The h Bar

Dawood ibn Kareem

8:00 AM

You'd have to ask one.

Kenshin

ok thanks

Who can now help me with some QM quesitons?

Dawood ibn Kareem

No seriously. You may think I'm being flippant. But if we ever make an AI machine that's conscious, how will we ever know? It might tell us it's conscious, but why would we believe it?

Kenshin

@DawoodibnKareem, but is consciousness merely saying that oneself is conscious?

The machine could be lying when it says it is conscious couldn't it be?

Dawood ibn Kareem

That's what I said.

Kenshin

Yes, so asking the machine as you initially suggested wouldn't really give us any new info about whether it is conscious or not

Dawood ibn Kareem

8:06 AM

So, again, how can we possibly know whether the machine is conscious? No other being can reliably tell us. And the machine itself can't be trusted.

Kenshin

So conciousness is a phenomenon that is real, but can never be proven to be so

reminds me of Godel's work a bit

Ok My next question is,

Dawood ibn Kareem

Sure. Theoretically, you can never be completely convinced of the consciousness of any being other than yourself.

Kenshin

@DawoodibnKareem, unless we one day figure out what causes conciousness, and maybe then we will have a means of measuring it

ok the next question is on space and time:

Dawood ibn Kareem

The philosophy called "solipsism" holds that the only conscious being in the universe is Kenshin, and the entire universe exists for Kenshin's benefit.

Kenshin

If there are no objects, does "distance" still exist?

Anonymous

8:11 AM

@Kenshin You can never figure that out unless there is an universally agreed upon definition of consciousnesses. As of now, there isn't any.

Kenshin

or is distance just a way of seeing how far objects are away relative to the sizes of other objects?

@Blue, as of now correct. But maybe in the future we will have a model of conciousness (which includes a proposed definition) that is consistent with the subjective phenomemon

Dawood ibn Kareem

Does time elapse in your hypothetical objectless universe?

Anonymous

@Kenshin Possible

Balarka Sen

Is this metaphilosophy going anywhere

If not I am not participating

Kenshin

My next question on distnace is going somewhere

the question is, does distance (or time for that matter) exist independently of the substances contained in the universe

Anonymous

8:13 AM

@BalarkaSen Metaphilosophy never goes anywhere ;)

Anonymous

It's just an alternative to weed

Balarka Sen

lmao

Dawood ibn Kareem

@Blue Sure it goes somewhere. There's just no distance to measure it with.

Kenshin

If you had a universe with no rulers or anything that could be used as a ruler, does space stille xist?

Dawood ibn Kareem

I suspect that an objectless universe probably violates the Uncertainty Principle.

Kenshin

8:15 AM

Ok I'll tell you where I'm going with this

Balarka Sen

@Kenshin Well, depending upon if you believe in the abstract concept of position, which is a place where an object may, but need not necessarily, stay.

Kenshin

In QM, we postulate the existance of a space and time coordinate system

from which we place the wavefunction within this

but if all objects are wavefunctions

then is it correct to even postulate that such a space and time coordinate system can exist itself, in which we can place the wavefunction?

Anonymous

@DawoodibnKareem That assumes that QM holds true in an objectless universe.

Dawood ibn Kareem

Oh, so now it's an objectless classical universe? Just how many laws of physics are we changing here?

John Rennie

@Kenshin yes, there is a perfectly good solution to the Einstein equations for a universe containing nothing.

Balarka Sen

8:16 AM

It's a perfectly consistent mathematical formulation.

Anonymous

@DawoodibnKareem It may be neither

David Z

@user626528 I think @DawoodibnKareem did a pretty good job of explaining to you why the question you asked isn't on topic.

To turn it into one that is on topic, you should make some progress on the problem yourself until you get stuck. Make sure to consult textbooks, websites, or whatever standard resources you have available to you. When you find yourself unable to figure out some detail of the solution and those resources don't help, then you ask here - but make sure to ask about the detail you can't figure out, don't just ask how to solve the problem.

