The h Bar

DIRAC1930

12:02 AM

Sorry my mistake. I don't require the determinant to be linear. I just require vector addition to be linear which restricts the $\mathbf{X}$ to be linear in coordinates

But I think my argument still holds. If $\det \mathbf{X}=0$, then the vector $X$ associated with the matrix $\mathbf{X}$ is isotropic

That's all that is needed I think

bolbteppa

In 4D, wouldn't the determinant be $-(t^2 - x^2 - y^2 - z^2)^4$ (off-diagonal blocks...)

You're saying there's a map from vectors to matrices for no reason, then saying there's another map from these matrices to the determinant which is the square of those matrices, then assuming the things those X matrices act on are called spinors. These assumptions are the whole thing we're trying to see without any guessing/foresight

DIRAC1930

@bolbteppa This is equal to $0$ for an isotropic vector

bolbteppa

Right but it's not the square, so already we see things are changing in higher dimensions, I don't think this determinant thing always holds or is taken as anything fundamental in general

DIRAC1930

So this det thing is only true for d=3?

bolbteppa

In D = 4 it holds when you split the spinor into the 2D projections and apply it to those 2 x 2 matrices in special relativity, but other things change when you do this projection, $X^2 = x^2 I$ no longer holds (if I remember correctly off the top of my head), this is why I wrote it with a power $4$ because you get two Weyl projections

DIRAC1930

12:14 AM

Yes that is true

bolbteppa

Cartan is trying to show all of this follows 'simply' from $F = 0$ and isotropic subspace thinking, the good thing about it is you derive these $\xi_{ij},...$ type coordinates as a by-product and the existence of a spinor, and he even derives the $X$ matrix and the fact $X^2 = x^2 I$ and the Clifford algebra as by-products. The bad thing is you have no idea what he's doing or where it's coming from, which is what I'm trying to figure out.

If you do it Dirac's way then everything makes sense except you have no idea what it means, though this quadric surface stuff above might give the whole picture, explaining Cartan's thing too. If you do it with creation/annihilation operators it's similar to Dirac's approach, but I think some other things are made clear this way as well

Dirac's trick is a good reason to postulate $X = x^{\mu} \gamma_{\mu}$ matrices directly, and you get the Clifford algebra, and then representation theory can tell you the dimension of a matrix representation is $2^{\nu}$, and so we see the vectors these matrices act on are $2^{\nu}$ component objects (called 'spinors'). Then you can basically write the creation operator stuff I did above and end up with Cartan's $\eta_A = 0$ equations before postulating anything about isotropy

DIRAC1930

I don't think you will be able to get anywhere without any assumptions. I think the whole point is that he observed that he found this structure for the $D=3$ case, then he tries to generalize it to higher dimensions which naturally involves using some assumptions from the construction in the $D=3$ case

bolbteppa

The Given's paper I linked to above is in some places pretty similar to Cartan, and the paper I linked to that says it motivated Cartan makes it worth trying to read/skim. I'd say that paper (talking about lines on quadrics and projective spaces) is the motivation, he also does Dirac's trick in the paper and uses it but it's an old paper so...

bolbteppa

12:31 AM

It seems like he's trying to say in that paper that the $\xi_{ijk}$-type tensors are basically ways to talk about lines, 2-planes, 3-planes, etc... living in a $\nu$-plane, by treating a line as being represented by two-points (vectors $A$ and $B$) in the plane (which the line passes through) and then saying the tensor $\xi_{ij} = 2 A^{[i} B^{j]}$ represents this line, and similarly with higher tensors, and then equations like $\xi_{ij} x^j = 0$ tell you that $x^j$ lives on the line,

something like that is going on

bolbteppa

1:04 AM

"In [3], we have also shown that Cartan’s spinor calculus has its foundation (and we think its origin) in Study’s and Beck’s work [3], and that these topics have to be treated as a subset of rational curves and advanced (projective) geometry (PG)"

antimony

1:22 AM

for someone who is pre-relativity. are there objects from the standard model who could classify as an observer in relativity?

i'm assuming no for the photon, because travels at c in freespace and no mass

More Anonymous

2:33 AM

Life on PSE

Felix Wotter

@RyderRude hi i wanted to say I found your question interesting as I have worried about similar matters, and have collected a small library of relevant papers. One thing worth mentioning here is that there is no measurement of a fraction of a global particle by a local measurement - a local measurement will measure th

the local number operator, related by a Bogoliubov transformation

Happy to discuss this general topic - a nice paper that helped me early on with my philosophical concerns in this area is arxiv.org/abs/math-ph/0607044

Relativisticcucumber

@ACuriousMind test

ah okay i see. thank you !!

