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02:26
0
Q: Spin Angular Momentum Quantum Mechanics

John.DA particle has spin $\hbar/2$. A measurement is made of the sum of x and z components of its spin angular momentum. What are the possible results of the measurement?

 
6 hours later…
08:20
in The h Bar, 25 mins ago, by Blue
So all these steps are justifed. After all we just need to find the constants such that RHS=LHS holds for all $x$. We don't really bother about the singularities as long as they are removable
09:02
0
Q: What decides the number of bands in one-dimensional tight-binding model?

SRSI was reading about the one-dimensional tight-binding Hamiltonian (TBH) $$H=E_0\sum\limits_{n}|n\rangle\langle n|-t\sum\limits_{n}\Big(|n\rangle\langle n+1|+|n+1\rangle\langle n|\Big)\tag{1}$$ where $E_0$ and $t$ denote the on-site energy and the hopping parameter, repectively. The Hamiltonian of...

09:18
0
Q: QM formulation in real vector spaces

NatanaelIs it possible to define quantum mechanics in real vector spaces instead of complex vector spaces and what would the dimensionality be of such a vector space? Can anyone referee me such a treatment of a 2 state system, say the spin system.

0
Q: Uncertainty principle on optical fiber communication

user129048 While the calculations seem fine, I can't make sense of the solution to this problem. How can the uncertainty principle be used to tell us what the shortest pulse is? Wouldn't using it tell us the error in the pulse lengths? And the pulse lengths then could be arguable any size we want it to be.

 
2 hours later…
11:09
0
Q: Commutation [x,p]=ih

benif we have position operator x and momentum operator p as follows: $$ \hat x = x \space ; \space \hat p = - i \hbar {∂ \over ∂x} \space $$ How do we show the commutator is: $$ [\hat x, \hat p] = i \hbar$$ I am stuck at: $$ [\hat x, \hat p] = x \space (-i \hbar {∂ \over ∂x}) - (-i \hbar {∂ \o...

11:58
0
Q: What does this quote by Niels Bohr mean?

doge"Everything we call real is made of things that cannot be regarded as real." And also can someone say what is the source of this quote.

 
2 hours later…
14:15
1
Q: How would the intersection of two uncountable sets be finite?

committedandroiderThis is a problem from Discrete Mathematics and its Applications Here is my book's definition on countable and definition of having the same cardinality The only example that my book gave of uncountable set was the set of real numbers. I understand that because if you try listing out all ...

2
Q: Can we apply a successor operation in Peano's arithmetic infinitely many times, is the successor well defined?

Dávid TóthIf we look at the axioms of Peano arithmetic, e.g. http://mathworld.wolfram.com/PeanosAxioms.html, they contain an axiom: If $a$ is a number, the successor of $a$ is a number. However, the axioms do not limit how many times we could apply this successor operation. So we could apply successor to...

googology.wikia.com/wiki/Introduction_to_BEAF
The notion of growth dimension:
There are many notions of comparing size, such as volume, duration, scope, scale, growth etc.
Rate of growth can be considered as like a dimension. It also has properties similar to both potential and actual infinity
Consider a growth of 100/s that occured together with a growth of 1000/s. Then the overall growth will be 1100/s
However if we had some growth $10^n/s$ and $10^m/s$ where $n>>s$, then the latter growth will be effectively drown out by the former, thus effectively, the system only felt the growth by th
1. There is a maximum, defined to be the unique number that has no successor.
2. There is a minimum, defined to be the unique number that has no predecessor.
3. Every number is either a successor/predecessor, the minimum or the maximum.
4. Between any two successors there are no numbers.
5. Every number except the minimum or maximum has both a unique successor and a unique predecessor.
6. The minimum has a unique successor, and the maximum has a unique predecessor.
1. $\not\exists n\exists! b[S(b)=n]$
2
Q: Verifying a Superior Integration Method for Step Functions

The Great DuckI should note that this is an integration algorithm and therefore intermediate steps DO appear to be unjustified. The purpose of this question is to justify or reject this algorithm as always giving correct solutions. Suppose we have some piecewise continuous function $f$ and that it can be writ...