Kenshin

@JohnRennie, yes that is non-quantum

Balarka Sen

This is pointless trash philosophy. I'm outta here.

Kenshin

@JohnRennie, I'm wondering in the quantum case, if all objects are wavefunctions, is it a bit contradictory to assume there is a well defined coordinate system (with known position and velocities) in which we can place these wavefunctions.

such a postulate @JohnRennie, kind of assumes that space and time in the quantum world exist independently of the particles it is trying to model yes?

John Rennie

8:18 AM

@Kenshin in GR coordinate systems are purely human inventions. The existence or otherwise of particular coordinate systems is irrelevant.

Dawood ibn Kareem

We can always measure things in Planck lengths, right? Even if there's nothing to measure?

Anonymous

@DawoodibnKareem That depends on the fundamental constants...

Dawood ibn Kareem

Frankly, if I were in an objectless universe, I'd be more worried about where my next meal's coming from, than in measuring its length.

Kenshin

@JohnRennie, when modelling the electron around the nucleus, we might take the nucleus to be well defined position and momentum, from which we derive the wavefunction of the electron? Is it correct to assume the nucleas can have such a well defined position and momentum?

Anonymous

@DawoodibnKareem You wouldn't have to. Because you wouldn't exist there. ;)

Dawood ibn Kareem

8:20 AM

@Blue Yes, but the fundamental constants must have values, even if they're different from the ones in this universe. So there'll always be a Planck length, even if it's an entirely different Planck from our one.

Anonymous

@DawoodibnKareem I don't have enough knowledge to validate that statement...

Dawood ibn Kareem

It would probably be a good deal smaller. I think the Planck in our universe would be the Max!

Anonymous

I'd like to see any source or references you have regarding that. I don't know if it's even necessary for an universe to have fundamental constants.

Dawood ibn Kareem

You'd prefer fundamental variables?

Anonymous

I don't prefer anything. I'd just like to see the relevant papers/calculations

Dawood ibn Kareem

8:26 AM

@Kenshin I don't think the nucleus can have a well defined position and momentum. Not at the same time.

Kenshin

@DawoodibnKareem, so how do we choose a frame of reference when modelling an atom in QM?

Dawood ibn Kareem

Do we have to?

Balarka Sen

It's easy to choose a frame of reference. There's just no canonical one.

Kenshin

well wavefunctions are always specified relative to some frame of reference

Balarka Sen

Mark off some point and call it the origin. Done.

Kenshin

8:28 AM

but how can i be sure of where "some point" is?

Anonymous

...

Kenshin

because if i use a physical object (e..g cornner of a lab) to be that point, and that point changes

Anonymous

@BalarkaSen Time to get back to weed

Balarka Sen

You're in a Euclidean space.

Kenshin

in classical physics, we can use physical objects to be our markers

Balarka Sen

8:28 AM

Any position can be marked off

I don't care about physics

Kenshin

i see that

you care about the maths

Balarka Sen

This is just purely a mathematical point.

Yes.

Kenshin

I want to know where the wavefunction is physically located

therefore I need a physical reference point for my coordinate system

but all physical points are subject to the uncertainty principle

Dawood ibn Kareem

You might be out of luck with that physical reference point thing.

Kenshin

so I can never know wher ethe wavefunction really is?

Dawood ibn Kareem

8:30 AM

What do you mean by "really"?

Kenshin

So not only is a particles position uncertain (due to the wavefunction), the location of the wavefunction itself is also uncertain since there is no physical coordinate system

@DawoodibnKareem, as in, I want to know in my laboratory where the wavefunction is located in physical space

@DawoodibnKareem, so to do that, I'll need some coordinate system to represent my lab

Balarka Sen

The wavefunction is not a physical object.

Kenshin

@BalarkaSen, that's debatable

Anonymous

Gosh, no. It's totally certain. Pick any reference frame and calculate the probability amplitudes w.r.t that frame.

