The h Bar

1:00 PM

@0celouvsky Wait, what exactly needs the accent to work? Is the 'R' pronounced differently?

@0celouvsky Slightly less than half Welsh and slightly less than half Northern Irish - the remainder being contributed by miscegenation with the English. However my accent is pure BBC English.

0celouvsky

@ACuriousMind yes

AccidentalFourierTransform

british r is way nicer

John Rennie

I didn't know there were different Rs, though the Scots are inclined to roll theirs.

John Doe

Whoever is of pure noble heart please have a look at my horrible QM post...Should be chicken feed for most of you.

John Rennie

1:04 PM

I was quite surprised the first time that someone told me the Scottish roll their arse.

0celouvsky

@AccidentalFourierTransform too bad the Bernice policy is a thing

ACuriousMind

@0celouvsky What's that?

AccidentalFourierTransform

the be-nice policy I guess

ACuriousMind

Oh.

AccidentalFourierTransform

Yep.

Yashas Samaga

1:24 PM

quora.com/…

AccidentalFourierTransform

1:38 PM

Awkward silence

An awkward silence is an uncomfortable pause in a conversation or presentation. The unpleasant nature of such silences is associated with feelings of anxiety as the participants feel pressure to speak but are unsure of what to say next. In conversation, average pause length varies by language, culture and context. An awkward silence may occur if a pause has exceeded, for instance, a length generally accepted for demarcating a subject change or the end of a turn. It may be preceded by an ill-considered remark or an imbalance in which one of the participants makes minimal responses. Alternatively...

0celouvsky

1:51 PM

@ACuriousMind my wife

Despite your sucking the energy out of me I think I did fine

I over studied the measure theory, it was more Lp heavy. Thankfully I like PDE so it was no problem

AccidentalFourierTransform

that boy needs therapy

0celouvsky

Which boy?

AccidentalFourierTransform

psychosomatic

crazy in the coconut

0celouvsky

Who?

@ACuriousMind can we say people need therapy or is that a medical diagnosis

John Doe

What is the basic protocol for finding the probability density of this type of sequential measurement. I only know the case where we want the probability of some measurement eigenvalue $a$ corresponding to operator $\hat{A}$, we project onto the eigenstate to get $\langle \psi|a\rangle \langle a | \psi \rangle$, the probability amplitude. In this case we start with $| + \rangle$ and end with $| - \rangle$. Also does it follow from the postulates of QM as in the simple case I outlined above?

0celouvsky

1:58 PM

@yuggib so if I get the first three Hormanders, I will use all of them?

@AccidentalFourierTransform is American english simplified vs. British english?

@JohnRennie need your expertise

Yashas Samaga

2:12 PM

What is that kind of determinant called?

I have never seen it before

0celouvsky

lol

where did you see that?

Yashas Samaga

I have seen the formula for area of a convex polygon in another form. It composed of a sum of 2x2 determinants.

@0celouvsky mathwords.com/a/area_convex_polygon.htm

I don't even know how to interpret it in that form.

0celouvsky

I've never seen it before

Secret

@ACuriousMind @yuggib Since the notion of entanglement is basis independent, hence suggesting some geometric property exists for that notion, and that any state in $\mathcal{H}_1 \otimes \mathcal{H}_2$ is a ray in said hilbert space (thus loosely speaking vectors of norm 1). What is this geometric property that distinguish between product states and entangled states?

Yashas Samaga

but I like that notation :D

0celouvsky

2:15 PM

@YashasSamaga it looks like you do diagonal sums going down

left -> right gives a +

Yashas Samaga

Yup.

0celouvsky

right -> left gives a -

no clue what that is supposed to be

John Rennie

@0celouvsky no, American English isn't simplified when compared to UK English.

0celouvsky

it's not something you'd see in actual math

Yashas Samaga

You can write it as a sum of 2x2 determinants taken in order. |[x1 x2][y1 y2]| + |[x2 x3][y2 y3]| + ...

[nums] represent a row

John Rennie

2:16 PM

Though Noah Webster attempted (but largely failed) to simplify American spelling. He is the origin of most of the spelling differences between American and UK written English.

ACuriousMind

@Secret The geometric description is by passing to the projective spaces. Then the entangled states are precisely those not in the image of the Segre embedding $P(H_1)\times P(H_2)\to P(H_1\otimes H_2)$.

