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/sci/ - Science & Math

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5834300 No.5834300 [Reply] [Original]

Gather round, everyfag. It's time to learn Quantum Mechanics, so you stop spewing nonsense on this board.

Imagine a particle of mass m moving in Euclidean space, subject to some force <span class="math">\vec{F}(\vec{x},t)=-\vec\nabla{V}(\vec{x},t)[/spoiler]

The program of Classical Mechanics is to determine its position vector <span class="math">\vec{x}(t)[/spoiler]. Once we know that. Solving Newton's 2nd law, <span class="math">m\vec{x}''(t)=\sum{\vec{F}(\vec{x},t)}[/spoiler], with appropriate initial conditions (typically the initial positions and velocities) suffices to determine <span class="math">\vec{x}(t)[/spoiler].

In Quantum Mechanics picture, we are looking for the particle's wavefunction, and we get it by solving the Schrödinger equation: <span class="math">i\hbar\frac{\partial { \Psi }}{\partial {t}}=-\frac{\hbar^2}{2m}\nabla^2 \Psi+V \Psi [/spoiler]. This role is logically analogous to Newton's 2nd law. Given initial conditions (typically <span class="math">\Psi(\vec{x},0)[/spoiler]), the Schrödinger equation determines <span class="math">\Psi(\vec{x},t)[/spoiler] for all future times, just as Newton's 2nd law determines <span class="math">\vec{x}(t)[/spoiler] for all future times in classical mechanics. When extended to multiple particles, this equation determines the (non-relativistic) behavior of every particle in the universe.

>> No.5834305

>>5834300

The beef:

We connect the wavefunction to classical mechanics through **Observables**. It's complicated and there are loads of important details, but I'll try and gloss over it. Observables correspond to classical quantities (e.g. position, momentum, Energy), and are represented by Hermitian **Operators**. These operators transform the wavefunction to eigenfunctions whose eigenvalues correspond to classical values, and the Expectation Value of an operator corresponds to the average value of repeated measurements of that observable. Experimentally measuring an observable of a wavefunction **collapses** that wavefunction to an eigenstate of the operator.

Pre-Decoherence Quantum Mechanics says absolutely nothing what actually happens during collapse: it avoids the issue entirely by postulating a non-deterministic "leap" of the wavefunction to this new eigenstate. This is a placeholder postulate, as we shall see. Here only, the wavefunction doesn't evolve **deterministically** according to Schrödinger's equation. The question of what actually happens during collapse is called the **measurement problem**.

However, the notion of collapse should set off alarm bells in your head. How can a Classical system interact with a Quantum system at all? The mathematics of wavefunction evolution never address these microscopic concerns! What about the (billion trillions of octillions of) wavefunctions that make up the Classical system that's doing the measuring?

>> No.5834307

>>5834305
Physicists long worried about this, and in the 1980's hammered out the mathematical framework of **Quantum Decoherence**. It's obscenely complicated, but it essentially pegs collapse down to the **einselection** of certain eigenstates, based on the instantaneous phase of the quantum state of the observer/measurer. The observer's quantum state is deterministic, following Schrödinger's equation, but ungodly large & well mixed, just as thermal movements of molecules are technically deterministic but so extremely well mixed as to appear almost-random! Quantum Decoherence is extremely experimentally successful, and its explanation of collapse and other predictions have been verified numerous times, primarily via Quantum Optics.

>> No.5834308

Oh look, it's the girl who spewed a bunch of crap in a previous thread claiming QM is deterministic. You're the one who needs to stop spewing bullshit.

>> No.5834309

>>5834307
Once decoherence is accounted for, collapse is easily seen as an approximation to environment effects. Consider the following example: there exists only an electron, inside a machine that can measure the electron's position, floating in space. If you consider just the wavefunction of an electron, it will exhibit collapse on interaction with the machine (e.g. when its position or momentum is measured by a machine). However if you consider the wavefunction of everything (the electron's wavefunction + the octillions of wavefunctions that make up the machine), they will always obey Schrödinger's equation and interact via Schrödinger's equation, and they will evolve deterministically. Decoherence is a natural consequence of Schrödinger's equation; non-deterministic collapse is discarded as an unnecessary, impotent assumption, now that we have a fundamental explanation for the epiphenomenon of collapse.