Ryder Rude

3:53 AM

@FelixWotter I will look into it. What is the local number operator btw? Is it the total energy operator whose sum is restricted to the discrete momenta values defined by the boundary of the region?

This operator is what @ACuriousMind 's answer mentioned too. I was thinking that, since in classical field theory we can in only measure local field values, our energy operator should be the integral of the energy density over the local region

Felix Wotter

5:12 AM

@RyderRude it is unfortunately more complicated than that because the Local and Global Fock spaces are unitarily inequivalent. A thorough discussion of this can be found here: arxiv.org/abs/1403.0073.

Also, keep in mind that in an eigenstate of the global energy/Hamiltonian, the local energy is not in an eigenstate. If you locally measure even the global vacuum state, you can end up detecting arbitrarily many particles (the Reeh Schlieder theorem).

And any physical measurement made for a finite time involves an on/off switching function, which inevitably pollutes the local region you want to measure with energy from the device. So the result of a local measurement of energy cannot be thought of as just passively evaluating a sub-volume integral from the globally prepared state.

Ryder Rude

5:32 AM

@FelixWotter i shouldn't have expected something like this to be simple. Much thanks for the link. But, even if the process is riddled with complications, we should still be unable to measure the global energy, right? So what would we measure when the global energy is in an eigenstate but the state is highly non-local?

Do we measure the energy operator which is the sum over the discrete wavelength sprectrum given by the boundary of the local region ( I assume this is what you meant by the local number operator)

If yes, then what I'm worried about is that the classical equivalent of this observable would end up measuring the global energy stored in those particular wavelengths

Felix Wotter

6:01 AM

@RyderRude the global vacuum is as non-local as it gets, and a local measurement of the global vacuum can return literally anything! Look into the Reeh-Schlieder theorem, which is how this is usually discussed. Th

Thr classical equivalent, I am not sure what that means

The local and global operators don't commute

It is like if you have a spin = 0 Bell state, you expect +/- 1/2 when measuring an individual qubit

Do you have a worry about a classical issue in this case?

Ryder Rude

6:59 AM

@FelixWotter i agree that the global and the local energies don't commute. This is problematic, yes. If we're not measuring the local energy, what are we measuring? Is it the total energy but restricted to the discrete wavelengths modes determined by the boundary of the region? As in $\sum \hbar \omega_p a^{\dagger}_p a_p lregio$ but $p$ is restricted to be in $\frac{2\pi k}{L}$ (in the one dimensional case), where $L$ is the length of the region

If it's not this either, what operator do we measure when the global state is non-local but is an eigenstate of total energy?

Felix Wotter

7:15 AM

@RyderRude a local measurement will return an eigenstate of the local energy, by definition, in that subregion. but local energy h

local energy has a tenuous connection to the global energy

Just as the "local" spin of oje qubit has a tenuous connection to the "global" spin of a GHZ state

*one qubit

The local modes of course have the boundary conditions of any finite box Fourier modes

But I feel like you are asking if this like a subset of modes pulled out of the global modes, which it is not

Ryder Rude

@FelixWotter yeah, thats what i was asking

Felix Wotter

It is a completely different basis, and for QFTs with infinite degrees of freedom, not even in the same Hilbert space

Ryder Rude

So if the measurement yields the eigenstate of the local energy operator, then is it really true that measuring the vacuum gives you arbitrarily many particles

Felix Wotter

Yes because the local number operator is maximally uncertain when the field is in a global eigenstate

Ryder Rude

@FelixWotter i was replying to this

Oh, i thought you were pointing it out as a problem rather than an experimental fact

Much thanks for this

Felix Wotter

7:24 AM

It isn't a perfectly balanced superposition of all possible outcomes, as long as the local region is not extremely tiny

Ryder Rude

So we're not measuring a fraction, but more like an uncertain number of particles

Felix Wotter

For a local region bigger than the Compton wavelength, the local and global basis are pretty close

Ryder Rude

How can that be when the global region is infinitr

I will have to read your links

Thanks

Felix Wotter

Sorry I am new to this chat app. How can that be referring to which point?

