Yeah \tr is infuriating, there's other ones like this I can't remember right now

@bolbteppa \diag is another one for me

no one thinks (La)TeX is a particularly great language

it's just that the effort that went into making its output typographically excellent is so much that no one is willing to do that again for a different language :P

and it doesn't help that the PDF standard is arguably even worse than TeX but somehow we've got pdfTeX, to the degree that in some contexts people prefer to generate LaTeX to then generate PDFs from rather than trying to directly generate PDFs

For years I'd seen constant shaming for not using bibtex for references, then I go read into it and it basically hasn't been updated in decades, has it's own general stupid rules you need to bypass, you're basically forced to double-check every single reference and even then there are stupid rules that I gave up on trying to fix...

another case where I think people generally use other tools to manage it?

like, use a citation manager like Mendeley and let it generate the bibTeX instead of trying to write it by hand

Yeah in the future I will try find an alternative, I'll check that out

I am little bit confused here. So it looks like we are writing the Hilbert space of all hermitian $2^n \times 2^n$ matrices since that will give us all possible density operators for a $n$ qubit system (with the generalized gell-mann basis). We then explicitly state constraints which pares the whole space down to genuine possible states of $n$ qubits. But then Martins (the author of this paper) seems to subtract the trace out of these resulting states? Is that accurate and if so for what reason?

@SillyGoose where in that except is a "subtraction" of the trace? I'm not seeing it

"The translated vector $\bar{\rho}$..." In the bottom paragraph. Perhaps i am misunderstanding

I don't see any trace there

Hm well I guess my question is just then what is this translated vector object.

I see a maximally mixed state being subtracted there, not a trace

and given that this is called "generalized Bloch vector", have you just tried to work out what this is for a single qubit?

well it'll just be linear combinations of pauli matrices right?

because the obvious interpretation to me would be that this means $\bar{\rho}$ is a vector inside a $4^n -1$-ball and the maximally mixed state is the center of the ball

@SillyGoose I mean that calling it a "generalized Bloch vector" should mean that this reduces to the usual Bloch vectors for $n=1$ and pure states

and then you could just see what happens if you don't subtract that maximally mixed state in that case and thereby understand why this is the correct generalization of a Bloch sphere

whenever someone calls something a "generalized X", the first order of business should be to understand in what sense this actually generalizes X, no?

heh i suppose so

and from what is written there this is actually immediate: you have $r_0 = 1/2^n$, and so subtracting the identity just removes those $0$-th components from the vector

i.e. the result is purely a $4^n-1 =3 $ vector $r_1, r_2, r_3$ in the 1-qubit case

and when the state is pure, this vector has Euclidean length 1, so it lies on the surface of the unit sphere in 3d space, i.e. the usual Bloch sphere, so everything matches up!

in contrast, the maximally mixed state maps to the zero vector

so this is a pretty neat representation: the nearer your vector $\bar{\rho}$ is to the surface of the sphere, the purer it is

Hm okay so is it the case that here we are dealing with 1) density operator representations of states that live in a Hilbert space, and then 2) translating these density operators into vectors (in Euclidean space) so that we can put them on generalized Bloch spheres?

I mentioned the trace business above because certainly these generalized Bloch vector representations are traceless precisely because we subtract out the only component of the matrix that contributes to the trace. Hence they are cannot be density matrices.