>> No.5834311

>>5834309
In short, the measurement problem asks,
>"By what physical mechanism does a classical system induce wave function collapse in a quantum system?"
As mentioned above, it's answered by decoherence.

I know this is counter-intuitive, but unfortunately there is a long-standing public misconception that Quantum Mechanics is non-deterministic. This stems from the fact that QM did not properly explain collapse until the 1980's, and before decoherence. But times change, and Copenhagen was replaced by Consistent Histories.

I cite Griffiths and Peskin.

TL;DR: Quantum Mechanics is deterministic.

>> No.5834312

>>5834308
>thinks I'm some random faggot

Lay off the cocksucking. Read the explanation.

>> No.5834314

>>5834312
I read your explanation the last time you posted it, it's still wrong.

>> No.5834316
File: 57 KB, 679x516, 1370874308060.jpg [View same] [iqdb] [saucenao] [google]
5834316

>>5834314
>faggot

Excellent, you're marginally less of a faggot now! You've moved from Name-calling to Contradiction.

>> No.5834321

>>5834314

Maybe someday, you'll explain what's wrong with it, instead of whining like a woman

>> No.5834322

>>5834300
strange Latex error, otherwise good

>> No.5834328

>>5834321
What's wrong with it is that quantum mechanics isn't deterministic. It predicts probability distributions of observables. Your misunderstanding of decoherence does not change this. You don't change the rules of the game by more complex play.

But there's not much to refute:
>It's obscenely complicated
is code for "I can't explain because I don't understand it."

>> No.5834329

>>5834309
Saying that the currently favored interpretation of QM is deterministic is not the same as saying that QM itself is deterministic.

>> No.5834332

>>5834328
Does anybody understand it enough to that level though?

>> No.5834333

>>5834329
The only interpretation of QM that's popular and deterministic is many-worlds, and that's not really deterministic because they bring back randomness via uncertainty in "which copy am I?" Bohm is truly deterministic but not at all popular.

>> No.5834334

>>5834328
>is code for "I can't explain because I don't understand it."
No, you raging faggot. It's code for "it doesn't matter for this discussion." 2 quarters of QM and 1 quarter of nuclear physics == kill yourself.

>It predicts probability distributions of observables.
Among other things. Your misunderstanding of time-evolution and wavefunction collapse cripple your understanding.

>> No.5834336

>>5834332
Yes, I do, as does anyone who takes undergraduate-level quantum mechanics. This is basic stuff.

>> No.5834338

>>5834329
>>5834333

This isn't about interpretations of quantum mechanics. Decoherence allows us to assert Quantum Mechanics itself is deterministic.

>> No.5834339

>>5834332
Yes. It's not that complicated. I'd have to review how einselection works to give a proper account of it rather than what OP's said, but most of what she's saying is like saying 2+2=5. If you just evolve a quantum state forward using the Schrodinger equation, you end up with states containing live and dead cat components and stuff like that.

>> No.5834340

>>5834336
It being deterministic or not is undergraduate-level?
I should've done Physics MSci.

>> No.5834341

>>5834328
> shitty, misguided counterargument
> then an Ad-Hominem followup

You're not even wrong

>> No.5834349

>>5834339
Not a she, faggot. No idea who the fuck you're talking about.

>If you just evolve a quantum state forward using the Schrodinger equation, you end up with states containing live and dead cat components and stuff like that.
Except you don't. That's not at all how einselection works.

>> No.5834353

>>5834340
If you've taken any course at all in quantum mechanics, you will know that it is not deterministic, at least in the standard formulation. There are deterministic interpretations, but all of these have to introduce random chance in one way or another to *exactly* reproduce QM predictions. (Bohmian mechanics approximately reproduces QM after a sufficient amount of time using uncertainty in initial conditions.)

>> No.5834355

>>5834333
Many-worlds is totally deterministic. If you make a measurement that would normally be said to collapse a wavefunction, what's really happening is that you're going into an entangled state. The question "which copy am i?" is answered by saying: there is no copy, you're in an entangled state.

