The h Bar - 2021-10-28

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2:37 AM

@DIRAC1930 Maybe these notes would help with getting the jist of representation theory in physics (not sure about the end but maybe it's helpful too)

2 hours later…

4:53 AM

@DIRAC1930 I enjoyed reading it, but it's not a physics textbook. I think it's aimed at physical scientists who are not theoretical physicists but who are interested in theoretical physics. That's me! :-)

5:09 AM

From this (mentioned here), pretty hilarious

5:37 AM

If any mods have a free moment please would you unfreeze this room. Thanks :-)

2 hours later…

7:50 AM

@ACuriousMind Thanks :-)

no problem

8:36 AM

"At the 1951 International Conference on Nuclear Physics and the Physics of Fundamental Particles, Eugene Wigner broached the idea that quantum mechanics (QM) must recognize situations in which the matrix elements of any observable taken between two states that belong to what would later be called different superselection sectors are zero.
According to Arthur Wightman’s recollection (Wightman 1995, p. 753), some members of the audience were ‘‘shocked,’’ presumably because Wigner’s proposal contradicts (as will be seen below) von Neumann’s (1932) assumption—an assumption that had become acce…

Shocked and appalled!

"Requiem was read by E. Teller who cited the apropos anecdote of a candidate for a doctor’s in philosophy who made a statement presumed to be true. Upon being asked by a professor on the examining board, ‘‘In which universe?’’ , he responded, ‘‘Which which?’’"

9:24 AM

Actually this relates a bit to what we were talking about yesterday : "It should be noted that such a statement implies that the set of (physically realisable) observables is strictly smaller than the set of all self-adjoint operators on Hilbert space. For example,
$$A = |\psi_1\rangle \langle \psi_2| + |\psi_2\rangle \langle \psi_1|$$
is clearly self-adjoint and satisfies $\langle \psi_1 | A | \psi_2 \rangle= 0$. Hence the statement of a SSR always implies a restriction of the set of observables as compared to the set of all (bounded) self-adjoint operators on Hilbert space."

3 hours later…

12:53 PM

Huzzah for finite-dimensional Hilbert spaces, where things are boring and easy

@Slereah SSR?

super selection rules

1:19 PM

and not selective serotonin reuptake

1:42 PM

I mentioned that on Monday:

2 days ago, by ACuriousMind

Also, for operators vs. observables, see https://physics.stackexchange.com/q/373357/50583, there's a pithy answer by me and a much more detailed one by Valter Moretti

1:55 PM

One thing I have never found out is how one checks that an apparatus corresponds indeed to the purported operator we assign it to be

it seems like a hard thing to do

How do we know what our prepared state looks like and what our apparatus acts on it

I guess maybe it's something you could try to work out in the 1D case

You have a State Preparator machine, say some electron gun with a knob to control the momentum they are shot at, and you have a Measuring Machine which performs some measure on them

although the hard part is that the electron canon only does eigenstates of the knob I suppose

You also need to test for superposition of states

It's probably doable for some dumb Hilbert space like for spin states but I want to see it for a real particle

I mean aren't some operators virtually by definition defined by how they change the physical system in a particular way e.g. parity reversal

lol, publishing: we've waited months to get the page proofs back, and they give us a week to get back to them with any corrections

If I have an operator $\hat{F}$, it will change a state $\hat{F}|\Psi \rangle \rightarrow |\Psi' \rangle$. Couldn't this possible then change the expectation value of a different operator e.g. $\hat{X}$. So the expectation value of $\hat{X}$ will change through $\langle \Psi | \hat{X} | \Psi \rangle \rightarrow \langle \Psi' | \hat{X} | \Psi' \rangle$

2:14 PM

you mean, like how time evolution works?

if $F$ is unitary that's probably fine

continuation of rant above: would be nice if they had a list of what changes they made in making the proofs.