Yashas Samaga

How Pi was nearly changed to 3.2 - Numberphile

That's funny.

0celouvsky

@JohnRennie I think I knew that

ACuriousMind

You can't describe entangled states nicely on the level of the vector spaces since entangled states don't form a subvectorspace.

0celouvsky

Algebraic geometry is needed for QM?

I'm triple out

Secret

2:21 PM

Wow, so that means my conception/knowledge that entangled states are vectors with 4 components (since dim (VW)=dim(V)dim(W)=2*2=4) are not entirely accurate

ACuriousMind

There's another geometric description if you are taking the space of all density matrices, i.e. include mixed states:

Then the pure states are extremal points in that space, and the separable states are the (closed) convex hull of the pure separable states, but this doesn't give you a geometric description of the pure separable states as such

0celouvsky

Interesting.

ACuriousMind

@Secret Exactly - that's the catch - for a 2-qubit system, the entangled states are a complex-two-dimensional subvariety of the complex-3-dimensional space of states (which you can see very straightforwardly by comparing how many parameters you need to describe an entangled state compared to a generic state)

Secret

Hmm so that means the usual bra ket way we compute unitary operations on entangled states that "rotate" it to a product state is in the strict sense, not really a rotation of vectors or rays.

I guess I have a lot to *relearn* about the nature of entanglement

Before today, I am always mysterified on how two vectors in a 4 dimensional space (since the basis has dimensions 4) give very different algebraic properties (entangled and products/separable). But I guess the above illuminate why I had the confusion

ACuriousMind

@Secret The points of the projective space are the rays, so yes, the operations do act on rays/vectors. I'm not sure how you got that from what I said.

It's just that the separable states don't form a subvectorspace - so if you thought of them as such a thing, then that's the source of your confusion

(That they aren't a subvectorspace is a very trivial observation that almost never gets spelled out: The tensor product is spanned by the pure states by definition, so if the separable states were a vector space, then there would be no non-separable states)

0celouvsky

2:35 PM

What's the difference between a separable and pure state?

Secret

Yeah, I think the notion of sub vector space (unless you mean subspace in linear algebra) never came to mind to me before. This is because in undergrad QM, we pick a basis to calculate 2 qubit systems, and since under this basis the coordinate vector is just a 4 tuple, this gives a naive picture of basically 2 different vectors in a 4 dimensional space, and then I don't see anything that makes them geometrically different that carries the notion of separable and non separable

ACuriousMind

@0celouvsky I tend to confuse the words too, but they're very different concepts: A pure separable state is a state of a combined system of the form $\lvert \psi_1\rangle\otimes\lvert \psi_2\rangle$. A pure state is a density matrix that can be written as $\rho = \lvert \psi\rangle\langle \psi \lvert$. A separable/non-entangled density matrix of a combined system is one that is a convex linear combination of matrices of the form $\rho_1\otimes \rho_2$.

Secret

Yes, the algebraic definition is straightforward. What I previously have spent a long time on is to translate and understand those into geometric terms

and it does not help at those time I only have that naive picture of two 4 tuples sitting in 4 dimensional space, which then suddenly change from separable to non separable by simply rotating it

0celouvsky

@ACuriousMind I still don't understand the connection between density matrices and state vectors

0celouvsky

2:57 PM

@ACuriousMind hello?

AccidentalFourierTransform

I hate the concept of 1st vs. 2nd quantisation

its just quantisation!

its the frigging same thing

0celouvsky

@AccidentalFourierTransform no

ACuriousMind

@0celouvsky I'm not sure what you mean by that

0celouvsky

@ACuriousMind you said a pure separable state is X, and a pure state is Y

X and Y look completely different to me

AccidentalFourierTransform

you shut up you hear me

0celouvsky

3:10 PM

so how is there a risk of confusion?

they don't even appear be remotely the same thing

ACuriousMind

There's no risk of confusion in the concepts

I just said that I also tend to confuse the words

I've called separable what is pure and vice versa several times

0celouvsky

@ACuriousMind and what is your $\psi$ there anyway

is it in the tensor product space?

ACuriousMind

It's a state vector in your Hilbert space - the concept of purity has nothing to do with combined systems as such, whether the space is composite or not is irrelevant

0celouvsky

@ACuriousMind and I said I don't understand why density matrices are called states

ACuriousMind

@0celouvsky Because they give you a notion of evaluating observables by $\mathrm{Tr}(\rho A)$. Abstractly a state is nothing but something that allows you to take expectation values

0celouvsky

3:13 PM

@ACuriousMind Of course, I know all of that

ACuriousMind

And they are the analogue to classical statistical states.