If it was a good measurement, the entangling process should have entangled your mental state with the state of the system. So if the measurement could be 0 or 1, and your mind could think "0" or "1", then you'll be in the state |0,"0">+exp(i*phi)|1,"1"> for some phi that depends on the system. If by "you" you mean "the wavefunction as a whole," then you're entangled, end of story. If by "you" you mean "the part of the wavefunction that corresponds to me thinking "0", then that is |0,"0">, and it makes up just half of the total wavefunction.

>> No.5834356

>>5834340
No, sorry, the mathematical formalism is undergrad level. That's things like the Hilbert space, observables, eigenfunctions of hermitian operators, the generalized statistical interpretation, etc.

Decoherence is hardcore grad level. That faggot who's making the ad-hominem arguments probably has never taken either before.

>> No.5834358

>>5834349
That is exactly what you get. The state is entangled with the environment, though, so it's not really

|live cat> + |dead cat>
but
|live cat, live cat environment> + |dead cat, dead cat environment>

It's no good for calculations to have to include a description of the whole environment, so what people do is work with density matrices. With density matrices you can trace over the environment and get something like this:

|live cat><live cat| + |dead cat><dead cat|

>> No.5834361

Faggots stop pretending to understand QM and stop espousing shit of an inherent order of the universe, reality would be retarded if it didnt contain the possibility of the impossible being probable, your feeble human minds are just unable to comprehend the contradictions of the posibilities of the impossible

>> No.5834362

>>5834353

No.

What you mean to say is that wavefunction collapse, caused by measurement of classical dynamical variable, from a Classical system, is nondeterministic.

Evolution via the Schrödinger equation IS deterministic.

>> No.5834363

>>5834361
0/10 faggot

>> No.5834365

>>5834362
Those are both essential parts of quantum mechanics.

>> No.5834368

>>5834353
I've done up to A2 Physics and also personal study. Recently I've been on /sci/ a bunch, and I've been looking at the resources in the sticky. Science is great.

>>5834356
Ah. Alright. This seems a pretty high-level argument.

>> No.5834369

>>5834362

>>5834328
No, OP makes sense. The evolution of any Quantum system is deterministic. If you measure it FROM a classical system, it quantum leaps.

But what if you treat the Entire system (quantum system + measuring system) as quantum? Then it HAS to obey the Schrödinger equation still!

>> No.5834370

>>5834369

Right, and you can extend that to say the entire universe must always obey the Schrödinger equation (in the non-relativistic limit)

>> No.5834371

>>5834369
That's essentially what many-worlders believe, but you absolutely need the probabilities or else quantum mechanics does not make predictions for experiments.

>> No.5834372

>>5834369
Why?

>> No.5834376

>>5834365
>>5834371
>many worlders do this
right

> you absolutely need the probabilities
Not at all, at least not the way you'd think.

Many-worlders use the experimental probabilities to generate multiple worlds.

Consistent Histories takes the sane approach, and argues that experiments give us information on the history of our quantum state. In consistent histories, the evolution of the universe is deterministic.

>> No.5834377

>>5834372
Why what?

>> No.5834379

>>5834376
Blatantly false. Consistent histories gives an explicit formula for the probability of a history. It is not by any means deterministic.

>> No.5834381

>>5834371
Yes it does. At the very least, there are some (idealized) experiments where you get a 100% prediction one way or the other. Besides, you can derive the probabilities independently in the many-worlds framework. Basically, what it gets down to is that if I (correctly) believe, say, a spin will be up with probability p on the classical level, than after the experiment the universe will be composed of an entangled state of which p is spin up and (1-p) is spin down. So if p is .95 and I say that's enough for me to be confident that p will happen, the future state has me right in 95% of it and wrong in 5% of it, which is just what you'd expect in classical probability. And since it looks to someone *in* the entanglement like his piece is the whole universe, it really does look like classical probability to you.

>> No.5834386

>>5834377
Why does it HAS to follow schrödingers equation thats not a given

>> No.5834388

>>5834381
Yes, but you still have to figure out what the right way of separating the state vector into "worlds" or "branches" is. This essentially gives you back the same problem you had in Copenhagen, with essentially the same not-quite-satisfactory methods for solving them.