Whats an induced representation

@DIRAC1930 Yeah but how does it work in practice

My machines don't have a $\Psi = f(x)$ button in practice

What are the rules of correspondance of QM, as epistemologists would say

I very roughly know how it works in GR, but QM seems a bit more vague

i suspect there's no "definitive" way, just ways which are good FAPP

like, suppose you had an a system which is consistently observed to give two outcomes 0,1

Don't we classically measure energy from the change in volume of mercury or something

Yeah but as I said

Those are the bullshit systems

If you're using spin hilbert spaces, there are essentially like 4 operators on it

2:20 PM

there's probably nothing stopping you from saying there's 'really' a third state $2$ which occurs with very very very small probability

It's a bit trivial to do

@Semiclassical Black swans are yet another bear I don't want to wake up :p

lol, fair

Maybe the change in expected value of something we definitively know what it is e.g. position is the only way

I dunno how else to measure energy classically

Well yes, but position isn't even a real measurement

The real machines to measure it are like cloud chambers or wire chambers

depends on which interpretation you use, but in the usual interpretation position isn't privileged

2:21 PM

It's kind of vague what operators those represent

@Semiclassical It's not priviledged in the formalism, but is it in the measurements

well, my point was alluding to pilot wave theory: it's very much privileged in that formalism

for better or worse

I think an experimentalist would know the answer to these questions maybe

e.g., the Stern-Gerlach experiment is 'really' just a position measurement

and the pilot-wave story for that very much embraces that

Isn't everything we do just a measurement of position?

(though arguably the Bohm story is not privileging "position" so much as it is "configuration", and that's dicier)

2:24 PM

Except we can infer energy from the wavelength of light emmited

@DIRAC1930 certainly some people have said that, e.g., John Bell has a quote to that effect

@DIRAC1930 I mean we have many senses

i'm not sure whether it's strictly true, but my own imagination does tend to default to that

We can see colors

We can hear noises

We can smell things

well. noise is arguably just a position measurement: interaction of sound waves with a membrane

2:26 PM

@Semiclassical You're confusing the formalism with the reality again!

now i want to see a scratch-and-sniff measurement of spin :P

The sensory input comes before the formalism

Smell doesn't usually come up a lot for QM, but it is very useful, if risky, in chemistry

Waft is a term meaning to "carry along gently as through the air." The term is commonly used to describe scents that have diffused into other parts of a room, or to describe smoke as being seen moving through the air. Wafting may be used for everyday substances, to make sure they are fresh, or consumable. In chemistry and other sciences, it is a term of laboratory safety. In "wafting" a person takes an open hand with the palm towards the body and moves their hand over the substance in a gentle circular motion so as to lift vapors of the substance towards the nose. This method allows for a lower...

spin up: smells like lilac
spin down: smells like burnt toast

Standard smelling protocol

Humans are just massive position operators

2:28 PM

There used to be a philosophical riddle

If you give a man blind from birth sight, and you show him a sphere, would he recognize that it's a sphere?

Now we can do it and it turns out that no, they cannot!

Touching a sphere and seeing a sphere are very different things

Can't infer that they are the same on gut instinct

Nakshatra Gangopadhay

Hey, can anyone help me by transferring a question I posted on Math SE to physics SE, as I don't have enough reputation, and came to realize that the question is more about physics than maths. Also, would it be terribly out of place, to ask if anyone can spare some time to help me clear a conceptual doubt that has been bugging me for a few days now ?

just ask, don't ask to ask

found the Bell quote, from his paper on 'the impossible pilot wave'

"The second moral is that in physics the only observations we must consider are position observations, if only the positions of instrument pointers. It is a great merit of the de Broglie-Bohm picture to force us to consider this fact. If you make axioms, rather than definitions and theorems, about the “measurement” of anything else, then you commit redundancy and risk inconsistency."

@Semiclassical True enough, but otoh imagine if papers were written like that

i don't take this as an uncontroversial 'moral', to be clear, but it's a position which Bell certainly took

"The little needle went all the way to the little circle symbol"

2:34 PM

lol

I'm not against using "temperature" as a shorthand

also as I said, a bit reductive

you can perceive all kinds of things!