Emilio Pisanty

@ACuriousMind OP is fighting for your record: physics.stackexchange.com/questions/322211/…

0celouvsky

@ACuriousMind what's that?

ACuriousMind

@0celouvsky A probability distribution, essentially. A density matrix represents classical probabilities $p_i$ to be in the state $\lvert \psi_i\rangle$.

@EmilioPisanty Heh

0celouvsky

@ACuriousMind on phase space?

Emilio Pisanty

3:18 PM

Before re-posting similar questions, please do take the time to read the guidance linked to in the on-hold message. — Emilio Pisanty 1 min ago

that should do it, I think

ACuriousMind

@0celouvsky On whatever your space of states is, but yes, commonly on phase space.

Emilio Pisanty

@0celouvsky because they are an informationally complete representation of the (informal notion of the) 'state' of the system

and plus, all other such complete representations are an absolute mess

0celouvsky

@EmilioPisanty what does that mean?

@BernardoMeurer Let $d$ be the Dirichlet function. Let $x\in\Bbb R$. Since both $\Bbb Q$ and $\Bbb I$ are dense in $\Bbb R$, you can find sequences $(q_n)\subset\Bbb Q$ and $(i_n)\subset\Bbb I$ with $\lim q_n=\lim i_n=x$. But $\lim d(q_n)=\lim 0=0$ and $\lim d(i_n)=\lim 1=1$, so $d$ is not continuous at $x$. This works for all $x$, hence it is nowhere continuous.

Bernardo Meurer

Ah

Interesting

John Doe

@EmilioPisanty How did you derive $p$ in your answer?

Bernardo Meurer

3:29 PM

But @0celouvsky How does this not contradict the Lebesgue theorem we proved?

0celouvsky

@BernardoMeurer On the other hand, this doesn't mean anything for the Lebesgue integral.

Bernardo Meurer

Ah

Nevermind

Got it

0celouvsky

It doesn't, we know that it's not integrable because of this.

Bernardo Meurer

Yeah, exactly

0celouvsky

For the Lebesgue integral you get $\int_a^b d(x)\,dx=b-a$.

Bernardo Meurer

3:31 PM

But since $\Bbb Q$ has measure 0 the integral will be over the measure of $\Bbb R\setminus\Bbb Q$

For Lebesgue

0celouvsky

Exactly, and on that set it's just $1$.

Bernardo Meurer

Yep

Cool!

0celouvsky

But you need a lot more machinery to show that, even though it seems fairly intuitive.

Bernardo Meurer

Thomae's function is so fucking weird

0celouvsky

Yeah I have no clue what that one is about.

@BernardoMeurer Do you know Lebesgue's singular function?

It is perhaps the strangest of functions.

Bernardo Meurer

3:35 PM

Nope, show me

0celouvsky

@BernardoMeurer $\psi:[0,1]\to[0,1]$ with $\psi(0)=0$, $\psi(1)=1$. $\psi$ is continuous, strictly increasing, and has $\psi'(x)\ne 0$ only on a set of measure zero.

That is, $\psi'(x)=0$ almost everywhere.

It violates the (naive) fundamental theorem of calculus even when written with Lebesgue integrals.

Bernardo Meurer

What the fuck

What's the naïve FTC?

0celouvsky

It's what led the old masters to define a stronger form of continuity for use in measure theory/functional analysis.

@BernardoMeurer $f(x)=\int_0^x f'(t)\, dt+f(0)$.

Bernardo Meurer

Ah

That is truly a weird function

0celouvsky

@BernardoMeurer Turns out the fundamental theorem of calculus is only true if $f$ is "absolutely continuous."

All $C^1$ functions are locally $\mathrm{AC}$, so it's fine most of the time.

Bernardo Meurer

3:41 PM

@0celouvsky Hm, that seems reasonable. How do you define absolutely continuous? Continuous everywhere?

0celouvsky

@BernardoMeurer $f\in \mathrm{AC}[a,b]$ means: $(\forall\epsilon>0)(\exists\delta>0)$ such that for $[a_1,b_1],\dotsc,[a_N,b_N]\subset [a,b]$, $\sum_{i=1}^N|b_i-a_i|<\delta\implies \sum_{i=1}^N|f(b_i)-f(a_i)|<\epsilon$.