>> No.5834389

>>5834388
>solving them
*solving it

>> No.5834391

>>5834376
|Dead cat</10

>> No.5834394

>>5834388
Not really, I think. If you had infinite computing power and sufficient prior knowledge of the wavefunctions in question, you could work out the exact evolution of the total wavefunction. Then you could give answers for the probabilities and the corresponding entangled states for every possible basis.

"Oh, you want a density matrix where your state of mind is traced over? Well then here are the probabilities for each state of mind and each corresponding external wavefunction." Something like that.

>> No.5834395

>>5834379
Blatently faggot

>Consistent histories gives an explicit formula for the probability of a history
I just said that.

>It is not by any means deterministic.
Yes, CH eliminates non-deterministic collapse. The interpretation of the reduction rule for wavepacket collapse is different. Collapse is valid, but no specific effect on the measured object can be held responsible for it. Decoherence in the measuring device is enough.

>> No.5834402

>>5834386

Because any quantum system obeys the Schrödinger equation, except during collapse.

>> No.5834404

>>5834395
What sort of doublethink is this? Having fundamental probability in your theory is the same thing as not being deterministic. That is what the word means.

>> No.5834415

>>5834358
You completely missed the point. foxnews/10

>>5834371
Not really. You don't know the current state of the Universal wavefunction. You can view measurements as yielding information about it; however, this comes at the expense of altering it, with your own part of the universal wavefunction.

>>5834376
Mostly correct. What consistent histories let's you do is assign probabilities to the histories of the Universal wavefunction, given your measurement's result.

The universal wavefunction still evolves deterministically; but we can only estimate its value.

>> No.5834418

>>5834395
>>5834404 here,
Okay, if you meant that one part of the interpretation is deterministic but which history is observed is non-deterministic, then I agree. But that still makes the interpretation as a whole non-deterministic.

>> No.5834422

>>5834404
>misunderstands what I'm calling deterministic.
I'm calling the evolution of the universal wavefunction deterministic.

CH doesn't determine what your measurement will yield.

But if you knew the universal wavefunction, you could obviously calculate this.

>>5834418
Yes, see above. I'd say this makes it deterministic, in that it allows one to view the evolution of the universe as completely deterministic.

I would say this makes the interpretation as a whole *deterministic.*

From wiki:
>[Determinism states] that for everything that happens there are conditions such that, given those conditions, nothing else could happen.

In other words, for one value of the universal wavefunction, one universe exists.

Contrast this with Copenhagen, which makes absolutely no such guarantees.

>> No.5834430

>>5834300
You said quantum mechanics not vector calculus
:(

>> No.5834432

>>5834339
dipshit, this is what OP was talking about:

>>5834415
>>5834422

>> No.5834433

>>5834430
Aw lol, sorry I presuppose you know Multivariable calculus, linear algebra, differential equations, and classical mechanics.

>> No.5834434

>>5834418
Well then how would you rewrite OP's explanation to include this clarification?

>> No.5834436

>>5834415
Are you pretending that a history in CH is the same as the universal wavefunction? It isn't.

>>5834422
>In other words, for one value of the universal wavefunction, one universe exists.
Substituting "history" for "value of the universal wavefunction," this would be correct. But by that same token you could call a collapse interpretation (which CH essentially is, just cleaned up) deterministic. You say, for a particular value of each measured quantity, one universe exists. But that's really missing the point. Your theory can't predict what those measured values were a priori. Equivalently, nor can it tell you which history is observed a priori. It only gives you probabilities for those things. Therefore, it is non-deterministic.

>> No.5834440

>>5834422
I might be completly wrong here (only had some very basic QM to this point) but wouldn't the uncertainty principle contradict your determinism? If there are some causes (e.g.energy ) which aren't defined at a certain timeperiod in the universe, how can the whole be deterministic? Determinism implies that every cause is known/possible to know (can be determined), yet there are missing informations in the universe.

>> No.5834449

>>5834440
Yes, completely wrong. This discussion is for adults only.

