagrams are actually not calculated at all

ACuriousMind

@JohnRennie If you mean the "vacuum bubbles" diagram, you're computing $\langle \Omega \vert \Omega \rangle$.

Slereah

17:07

They make the calculations diverge

There's only one diagram for the vacuum state

The empty diagram

All other diagrams, the vacuum bubbles, make things diverge

That's where you have to use renormalization

Basically the vacuum bubbles correspond to the vacuum energy

John Rennie

@ACuriousMind silly question but what is the physical significance of $\langle \Omega \vert \Omega \rangle$?

Slereah

That's the probability of transition from the vacuum to the vacuum

Which should be 1 in most theories

ACuriousMind

@JohnRennie Naively, it's the zero point function, which has no evident meaning. However, if you look closely at the expression I used to define the vacuum, one sees that the diagrammatic series in the vacuum bubbles computes the vacuum energy.

John Rennie

So when physicists say the vacuum isn't fluctuating they mean $\vert \Omega \rangle$ isn't time dependent? And the bubble diagrams are just part of the computation of $\vert \Omega \rangle$?

DanielSank

@JohnRennie no

er, maybe

John Rennie

17:13

Ah, dash it I thought this seemed to easy :-(

Slereah

Well measurements are still probabilistic

DanielSank

"Zero point fluctuations" are in a logical/semantic superposition of time dependent and not time dependent.

Slereah

If you make two measurements on the vacuum you might get two results

DanielSank

^ That

Slereah

That is what the fluctuation means

DanielSank

17:13

If you make measurements, you get time dependent fluctuations even in the vacuum state.

John Rennie

Yes, but that doesn't mean anything is fluctuating in the sense of bobbling about as a function of time.

DanielSank

Well, note that immediately after each measurement, the system is not in the vacuum state.

The fact that you're making measurements is a critical part of why there are fluctuations in time.

John Rennie

Independent measurements of any quantum system that isn't an eigenfunction will return different values ..

DanielSank

Note, however, that even if you let the system decay back to the ground/vacuum state following each measurement, you still get a fluctuating sequence of measurement results.

Slereah

Well the vacuum, by definition (at least the free vacuum) is Poincaré invariant

Hence it is time independant

ACuriousMind

17:15

@JohnRennie Yes. That's why I have repeatedly said that "vacuum fluctuation" is an unnecessarily mysterious term for that elementary property of quantum mechanics.

DanielSank

The state that you measure each time can be exactly the same, i.e. $|0\rangle$, but your sequence of measurements of, say, $\hat{x}$ is stochastic, a.k.a. "fluctuating".

In this case, it's not the state that's fluctuating, it's just the measurement results that are fluctuating.

John Rennie

@DanielSank Yes, that^

DanielSank

People call that "vacuum fluctuations" or "quantum noise".

Slereah

Are interacting vacua assumed to be Poincaré invariant?

In flat space, anyway

DanielSank

As m y colleague and I remind each other often: "quantum noise is not noise"

Of course, it is sort of noise, but not in the usual sense.

ACuriousMind

17:17

@JohnRennie The main problem is that there is a pop-sci meaning of "vacuum fluctuation" that somehow has come to mean those vacuum bubbles we just talked about, and from which people derive all sorts of silly claims. I'm 99% certain you won't ever decode what that's supposed to mean rigorously.

Slereah

I'm guessing if you have degrees of liberty for the vacuum maybe not

DanielSank

@ACuriousMind That's just one manifestation of vacuum fluctuations where the operator you care about is $\hat{n}$.

ACuriousMind

@DanielSank I don't know what the operator $\hat{n}$ is when trying to act on the interacting vacuum.

John Rennie

@ACuriousMind if we can sort of write the inteacrting vacuum as a sum of Fock states and hope no-one looks too closely then isn't $\hat{n}$ defined by the way it acts on that sum?

DanielSank

@ACuriousMind Why not?

ACuriousMind

17:21

@JohnRennie Well, yes. The problem is that you can't evaluate $\langle n \rangle$ on that state, at all. No one knows how the "sum of Fock states" actually looks, and the whole machinery of perturbative QFT is cleverly set up so that we can't evaluate operators that "shouldn't" be able to act on the vacuum. Basically, you can only compute the expectation of operators that are polynomials in the fields.

John Rennie

Ah OK. Would evaluating the number operator require a non-perturbative approach?

ACuriousMind

@DanielSank Because the interacting vacuum is not part of the Fock space. We pretend it is to derive the LSZ formalism and that works mysteriously well, but I don't know how to write down the number operator in an interacting theory - you are not allowed to use the creation and annihilation operators, as those only are for the free theory.