Bernardo Meurer

Hm, nice

0celouvsky

@BernardoMeurer basically... the function is continuous and doesn't oscillate too much

that's in the realm of "not even wrong" because I don't have a better explanation

Bernardo Meurer

That's a pretty restrictive requirement :/

0celouvsky

It's not oscillations, sorry. That's $\mathrm{BV}[a,b]$.

"bounded variation"

BV also has a nice measure theoretic definition in terms of distributional derivatives

Bernardo Meurer

3:46 PM

I dunno measure theory

0celouvsky

@BernardoMeurer Aha

Bernardo Meurer

and, surprise, it's not on my course :P

0celouvsky

A function is absolutely continuous if and only if it is continuous and maps sets of measure zero to sets of measure zero.

That's a better definition.

@BernardoMeurer lame-o

Emilio Pisanty

@JohnDoe I don't

that's what the question asks for

0celouvsky

savage

Emilio Pisanty

3:48 PM

if you want a sharper answer, ask a sharper question

Bernardo Meurer

@0celouvsky That is a nice definition, yeah

0celouvsky

@BernardoMeurer of course, that $m(f(A))\ne 0$ for $m(A)=0$ is even possible is perhaps surprising

Bernardo Meurer

It is weird, to say the least

0celouvsky

measure zero is strange

in a sense, sets of measure zero can be very large :D

Bernardo Meurer

Measure 0 is a really weird property

Emilio Pisanty

3:50 PM

@0celouvsky not true

Bernardo Meurer

it's counter intuitive that an infinite set can have zero measure

0celouvsky

@EmilioPisanty what?

Emilio Pisanty

@0celouvsky by any reasonable definition of "large", measure-zero objects cannot be large

anyways, here's one for y'all

> SuitSat-1 inside the ISS

0celouvsky

I'm not going to get into an argument here because it was just a comment.

Bernardo Meurer

@EmilioPisanty Large element count

0celouvsky

3:52 PM

But there are topological notions of size, such as Baire category.

Emilio Pisanty

@0celouvsky you trigger way too easy

0celouvsky

There are sets which are large in the sense of Baire but have measure zero.

@EmilioPisanty No

Yes.

be quiet

No.

3:54 PM

> Suitsat-1 in orbit after being deployed from ISS.

0celouvsky

@skillpatrol yes

John Doe

@EmilioPisanty In the simple case where we want to find the probability of say measuring some eigenvalue $a$ corresponding to some operator
$\hat{A}$, we then project onto the eigenstate to get
$\langle \psi|
a
\rangle
\langle a|
\psi \rangle$ which is the probability amplitude of the measurement yielding $a$, so this follows from one of the postulates of QM. In this case I guess we want to find the probability of the measurmement $\hat{S}$ yielding $\frac{\hbar}{2}$ after a sequence of measurements. Hence I assume that we must adapt the simple case that I gave. Am I right?

Emilio Pisanty

@JohnDoe thus far, yes

0celouvsky

@BernardoMeurer By taking a union of cantor sets + $\Bbb Q$, you can get an unbounded, uncountable, dense subset of $\Bbb R$ with measure zero

Bernardo Meurer

Wat

What the fuck

How can an uncountable unbounded set have measure zero??

And it's dense too wtf

Wat

0celouvsky

3:56 PM

@BernardoMeurer magic lol

Emilio Pisanty

@0celouvsky that's pretty cute

0celouvsky

Cantor set is the ultimate fuckery

Emilio Pisanty

@BernardoMeurer easy, the union of the Cantor set and the rationals

oh, wait

@0celouvsky what exactly do you mean by Cantor sets?

0celouvsky

I want to translate Cantor sets around to all intervals $[k,k+1],k\in\Bbb Z$

then it has the property that any intersection with intervals is uncountable

Emilio Pisanty

@0celouvsky Why not just $C\cup \mathbb Q$?

0celouvsky

3:58 PM

so it's "nowhere countable" in a sense

@EmilioPisanty Call my set $K$. Then $K\cap [a,b]$ is uncountable for any $a,b$.

Emilio Pisanty

@0celouvsky if $k\in\mathbb Z$, then no

0celouvsky

It's so uncountable that no reasonable subset is countable.

@EmilioPisanty Cantor set is uncountable.