Have you even tried tasting uranium

Maybe some data to extract there

Nakshatra Gangopadhay

Suppose I have some wavefunction represented in terms of x. I've been given some u=g(x) function, and I want to represent the wavefunction in terms of this new variable u. Normally I would just proceed by change of variables or substitutions. $$\psi(x)=\psi(g(x))=\phi(u)$$. My confusion is, whether this simple variable substitution, is the same as changing the basis of the abstract wavefunction in Bra ket notation. Like is \psi(x)=<x|\psi> and \phi(u)=<u|\phi> ?

ah. almost certainly not

if only b/c there's no guarantee that $\phi(u)$ will actually be normalized

as in, suppose $\phi(u)$ is normalized so that $\int_{-\infty}^\infty |\phi(u)|^2\,du=1$

if we assume $u=g(x)$ is a bijection, we have $\int_{-\infty}^\infty |\phi(u)|^2\,du=\int_{-\infty}^\infty |\psi(x)|^2 g'(x)\,dx=1$

in which case the wavefunction, considered as a function of $x$, isn't normalized

Nakshatra Gangopadhay

Thank you so much for confirming this, but can you please elaborate on this ? For example according to Wikipedia, if I have a function f(x), and I want to change its basis to some $u=g(x)$, then I should just substitute x by u. So, in those functions substitutions and change of basis are the same. Here, it isn't

the only way to avoid this is to assume $g'(x)=1$, i.e., $u=x-x_0$, so all you're doing is translating the wavefunction

@NakshatraGangopadhay a point of comparison: have you seen the Planck distribution for black-body radiation?

Nakshatra Gangopadhay

2:42 PM

yeah I have

okay. then one weirdness you may remember from that is that, if you want to go from the wavelength distribution to the frequency distribution, you can't simply substitute $\nu=c/\lambda$

that would give you the wrong distribution function

a similar point applies here, to the probability density $\rho(x)=|\psi(x)|^2$. it's a distribution function, so you can't change variables simply by substituting. you have to consider their effect on integrals

Nakshatra Gangopadhay

If I have a function sin(x), and u=x^2, then I can write this function as a function of u, such that sin(\sqrt{u}), then this new function is NOT the u basis representation of the old abstract ket vector. In fact sin(x) and sin\sqrt{u} are x and u representations of two different abstract kets. Is that what you are trying to say ?

you really cant, $sin(|x|)$ and $sin(x)$ are not the same function

i mean, that's not a big deal. use $x^3$ instead

it's not clear what the "$u$-basis" is supposed to be

2:48 PM

it should be noted that expressions like $|x\rangle$ are, in the standard formalism, a bit nonsense

there's mathematically no such thing as an x-ket

and therefore it shouldn't be shocking that there needn't be such thing as an $|x^3\rangle$ ket

What's wrong with $\psi(x(y)) = <x(y)|\psi>$?

@bolbteppa the question here is more like "can you interpret $\psi(x(y))=\langle x(y)|\psi\rangle$ as $\langle y|\phi\rangle$"

and aside from the case of $x(y)=y-a$, i.e., translation, i'd say no you can't

if only b/c normalization is going to smack you

Right, well a change of variables can be 'non-linear' so while $|x> = a|y>$ seems plausible, beyond that...

Yeah

Nakshatra Gangopadhay

If a certain basis |u\rangle is not normalized, \langle u|\psi\rangle doesn't make any sense does it, we have to normalize the basis first I suppose

But what about $\psi(x,y,z) = <x,y,z|\psi> \to \psi(r,\theta,\phi) = <r,\theta,\phi|\psi>$ i.e. a simple change to spherical coordinates...

2:54 PM

i don't think one typically labels the kets that way tho

@Semiclassical if you change variables the measure is not $\mathrm{d}u$.

You know

what? i'm saying to start about by assuming that $u$ made sense and gave a normalized distribution

String theorists sometimes say "Oh string theory is so simple, there is only the string to consider!"