Ryder Rude

There's an arrow beside that comment of mine, which takes you to the message i replies to

Felix Wotter

7:30 AM

When you said "how can that be" I don't see an arrow

Ryder Rude

Oh, that was only a reply to ur previous message

@FelixWotter this one

So if we measure in a region larger than the Compton wavwlength, it serves as a good approximation to the global energy measurement?

Felix Wotter

Correct

Assuming a one particle state I mea

If you prepare a one particle state in a box of 2 Compton wavelengths, and you measure each half locally, very often you find one particle in one region, 0 in the other

Ryder Rude

This is very unintuitive to me as the expected energy density of a single particle state can, in principle, be spread out over the whole universe. How can we expect to measure the total energy in a small laboratory?

Felix Wotter

It is just collapse of the wavefunction

That spread out state is a superposition of localized Gaussians, peaked at different points

Ryder Rude

That idea works for a position measurement, as different positions are different eigenkets of the position operator. But does it hold for energy measurements too?

Felix Wotter

7:46 AM

Approximately, yes

Ryder Rude

With a position measurement, most of the time, one would detect nothing at all in the case of delocalised states

But with an energy measurement, the entire energy spread out over the universe would have to get concentrated immediately in the laboratory

So im finding it very weird lol

Classically, we wouldnt be able to do this

Felix Wotter

You can't reason from the spread of the wavefunctional to the spread of the energy

Ok lets simplify to a spin chain

Ryder Rude

I dont know about spin chains..

Never studied

Felix Wotter

Consider |0001 + 0010 + 0100 + 1000>

Ryder Rude

That works. Sorry

Felix Wotter

7:49 AM

I just mean an entangled state of the spin of 4 particles here

Ryder Rude

Go on

Felix Wotter

So the "global state" here is like each particle has 25% of the |1> value

Ryder Rude

Yeah

Felix Wotter

Now, if the 1 configuration has a quanta of energy while the zero has none

This is like a global one particle state, expressed as a linear combo of 4 possible localized states

This is approximately how local energy measurement works for large enough discretization of space

Ryder Rude

So the wavefunction is de-localised but it collapses to a local state?

Felix Wotter

7:57 AM

Yes, ignoring the difficult but important issues with the exponential tails of the "localized" state, discussed at length in the papers I linked

Ryder Rude

I will have to read about this more. Thanks.

Felix Wotter

@RyderRude Sure ping me if you want to discuss further. Another excellent intro-ish paper that gets into these issues is arxiv.org/abs/quant-ph/0112148

Ryder Rude

Thanks

Relativisticcucumber

8:16 AM

hello, i was wondering, in quantum mechanics, we are introduced to spin states being spinors. is there a point in which spinors have different mathematical properties than vectors that i should take note of? in sakurai and griffiths, it is mentioned that they are spinors, but this is never discussed further, and the wikipedia page for spinors goes down the rabbit hole very quickly (for me at least ;P )

@Mr.Feynman regarding the question about the hamiltonian that you and ACM addressed, i think i maybe you touched on something of interest but skipped over it? so what i am wondering is how to know that this extra term belongs to the hamiltonian? they say we have the normal KE term on left and normal potential term on right, so rest is hamiltonian, but why couldnt we take this to be a correction to, say, the potential?

@Mr.Feynman this is what i am referencing

Mr. Feynman

9:22 AM

@Relativisticcucumber I think that at an elementary level all you need to know is that they are (in this case) two-complex components vectors and how they transform after a physical rotation of the system, which Sakurai addresses

@Relativisticcucumber The extra term arises as soon as you make the minimal coupling substitution $p\to p-eA$ (and calculate the square), so this is why it's a piece of the hamiltonian. Are you asking why we make this substitution?

Arsenal Creation

Is this real feymann?

or someone else with same name

Relativisticcucumber

@Mr.Feynman and when do I leave "elementary level" XD i mean when do these become important haha

@Mr.Feynman ok ok i think i see.

Mr. Feynman

@ArsenalCreation Well...

Jul 13, 2022 at 19:14, by Feynman_00

And I'm not the real Feynman either...