@SillyGoose no one claimed that they should be density matrices

they explicitly say it's a vector representation of the density matrices

I think they're expecting you to be familiar with the usual Bloch sphere and how to represent mixed states there

because it is pretty standard to do the 1-qubit version of this, with the maximally mixed state indeed corresponding to the zero vector

Right but the original objects they are constructing are meant to be density matrices (at least to my understanding by the constraints they write)

Hm I see

@SillyGoose yes, and the claim here is that you can usefully represent all possible density matrices as $4^n - 1$-vectors inside the $4^n$-sphere

okay now I see

how convenient :D

I mean this should not be that surprising: $n$-by-$m$ matrices form a vector space of dimension $nm$

so all complex matrices on a space of dimension $2^n$ are vectors of dimension $4^n$, self-adjointness imposes a reality condition so this is actually $4^n$ real dimensions

the trace is a metric on the space of matrices, and a condition for density matrices is $\mathrm{tr}(\rho^2) \leq 1$ with equality for pure states, so that there should be a map from the space of density matrices to the space of vectors inside a $4^n$-ball with the pure states lying on the surface of the ball could be argued on purely abstract grounds

oh mayn okay that does sound pretty reaonsable

wait now is trace a metric as defined by $d(x, y) = Tr(xy)$?

@SillyGoose it's $\langle A,B\rangle = \mathrm{tr}(A^\dagger B)$ (I should have said inner product rather than metric) and it's called the Hilbert-Schmidt inner product on linear operators

ah

but inner products induce metrics, so....

what is the origin of the fact that if you want to generate a tensor product of Lie Groups you exponentiate the direct product of the Lie Algebras? Is it encoded in the exponential map?

(pic in the previous message)

yep, should be pi

a'ight

@Slereah I am glad that it's Sunday.

@SillyGoose the idea is just that exponentiation turns addition into multiplication!

$\mathrm{e}^{x+y} = \mathrm{e}^x\mathrm{e}^y$ when $x$ and $y$ commute

also: You're not taking a "tensor product of Lie groups", that doesn't even really exist

I'm 99% certain you're talking about the direct product, not a tensor product

since in $G\times H$ you have that $(g, h) = (g, 1_H)(1_G, h)$ and the two factors on the r.h.s commute, you have that for algebra elements $t,s$ with $\mathrm{e}^t = g, \mathrm{e}^s = h$ that $\mathrm{e}^{t+s} = (g,h)$

I'm not 100% sure why so many physics texts are confused about what a tensor product is, but I think the problem here is that the representations of $G\times H$ are given by tensor products of representations of $G$ and representations of $H$

@ACuriousMind every time I open a Physics book and read $\mathrm{SU}(2)\otimes\mathrm{SU}(2)$ my heart skips a beat

@ACuriousMind my theory is that they get confused with the direct in "direct product" because the direct sum is a circled plus

That's the only possible stretch I can think of. They sure know groups are not vector spaces

A group can have a vector space structure

@Amit Not a lot of them

A lot numerically or a lot usefully? 😋

@Amit not in general and also, that's irrelevant to the group structure

$\mathbb{R}^n$ is important but maybe not the best representative of groups in physics

All vector spaces are groups wrt sum, yeah

@Amit what exactly do you mean by that? Vector spaces are Abelian groups, i.e. $\mathbb{R}^n$ is both an Abelian group and a vector space, but do you have any other examples?

$7$ is a group too

what

No I only said it's possible, I understand the point that the tensor product is not supposed to come between two groups which are definitely not vector spaces like su(2)

@ACuriousMind Nonsense

I didn't mean to deny what @Mr.Feynman said only point out that it is possible there are exceptions

@Amit do you mean the group $\mathrm{SU}(2)$? That is not a vector space

Yeah

Oh ok, I read your message again. A comma can make a difference :P

I understand we're not that big on using capitalization these days but if you're not using mathJax I beg you to at least distinguish between the group SU(2) and the algebra su(2)

SU(2) is a group, but not a vector space. su(2) is not a group, but a vector space!

Right, sorry

Also today we got some group theory talk

@Mr.Feynman Sorry again. Mobile typing makes me a lousy chatter

Not that it's an excuse

@Amit why are you saying sorry? No need to be sorry :P

Ah, okay :) I just don't like saying dumb stuff out of carelessness, I do enough of it anyway 😂

I say dumb stuff all the time here

Thank you sir for the encouragement

@SillyGoose hilarious

Hi

If I give you $\langle \hat{T} \hat{S}_z(X) \hat{S}_z(Y) \rangle$, what is the physical interpretation? It seems to only make sense in the non-interacting regime

(or replace the above with field operators)

Is it best just to think of the correlation function as giving the excitation levels of the system by it's poles?