Emilio Pisanty

@0celouvsky $C$ has a big gaping hole in the middle

i.e. its intersection with $(1/2-\delta,1/2+\delta)$ is empty

John Doe

.

Emilio Pisanty

where $\delta$ is 1/12 or something

0celouvsky

4:00 PM

@EmilioPisanty Ah. Take more Cantor sets then.

Emilio Pisanty

@0celouvsky then yes

0celouvsky

Union cantor sets so that there are no holes.

That's my idea.

Emilio Pisanty

so you mean $C+\mathbb Q$ or something

0celouvsky

Exactly.

That has measure zero too.

John Doe

Let me rethink that question.

Emilio Pisanty

4:02 PM

@JohnDoe frankly, you need to push a bit more on this one

you're kinda asking to be spoon-fed a solution, when it just boils down to straightforward applications of principles you already know

0celouvsky

That should contain $\Bbb Q$

John Doe

@EmilioPisanty Okay I will try to push a bit more.

0celouvsky

@BernardoMeurer Yep, $C+\Bbb Q$ is dense, uncountable, and every $(C+\Bbb Q)\cap(a,b)$ is uncountable. But, it has measure zero.

Emilio Pisanty

@0celouvsky Here's a challenge for you

> A set $A$ for which there exists no function having $A$ as its set of points of discontinuity

Bernardo Meurer

What does $+$ mean when talking about sets?

Emilio Pisanty

4:05 PM

@BernardoMeurer $A+B=\{a+b:a\in A,b\in B\}$

ooooh, here's another nice one

Bernardo Meurer

Intuitive, cool

Emilio Pisanty

> Two disjoint nonempty nowhere dense sets of real numbers such that every point of each set is a limit point of the other

0celouvsky

Irrational numbers @EmilioPisanty

Emilio Pisanty

@0celouvsky that the first one or the second one?

0celouvsky

First

skill patrol

4:08 PM

in Mathematics, 41 mins ago, by Semiclassical

But then again, anything which has to address the Cantor set is going to be weird :)

Emilio Pisanty

@0celouvsky ok, yeah, fair enough

I'm assuming you know

> A continuous monotonic function with a vanishing derivative almost everywhere.

0celouvsky

Yeah, although I admit I haven't gone through the proof in the discontinuous case.

@EmilioPisanty oh Christ

I don't like analysis trivia enough for that one :D

Emilio Pisanty

oooooh, nice

> A function defined on $\mathbb R$, equal to zero almost everywhere and whose range on every nonempty open interval is $\mathbb R$.

man, that book is a jewel

@0celouvsky what about strictly monotonic, though?

0celouvsky

@EmilioPisanty we did it in class for continuous ones, he put the general case in the notes but I didn't check

Emilio Pisanty

@0celouvsky wait, why do you think that discontinuous is harder than continuous?

"A continuous monotonic function with a vanishing derivative almost everywhere" is just the devil's staircase

0celouvsky

4:14 PM

@EmilioPisanty I told Bernardo about that an hour ago

Emilio Pisanty

@0celouvsky so now make it strictly monotonic

0celouvsky

@EmilioPisanty one of the intermediate steps strongly requires continuity

There might be an easier proof that doesn't use it

Emilio Pisanty

@0celouvsky what?

what is it you're proving?

0celouvsky

@EmilioPisanty I'm confused, what am I supposed to be doing.

Emilio Pisanty

this is a counterexample

0celouvsky

4:15 PM

We're having very different conversations I'm afraid.

Emilio Pisanty

i.e. provide a function with those properties

John Doe

@EmilioPisanty I just want to ensure my basic idea (which is supposed to be an anology to the situation in the post) is correct:
So say I start with state $| \psi \rangle$ and want to find the probability of measuring $a'$ (corresponding to operator $\hat{A}$) after measuring $b'$ (corresponding to some operator $\hat{B}$) then would the probability be $\langle b' | a' \rangle \langle a' | b' \rangle\langle \psi| b' \rangle \langle b' | \psi \rangle$?

Emilio Pisanty

@JohnDoe yes, but never ever ever ever order it like that

John Doe

@EmilioPisanty What's wrong with it?

