It doesn't really matter

I mean there are standard dimensions

but none of them are particularly fundamental

I understand, but

I'm trying to argue purely from the basic QM equations why we are already done

if we've specified mass, length, time

however, it seems like I'm missing a formula

if we were to use $E=1/2mv^2$, then it's clear that we can get the dimensions of energy now

and therefore of $\hbar$

a ballot?

I mean

but I didn't want to use that formula

If you want an example on how to relate energy, position and momentum

so which more basic QM formula can we use, to derive the dimensions of either $\hbar$ of $E$

Try $[H, x] = p/m$

@NovaliumCompany a survey?

and $[x,p] = \hbar$

That should be enough to go on

You can't even define the Schrodinger equation unless the units are right from the beginning

@Slereah is it though? because $p$ itself also depends of $\hbar$. So we have two equations, and two unknowns

@JohnRennie no, not survey. In the Justin Bieber case, you are not doing it for research or similar. I think ballot best describes the thing people vote in, where it's main purpose is to show in percentages which one has gotten the most votes.

nvm

it's not ballot

oh, maybe with $\Delta E\Delta t\geq \hbar/2$ it is enough then

I mean the question I really want to know is

What assumptions do you make?

@JohnRennie a form?

I don't know what you consider as obvious as far as dimensions go

at this point, I don't even have a clue what I'm doing tbh

@NovaliumCompany ballot is probably the best word so far.

the reason for that being is, that people always grab random formulas like $E=\hbar f$ to argue things about dimensions

so I never know what is being assumed

@Slereah I guess this is my question as well

@JohnRennie Yeah but ballot is for secret voting. If you display the results publicly?

but I guess working with commutation relations should be a good start?

I don't know if I should include uncertainty principles as well, but I guess that's fine as well

@NovaliumCompany that would be a survey

@Slereah I feel like that wasn't enough:x I wasn't able to derive that the dimensions of $\hbar$ are mass * length^2 * time^-1

@JohnRennie Yeah, but the word survey is for research or to examine something, not for a decision to be taken based on the results (Such as the future state of Bieber's pants)

Uncertainty will give you the same informations as the commutation relations

yea, I realised that as well just now

I would probably just say a vote

though the thing is, I don't know of a commutation relation with $H$ and $t$

Vote on whether Justin Bieber is pants.

so that's why I invoked $\Delta H\Delta t\geq \hbar/2$

Also part of the issue is that you're assuming a representation :p

$p = -i\hbar \partial_x$ is certainly not a fundamental relation

but is it possible to argue using that?

@JohnRennie Yeah, makes sense. Why would form be incorrect?

because I'm fine with not working with the most fundamental formulas

I am just trying to reason from the definition of the hamiltonian and the definition of the operators

is that sufficient?

@NovaliumCompany A form is normally a way of recording information, not part of a vote/ballot/poll/whatever

@ShaVuklia Well, define what you mean by "mass", "length" and "time"

@JohnRennie Alright. Thanks.

well, I'm not really defining things, right? I'm just trying to label them so that I can do dimensional analysis. I was just trying to find an expression for the dimension of $\hbar$ in terms of the dimensions for mass, length and time, without resorting to arbitrary formulas

I feel like that would be a completely different and much deeper topic?:x

@JohnRennie selection!

if I didn't know any classical physics, how would I know the dimensions of $\hbar$ is my question:x (so if I didn't know the formula $E_{kin}=1/2 mv^2$

and $v=dx/dt$

A selection on whether JB pants should be up or down.

Well as I said

$[H, x] = p/m$

Is the quantum version of that

but $p$ is itself in terms of $\hbar$ :x

Well look

in the case of $E_{kin}=1/2mv^2$, you can express energy in terms of you fundamental quantities

It basically means, dimensionally, $[H] [x] = [p] [m]$

If you didn't know any classical physics you couldn't do QM because QM inherently depends on the existence of CM for it to even exist as a theory

If you really want to insist on using the $L^2$ representation, that means

p is not a fundamental quantity:x

Hush

we only have mass, time, length right:x

let me finish

@JohnRennie a selection?

sry:x

This is one of the most fundamental claims of (Copenhagen) QM

$$[\hbar] [T^{-1}] [L] = [\hbar] [L^{-1}] [M]$$

e.g. the notion of measurement depends on a classical world existing

Errr wait

@NovaliumCompany not really.

Of course, that wouldn't mean that QM behavior couldn't exist if we didn't know CM. But it would mean we'd have no idea WTF we're doing

If Schrodinger is an axiom you know $\hbar$ has to have the units of $H$ and $T$ by definition in $i \hbar \partial_t \psi = \hat{H} \psi$

@NovaliumCompany Selecting something doesn't necessarily imply a vote on it.

Ah yes, there we go

@JohnRennie pff.