Instead of giving it a physical interpretation

Or is it best to think of it with the physical interpretation given in the non-interacting regime + corrections

which is how it's calculated with an expansion in $S$

There is a state $| \psi_i \rangle$ with an energy $E_i$ (given by the pole in the KL spectral rep) and looking at system near that energy, the only contribution will be given by the singular part which looks like a free particle

Is that a coherent way of looking at it?

Anyone?

@DIRAC1930 idk the answer to ur question but what is the physical interpretation of correlation function in case of the free theory?

Well we define the field operator as $\hat{\Psi}^\dagger(X)= \sum_i \hat{a}_i^\dagger \psi^*_i(X)$. We have $$|X \rangle = \sum_i | i \rangle \langle i | X \rangle = \sum_i | i \rangle \psi_i^* (X) = \sum_i \psi_i^* (X) \hat{a}^\dagger_i | 0 \rangle = \hat{\Psi}^\dagger(X) |0 \rangle$$ i.e. the feild operator creates a one particle state at $X$

The interpretation of the non-interacting correlation function follows from this

Oh u mean the business about "the probability of a particle to travel from x to y"

Yes

Hey guys does this make sense, $$\mu = \frac{\sum_i \overline{X}_i}{(N/n)}$$ where $\mu$ is the population mean, $\overline{X}$ is the sample mean, $N$ is the population size, $n$ is the sample size?

That stuff is not accurate tho becuz there's no position basis.

I personally think of a correlation function as just a correlation function. The general concept of correlation between two quantities is defined as the expectation value of the product @DIRAC1930

The above does not require any particle interpretation. In fact, the above also applies to classical field theory

Okay but then what do you interpret as the correlation function calculating?

I think we're really interested in the S-matrix element. We're calculating the correlation function only becuz of the LSZ theorem. So just think of it as useful math.

The S-matrix element has physical interpretation

Perhaps but in cond mat, an expression like $\langle T S_z(X),S_z(Y) \rangle$ must have some sort of physical interpretation since I doubt they are using LSZ for that

I dont know about the applications of correlation func. in condensed matter. U shud look at what they're using it for

At the very least, a correlation func. can b thought of as just that.... as something that quantifies the correlation between the field "observables" in the vacuum state. This interpretation also holds for classical field theory. It provides some interesting contrast between QFT and CFT @DIRAC1930

Becuz the derivation of the Schwinger-Dyson equation yields different result in case of CFTs. The delta term is absent in the final equation. This leads to the absence of loops diagrams in the classical theory @DIRAC1930

@Obliv i think it's correct

Hav u divided the population exhaustively into samples of sizes n?

Perhaps the spin-spin (in the z direction) correlation function has a physical interpretation because of $S_z$ being a conserved quantity (I don't know)

Would a number operator - number operator correlation function have a definite interpretation in rel QFT?

what is a spin-spin correlation function in a general QFT?

in cond. mat. where you have a lattice with spin operators $S_i$ at every lattice site I know what that means, I do not know what that means in a generic QFT

My confusion is of the interpretation of $S_i$ in an interacting theory

Hmm maybe it is the same in this case

and that the states become complicated superpositions once the interaction is switched on

what is $S_i$

Sorry I means $\hat{S}_z$

what is $\hat{S}_z$, then :P

The projection of the spin in the z direction

is that just supposed to be the z-spin operator acting on all particles?

Yes so it would be $\sum_i \hat{S}{}^z_i$ written in 2nd quantized form

because usually interesting spin-spin correlation functions are like the ones in the Ising model where you have that $\langle \sigma_i \sigma_j\rangle$ is an order parameter for a phase transition and $\sigma_i$ is the z-spin operator at the i-th lattice site

Yes sorry I meant that it is the spin at the ith lattice site

But my issue is that doesnt $S_z$ only take that interpretation in the free theory

Doesn't it become dressed when interactions are introduced?