Anonymous

I don't know what you're on about

John Rennie

8:32 AM

There's nothing uncertain about the wavefunction. If you choose some coordinates $\mathbf r$ then you can write $\psi(\mathbf r)$ with no uncertainty. The uncertainty comes in only when we try and compute observables.

Kenshin

@Blue, the trouble is, how do I map the reference frame to my lab, if the lab's position is not defined

Anonymous

......................

Kenshin

@JohnRennie, yes relative to a reference frame, the wavefunction isn't uncertain I agree with you there

Balarka Sen

@JohnRennie He wants to chose a specific particle as the origin in his reference frame. It's a totally trash endeavor

Kenshin

but @JohnRennie, the reference frame itself however is mathematical, not physical

To map the reference frame to physical space is difficult, since all objects in physical space have uncertainty

Slereah

8:34 AM

mornin

Kenshin

@BalarkaSen, what's the point of a reference frame if i can't map it to physical space tho

Balarka Sen

@Kenshin What the fuck does that even mean

Kenshin

well to give a classical analogy

I could say let the corner of my desk be the origin of the coordinate system

I can then do calcs using this reference frame, and ultimately any answer to the calc I know is relative to the corner of my desk

I have a mapping from my mathematical reference frame to physical space

John Rennie

@BalarkaSen well that's the Born Oppenheimer approximation. A perfectly reasonable thing to do, but still only an approximation.

Kenshin

But in QM, since there is no way to set the reference frame to say a particular point in physical space, I can still do the maths in my reference frame, but I cant map the results back to the physical world

Anonymous

8:36 AM

@Kenshin You can. Approximately. And even mathematicians are happy doing that kind of practical approximations.

Balarka Sen

@Kenshin Yes, similarly you model the origin as the nucleus and compute the corresponding differential equation satisfied by the wavefunction of the atom or whatever. There are two justifications for this (1) you completely forget the uncertainty in choosing the origin as the position of the nucleus, that has an infinitesimal contribution to the calculations (2) moreover, the uncertainty is not present unless you choose a coordinate system beforehand

Kenshin

ok thanks guys, I will have to look more into the Born-Oppenheimer approximation I think

@BalarkaSen, what do you mean by "the uncertainty is not present unless you choose a coordinate system beforehand"?

Ok next question:

Balarka Sen

The "position" and "momentum" of a particle are undefined quantities before choosing a coordinate system

So the uncertainty principle is nonexistent.

Kenshin

If I have a wavefunction/state and I measure it's position

Then immediately after I measure the position again

will the second position measurement match the first?

Dawood ibn Kareem

What do you mean by the position of a wavefunction?

Kenshin

8:41 AM

Sorry for all the questions guys but since Hawking is gone now someone has to take his place

Dawood ibn Kareem

So you used to ask Hawking all these questions?

Kenshin

@DawoodibnKareem as in you measure the position of the "particle"

@DawoodibnKareem lol no, I meant the other interpretation

Dawood ibn Kareem

So you collapse the wave function. Then after a delay you do it again. Now why would you expect the position to come out the same?

Balarka Sen

The wavefunction has nothing to do with position of a particle. What you said makes 0 sense. The amplitude squared of the wavefunction at a point in your coordinate system gives the probability that the particle takes that position.

Kenshin

I gotta learn the physics to take his palce

Balarka Sen

8:43 AM

The particle is modeled as a random variable.

Kenshin

@BalarkaSen the wavefunction represents all knowable info about the particle, so yes the wavefunction contains position info

Balarka Sen

Read what I said again.

Kenshin

Yeah I wasn't sure why you thought was I said made 0 sense, because I understand what you said, and it doesn't contradict my quesiton

basically you have a wavefunction representing a particle, you then measure the particles position, then you do it again, will the two position measurements match?

John Rennie

No

Balarka Sen

The "position of a particle" is not a well-defined quantity. There is no specific position

There is only probability associated to each position

John Rennie

8:45 AM

You get the position from the wavefunction by applying the position operator.