Typhon
8:35 AM
@Secret I believe F is constructed within the algorithm and returned as the resukt but i may be mistajen or made a typo
no i dont
just replace floor(g(x)) with c
if you have infinite floor terms to replace
then youve violated my close form stipulation
yeah step 3 actually constructs F
it doesnt exist until then
@Secret finly after literally two years and three months i am about to write a proof of that algorithms validity
heh
heh heh heh
pastebin.com/PgL0GtG1
^^ see?
if you have a suggestion so far let me know
it isnt done of course
once i prove the algorithm is equivalent to the continuous subet of the implied antiderivatives
i have to show that is equivalent to the integral
that will be the part ill have trouble with
but I WILL prove it
@Secret I'm afraid this remark seems to be a joke. I could not find any such thing as "transfinite metachemistry" on Google.

Secret
Secret
yeah alessandro is joking, though it is still fun to ponder what happens if it is assumed to be serious and see where such rabbit hole leads
nevertheless, for this example, it is not going to be useful for a very long time, because we don't even have countably infinite entities in the real world known so far, let alone a chemistry based on them

user21820
user21820
I'm not saying it's feasible to have such stuff in nature, even in the sense of finite approximation.
A crystal structure or quasi-crystal structure will have local rules.
This kind of structure has non-local rules that reek of arithmetic, and nature just does not have complicated discrete arithmetic.

Secret
Secret
1:20 PM
It's still fun to ponder about that, given that in organic chemistry we do check reactivity trends as the carbon chain gets longer. If a plateau behaviour is observed, we can perhaps say that at the limit of infinitely long carbon chain, we get such and such value.
gg rationals
Attempt 2:
Start with the reals R, construct the irrationals I=R∖Q
Now, pick one irrational, say π and then construct the set πQ={q∈Q:π+q}
...
ok gg, does not work, as for every such shift, there will be some irrational r<π which then becomes r>π thus ordering was flipped for these entries
Leaky Nun
6:52 AM
@Secret are you trying to order-biject $\Bbb I$ and $\Bbb R$?
hmm, I need to think about it
I think you would need a closed subset of $\Bbb I$

Typhon
Typhon
@LeakyNun whats up?

Leaky Nun
Leaky Nun
the sky

Typhon
Typhon
indeed
and that is the only correct answer
know why?
@LeakyNun up is relative. therefore the only thing you know is up is something in all directions such as the sky which surrounds the Earth.
so when someone says the ceiling feel free to politely correct them
in Logic, Jan 6 at 15:32, by Leaky Nun
busy oracle
The busy beaver game consists of designing a halting, binary-alphabet Turing machine which writes the most 1s on the tape, using only a limited set of states. The rules for the 2-state game are as follows: the machine must have two states in addition to the halting state, and the tape starts with 0s only. As the player, you should conceive each state aiming for the maximum output of 1s on the tape while making sure the machine will halt eventually. The nth busy beaver, BB-n or simply "busy beaver" is the Turing machine that wins the n-state Busy Beaver Game. That is, it attains the maximum number...
In the computer science subfield of algorithmic information theory, a Chaitin constant (Chaitin omega number) or halting probability is a real number that, informally speaking, represents the probability that a randomly constructed program will halt. These numbers are formed from a construction due to Gregory Chaitin. Although there are infinitely many halting probabilities, it is common to use the letter Ω to refer to them as if there were only one. Because Ω depends on the program encoding used, it is sometimes called Chaitin's construction instead of Chaitin's constant when not referring to...
What is the class of most indescribable number
in Logic, Jan 6 at 15:34, by Alessandro Codenotti
That's a problem of bounded sets, not of $\omega$, it has no accumulation points in its natural topology
$\omega$ literally give us a glimpse on what a proper class would have looked like had high infinities don't exist
specifically, it only requires to go forever, no limit points necessary need to be within the set

user21820
user21820
@Secret In case it wasn't clear, LeakyNun was joking about the oracle being busy, and nothing to do with the BB function.