Or are we always viewing it as 'spin as described in in the non-interacting regime' plus corrections?

Omg look up "Ames Window" on utube

I've seen illusions but this stuff is...

On second thoughts maybe I can phrase the question differently. The energy operator $\hat{H}$ takes a different from in the free case and the interacting case but we use the same term 'energy' for both. The spin operator $\hat{S}_z$ must do the same thing

@DIRAC1930 why would it? why would the spin operator care that the Hamiltonian is different?

Because in the Heisenberg picture, I will get $e^{\imath H t}\hat{S}_z e^{-\imath H t}$ (and $S_z$ won't always commute with $H$)

but why are you using the Heisenberg picture? :P

That is true

Everything seems to make sense in the Schrodinger picture

So for physical interpretation of operators, is it best to think in terms of the Schrodinger picture?

So in the schrodinger picture, the $S_z$ operator would always just be the free one

Is that Albert Einstein as Richard from richard and mortimer @ACuriousMind

no, it's Andre from Disco Elysium

Ah, I thought it was an artistic rendition of einstein, maybe even crafted by you.

Well now I can't unsee it

@ACuriousMind According to transcendental idealism, do the properties of mind transcend the mind to the outer world?

Surely evidence of the adiabatic switch on/off would be in the momentum space propagator

@RyderRude I have no idea what "do the properties of mind transcend the mind to the outer world" is supposed to mean

like, that doesn't even make grammatical sense to me

When doing the fourier transform over time from the position to momentum representation, how do we know that the integral over time (which will include the switch on/off ramp) is not affected by the adiabatic switch on/off?

Is there something semi-rigourous on why the adiabatic hypothesis works?

@DIRAC1930 if you want rigor in perturbation theory, you need to go to causal perturbation theory/Epstein-Glaser renormalization

they construct S-matrices that depend on the adiabatic switching function and, if you can cure or ignore the infrared divergences, you can take the limit of taking the switching function to $1$ (i.e. "always switched on") in the end

Ah okay thanks

@Amit things might change in a few weeks

There's a lot to think about in QFT that's all swept under the rug

When in GR we say that there is no preferred coordinate system, isn't that a property intrinsic in the definition of a manifold, i.e. the mathematical structure we use to model spacetime, instead of a property of GR itself?

@Mr.Feynman yes

You have to place this in the context of the time where nobody was using manifolds for physics

People just wrote physics as a loose collection of equations

I feel like all that beating around the bush physics books make about coordinates being meaningless make a simple concept more complicated

@Slereah Yeah, I mean, I'm not reading 1920 books

@Slereah that's an excuse for Einstein, but not an excuse for modern physicists ;)

I'm talking about 2000 and later books :P

@ACuriousMind I mean are people not doing this still today

Nobody learns Newtonian mechanics on a manifold

We learn Hamiltonian mechanics on manifolds

(no, I didn't :c )

@Slereah sure, not in a first course, but generally the abstract formulations of all the mechanics are more or less well-known

> he doesn't learn Hamiltonian mechanics on the category of smooth spaces

but for some reason textbooks insist on pretending coordinate invariance or "diffeomorphisms" are somehow a GR topic

It makes it much more difficult for those who already know differential geometry as you suspect there's something more subtle you don't get

Is anyone a fan of Eric Weinstein?

I mean those who know some differential geometry and no GR

This guy is a Julian Schwinger impersonator

I'm reading Rovelli's latest booklet on GR to get more interested in the topics (it's really a booklet with few pages on each topic but I think it's enjoyable). There are some mistakes here and there (like the one of $\pi/2$ rotations and spin $2$ or the same thing for spin $1$ and $\pi$), though

Anyways, in a paragraph he basically says "dark energy" is bullshit

It seems quite a strong statement, is this a common idea in the scientific community?