I mean, it's up to you whether to include the right factors in the measures or in the 'new wavefunction'

2:55 PM

But what of the $B$ field

and then show that this wouldn't make sense in the original variables

Are they lying

Also branes are basically independent from strings, no?

The Schrodinger equation for the Hydrogen atom can be written in terms of spherical and Cartesian coordinates directly, so that's probably the way to go in understanding this

mostly i just didn't want to hassle with $g'(x)|_{x=g^{-1}(u)}$

@Semiclassical eg in bolbterra's example the measure would be $r^2\sin{\theta}dr d\theta d\phi$, not just $dr d\theta d\phi$

2:56 PM

The $B$ field arises as part of the study of the spectrum of the strings

@bolbteppa I thought the B field was applied to the string itself?

@bolbteppa well, when you do the hydrogen atom and have to normalize the radial parts of your wavefunctions, the condition is $\int_0^\infty |\psi(r)|^2\,r^2 dr=1$

That it was the gauge field of the string

acting on the string, I should say

@Slereah no

Hm

2:58 PM

the Kalb-Ramond field is just a field that arises from the spectrum of the string excitations - it lives in the effective QFT, not in string theory

But then why is it a $2$-gauge field

it is a little strange

or is it like an effective action thing?

so while we like to write $\Psi(x,y,z)=\varphi(r)\psi(\theta,\phi)$, that doesn't mean we've suddenly got an $r$-representation

we've still just got position, merely expressed in another form

Branes arise from talking about open strings, of course m theory (whatever it is) is where direct contact with strings is less direct

@Slereah yes - there are states in the string excitations that are the excitations of such a field

I see

3:00 PM

the string spectrum has three different kinds of massless excitations - dilatons, gravitons, and whatever you want to call the excitations belonging to the B field

the bagnetic field

@Semiclassical if you change coordinates, the scalar product also changes. The integral gets precisely the right factor to preserve normalisation

it's a 2-gauge field because it's a massless excitation with two antisymmetric indices

the bhotons?

$\hat{H} \psi(x,y,z) = E \psi(x,y,z)$ i.e. $<x,y,z|\hat{H}|\psi> = <x,y,z|E|\psi>$ i.e. $\hat{H}|\psi> = E|\psi>$ i.e. $<r,\theta,\phi|\hat{H}|\psi> = <r,\theta,\phi|E|\psi>$ i.e. $\hat{H}\psi(r,\theta,\phi) = E \psi(r,\theta,\phi)$ :\

bolbtons

3:02 PM

i'd still insist that $|r,\theta,\phi\rangle$ only makes sense to the extent that it's another name for $|\vec{r}\rangle$

Right

what does the effective string action look like?

let's see

but you're not completely wrong - if you think about strings propagating in 10d target space with a background B-field, people like to talk about this field coupling to the strings

by contrast, $\langle x_1,x_2,x_3|$ makes more sense to me

I don't think $|x,y,z> = \sum <r,\theta,\phi|x,y,z> |r,\theta,\phi>$ makes that much sense beyond the $|r> = <r|r>|r> = |r>$ interpretation

3:04 PM

b/c you could imagine doing a Fourier transform w/r/t to just $x_1$ and not the other two

"Differential cohomology in a cohesive $\infty$-topos"

Nakshatra Gangopadhay

Another question - does unitary transformation preserve the normalization, and if so, then is the change of basis an unitary transformation ?

That's the most nlab title of them all

I've never 100% understood what they mean by that, but there's lots of stuff I don't understand when they start talking about a ostensibly classical description (strings floating around) and then also talk about quantum stuff like the resulting particles

@Slereah i don't have a counterexample, but somehow i doubt that's peak nlab

like, "differential cohomology" at least sounds somewhat possible to understand

3:06 PM

have you ever seen a cohomology in the night sky

@Semiclassical you can do a Fourier transform in $\theta$ and not $r$, no? spherical harmonics are almost that