Emilio Pisanty

$$ \langle \psi| b' \rangle \langle b'|a' \rangle \langle a' | b' \rangle \langle b'|\psi\rangle$$

@JohnDoe it is completely opaque, which makes it much harder to analyze, and much easier for you to make a mistake when you're manipulating it

Anyways @0celouvsky, this is the book

amazon.com/Counterexamples-Analysis-Dover-Books-Mathematics/dp/…

free copy here

0celouvsky

4:25 PM

I have it

Emilio Pisanty

@0celouvsky figures

0celouvsky

@EmilioPisanty is there a "counterexamples in quantum mechanics" book out there :D

Emilio Pisanty

@0celouvsky as in?

0celouvsky

Hamiltonian with continuous spectra, empty point spectrum, weird perturbation series

That kind of stuff

Emilio Pisanty

@0celouvsky oh, that's boring

that's not even QM

that's just functional analysis

I mean, interesting enough in a functional-analysis sense

0celouvsky

4:29 PM

@BernardoMeurer A+B means you translate A by every element of B and union them up

Emilio Pisanty

but there's nothing specific to QM about any of those

0celouvsky

@EmilioPisanty speaking of FA, any infinite dimensional Banach space has only uncountable bases.

Emilio Pisanty

A book that had all the uncomfortable experimental truths why your cousin's fantastic new interpretation of QM is necessarily bullshit, though, now that'd be interesting

2

0celouvsky

it's pretty strange

and in an infinite dimensional Hilbert space, an orthonormal (Hilbert) basis is never a basis

Emilio Pisanty

@0celouvsky what do you mean by basis there?

0celouvsky

4:31 PM

@EmilioPisanty Hamel basis

@EmilioPisanty Do you have such a cousin :P

Emilio Pisanty

@0celouvsky Hamel bases are (ultimately) boring

ACuriousMind

@EmilioPisanty I'd buy that one.

2

Emilio Pisanty

@0celouvsky ^ proof

0celouvsky

ACM is your annoying cousin?

Emilio Pisanty

@0celouvsky ring me up when you can construct one, though

@0celouvsky blast, we've been discovered

0celouvsky

4:33 PM

@EmilioPisanty they are useful in certain proofs

Emilio Pisanty

no, thankfully my family tends not to say stupid shit like that

Bernardo Meurer

My mom thinks QM is about the power of positive thought

Emilio Pisanty

@0celouvsky those proofs are likely equivalent to Banach-Tarsky

0celouvsky

but I don't accept C so you're right

they're worthless

@EmilioPisanty are you a fellow C denier?

Emilio Pisanty

@0celouvsky I'm on the fence on C

Bernardo Meurer

4:34 PM

Some people don't believe in C, and that's okay

Emilio Pisanty

thus far I'm undecided about C

Bernardo Meurer

I go a step further, I think ZFC is a lie

Emilio Pisanty

Banach-Tarsky is bizarre

but then again so is "$\exists$ a nonzero vector space whose dual is zero"

0celouvsky

even with C there are topological vector spaces with trivial strong duals, btw

@EmilioPisanty do you know what the canonical example for that is?

Emilio Pisanty

@0celouvsky by trivial strong duals, you mean there's no nonzero continuous linear functionals?

that's fine by me

continuity is a strong condition

0celouvsky

4:37 PM

I think that's one of those things (like Banach Tarsky) that we all "know" but have never cared to check the proof

John Doe

@EmilioPisanty Following from my simple case analogy then I would write: $$\langle \psi|+ \rangle \langle \hat{n}; +| - \rangle \langle-| \hat{n} ;+ \rangle \langle+ | \hat{n};+ \rangle \langle \hat{n}; + \rangle \langle + | \psi \rangle$$ would be the probability of measuring $s_z = \frac{\hbar}{2}$ starting with state $| \psi \rangle$.

0celouvsky

@EmilioPisanty I know it's fine, I'm just giving you an FYI

Emilio Pisanty

@0celouvsky there's an MO question about that, let me look for it

@JohnDoe mind your notation there

0celouvsky

I'm 99% sure every locally convex space has at least one nontrivial continuous functional

every normed vector space does

Emilio Pisanty

there we go

37

Q: Is the non-triviality of the algebraic dual of an infinite-dimensional vector space equivalent to the axiom of choice?

If $V$ is given to be a vector space that is not finite-dimensional, it doesn't seem to be possible to exhibit an explicit non-zero linear functional on $V$ without further information about $V$. The existence of a non-zero linear functional can be shown by taking a basis of $V$ and specifying th...