Oh that's my old nickname

Arsenal Creation

9:38 AM

I will name myself Einstein

Relativisticcucumber

there's also a dirac in here

Mr. Feynman

@Relativisticcucumber As soon as one discusses quantum mechanics in terms of representation theory, I guess? Also, in QFT you'll definitely need to discuss spinors

Relativisticcucumber

@Mr.Feynman okay thank you

Mr. Feynman

Not Pauli spinors but I don't think I'm able to address the question further, ask me 6 months from now :P

Arsenal Creation

CAn i ask anything about thermodynamics. sorry to interrupt

Relativisticcucumber

9:39 AM

@Mr.Feynman setting a calendar reminder now

Mr. Feynman

*Don't ask about asking, just ask.*
If someone is interested they will answer your question

Arsenal Creation

i will remeber it

I was studying about Enthalpy. the equation was H= U +PV.
What does P mean here. Is it internal pressure or external

Relativisticcucumber

4:03 PM

hello again. i have a question about the quantum numbers related to spin. so in sakurai, the discussion of spin originally kicks off with stern gerlach, which motivates defining states and the spin operator, and then from this, more formalism is developed. however, in the discussion of the spin 1/2 system, i was confused how we know the values in stern gerlach correspond to spin 1/2. [...]

[...] we can say that we know these are electrons and that this is their spin, but i am wondering if we notice that, in stern gerlach, there are two values we see, and by theory we decide this is a 1/2 system? this is question 1, and it prompted me to return to griffiths. i am also confused about the way griffiths derives these quantum numbers. for angular momentum, he uses the functional form to find l and m, which is fine to me. [...]

[...] then he says since the spin algebra is the same as that of angular momentum, we use these values but do not restrict to half integers. since the functional form of spin is not the same as that of angular momentum, i am having trouble buying this argument. am i missing something? how can i find out more about how to rigorously derive these quantum numbers?

ACuriousMind

@Relativisticcucumber Yes, we decide that the Stern-Gerlach atoms must be spin-1/2 because spin-1/2 objects are only objects with exactly 2 spin states

@Relativisticcucumber rigorously this is about the representation theory of $\mathfrak{su}(2)$; I have an alternative explanation of what's going on in the Wiki article here

bolbteppa

9:44 PM

God damn spinors

2

Silly Goose

10:03 PM

is the partition implied by the labeling of the pauli matrices? i.e. $\sigma_1$ would correspond to an operator only acting on subsystem 1?

ACuriousMind

yes

Silly Goose

ah okay

how did you learn about lie groups and the such :P @ACuriousMind

ACuriousMind

I took a course called "Lie groups" :P

Silly Goose

XD

up to this point...I have thought that a statement like "in general...[insert mathematical claim]" means that in all cases. but does in general mean in almost all cases, but not all?

Mr. Feynman

@bolbteppa Describe the mood of the chat during the last three months: mission accomplished

ACuriousMind

10:18 PM

@SillyGoose I'm sorry to inform you that it depends on who's talking

Silly Goose

lol

Mr. Feynman

I'll need time to get used to the new propic, until then I can't consider you the real ACM :P

bolbteppa

Klein quadric

In mathematics, the lines of a 3-dimensional projective space, S, can be viewed as points of a 5-dimensional projective space, T. In that 5-space, the points that represent each line in S lie on a quadric, Q known as the Klein quadric. If the underlying vector space of S is the 4-dimensional vector space V, then T has as the underlying vector space the 6-dimensional exterior square Λ2V of V. The line coordinates obtained this way are known as Plücker coordinates. These Plücker coordinates satisfy the quadratic relation p 12...

This is part of the secret...

Silly Goose

why are isomorphisms between hilbert spaces necessarily unitary maps?

or why are the concepts equivalent

ACuriousMind

how did you define an isomorphism of Hilbert spaces?

Silly Goose

10:28 PM

hm i guess im not sure but probably to preserve the vector space operations...then i guess if you also include preserve norms it must be unitary?

is that a usual condition on isomorphisms between inner product spaces?

ACuriousMind

yes

Silly Goose

oh hm okay i guess that makes sense

so it being a linear map preserves the vector space operations, then being unitary preserves inner products?

ACuriousMind

yes

Silly Goose

by presenting the definition of $TOT^{-1}$ being local to $\mathcal{H}_i$, the idea in mind is that okay if $T^{-1}$ is mapping a vector from $\mathcal{H}_i \rightarrow \mathcal{H}$, then $O$ maps it to another vector in $\mathcal{H}$, then $T$ maps it again back to $\mathcal{H}_i$, the net effect of the operator is sending a vector from some subsystem to another vector in the same subsystem?