You must have taken at least one year of college QM to know what we're talking about.

>> No.5834460

>>5834436
>Are you pretending that a history in CH is the same as the universal wavefunction?
No.

>Equivalently, nor can it tell you which history is observed a priori.
Yes.

>Therefore, it is non-deterministic.
No, that's not the definition of determinism. If you could write down the wavefunction of both your measuring device and your "measuree", you could calculate what your measuring device will record.

>> No.5834464

>>5834358

this: >>5834460

>> No.5834469

>>5834358
Let's give an example to illustrate how "tracing out the environment" works.

I prepare two spin one have particles in the state

(1/sqrt(2)) (|up,down> + |down,up>)

Now Alice measures one particle and Bob the other. Alice measures whether her particle is in the state |A1> or |A2>, and Bob measures whether his particle is in the state |B1> or |B2>.

The probability for Alice to find A1 and Bob to find B1 is

(1/2) <A1,B1| (|up,down> + |down,up>) (<up,down| + <down,up|) |A1,B1>

and the probability for Alice to find A1 and Bob to find B2 is

(1/2) <A1,B2| (|up,down> + |down,up>) (<up,down| + <down,up|) |A1,B2>.

If we add the probabilities together, we can rewrite this as:

(1/2) tr[(|A1,B1><A1,B1| + |A1,B2><A1,B2|) (|up,down> + |down,up>) (<up,down| + <down,up|)]

or

(1/2) tr[(|A1><A1| x (|B1><B1| + |B2><B2|)) (|up,down> + |down,up>) (<up,down| + <down,up|)]

But |B1><B1| + |B2><B2| is just the identity operator on B. So we can write the total probability of Alice finding A1 as:

(1/2) tr[(|A1><A1| x 1_B) (|up,down> + |down,up>) (<up,down| + <down,up|)]

which evaluates to

(1/2) <A1| (|up><up| + |down><down|) |A1>

So without knowing what measurement Bob actually performed, Alice can average over Bob's possible results and obtain an expression giving her the probability of her measurements. And the expression is equivalent to what she'd get if she assumed that Bob collapsed the state randomly to either up or down, even if that's not actually what Bob measured.

The (1/2) (|up><up| + |down><down|) expression is called the reduced density matrix.

>> No.5834475

>>5834436
that's not what OP is saying.

>> No.5834482

>>5834475
Well, no, OP said that QM is deterministic, which is wrong.

>> No.5834483

>>5834469
point?

>> No.5834488

>>5834482
No, he said something like this: >>5834460

>> No.5834489

>>5834483
Just trying to give an explanation for some of the ideas behind decoherence, which at least OP doesn't understand because she thinks it makes QM deterministic.

>> No.5834493

>>5834489
Ok, so what is the net effect of this tracing then?

>she thinks it makes QM deterministic
this one doesn't sound like the one from the other thread...

>> No.5834496

>>5834460
>If you could write down the wavefunction of both your measuring device and your "measuree", you could calculate what your measuring device will record.
Care to give an example (or a link to an explanation) of how this is supposed to work? I think you are confused, because that's not the case in any formulation of QM I've ever heard of (assuming by "wavefunction" you mean a quantum state, a vector in a Hilbert space).

>> No.5834501

>>5834449
what you basicly say is that if one could gather all infromation (->knowing the wavefunction of the whole) the universe would be deterministic.

>> No.5834504

>>5834404
>Having fundamental probability in your theory is the same thing as not being deterministic.
But the probability is not fundamental, it is emergent. That's literally the whole point.

>> No.5834508

>>5834496
Not him, but if you have a spin in state |up> + |down> and a measuring device that registers "u" for up and "d" for down, the output state of the measurement will be [neglecting phase and normalization]: |up,"u">+|down,"d">.

>> No.5834511

>>5834493
In general, quantum systems interact with the environment -- other stuff in the experiment with a huge number of degrees of freedom which we're never going to measure. In order to make any sort of calculation that predicts the results of your experiment, you need a way to get rid of those degrees of freedom, and that's what tracing out the environment does for you. You describe your quantum system not with a state vector, and a density matrix, and you can now reasonably include the interaction with the environment in your evolution of the description without including the whole environment. It's a tool.