Kenshin

@BalarkaSen that doesn't prevent measuring position, it merely causes the wavefunction to collapse

@JohnRennie, I thought you get position by measuring it, then the wavefunction collapses to a position?

Dawood ibn Kareem

@Kenshin You seem to be assuming that the particle's momentum is zero. That seems to be an odd thing to assume while you're measuring its position.

John Rennie

But the position operator has no eigenfunctions, so applying the position operator always changes the wavefunction.

Kenshin

I know

You measure the position, it causes the wavefunction to change

then you measure the position again

so I'm asking in that second measurement, how does it compare to the first?

John Rennie

@Kenshin the position is given by $$ x = \hat{x}|\psi\rangle $$

Kenshin

8:47 AM

yeah

John Rennie

But since $\hat{x}$ has no eigenfunctions the operator always changes $\psi$

Dawood ibn Kareem

@Kenshin Why would you expect it to be the same? If your answer is "because the particle's not moving", then you have thrown out the uncertainty principle.

Kenshin

@DawoodibnKareem I never said I expected to be the same

I'm just asking if it would be the same

Dawood ibn Kareem

@Kenshin OK, rephrasing that - if it would be the same, in every case, then the Uncertainty Principle would be contradicted.

Therefore, you should expect it NOT to be the same.

Don't need any fancy Greek letters or bras and kets to work that out!

Kenshin

let me quote a QM textbook called griffiths for yo uguys

"What if I made a second measurement, immediately after the first? Would I get C again or does the act of measurement cough up some completely new number each time"

then it goes on to say

"On this question everyone is in agreement: A repeated measurement (on the same particle) must return the same value."

"Indeed, it would be tough to prove that the particle was really found at C in the first instance, if this could not be confirmed by immediate repetition of the measurement"

@DawoodibnKareem I agree with your reasoning, that is why I was confused by griffith's remakrs above

John Rennie

8:50 AM

When yu measure $C$ the observation collapses the wavefunction to an eigenfunction of $\hat{C}$. So reapplying $\hat{C}$ returns the same eigenvalue.

Balarka Sen

@Kenshin You're not measuring position of the particle. You're measuring if the particle is on that specific position: that's the content of the position operator as John said.

John Rennie

But there are no eigenfunctions of $\hat{x}$ because that woud require infinite energy.

Kenshin

@JohnRennie, but shoudln't the particle have uncertain momentum, thus have moved between the first and second measurement?

Anonymous

@Kenshin You're misinterpreting HUP....

Kenshin

ok please explain then

John Rennie

8:52 AM

However a second application of $\hat{x}$ will return a very similar value to the first application, if you make the the second measurement quickly enough.

Kenshin

if the position never changes between measurement, that to me seems like momentum is 0?

Dawood ibn Kareem

Hang on - is the second measurement "where is the particle now", or is it "where was the particle when we made the first measurement"?

Kenshin

so that menas both position and momentum seem well defined? where is my reasoning error?

Balarka Sen

@Kenshin Position is not a well-defined quantity. It doesn't change.

I am not sure what "position changing" even means.

Anonymous

Position and momentum are well-defined. Just not simultaneously.

Kenshin

8:53 AM

@BalarkaSen, well the griffith's textbook implies that you can measure position to be at C (or at least in the near vicinity)

Anonymous

Also I don't recommend Griffiths for learning QM

Balarka Sen

@Kenshin Well, that means you can check if that particle is on a position P. That's restricting the wavefunction to P.

Kenshin

@Blue, but griffiths says if you measure the position two times in a row you get the same answer. Doesn't that mean you (1) know the position and (2) can deduce velocity to be0 thus also know momentum simultanously?

Balarka Sen

Because the point P is well defined in your coordinates

The particle is just a random variable which takes value P with some probability

Kenshin

seems to be some variance of opinions, maybe I will post this to the main site and see what the consensus is

Bernardo Meurer

8:55 AM

Don't bother

QM is a lie

It's a made up theory to trick students

Kenshin

fark i suspected so

Bernardo Meurer

In reality @JohnRennie Is orchestrating the movements of all the particles in the universe

Secret

what have a i missed reads transcript

Anonymous

@Kenshin (1) Yes (2) No

Kenshin

why (2) is no?

the particle hasn't moved over some time period

so that means the velocity was 0, or could velocity still be non 0?