DanielSank

@ACuriousMind You've mentioned this several times before and I don't get it.

ACuriousMind

@JohnRennie No, because you can't define it in the interacting theory. There is no notion of "particle" in the non-perturbative interacting theory.

DanielSank

If I have a single mode system, I can define the number operator.

If I have a two mode system, I can define the number operators for each mode.

Where does this all go wrong?

ACuriousMind

17:23

@DanielSank You don't have "modes" in the interacting theory. The quantum field is not a collection of harmonic oscillators anymore

John Rennie

But you don't have modes in an interacting theory - do you??

Slereah

Well you can have notions of particles in interacting QFT

But it becomes complicated

You have like weird non-linear superposition of solitons

DanielSank

@ACuriousMind I don't believe you.

Why don't I have modes?

I can list the modes.

Slereah

Define "mode"

DanielSank

Put a box around the universe. Bam. Modes.

John Rennie

17:26

@DanielSank same reason you don't have normal modes in anharmonic oscillators

DanielSank

@JohnRennie What? Sure you do.

I can have an anharmonic mode.

Slereah

The basic problem is that you can't write field solutions as sums of existing solutions

Since the EoM is nonlinear

DanielSank

Particle-in-a-box-is a good example.

ACuriousMind

@DanielSank Because the quantum field no longer fulfills the Klein-Gordon equation, but a modified equation, which also changes the canonical momentum, so that the Fourier modes no longer fulfill the commutation relations of creation and annihilation operators.

The whole reason free QFT has modes is because Fourier transforming the fields leads to the Fourier modes having those nice relations.

DanielSank

@ACuriousMind, suppose I have two particles-in-boxes.

And I couple them through a spring.

Does that system have modes?

ACuriousMind

17:28

Probably? But we're not putting particles here anywhere. We're starting from an object $\phi(x)$ that fulfills an equation of motion, and there's nothing inherently oscillating about it (while there is about a spring)

Slereah

I mean

I guess technically you have eigenstates of the momentum operator

If you want to consider those modes

DanielSank

@ACuriousMind Ok, but as usual, I need to understand where the mode idea breaks down instead of just talking about QFT where it has already broken down :)

John Rennie

Just when it's getting interesting I have to go and get dressed up in preparation for a night's drinking ...

DanielSank

@JohnRennie Priorities, man!

Slereah

@JohnRennie is gonna get some scrumpy

DanielSank

17:29

scrumpy?

John Rennie

@Slereah if I never drink scrumpy again it will be too soon!

Slereah

It is some kind of british hillbilly cider

John Rennie

@DanielSank we were taking about life in rural Somerset in the 70s.

Why was alcoholism such a major problem? Because there was frak all else to do!

ACuriousMind

@DanielSank I think it would be better if you told me why you so easily believe it for the free field. I only believe that there are particle modes for the free field because I have seen how you can transform it to make it look like an infinite collection of harmonic oscillators. I don't know how to make an interacting field look like a collection of systems that have modes, so I don't know what a "mode" would be for it.

DanielSank

@JohnRennie Indeed.

Slereah

17:32

Define what you mean exactly by "modes", for a start

If you mean Fourier modes then certainly not

ACuriousMind

And I'm afraid I also have to go soon-ish

Slereah

Since those can only be defined for linear systems

DanielSank

Ok, let's continue this later.

I have to go too.

@Slereah Yes, we should do that.

John Rennie

On that note I'm off to drink some of Chester's finest real ale - and no scrumpy (though Chester has some very nice dry ciders)

Emilio Pisanty

@JohnRennie have some Old Rosie =D

Slereah

17:42

Interacting field theory is not a fun field because I'm 90% sure that literally none of them are fully solved

MAYBE massless Thirring model

koolman

How can I turn on mathjax

On chat

Slereah

17

A: Any chance of MathJax in chat?

As a workaround while this request is pending, there exist several client-side workarounds that can be used to enable LaTeX rendering in chat, including: ChatJax, a set of bookmarklets by robjohn to enable dynamic MathJax support in chat. Commonly used in the Mathematics chat room. An altern...