If you conduct the Schrodinger's cat experiment, you would never try to figure out the details of

|live cat, live cat environment> + |dead cat, dead cat environment>

but you would work with

|live cat><live cat| + |dead cat><dead cat|.

You can treat the cat as if it has already collapsed randomly into a live or dead state before measuring it. This is independent of what actually happens to the cat, which is very interpretation-dependent.

>> No.5834512

>>5834496
>>5834501

Got to go to dinner. Back later. BTW I'm OP you've been talking to. Not a fucking woman. Will post revised explanation + dickpick then, LOOK FOR ME AT THE FIRST LIGHT OF THE FIFTH DAY.</gandalf>

>> No.5834513

>>5834508
>the output state of the measurement will be [neglecting phase and normalization]: |up,"u">+|down,"d">
No it will not. A measuring device that displays human-readable input is necessarily classical. What you are effectively saying is that Schrodinger's cat is both alive and dead.

The machine will output either u or d, but it is not humanly possible to predict which one.

>> No.5834514

>>5834511
>This is independent of what actually happens to the cat, which is very interpretation-dependent.
So in your understanding of QM, cats can be both alive and dead, and the human mind has the magical property of randomly collapsing the universe when it thinks about stuff?

>> No.5834515

>>5834501
>what you basicly say is that if one could gather all infromation (->knowing the wavefunction of the whole) the universe would be deterministic.

yes

The thing is that we are a part of the universe, so we only experience one of the branches. The branch we end up on seems to be non-deterministic.

But on a theoretical level (assuming many worlds), there is determinism.

>> No.5834518

>>5834515
>assuming many worlds
You don't need many worlds for CH.

In many worlds, you aren't on "one branch," you (or copies of you) are on infinitely many branches.

>> No.5834519

>>5834518
but what you EXPERIENCE is one branch.

>> No.5834522

>>5834514
That is one interpretation of quantum mechanics, and not one I find particularly likely. But decoherence (evolving quickly from a pure state like (|live>+|dead>)(<live|+<dead|) to a mixed state like |live><live|+|dead><dead| due to interaction with the environment) works no matter what (valid) interpretation of QM you pick.

>> No.5834528

>>5834513
I am not saying Schodinger's cat is alive and dead. I didn't say what "u" and "d" were. They could be very complicated states. Or they could be very simple. With sufficient (read: not coming anytime soon) technology, a single spin could be flipped in a quantum memory, then held without further decoherence for 1000 years for future readout.

As long as the memory spin is not allowed to be further interfered with the measured spin, the amount of decoherence would be the same as if we had measured it registered it in our memory. But if both were perfectly preserved, then 1000 years later we could do an appropriate experiment to show the entanglement. (we could predict a particular beamsplitter output with certainty.) This all is just practically impossible, but theoretically fine.

>> No.5834533

>>5834519
>but what you EXPERIENCE is one branch.
For the appropriate definition of "you".

>>5834522
>That is one interpretation of quantum mechanics
That is a nonsense first grader interpretation of quantum mechanics that nobody has ever subscribed to. Magic has no place in physics.

>But decoherence works no matter what (valid) interpretation of QM you pick.
Yes, and the formalism in the context of CH renders collapse an unnecessary postulate. The fact is that decoherence occurs, and that decoherence alone is sufficient to explain the appearance of collapse.

What you are saying is like a classical physicist saying, "Yes, electromagnetism is real, but Brownian motion is a perfectly random and independent process."

>> No.5834535

>>5834528
Well sure, I would agree with that. In principle, if you could keep a quantum system in isolation for sufficiently long, you could observe quantum phenomena on a macroscopic scale. Of course, uncertainty still applies, in the sense that the system would not have defined classical observables.

>> No.5834538

>>5834533
>CH renders collapse an unnecessary postulate
Consistent histories is the collapse postulate in slightly different language.