Slereah

9:01 AM

You can take two positions and then try to deduce the velocity $(x_2 - x_1) / \Delta t$, but that will not really give you any relevant information on the instantaneous velocity

Kenshin

@Slereah, so the particle may have jiggled back and forth, even in the absense of external forces?

Slereah

Yes.

Kenshin

I thought it would have travelled at a constant velocity unless acted upon by an external force

Anonymous

You know the position at a certain instant. If you measure at another instant (very quickly as JR mentioned), you'll get almost the same value. You can say that the average velocity is $\approx 0$. Not the actual velocity between those two instants of measurement.

Kenshin

But quantum particles can accelerate without force?

Slereah

9:02 AM

if you insist on trying to find the trajectory of a quantum particle, the trajectory will be a Holder-continuous path

Which is a fractal of Hausdorff dimension 2

hence a lot of "jigglign"

Kenshin

OK guys, I've posted my question here too so that I can be confident that the answer has community consensus:

0

Q: Repeated Position Measurements

I'm reading Griffith's Introduction to Quantum Mechanics and came across the extract below: "What if I made a second measurement, immediately after the first? Would I get C again, or does the act of measurement cough up some completely new number each time? On this question everyone is i...

quantum-mechanics

Slereah

this is by the way the reason behind the time-energy uncertainty

Kenshin

wow it is all connected

I hope I can wrap my head around this soon

Anonymous

I actually don't understand why people insist on using Griffiths in the first QM course. It's terribly handwavy

Kenshin

I'm really suprised that QM particles jiggle about without external forces

Slereah

9:06 AM

Physicists love handwavy QM :p

Kenshin

@Blue do you have some other recommendations?

Slereah

Well that's the path integral description

Kenshin

true

Slereah

The wavefunction description simply doesn't have paths

Anonymous

@Slereah I've come to the conclusion that people should study QM from mathematicians and then the experimental aspects from engineers :P

Anonymous

9:08 AM

@Kenshin Shankar is slightly better. MIT OCW has a good three part course. Also check out Vazirani's lectures

Dawood ibn Kareem

@Kenshin So you're now interested in a universe with only one particle? Because as soon as you have more than one particle, there's no such thing as "without external forces".

Slereah

this is also true in a universe with one particle, though :p

Balarka Sen

they don't really "jiggle about"

that's dangerously close to keeping the newtonian picture in mind

Anonymous

@Kenshin I actually recommend properly studying the lagrangian and hamiltonian formulations first.

Anonymous

Maybe spend a few days on that

Balarka Sen

9:10 AM

Look up Hardy's uncertainty principle. There's nothing special about the physics bullshit.

Anonymous

youtube.com/playlist?list=PL0F530F3BAF8C6FCC

Balarka Sen

It's just a fact about moments of a function and it's Fourier transform

Anonymous

Oh, and yeah. This is THE best set of QM lectures on the whole internet ^

Anonymous

Missed that

Slereah

yeah it is also true for classical waves

Kenshin

9:12 AM

@BalarkaSen I understand the uncertainty principle, but QM isn't just the wavefunction. It is a combination of (1) the wavefunction and (2) the measruement process

Slereah

a similar relation exists for classical waves

Kenshin

the evolution of (1) and (2) are two distinct dynamics

Dawood ibn Kareem

@BernardoMeurer I think all QM students half-expect that on the last day of the course, the professor will show up and say "ha ha, it was just a joke - the universe is Newtonian after all".

Balarka Sen

@Kenshin The measurement process is a purely mathematical idea.

Slereah

@DawoodibnKareem that's the idea behind the Bohm interpretation :p

Kenshin

9:13 AM

@Slereah does Bohm work for QFT?