@Mr.Feynman this is one of these topics where a lot of people have a lot of strong opinions in the complete absense of evidence :P

I don't think anyone denies that "it's just a cosmological constant, just write it into the equations as another parameter" is a possible solution

I love the kind of situation in which scientists get all heated up about unresolved problems because of their opinions

I miss the days when I thought that physics was just condensed matter

but there's at least one faction who thinks that QFT predicts a larger cosmological constant than we observe (an argument I have never understood because it's just silly dimensional analysis), and another faction who thinks that QFT/string theory predict zero cosmological constant, requiring other mechanisms to arrive at the observed value for the "dark energy constant"

> You have to place this in the context of the time where nobody was using manifolds for physics

That. Even vector analysis was radical new mathematics in 1900. en.wikipedia.org/wiki/Vector_calculus

I'm waiting patiently for this book to be released

cambridge.org/core/books/quantum-phases-of-matter/…

@DIRAC1930 truly the greatest physics thinker of the modern era

I'm not familiar with his research but I wish I understood how interesting cmt is in it's own right

@NiharKarve Do you work in condensed matter?

@DIRAC1930 I'm around 95% sure Nihar Karve was being sarcastic :P

(also note that that was a reply to the question about Weinstein, not to the mention of the book on CMT)

Oh right

@RyderRude Here's a GIF that Google found on Imgur

Hmm it doesn't want to upload. Oh well

imgur.io/j9h1yIW?r

Ah. It's a mp4. Stupid Google...

@DIRAC1930 lol

I think that a lowercase "lol" is the most savage possible reply on the internet

@DIRAC1930 Eric Weinstein is one of his biggest fans. ;)

And also Witten's

I was watching a lecture from Ed Witten and the comment section was full of people saying this guy Weinstein said he was scared of Witten

Not the first time these kind of people have tried this schtick

I've recently been browsing Scott Aaronson's site. I'm not that interested in quantum computing, but Scott is a great communicator, and he seems to be a pretty nice person. He's been blogging since 2005 scottaaronson.blog And the commenters on his blog tend to be intelligent and well-informed. I quite like this quote from his lectures (and book): scottaaronson.com/democritus/lec9.html

> But if quantum mechanics isn't physics in the usual sense -- if it's not about matter, or energy, or waves, or particles -- then what is it about? From my perspective, it's about information and probabilities and observables, and how they relate to each other.

That quote became famous in 2007, when it got plagiarised for a TV ad for Ricoh printers. scottaaronson.blog/?p=277 youtu.be/saWCyZupO4U

I also like this, from a poem he wrote. scottaaronson.blog/?p=6534

@Mr.Feynman If you know about Schuller's online GR course he gives a very differential geometric take. Someone also compiled lecture notes which aren't half bad for these lectures. But actually if you are very thoroughly familiar with DG you may get bored, he takes a lot of time to develop the math

@PM2Ring I would add: If quantum mechanics isn't Physics in the usual sense, the usual sense of Physics needs to change :)

Is it okay to interpret a direct sum of Hilbert spaces as representing a single system? For tensor product of Hilbert spaces, I feel it is easier to interpret as two different subsystems you are bringing together to make a composite system, but I am not so sure for the direct sum

@SillyGoose the first question I have is why you're taking the direct sum?

if you have two systems and want to represent the combined system, the tensor product is the correct choice

well i am thinking about the situation where we decompose a tensor product space into the direct sum of invariant subspaces (w.r.t. some action).

is this just convenient mathematically and its physical interpretation doesn't matter so much?

@SillyGoose okay, that's not a decomposition into subsystems, that's a decomposition of the state space of a system into "regions with the same value for some conserved quantity"

you're usually decomposing into the eigenspaces for some operator(s) here

in the case where the action is say $SU(2)$ on the space, is the conserved quantity that is conserved in these various invariant subspaces total angular momentum?

