Recent content by SergejMaterov

Are you familiar with these experiments? https://doi.org/10.1038/nature15759

Exactly. All the physics lives on the projective space ##\mathbb C\mathrm P^1## (or Bloch sphere); ##\mathbb C^2## is just a convenient, basis‑dependent coordinate chart.

Exactly right. You’re drawing the perfect analogy. Point [a:b] in ##\mathbb C\mathrm P^1## isn’t literally the qubit state, but represents the unique physical pure state (ray) in the abstract ##\mathcal H##. All the physics lives in the projective geometry; picking ##\mathbb C^2## is only a...

$$H \rightarrow \mathbb{C}^2,\quad a|0\rangle + b|1\rangle \mapsto \begin{pmatrix}a \\b\end{pmatrix}$$ This is already the realization of an abstract space, and not the physical state itself. The physical state of a qubit is a class of vectors {λ(a,b)} = point ([a:b]) in ##\mathbb{CP}^1##

The vectors ∣0⟩ and ∣1⟩ are not cubit states in themselves, but only their representatives. The only physical content is the rays, that is, the points of the project space.

Gravity’s coupling G is dimensionful, so you only get a meaningful gravitational fine‑structure constant once you specify masses or energies. In hydrogen: ##\alpha_g=\frac{Gm_em_p}{\hbar c}\sim3\times10^{-42}## In particle scattering at energy E: ##\alpha_g(E)\sim\frac{GE^2}{\hbar c}## This lets...

The chosen basis provides a convenient coordinating, but does not change the geometry of the state space itself

You can of course be skeptical about my previous post, but: If it does exist, then formally the "graviton" should be very similar to the photon, and many techniques (Feynman rules, soft theorems, double copy) can be transferred. They share the same quantum‑wave duality, but that’s where most of...

Without convexity and canonical form you will not get a powerful geometric tool for calculation or classification. It is possible to make a "negative EFT-hedron", but its usefulness is unlikely to be comparable to the positive one: the main properties are lost. I have not yet seen a single fully...

why is CP violation in baryons so small compared to mesons? paradox?

Trotter splits ##e^{−i(T+V)ε}## into “purely kinetic” and “purely potential”. The Gaussian integral yields the leading term ##exp[im(x−y)^2/(2ε)]## and normalization. Taylor potential around the midpoint 𝑧=(𝑥+𝑦)/2 yields −iεV((x+y)/2) plus ##𝑂(𝜀(𝑥−𝑦)^2)##. All remaining inconsistencies are...

Exactly so—multiplication operators fit the same pattern as the differential ones: Multiplication by an unbounded function is fundamentally not everywhere defined, so it doesn’t contradict Hellinger–Toeplitz. Any physically meaningful unbounded operator—differential or multiplicative—lives on a...

Differential operators are unbounded because they’re only defined on a proper dense subset of the Hilbert space, not on every vector. If you insist on “full” domain, the only unbounded ones you get are pathological, non‑continuous algebraic beasts with no symmetry or spectral story.

If you drop symmetry, you get pathological, discontinuous, unbounded maps—but they’re of almost no use in analysis or physics.

I think it is too early to talk about "graviton" at all for obvious reasons: 1. quantum gravity has not been built 2. "graviton" has not been discovered Quantum field theory predicts a lot, but creates more problems.