The h Bar

12:00 AM

And (from Cotler) instead of actually changing this tensor product factorization, we instead unitarily vary the hamiltonian with arbitrary unitary operators parameterized by d^2-1 parameters, so 15 in this case.

I then define a function that computes the linear entropy of a time evolved initial candidate pointer state parameterized by the d^2-1 parameters. this outputst the "score" of the unitarily varied hamiltonian and hence tensor product factorization.

i find the set of d^2-1 parameters which minimizes the score. there are many such sets of parameters.

ACuriousMind

ah, I see

Silly Goose

and to clarify each set of 15 (in the case of 4 dimensions) parameters corresponds to the unitary transformation which sends our initial hamiltonian to a new unitarily varied hamiltonian and hence potentially new tensor product structure

of course the difficulty is thatt some of these unitary transformations stay within the same tensor product structure

ACuriousMind

you're trying to find a factorization/Hamiltonian for which your non-entangled state becomes the "least" entangled over time

since that's the most convenient way to think about the system

Silly Goose

yes

most convenient in terms of writing code at least

well actually it is also just as much about finding if the minimum factorization is unique

because i can find many minimum factorizations

ACuriousMind

that's the dual business we discussed a while back, I remember

Silly Goose

12:05 AM

yes

!

ACuriousMind

!

Silly Goose

i am thinking that i would only be able to maybe statistically justify that "most of these minima are part of the same TPS" but that seems kind of not rigorous enough

also i guess to be more precise, I should have said tensor product structure everywhere i said tensor product factorization

ACuriousMind

I mean of course "most" of them are the same TPS because there are infinitely many unitaries but usually only finitely many duals :D

Silly Goose

XD heh

well i guess at least i can find minima XD there is one step in the right direction

previouslyt i had been trying to use gradient descent but it turns out one line of scipy does the trick lol

Slereah

I think the way to get GR from some gauge theory is basically 1) have some Lorentz gauge theory 2) have the bundle be natural 3) have an AVB in the standard rep of the group

I think it's all you need

Maybe the AVB is also natural idk if that naturally flows otherwise

Though I'm not sure if just naturalness is enough to guarantee some relation to the tangent bundle

Silly Goose

12:11 AM

Also, I think I have finally found peace with the current level of my understanding of the Cotler business lol.

maybe in 10 years i will understand it fully but that is for 10 years

parvin

2:16 AM

Hey! I got a question,
If we add two white noise together would the result amplitude or sound pressure level of the sum be different than when we add sine waves of the same amp and SPL?

Also, in adding the sine waves, if the speakers are in front of eachother, should we substract or add?

Manas Dogra

6:46 AM

Why is the conservation of any four vector other than the 4 momentum forbid all except forward collisions?

This is from a footnote in Weinberg's QFT book---
""The only conserved tensors are the scalar
`charges' associated with various continuous symmetries, and the energy-momentum four-vector .
The conservation of any other four-vector, or any tensor of higher rank, would forbid afl but
forward collisions .""

Silly Goose

7:46 AM

Consider a density matrix represented as an nxn matric $M$. Let's say we are working in a Hilbert space $\mathcal{H} \cong \mathbb{C}^2 \otimes \mathbb{C}^2 \otimes \mathbb{C}^2$, where each factor corresponds to a qubit.

Is it true that in general the decomposition of $M \mapsto M_1 \otimes M_2 \otimes M_3$ is unique?

Mr. Feynman

8:56 AM

Is there any resource (that is not L&L as I've read that part) that makes clear the distinction between the magnetic fields $\boldsymbol{H}$ and $\boldsymbol{\mathfrak{H}}$?

From "standard" E&M I only know $B$, $H$ and $M$ as macroscopic quantities

bolbteppa

9:55 AM

@ManasDogra he's probably referring to this

It's explained better (e.g. the conservation of higher rank tensors part) in the first chapter of Freund's supersymmetry book

I cannot believe how simple it is to see where the LRL vector comes from after that post above, it was staring everyone in the face the whole time

@Slereah There's more ways than this, there's also a Poincare version for example, confusing stuff

Slereah

10:30 AM

@bolbteppa yeah with affine connections

But one step at a time

ACuriousMind

10:59 AM

@SillyGoose What exactly does the $\mapsto$ in your "decomposition" denote?

because in general not every $M$ is equal to such a simple tensor; the point of entanglement is precisely that not all states/operators in a tensor product are of this form

and when such a decomposition exists it is not unique without further constraints on the $M_i$ - consider that $(ca)\otimes b = a\otimes (cb)$ for any constant $c$

user2236

11:29 AM

How goes your recovery from CoV pal @Slereah

Slereah

11:45 AM

Still not doing amazing but better than in february

user2236

Good to hear.

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