Secret
Secret
Well, the main reason I post this in rambles and not here (as otherwise I would have post a condensed version of the above in a sentence in the logic room) is the above is really just a stream of consciousness. Specifically:
My thought process in a nutshell when not required to be in analytic mode:
Jotting down the process of generating an idea which is the the first thing that came to mind when reading something, and then repeat the process until the final product is reached
It's a ... semiconscious thing, probably because I spend too much time brainstorming and mixing ideas that the whole thing becomes second nature and it just spontaneously recall and mix ideas while I am busy on other tasks
but as you can see, most ideas are junk though, only the few ones that are promising will then be further polished in
in Logic, Jan 7 at 7:19, by Leaky Nun
we posted the same thing at the same time in different rooms
lol that delocalisation
The hallmark of The Weirdness
@Secret Yes it's good to keep notes of all your ideas, even if they are just ideas. I just thought you may potentially have been confused by his joke. =)

Secret
Secret
an actual busy oracle can be potentially interesting though, despite its sheer uncomputability

Secret
Secret
(Some other notes)
I like to mix ideas, to the point it is part of my purpose of life "to unlock the mystery of nature"
Because of that, I have a lot of sources to obtain ideas such as:
1. My friends, my collegues, speakers in conferences and seminars, my professors and teachers as well people met transiently in life
in Mathematics, Jan 17 at 4:45, by Secret
(To be done later) Think of a geometry such that if the ant walked at some velocity < n, then it got stuck, otherwise it walked some distance x > n
Requires an infinite set that becomes infinite when it involves finite union of finite sets, but finite when the finite union exceeded a certain threshold
In topology, the Tychonoff plank is a topological space defined using ordinal spaces that is a counterexample to several plausible-sounding conjectures. It is defined as the topological product of the two ordinal spaces [ 0 , ω ] {\displaystyle [0,\omega ]} and [ 0 , ω 1 ] {\displaystyle [0,\omega _{1}]} , where ω {\displaystyle \omega } is the first infinite...
to be visualised later
in Mathematics, Jan 26 at 7:44, by Secret
because that will mean there exists a finite set that has infinite proper subsets
in Mathematics, Jan 26 at 7:14, by Daminark
Think about it, infinity divides no finite number, so by Lagrange's theorem it doesn't work
Construct a foundation system such that finite cardinals are divisible by infinite cardinals (or any notion of size)
S+S+S+S+S=5
may be related to the notion of quasifinte sets rambled earlier
in The Factory Floor, Jan 26 at 10:15, by Secret
logically speaking, however it may not because it could be that there really exists technology that can be compatible with it, but the universe conspires in a way such that this can never be shown, hence making Clark's Law vacuously true for such case
Considering vacuous truth cases for a problem may be useful, thus do not ignore them in the future
4
Q: Uncountable orderings

Jacopo BelboLet $P$ be an uncountable linear ordering. Is it true that either $P$ contains an order-copy of $\omega_1$ or there is $x_0\in P$ such that there exist uncountably many distinct $y\in P$ with $y< x_0$? If so, where can I find a reference for this? Thank you.

1=L,0=R
00=0
q=L
q00=q0
10=1
0=1
(L,R,L) = false
(L,R,R)
00=0
q=R
00q=0q
01=1
01q=1q
0q=q
1=q
(L,R,R) = false
(L,L,L)
00=0
q00=q0
mistake
(q01)=(LLL)
00=0
q00=q0
10=1
0=1
(LLL)=false
(LLR)
00=0
q00=q0
10=1
q10=q1
q0=q
1=q
(LLR)=false
mistake
(q1)=(LL)
00=0
q00=q0
10=1
0=1
(LL)=false
(LR)
00=0
q00=q0
10=1
q10=q1
q0=q
1=q
(LR)=false
(RL)
00=0
00q=0q
01=1
01q=1q
0q=q
1=q
(RL)=false
(RR)
00=0
00q=0q
01=1
0=1
(RR)=false
therefore 00=0 => q=1
in Mathematics, 6 mins ago, by Secret
Theorem a,b,c
in Mathematics, 6 mins ago, by Secret
a: $00=0 \implies q,1$ has opposite sideness and $q=1$
in Mathematics, 3 mins ago, by Secret
b: $\exists x \neq 0, 00=x \land q,1$ has same sideness $\implies q$ is a zero divisor of $x$
in Mathematics, 1 min ago, by Secret
c: $\exists x \neq 0, 00=x \land q,1$ has opposite sideness $\implies $(compiling...)
in Mathematics, Jan 28 at 8:48, by Secret
Theorem a,b,c
1
Q: Are two functions $f$ and $g$ with linear independent derivatives neccessarily linearly independent and vice versa?