0celouvsky

4:38 PM

aha

I've seen that MO post about 10 times

but I can't read that proof

Bernardo Meurer

@0celouvsky What's a dual?

0celouvsky

if you can...I'm very impressed

Emilio Pisanty

@0celouvsky I can't

0celouvsky

@BernardoMeurer the dual of a vector space $V$ is the set of all linear maps $V\to\Bbb F$

Emilio Pisanty

anything technical on AC is beyond me

John Doe

4:39 PM

@EmilioPisanty $$\langle \psi|+ \rangle \langle \hat{n}; +| - \rangle \langle-| \hat{n} ;+ \rangle \langle+ | \hat{n};+ \rangle \langle \hat{n}; +| + \rangle \langle + | \psi \rangle$$

0celouvsky

if $V$ has a topology we can require these to be continuous, and get a (perhaps) smaller continuous dual

ACuriousMind

In the end, denying choice means that you claim that there are surjections of sets that don't have right inverses, which is a statement one should perhaps dwell on before going to all the paradoxes.

Emilio Pisanty

@JohnDoe I'm going to refuse to read any such expressions henceforth unless you order them correctly

0celouvsky

and in many cases you want to topologize the dual, which gives the strong dual, wk* dual, and more

Emilio Pisanty

really

0celouvsky

4:40 PM

@ACuriousMind I don't claim to understand infinite sets

and I don't think you should either

Bernardo Meurer

What was the first SE site after SO?

0celouvsky

As Dr. Duffield famously said, "there ain't no infinite cardinalities in the night sky"

Emilio Pisanty

@BernardoMeurer SU?

possibly SF

0celouvsky

Why not U

Surely people thought of U before SU

Bernardo Meurer

SF?

Super User

0celouvsky

4:41 PM

or maybe SL

Emilio Pisanty

@0celouvsky ain't no Dr. behind Duffield, I should think

@BernardoMeurer serverfault

Bernardo Meurer

Ah

@EmilioPisanty I believe you are correct

He's an EE IIRC

0celouvsky

@EmilioPisanty honorary degree from the Ocelot University

John Doe

@EmilioPisanty I'm not ordering incorrectly on purpose. I'm still figuring the subleties like ordering out. As long as the idea is right for now.

Bernardo Meurer

lol

Emilio Pisanty

4:43 PM

@JohnDoe As I said before, if you put them out of order they become much harder to read

ACuriousMind

@0celouvsky What I mean, is, think about how these statements can fail in other contexts: E.g. not all ring epimorphisms have right inverses, but that's because the naive inverse fail to be a homomorphism. That is, the failure of an epimorphism to split should be related to some structure on the objects you're considering. But a set has no structure - what is the obstruction for the inverse to exist?

0celouvsky

@ACuriousMind can you write this in symbols

Emilio Pisanty

I can look them over but I'm not going to decode stuff that's been incorrectly put together

0celouvsky

I know what you're saying but I can't see why it's bad

ACuriousMind

The obstruction in ZF is simply that it doesn't allow you to "construct" that function from the axioms, but that strikes me as an inherently silly obstruction. I'll admit that's a purely subjective argument, but it's really not the notion of "set" that obstructs us here, but the particular axiomatization

Emilio Pisanty

4:44 PM

-6

Q: Modern Formalism

Modern formalism Question on Bloch Vector, Hermiltian matrix

quantum-mechanics homework-and-exercises

oh, hell, no

0celouvsky

I know what you're saying, Bajoran

But to me ZF formalizes what I can write on paper

I can't write an infinite set

or a counting function

@EmilioPisanty don't gold badge it

Emilio Pisanty

how the hell does something like that get an upvote?

@0celouvsky that's... a weird understanding of how goldbadging works

you can't dupehammer stuff that's not a duplicate

0celouvsky

oh, it's just a dupehammer?

til

Bernardo Meurer

Turing machines will replace set theory

0celouvsky

@ACuriousMind you should add turing machines to your starred quote

Emilio Pisanty

4:47 PM

@0celouvsky dupehammer plus editing the list of duplicates

Bernardo Meurer

A question that gives insight on the creepy stuff I develop sometimes

10

Q: Filtering a list of citizens

I have a filter method that receives a list composed of dictionaries. You must discard dictionaries that have their handle or their citizen_number elements off the filtered result as well as concatenate any lists inside that dictionary. In addition to that, in case an element on that main list is...