ACuriousMind

yes :)

Silly Goose

10:39 PM

:D

Silly Goose

10:52 PM

in the same screenshot i sent... so even though in say a math course we do not care so much to distinguish between isomorphic spaces, here we are defining an isomorphism between hilbert spaces and defining an addition criteria (the hamiltonian being local) which artificially prefers an isomorphism and thus tensor product structure.

however when we talk about the second to last sentence

we are referring to unitary transformations acting only on subsystems, which to me sounds like an automorphism defined on a subsystem; we treat these automorphic spaces as physically equivalent because we have not defined such an artificial criteria to prefer one subsystem over the other, right? i.e. we have no want to distinguish between automorphic subsystem spaces in a particular tensor product structure

ACuriousMind

I don't see anything in that quote treating the different $H_i$ as equivalent

they're just saying each of the $U_i$ is a unitary map $U_i : H_i\to H_i$

Silly Goose

hm i guess i am confused by what "notions remain unchanged" means

ACuriousMind

"these notions" refers to the concepts of local operators and entanglement entropy

An operator that is local to $H_i$ will remain so after application of $\otimes_i U_i$

Silly Goose

ah okay yes yes

ACuriousMind

and the entanglement entropy stays the same too

Silly Goose

10:58 PM

so then with respect to entanglement entropy and the locality of operators, automorphic subsystems of a given tensor product structure are physically equivalent? if those two criteria are the only physical criteria we care about?

ACuriousMind

what does "automorphic subsystem" mean?

automorphic as an adjective does not make sense to me

Silly Goose

ohh lol

i see

are you saying because the two spaces are just the same spaces?

ACuriousMind

which two spaces?

Silly Goose

so an automorphism $U$ defined on subsystem $\mathcal{H}_i$ is a map $U: \mathcal{H}_i \rightarrow \mathcal{H}_i$?

which means the domain and codomain are the same space? (those are the two spaces i am referring to)

ACuriousMind

yes, an "automorphism" is just any isomorphism from a space to itself

so I don't know what "automorphic space" means because every space has at least one automorphism, the identity map

Silly Goose

11:03 PM

i see i see

i think i was saying automorphic spaces (plural) to describe two spaces related by an automorphism

but i guess that doesn't really make sense because by definition of automorphism the spaces are genuinely the same

bolbteppa

@DIRAC1930 this has something interesting:

For $\nu = 1$ it reads: "Consider in an $\mathbb{R}^3$ the $(3-1)=2$-dimensional unit sphere with center in the origin. On this sphere there is lying for $n = 2 \cdot 1 + 1$ two systems of $\infty {1 + 1 \choose 2} = {\infty}1$ $E_{1}$'s."
For $\nu = 2$ it reads: "Consider in an $\mathbb{R}^5$ the $(5-1)=4$-dimensional unit sphere with center in the origin. On this sphere there is lying for $n = 2 \cdot 2 + 1 = 5$ two systems of $\infty {2 + 1 \choose 2} = {3 \choose 2} = {\infty} 3$ $E_{2}$'s."

For $\nu = 3$ it reads "Consider in an $\mathbb{R}^7$ the $(7-1)=6$-dimensional unit sphere with center in the origin. On this sphere there is lying for $n = 2 \cdot 3 + 1 = 7$ two systems of $\infty {3 + 1 \choose 2} = \infty {4 \choose 2} = {\infty} 6$ $E_{3}$'s."
For $\nu = 4$ it reads: "Consider in an $\mathbb{R}^9$ the $(9-1)=8$-dimensional unit sphere with center in the origin. On this sphere there is lying for $n = 2 \cdot 4 + 1 = 9$ two systems of $\infty {4 + 1 \choose 2} = \infty {5 \choose 2} = {\infty} 10$ $E_{4}$'s."

(In the odd case)

What is it trying to say...

Then you have that Klein thing from the above wiki link in the comment after this...

("$E_{\nu}$ =$\nu$-dimensional flat affine space. In an $E_{\nu}$ there exist parallel flat subspaces but no lengths and no angles")

Silly Goose

11:32 PM

where does the mathematical formulation of quantum mechanics begin (or at least one way to see it)?