Alex

18:11

@ACuriousMind For the operator $\hat{S}_z$, which is the observable of the z component of spin operator, I have in Griffiths that the eigenvectors of this operator are given as
\begin{align}
S_1 &= \begin{bmatrix}
1 \\
0
\end{bmatrix}
\end{align}
and
\begin{align}
S_2 &= \begin{bmatrix}
0 \\
1
\end{bmatrix}
\end{align}
I understand that this obervables only has two nondegenerate eigenvalues, so it therefore has two eigenvectors, which are also theorefore a basis for the two dimensional vector space. How do we know that the eigenvectors are of this form? Does it not matter since we are in a …

Physics Meta

19:24

Recent Questions - Physics Meta Stack Exchange: Trying to learn how I lost reputation

posted on November 18, 2016 by Ghotir

I'm trying to understand how I lost reputation. (Note that I am not necessarily disputing it - I'm more confused than anything else.) According to my reputation for today, I lost one point: -1 Why is nuclear waste more dangerous than the original nuclear fuel. I did not write the question, nor did I write an answer, nor did I even write a comment (which may be th…

Kyle Kanos

19:34

FWIW, the Louisville-Houston (American) football game was quite entertaining

Emilio Pisanty

@PhysicsMeta you big tease

heather

21:27

@ACuriousMind, thank you, I think I understand better now what Daniel Sank meant. I'll start to do that =)

heather

21:47

@Mew, I realized I calculated the problem incorrectly; I think the right answer is 17690.4

DanielSank

22:07

@heather w00t

heather

@DanielSank, I did a bunch of practice integration problems and I think I understand what I'm doing a whole lot better. =)

DanielSank

22:21

@heather that's great!

heather

@DanielSank, What all do I need to know of calculus? Do I need to know multivariable calculus, differential equations, vector calculus...? After I finish practicing/learning single-variable calculus, I mean.

DanielSank

22:44

@heather I think you should make sure you can actually solve problems using single variable calculus.

Solve some physics problems.

Then probably learn two/three variable calculus and again solve problems

Then maybe learn basic differentia l equations. Just the simple stuff.

heather

@DanielSank, okay. Where could I find some good physics problems to solve?

DanielSank

23:09

@heather Traditionally, a physics book.

heather

@DanielSank, good point =)

Mew

@heather, I didn't get that answer

heather

@Mew, oh...what did you get?

Mew

no

i did get ur answer

sorry

good job

@heather, do you know the difference between definite and indefinite integrals?

heather

@Mew, sure, definite are ones with the limits, and you don't need to put in $c$, the answer is certain, and indefinite integrals you do need the $c$, because one isn't actually sure of which of several lines it could be.

Mew

23:12

good

for your next problem, please solve the following indefinite integral

$\int 1 + x + \frac{x^2}{2} + \frac{x^3}{6}$

hmm

yep try that one

and add $dx$ at the end ;P

heather

okay =)

DanielSank

After that, let's do a physics problem.

Mew

Also if we call the answer to the above integral I, and we say that I = 1 when x = 0 solve for the value of c

heather

I think I messed up the $c$'s, but:

$\frac{x^4}{24}+\frac{x^3}{6}+\frac{x^2}{2}+x+4c$

is what I got.

Mew

not bad

@heather, the 4c bit is wrong but the rest is correct

heather

23:17

is it just $c$?

Mew

it looks like you split the question up into 4 integrals and got 4 constants

this is a valid approach

heather

yeah, that's what I did

Mew

but the value of each constant isn't necessarily equal

so you can't say it is 4c

heather

oh...

Mew

but you can say it is a + b + c + d for instance

heather

23:18

ah, gotcha

Mew

and what is the sum of several constants?

just another constant

so we can replace that big mess by just a single $c$

heather

okay.

Mew

so the answer becomes

$c + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}$

now I said that the integral has a value of 1 when x = 0

so it can be easily seen by substituting $x =0$ that $c$ must take on the value of 1

so the final answer becomes

$1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \frac{x^4}{24}$

heather

wow, that makes tons of sense!

Mew

now look back at the original itnegral I asked you to solve

do you notice any similarities between the answer and the integral?

they are the same (except for an additional term in the answer)

heather

23:21

there's a term with the same power for every one except the $x^4$ term. The denominators are all multiples of $2$. The constant is 1 in both cases.

Mew

yes, they are exactly the same if not for the x^4/24 bit

heather

maybe the denominator is related to the power too?

Mew

@heather did you know that $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + ....$

where the donominator is the factorial of the power

heather

no, I didn't

that's cool!

Mew

so this approach can prove that $\int e^x = e^x$

a very useful result in calculus

heather

23:23

the derivative of $e^x = e^x$ as well, right?

Mew

correct

heather

which makes sense because integrals and derivatives are inverses

Mew

yep and you can prove it both ways by looking at 1 + x + x^2/2 + ...

DanielSank

@heather yes, and it's really important.