@bolbteppa we don't take energy as a fundamental quantity

Yeah, it could all be dancing turtles behind the curtain

only time, mass, length for now

Consider the free Schrodinger equation

Without worrying too much about factors, it's

$$\hbar \partial_t = \frac{\hbar^2}{m} \partial^2_x$$

ahh, there's a square:0

If you assume the Schrodinger equation as an axiom, then from $i \hbar \partial_t \psi = \hat{H} \psi$ you know by dimensional analysis that $\hbar$ HAS to have the dimensions $[H][T]$ so the units agree on both sides, whatever the units of $\hat{H}$ actually are is another question

So $$[\hbar] = [T^{-1}] [M] [L^{-2}]$$

okay, that solves it:D

on the note of QM, a long-ass philosophy/history paper on quantum foundations with my name on recently went online

I had completely forgotten that we had $\hbar^2$

@Semiclassical link

@Semiclassical Not hard when your name is semiclassical

omg thanks for the patience

lol

top entry here: philsci-archive.pitt.edu/cgi/latest_days2?n=30

(I link that page b/c I'm inordinately pleased to be at the top of it)

like, really, thanks for your patience:D @Slereah

it should show up on arxiv tomorrow as well

That will be 30 bucks

Nice

171!!!

lel~ if I see you irl one day, I'll treat you on an expensive coffee~

like I said, long-ass

now, I didn't write -all- of that

@JohnRennie I think the word poll works fine enough because when you type online polls in google there are a few websites where you create your own poll on any topic. Wachya think?

my major contribution was sections 3,4 along with some of section 2 (mostly the last subsection)

@NovaliumCompany I think you're worrying unnecessarily about the exact wording :-)

Poll or vote or ballot would probably all do.

@JohnRennie Oks, let it be a poll.

the sections of prose were the work of people who actually know how to write philosophically and historically sensible prose :P

"We use these results to advocate for Bubism (not to be confused with QBism), an interpretation of quantum mechanics along the lines of Bananaworld."

I guess it's bananas all the way down, not turtles :p

Lots of math in philosophy papers apparently!

'was and remains a Bohm sympathizer' :(

Why did the mathematical physicist blow down the singular circles in anti de sitter space?

Alice and Bob in that paper is good to see

Why people interpret Schrodinger's equation differently? It can't be derived doesn't make any sense, then they end up deriving them.

$$ -\dfrac {\hbar^2} {2m} \nabla^2 \psi + V \psi = T \psi $$

@Semiclassical vaguely what is the main idea of bub'ism

@AbhasKumarSinha That $T$ should be $\hat{H}$

@bolbteppa That is $$\hat H \psi = E \psi$$

$\hat H$ is an operator for the sum of KE + PE

$\hat H = - \dfrac{\hbar^2}{em} \nabla^2 + V$

@bolbteppa What is the intuition behind that equation? Why it is said that it can't be derieved whereas there are a lot of derivations on the internet?

there are plenty of ways to derive the schrodinger equation

but of course that all depends on what axioms you choose

@Slereah That makes the thing more confusing...!

Where did we get that from? It's not possible to derive it from anything you know. It came out of the mind of Schrödinger.

I'm not sure I could explain the intuition without advanced math ideas

@bolbteppa oh okay...

how is changing velocity like rotating in 4 dimensional spacetime

Schrodinger came up with it via just basic Hamiltonian mechanics and the idea of matter waves

@Semiclassical this thing looks like a magnum opus

can spacetime rotate?

locally and\or globally?

@Slereah Some people intrepret it as $$ v^2 \nabla^2_x = \nabla^2_t $$ while some intrepret it as $$ KE + PE = TE $$ for KE operator $-\dfrac{\hbar^2}{2m} \nabla^2 $ and using it by multiplying $\psi$ both sides to get the equation.

Why so different interpretations...?

Ah $T$ means Total

@bolbteppa yes, total energy, some also use E, but I prefer T... (just like that).

@Ultradark what do you mean

There is frame dragging

If that is what you mean

yeah i looked it up and saw "frame dragging"

but the universe as a whole can't rotate (godel's model of the universe) right?

@bolbteppa I was talking of this: physics.stackexchange.com/q/102771

might be the best answer

@bolbteppa its an awesome paper, at the very least in the sense of its size

Hopefully it justifies its length

what is more effective at knocking something over? velocity, or acceleration?

@bolbteppa How to show that By looking at level surfaces S=const. we see that particle's trajectory is orthogonal to the level surfaces. what does this mean?

The gradient $\nabla$ of a function $S = \text{constant}$ is a vector orthogonal to that surface

@bolbteppa It means S function is flat everywhere?

This is a multivariable calculus result

oh okay...

ocw.mit.edu/courses/mathematics/…

@bolbteppa The answer makes me think that all classical mechanics are an approximation of waves...

one last question about units. if we consider the hamiltonian $-\hbar^2/2m\partial^2/\partial x^2-e^2/(4\pi\epsilon_0 r)$, why doesn't it follow then that $e$ is a derived unit, instead of a fundamental unit? because I would think that if we specify mass, length and time, then we can derive the units of $\hbar$ and energy, but also of $e$

I would think that the unit of charge is then simply (mass*length^3*time^-2)^1/2

I mean, it works

@ShaVuklia $\varepsilon_0$

ohhh right..

Although to be fair

so... do we specify $\epsilon$ or $\epsilon_0$?

I don't think charge really needs to be its own dimension

I guess charge then

since charges are all discrete

hms

@Slereah I’m busy idk when I can scan it

Next week

@RyanUnger thx

No classes so should have time during the day

I need to find a scanner

This thesis better be worth it