@SillyGoose yes

okay I see that helps so we are just reorganizing the space in a neat way

but what should not be lost is that the original system is a tensor product space (presumably because you were composing subsystems in the first place)

yes, if you care about the subsystems you must not forget that structure

Hm so is this decomposition is not really fruitful in all cases? Say if we composed two harmonic oscillators together and we wanted to decompose w.r.t. to the action of the hamiltonian

depends on what you mean by "fruitful"!

but indeed, you usually will have to switch between various descriptions of the state space depending on what you're interested in

:0 I guess in the spin case, it seems quite nice because the spin spectrum of a given total angular momentum particle has a very clear start and end, whereas for the hamiltonian of a harmonic oscillator...has a countably infinite spectrum where you can keep on raising and raising?

but maybe i am attributing the origin of the niceness of the spin decomposition wrongly

the reason Clebsch-Gordan coefficients, for instance, pop up so often is because for systems composed of individual particles you want to switch between the idea of "definite total angular momentum" and "definite angular momentum of each constituent particle" depending on what you're looking at

the former being using the direct sum decomposition and the latter being looking at the OG tensor product space?

yes

@Amit I want a physical understanding of GR first, handwaving is something I have to bear

Schuller's course is in my to-do list, which is expanding faster than the universe

Now wait so should I have said the "action of $\mathfrak{su}(2)$ on the space" here and anywhere previously I wrote the $SU(2)$ group instead of the algebra

@SillyGoose the invariant subspaces under the group and algebra are the same

doesn't matter

representations of a Lie algebra and its simply-connected Lie group are equivalent

i see so in the spin case, you can think of the invariant subspaces as states that rotate into each other or state that have the same total angular momentum, which should intuitively be equivalent

@Amit Maybe. But see the lecture that the quote comes from. The passage immediately preceding it is:

> So, what is quantum mechanics? Even though it was discovered by physicists, it's not a physical theory in the same sense as electromagnetism or general relativity. In the usual "hierarchy of sciences" -- with biology at the top, then chemistry, then physics, then math -- quantum mechanics sits at a level between math and physics that I don't know a good name for.

> Basically, quantum mechanics is the operating system that other physical theories run on as application software (with the exception of general relativity, which hasn't yet been successfully ported to this particular OS). There's even a word for taking a physical theory and porting it to this OS: "to quantize."

Scott admits that this is "very much a computer-science point of view".

@PM2Ring Yeah, I understand now where he's coming from. But actually if it is really true that QM is the "OS" in that way, it should be considered even "more physics than physics", e.g. closer to a fundamental theory of nature than EM or even GR

Emphasis on if

:)

@Mr.Feynman I see, completely makes sense

@bolbteppa to the extent that GU is well defined, one issue is that it's not actually radical enough

it's a classical field theory that has an immediate gauge anomaly (lol)

and so even if you were to somehow fix that, you would run into the usual no-go theorems

To be clear, you are taking a work of entertainment provided by an entertainer seriously

Yeah as far as I remember, I couldn't see where the quantum mechanics came up in this joke of theory

I mean it doesn't hurt to put forward a serious rebuttal

anything that can be misconstrued as "character assassination" only helps his populist rhetoric

"entertainment" is a good description tbh, like in a span of 3 pages he will drop some interesting geometry construction, say some poetic nonsense to stir up the non-technical crowd, then hit you with one of these:

It only would if his nonsense hadn't been debunked even before his woe 'paper' came out

The guy literally talks about how he only partially remembers this stuff, talks about forgetting things, how he can't find his old notes, you don't even find this level of incompetence on vixra

Remember this is a theory of everything, the fate of the universe potentially depends on him finding his old notebook and becoming 'reconversant' in the language of highest weight representation theory, what is humanity going to do until this guy looks through his closet...

haha yeah