Mew

@heather you asked someone about what to do after being able to calculate integrals

in my opinion after you can calculate them, you MUST understand exactly what they are

have a physical idea of what the calculation is actually doing

in physics this is more important than being able to compute the results

many integrals in physics are unsolvable but it is essential to udnerstand what they mean

Kyle Kanos

23:27

Simpson's rule is pretty useful for integration

heather

isn't what's happening is riemann sums? like where you pick a certain rectangle width and set them side by side under the curve representing the function you are integrating, such that their height is just under the curve, and then do the same thing with the height just above the curve, and find the area of both sets of rectangles, and then find the area between the two? And then you can make the width of the rectangles smaller for increased accuracy? @Mew

Bernard Meurer

@KyleKanos My rule for integrals is looking at them really hard until God shows me the way

Mew

@heather some of what you said is confusing but I think your on the right track

that's definitely enough of a understanding for now

heather

@Mew, okay

Kyle Kanos

@BernardMeurer You're better than me. I immediately plug it into either Wolfram Alpha or wxMaxima and if no answer appears, use Programming

DanielSank

23:29

It's just the area under the curve.

Why make it more complicated than that?

Bernard Meurer

@KyleKanos Tell that to my prof when he's correcting my exams

zounds

how are you guys exposed to lagrangian/hamiltonian mechanics for the first time?

heather

@DanielSank, yeah...that too =)

Kyle Kanos

@zounds In a classroom?

zounds

i have a bachelor's in math, and i took a classical mechanics course

Bernard Meurer

23:30

@zounds He told me it wouldn't hurt, and that I would be fine, but then there was blood everywhere

zounds

they just threw lagrangians at us without any motivation, and i decided physics was not for me an dropped the course

Bernard Meurer

@DanielSank Dood

I found out

Mew

@zounds they were motivated as an alternativive to Newtonian mechanics that produces the same results

Bernard Meurer

That if the set of points for with a function is not continuous is countable (note, not finite, but just countable) then the function is integrable and the integral is the same as if it were continuous

because they'll have 0 measure

Mew

by minimizing a lagrangian laws of physics can come out

Bernard Meurer

23:31

It's like some magic shit

DanielSank

@BernardMeurer Yes.

Mew

and you can choose the lagrangian so that NEwton's mechanics comes out

so it's just a usefult ool

The ends justify the means

heather

@DanielSank, what was the physics problem?

Mew

LIke chess, how do you explain to a kid why you make a certain move

WEll you have to be able to see the end game to explain why you did what you did

Kyle Kanos

No way, reactionary chess is the way to go

Mew

23:34

lol

Bernard Meurer

I need a chess board

Anyone wants to sell me theirs if they happen to have a good one?

Kyle Kanos

Ebay

Craigslist

Home Depot / Lowes

heather

game stores

target

walmart

Kyle Kanos

Actually, now that I've said it, I think I do want to try making a good chess board

I have some nice stain and paint, might be a nice project for a Saturday

Bernard Meurer

@KyleKanos No craigslist here

Or Home Depot

or Lowes

or Target

or Walmart

Kyle Kanos

23:45

@BernardMeurer Ooo...become a millionaire by inventing it then

Bernard Meurer

Ebay is a good idea though

@KyleKanos lol, it exists, craigslist, but no one uses it

Kyle Kanos

Ah.

Well Brazilians suck then (no offense)

Bernard Meurer

@KyleKanos I live in Portugal :P

Kyle Kanos

Even worse :P

I had some Portuguese friends back in HS.

Bernard Meurer

@KyleKanos Yeah, I don't like it here to be honest

Kyle Kanos

23:49

Tried dating one of them; she didn't even respond.

Bernard Meurer

@KyleKanos Try Slovenians

Kyle Kanos

Don't know any

Bernard Meurer

@KyleKanos Yes you do, Trump's wife is Slovenian

Kyle Kanos

Okay, I don't know any personally

Bernard Meurer

Call 1800-SLOVENIA

Kyle Kanos

23:52

Not sure if real

Also not sure if necessary

Bernard Meurer

Sadly no

theswedishnumber.com

But this is

Kyle Kanos

I've got a nice Germano-Irish wife

Bernard Meurer

and it's absolutely amazing

I use it sometimes when I'm lonely, the swedes are always nice

@KyleKanos You have a good job, you have a platoon of children, and you married a good looking girl?

Go away

Kyle Kanos

@BernardMeurer Check, Check, and Check

Bernard Meurer

@KyleKanos I'm calling a Swede and talking shit about you

Oh no

They closed

Dammit

Kyle Kanos

23:57

Well I'm going to bathe the kids

Adieu!

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