@Slereah yes

it was beautiful

hmm

that does sound halfway plausible. something-something spherical wave

i don't mean that they are literally a Fourier transform, but solving a central potential with spherical harmonics is the spherical equivalent of solving the free particle with plane waves

it's an expansion in the eigenbasis of a Laplacian

yeah

all of this does make sense when you look at the integrals you do, of course. then it's just good old-fashioned calculus

it's whether it means much at the level of bra's and ket's that i'm more dubious

3:14 PM

the $|x\rangle$ kets don't exist anyway so I am not sure what the point is

yes :P

but yes, Dirac notation is probably not great for this, unless one has very clearly in mind what everything means

@Semiclassical in general for it to make any sense, this "change of basis" is actually a relabelling of the same basis, e.g. for $u=x^3$, $|x=2\rangle$ and $|u=8\rangle$ must be the same thing. From that you can derive scalar products between the two basis, depending on how you rationalise stuff like $\langle x_1|x_2\rangle$

$$<\psi|\psi> = \int d^3 \mathbf{r} <\psi|\mathbf{r}><\mathbf{r}|\psi> = \int d^3 \mathbf{r} |\psi(\mathbf{r})|^2$$
is independent of the 'variables' you use for $\mathbf{r}$ whether Cartesian, spherical polar etc... I think it just comes down to recognizing a vector is invariant of the coordinates you use (because the basis changes to compensate)

Maybe that should be phrased with a $\sqrt{-g}$ type term in the integral in general

yeah it's hidden in the measure

3:32 PM

The Dirac equation seems like one of Diracs least impressive accoplishments

The analysis of it that he did seems way more impressive

@Slereah The idea is in e.g. Polyakov, you usually write it as $\sqrt{-h} h^{\alpha \beta} \partial_{\alpha} X^{\mu} \partial_{\beta} X^{\nu} \eta_{\mu \nu}$ then for a closed string in the massless spectrum you find gravitons i.e. the massless carriers of the thing that causes perturbations away from $\eta_{\mu \nu}$,

but you also find more (trace and the anti-symmetric $B$) so if we promote $\eta_{\mu \nu}$ to $g_{\mu \nu}$ to go from SR to GR in the above action, why ignore the other contributions

Really why ignore the rest of the spectrum, but even if we just want to talk about gr this arises

Can a certain group be a limit of another group?

depends on what you mean by "limit"

Group contraction

In theoretical physics, Eugene Wigner and Erdal İnönü have discussed the possibility to obtain from a given Lie group a different (non-isomorphic) Lie group by a group contraction with respect to a continuous subgroup of it. That amounts to a limiting operation on a parameter of the Lie algebra, altering the structure constants of this Lie algebra in a nontrivial singular manner, under suitable circumstances.For example, the Lie algebra of the 3D rotation group SO(3), [X1, X2] = X3, etc., may be rewritten by a change of variables Y1 = εX1, Y2 = εX2, Y3 = X3, as [Y1, Y2] = ε2 Y3, [Y2, Y3]...

That's one notion of taking a limit of a group to get another group at least

there's no "space of groups" in which you could take any ordinary notion of limit, but Wigner-Inönü contraction is e.g. how you can view the Galilean group as a limit of the Poincaré group

3:47 PM

This is a fascinating little history of how it came about

Weird that it took QM for it

You can do group contraction just from SO(3) to SO(2)

it is weird? you can do it, but people did not care much about group theory in physics before QM

Rotations are also a thing for mathematicians!

given how many examples there are of physicists thinking they've invented something new and subsequently finding out someone did it a hundred years ago

fair enough

3:58 PM

it's nice to see the opposite scenario once in a while

yes, the early 20th century was a great time, physicists could do something new and not expect to find a russian paper from the 60s-70s with the exact same thing or a more general case

heh

I mean sometimes I do find original ideas to do, but then I realize that it's hard!

4:19 PM

@bolbteppa no real reason except practicality to integrate out all the way down to the massless fields, I guess

you can find the correct contribution to the action of the higher-order members of the spectrum through the state-operator correspondence of course

4:33 PM

the procrastinator's folly: "hmm, i've got X,Y,Z to do today...i can't decide where to start, so let's do something else until I do."

3

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