Okay the answer to 1, 3 and 4 are good.In post 2292, Scorpious wrote:#1 For a given system, the choice of which commuting degrees of freedom to use is not unique, and correspondingly the domain of the wave function is also not unique. For instance, it may be taken to be a function of all the position coordinates of the particles over position space, or the momenta of all the particles over momentum space; the two are related by a Fourier transform. Some particles, like electrons and photons, have nonzero spin, and the wave function for such particles include spin as an intrinsic, discrete degree of freedom; other discrete variables can also be included, such as isospin. When a system has internal degrees of freedom, the wave function at each point in the continuous degrees of freedom (e.g., a point in space) assigns a complex number for each possible value of the discrete degrees of freedom (e.g., z-component of spin) – these values are often displayed in a column matrix (e.g., a 2 × 1 column vector for a non-relativistic electron with spin 1⁄2).In post 2282, Frogsterking wrote:Scorpius please answer theseIn post 2186, Frogsterking wrote:Hey guys! I just worked on this questionnaire with my friend to help with scumhunting! It was built around the idea that mafia forums are a low scope low testability domain.
Here's the "low-scope low-testability scumhunting questionnaire:"
#1 How does the wave function evolve?
#2 Does wave function imply many worlds?
#3 Have we used all the available technology?
#4 Do we live in a simulation?
#2 As has been demonstrated, the set of all possible wave functions in some representation for a system constitute an in general infinite-dimensional Hilbert space. Due to the multiple possible choices of representation basis, these Hilbert spaces are not unique. One therefore talks about an abstract Hilbert space, state space, where the choice of representation and basis is left undetermined. Specifically, each state is represented as an abstract vector in state space.[41] A quantum state |Ψ⟩ in any representation is generally expressed as a vector
|
Ψ
⟩
=
∑
α
∫
d
m
ω
Ψ
(
α
,
ω
,
t
)
|
α
,
ω
⟩
{\displaystyle |\Psi \rangle =\sum _{\boldsymbol {\alpha }}\int d^{m}\!{\boldsymbol {\omega }}\,\,\Psi ({\boldsymbol {\alpha }},{\boldsymbol {\omega }},t)\,|{\boldsymbol {\alpha }},{\boldsymbol {\omega }}\rangle }
where
|α, ω⟩ the basis vectors of the chosen representation
dmω = dω1dω2...dωm a "differential volume element" in the continuous degrees of freedom
Ψ(α, ω, t) a component of the vector |Ψ⟩, called the wave function of the system
α = (α1, α2, ..., αn) dimensionless discrete quantum numbers
ω = (ω1, ω2, ..., ωm) continuous variables (not necessarily dimensionless)
These quantum numbers index the components of the state vector. More, all α are in an n-dimensional set A = A1 × A2 × ... An where each Ai is the set of allowed values for αi; all ω are in an m-dimensional "volume" Ω ⊆ ℝm where Ω = Ω1 × Ω2 × ... Ωm and each Ωi ⊆ ℝ is the set of allowed values for ωi, a subset of the real numbers ℝ. For generality n and m are not necessarily equal.
Short answer yes
3.No, we’re being fed technology at a pace that appeases the overlords.
4. Only an overlord has permission to ask that question.
The answer to 2 is good in its conclusion (arguably) and weak in its demonstration of how to improve testability of
said conclusion.
Overall I'm thinking that Scorpius is a townlean here guys!
Question 3 needs to be revised for future low info low testable players (aka Eyes slot replacement):
3 With sufficient technology can all low scope low testability questions be made testable?
