Interesting article

@MahNeh

@Obliv It is just the implementation. The writing is much more focused upon chemical analysis than is able to properly teach physics. It succeeds at neither thermo nor stat therm.

@ACuriousMind Got it.

I have a new question related to gravitational path integrals.

A standard argument by Gibbons, Hawking and Perry shows that a Weyl rescaling of the metric can render the Euclidean Einstein-Hilbert action negative definite which makes the grav. path integral more pathological. What is the idea behind making a transformation which is not even a symmetry of the classical theory in the first place? In string theory, there's a Weyl symmetry on the worldsheet but in GR there's no spacetime Weyl symmetry, right?

I mean why would I want to do the Weyl transformation in the first place?

@Sanjana The point is that the Weyl transform is also part of the configuration space

Since you have to sum over all configurations it means that there are terms in that sum that are arbitrarily negative

@Slereah Okay, got it

It would be the opposite of the argument if it was an actual gauge symmetry since those you can actually discard

Also if it was a gauge symmetry they would have the same action, not be arbitrarily smaller :p

Yes, true.

Rovelli says that his interpretation supports mild panpsychism philarchive.org/rec/ROVRAP-2

i think Rovelli's relational psi interpretation is almost there but it would need modifications due to general relativity

@ACuriousMind i have a few follow up questions on this answer physics.stackexchange.com/a/721726/50583. the first one is that you say that ordinary qm has a fixed hilbert space $L^2(\mathbb{R^n})\otimes S$ where $S$ is a representation of the rotation group. i noticed that OP has in their first bullet no rotation group, then $\mathbb{C}^2$ and $\mathbb{C}^4$ in their second and third bullets respectively. in what way are these complex vector spaces rotation groups?

@Relativisticcucumber the $\mathbb{C}^2$ is the spin-1/2 representation of SO(3), the $\mathbb{C}^4$ the $(1/2,0)\oplus (0,1/2)$ representation of SO(1,3)

"no rotation group" is just having $S$ be the trivial representation, since $V\otimes \mathbb{C} = V$ for any complex vector space $V$

i see. i remember learning that the representation of $SO(3)$ is the pauli spin matrices -- is this related to $\mathbb{C}^2$? they live in $SU(2)$ i think, but is there a more proper way to say this? @ACuriousMind

okay i see. and then in the wiki page you linked for SVnT, it says that the commutation law was observed by weyl to be impossible to satisfy for linear operators acting on a finite dimensional space unless $\hbar$ vanishes. but im confused why finite versus infinite matters here. the point raised that taking the trace of both sides gives a contradictory result seems problematic to me by itself, so i think im missing something? @ACuriousMind

@Relativisticcucumber in the infinite-dimensional context, the trace is not well-defined for all operators, so it produces no contradiction there if $x$ and $p$ are operators for which it is not defined

i.e. $x$ and $p$ are not trace class