Slereah

@Kenshin eeeeeeeeeeeeeeeeeeeh

I've seen some attempts but it's not terribly convincing

Kenshin

I think i'll direct my research in that area

Slereah

Good idea, go for the approach that made it into a book called "Lost causes in physics"

Anonymous

XD

Kenshin

lol

Slereah

9:15 AM

springer.com/fr/book/9783540365815

a fine book

Kenshin

cool quantum theory of the brain sounds interesting too :p

Slereah

Well there is a reason why it's in the lost causes

Kenshin

Guys someone answered my question here, but his answer was that it was because 0 time has elapsed between measurements, not because the particles jiggled about and moved back to the same spot between measurements:

0

Q: Repeated Position Measurements

I'm reading Griffith's Introduction to Quantum Mechanics and came across the extract below: "What if I made a second measurement, immediately after the first? Would I get C again, or does the act of measurement cough up some completely new number each time? On this question everyone is i...

quantum-mechanics

I'd appreciate if someone can please post the jiggling answer too so the community can vote onthe more correct answer

Dawood ibn Kareem

Almost 30 years ago, I attended a lecture by Roger Penrose where he presented the idea that since a classical universe is entirely deterministic, free will must be a quantum phenomenon. Somehow from there, he jumped to the idea that consciousness must also be a quantum phenomenon. I happened to be sitting next to a leading expert in AI, who spent the entire lecture with his head in his hands muttering "you can't say that" over and over.

Anonymous

@DawoodibnKareem The AI person did have a point.

Anonymous

9:20 AM

And a very reasonable one

Dawood ibn Kareem

Well, yes.

Kenshin

I'm not an expert in QM tho so I think I should trust Penrose until I have enough knowledge to critisize him

Slereah

Don't you fool!

Anonymous

You shouldn't trust anyone till you have enough knowledge to claim they're right.

Bernardo Meurer

That's a dangerous game

Kenshin

9:25 AM

ok

SO i shouldn't believe in global warming then?

Dawood ibn Kareem

You need to trust Blue on this matter.

Kenshin

cos I haven't seen first hand proof global warming is real

Bernardo Meurer

Oh no

@BalarkaSen Look how low we've sunk

Slereah

If you haven't seen any proofs you shouldn't indeed

Dawood ibn Kareem

Global warming was invented by the Chinese. We know this for certain because the President said so.

Bernardo Meurer

9:27 AM

The earth is hollow and inside it live the ten lost tribes of israel

The earth is a hollow donuts

with the tribes inside the donut

And all the riches are inside the donut too

Balarka Sen

Lol

Bernardo Meurer

And since donuts don't heat

There is no warmin

Balarka Sen

oof Bernardo redpilling us hard

Bernardo Meurer

I know my shit

ourhollowearth.com/ourhollo/index.html

No joke

Kenshin

yes well feynman said that if two theories fit the data, then both theories should be kept in mind

Balarka Sen

9:28 AM

Hmm, on an unrelated note, I think I might have gotten addicted to tranquilizers

I haven't taken one today and I felt groggy all morning and felt I should take one today

Fucking hell

Bernardo Meurer

Tranquilizers?

Like Xanax?

Balarka Sen

I have been experimenting with a few. Clonazepam made me a little sick but diazepam worked wonders.

I have only take one though

one each

(so few = 2)

Bernardo Meurer

Meh, I wouldn't mess with those

Balarka Sen

I think I am going to force myself to not feel groggy.

Bernardo Meurer

virtualcityoflight.org

Balarka Sen

9:31 AM

I don't really want to get addicted to this shite

Gonna listen to this

Aesop Rock - Float [Full Album]

►

Secret

Random philosophy question:

It is easy to understand what timelessness is like (e.g. throw yourself into any steady state system or the heat death), but what does spacelessness is like?

I can certainly model this by grabbing some metric such that any spacelike separated events have vanishing squared interval, but what does that mean physically?

Balarka Sen

9:52 AM

\o @CooperCape