The Great DuckSuppose that there exists functions $f$ and $g$ defined on the real numbers and differentiable everywhere. If their derivatives $f'$ and $g'$ are linearly independent on some nonzero interval then are $f$ and $g$ linearly independent on the same interval? Similarly if $f$ and $g$ are linearly ind...

exhaustive sets
Some stuff on $\omega_1$
$\exists \alpha \forall \beta [\beta \hookrightarrow\Bbb{N} \implies \alpha \not {\hookrightarrow} \beta]$
$\forall \beta \exists \alpha [\beta \hookrightarrow\Bbb{N} \implies \alpha \not {\hookrightarrow} \beta]$
$[]_{0}$
$[0]_{0}$
$[0,1,2,...,n]_{n}$
$[0,1,2,..._{huge}$
$[0,1,2,...]_{\omega}$
$[0,1,2,...,\omega]_{\omega+1}$
$[0,1,2,...]_{\omega_1^{CK}}$
$[0,1,2,...]_{\aleph_0}$
$[0,1,2,...]_{n}$

Secret
Secret
8:33 AM
$\forall x : \infty_n + x = \infty_{n+1}$
$0 + x = \infty$
$A \cap B = \{c : c \in A \land c \in B\}$
$A \cup B = \{c : c \in A \lor c \in B\}$
$A - B = \{c : c \in A \land c \not\in B\}$
$A \triangle B = \{c : c \in A - B \lor c \in B - A\}$
$A \times B = \{\{a,\{a,b\}\} : a \in A \land b \in B\}$
$A + B = \{a + b : a \in A \land b \in B\}$
$A (R) B = \{aRb : a \in A \land b \in B\}$
$A \sqcup B = \{\{a,\{a,0\}\} : a \in A\} \cup \{\{b,\{b,1\}\} : b \in B\}$
Cardinality basics:
$|A \cup B| + |A \cap B| = |A| + |B|$
$|A \cup B| = |(A \triangle B) \cup (A \cap B)|$
Null set axiom:
$\forall A\exists Z [A \cap Z \neq \varnothing \land |A \cap Z| = 0]$
Results:
$|A \cup Z| + |A \cap Z| = |A| + |Z|$
$|A \cup Z| = |A| + |Z|$
$|A \cup Z| = |A \triangle Z | + |A \cap Z| = |A \triangle Z|$
$|Z \cup Z| + |Z \cap Z| = |Z| + |Z|$
$|Z \cup Z| = |Z| + |Z|$
$|Z \cap Z| = |Z| = 0$
$|Z \cup Z| = 0$
$|Z \triangle Z| = |\varnothing| = 0$
$|Z \cup Z| = |(Z \triangle Z) \cup (Z \cap Z)| = |\varnothing \cup Z| = |Z| = 0$
$A \triangle B = (A \cup B) - (A \cap B)$
if $A \cup B = \varnothing$
$A \triangle B = \varnothing - (A \cap B) = \varnothing$
$|A \cap B| = |A| + |B| = |A \cup B| = 0$
$|A| + |B| = 0$
$A \triangle B \implies A = B$
$|A|=|B|=0$
if $A \cap B = \varnothing$
$A \triangle B = (A \cup B) - \varnothing = A \cup B$
$|A \cup B| = |A| + |B|$
$\therefore A, B$ disjoint or $|A|, |B|$ is infinite
if $A \triangle B = \varnothing$
$A = B$
$\varnothing = (B \cup B) - (B \cap B) = B - B = \varnothing$
if $A - B =\varnothing$
One naive guess does not work: Isomorphism classes of sets (aka cardinals) under disjoint union form a commutative monoid - for size issues one should first choose a Grothendieck universe (or something similar) and consider only sets with smaller cardinality. A commutative monoid can be group completed. But by a naive variant of the Eilenberg swindle, every cardinal becomes 0 in the group construction: For example $1+\aleph_0 = \aleph_0$, so $1 = 0$ in the group completion. Note that this problem does not occur for finite cardinals and here we get just $\mathbb{Z}$. — Lennart Meier Jul 11 '13 at 15:58
Preliminary conclusion: Axiom of choice allow the partition of an interval into countably many uncountable dense sets that has no intervals. Then since there are no infintesimal elements in the Lebesgue measure, countable additivity of countably many nonzero measure cannot give a finite value
Thus now the question is then:
How to show that it is impossible to construct a countable partition of an interval that contain no intervals?
in Mathematics, 6 mins ago, by Secret
Given an non archimedian ring, a finite element $c$ is interfinite if there exists finite $d$ such that $cd=e$ where $e$ is an infinite element
Interfinite elements, the gap between the finite and the infinite
to be analysed
14:58
in Mathematics, 3 mins ago, by Alex
@Secret Dennis and Farb, or Lam
These guys cover noncommutative algebraic structures
Division by zero algebra Theorem 1: Given any algebraic structure $(S,\cdot,+)$ such that $\cdot$ left distribute over $+$ and the underlying set $S$ arbitrary, and that there exists a left or right multiplicative inverse to a left or right additive identity, the addition structure must have the following form:
Proof:
For all $x \in S$
$0+x=x$
Multiply both sides by the zero inverse on the left
$q0+qx=qx \implies 1+qx=qx$
Since $q$ is injective by definition of an inverse $X = qS \subseteq S$. By the above equation, all elements of $X$ dominates 1 on the right.
Now multiply both sides with some $y \in S$
$y1+y(qx)=y(qx)$
The morphisms $y\cdot$ can be injective, surjective, bijective or neither, thus $y(qS) \subseteq qS$
Now, note that the term on the RHS dominates $y$. Repeat the above argument, we end up with a linearly ordered decreasing chain of subsets of $qS$ that is the images after multiplying by $y$, thus the products in the structure $S$ that right dominates some given element under addition, when arranged in descending order of the images, will give the resulting diagram.
In particular, if all $y$ are permutations in $qS$, then the gray area where all products right dominates will be bounded by $f_{\text{bijective in S}}$
and if $q$ is also a permutation, then the offset is empty and thus the addition structure will be a right null semigroup, as required
15:57
-1
Q: Does observation require understanding to collapse the wave function in the double slit experiment?

Don IvieWhat if observation does not include "understanding"? For example, what if a 9 year old boy who knows nothing about QM is the observer? Does the wave function collapse? Further, if the observer has no understanding of exactly what they are observing but can relate their observation to someone who...

16:43
0
Q: Is this specific type of 'Time Travel' possible?

PhysicallyStupidI'm new here and I found out that you can't post duplicate questions, but mine is a bit obsure but I hope it is ok. I was wondering if this type of time travel: http://tvtropes.org/pmwiki/pmwiki.php/Main/GroundhogDayLoop is possible. You see, as this type of time loop (not to be confused with a ...

17:09
0
Q: Basic question from quantum mechanics

user176263Can a particle, having zero kinetic energy but with rest mass $m_0$ ($\sim E_0$) > V (V is potential energy), penetrate the potential wall V ?

 
3 hours later…
19:50
0
Q: How does an electron qubit work?

MathisImagine a qubit, who's physical support is an electron. If its information is the spin, how could we change the spin of the electron? It can have two spins, so I thinks that to change it, you have to make it absorb a photon. Also, and most importantly, how do you measure its spin not only once, ...


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