>> No.5834544

>>5834538
The only postulate required is the probability assignment, and even that is justified. It provides an actual description and explanation of the collapse process, unlike Copenhagen which explicitly avoids the issue, and it is the best candidate for calculations in mesophysics (at the quantum-classical boundary), which is already beginning to be researched.

>> No.5834550

>>5834535
>In principle, if you could keep a quantum system in isolation for sufficiently long, you could observe quantum phenomena on a macroscopic scale

Yes, but the real point comes from the next logical step: if we were technologically advanced enough to account for the crazy behavior of 10^24 or so particles, we could show that your brain really went into an entangled state just like those particles. It's just a practical matter that we cannot. Either this contradicts the collapse postulate or we must agree that my observation of some state does *not* cause wavefunction collapse. (But then when does it happen? Has it ever happened? How would we know?)

>Of course, uncertainty still applies, in the sense that the system would not have defined classical observables.

I'm sorry, I don't understand what you mean by this. The state certainly has observables, we just haven't personally done any "observing" yet.

>> No.5834556

>>5834550
>Yes, but the real point comes from the next logical step: if we were technologically advanced enough to account for the crazy behavior of 10^24 or so particles, we could show that your brain really went into an entangled state just like those particles.
It's not just a question of technological advancement but of decoherence time. A system that large will decohere much faster than the Planck time, and so cannot be observed even in principle.

>The state certainly has observables, we just haven't personally done any "observing" yet.
Well, it doesn't have unique observables. The state is well mixed and has many eigenvalues. When the state decoheres, one eigenvalue is selected by the environment (einselection), but it is not generally possible to predict which one.

>> No.5834567

>>5834556
>It's not just a question of technological advancement but of decoherence time. A system that large will decohere much faster than the Planck time, and so cannot be observed even in principle.

Wait, are you a subscriber to one of the Objective Collapse theories? I don't think they are true, but I respect them for actually making a meaningful prediction.

In my view, decoherence really *is* just a complicated entanglement. So if we could, somehow, keep track of every entangling interaction and reverse them all [like in a spin-echo experiment], we could reverse even a macroscopic "decoherence" and show the whole, vast entangled state.

>> No.5834580

>>5834567
>Wait, are you a subscriber to one of the Objective Collapse theories?
No. I'm not OP, but I'm still advocating Consistent Histories.

>>5834567
>In my view, decoherence really *is* just a complicated entanglement
Yes.

>So if we could, somehow, keep track of every entangling interaction and reverse them all [like in a spin-echo experiment], we could reverse even a macroscopic "decoherence" and show the whole, vast entangled state.
Yes you could, and the resulting quantum state would be just that--quantum. It could be described by the universal wavefunction, but not by classical variables like "momentum," because these correspond to just one eigenstate.

>> No.5834668

>>5834504
In my understanding of consistent histories, the probability assignment to histories is postulated, not emergent.

>> No.5834689

so many misconceptions

>> No.5834731

Am I getting this much right?

"Let's think of light as a wave in a Maxwellian field that nevertheless only transmits with discrete packets of energy, so it's basically the Maxwell model but restricting which waves can appear in that model because only certain energy quanta are allowed."
= old quantum theory, 1900, Planck's Law

>> No.5834745

>>5834731
>tfw no one can answer even this question

>> No.5834753

>>5834731
Planck didn't think of the electromagnetic field as quantized, but rather thought the energy of the objects emitting and absorbing the radiation was quantized. The Wikipedia article is probably better than my knowledge:

http://en.wikipedia.org/wiki/Planck%27s_law#Trying_to_find_a_physical_explanation_of_the_law

This is really a history of physics question, not a physics question.

>> No.5834756

>>5834753
fact: people are not going to be able to jump into QM without getting at least some of the early history leading up to it

>> No.5834760

>>5834756
fact: people are not going to be able to jump into QM from some descriptions of 19th century speculations on 4chan.

>> No.5834772

>>5834756
Well, yeah, it's good to have a rough understanding of where this stuff came from. But the treatment of history given in physics courses is often inaccurate. There's not much concern for the details, just the bits and pieces that made it into real quantum theory. But yeah, Planck was the guy who proposed light was emitted and absorbed in discrete energy packets. He used this to explain the observed blackbody spectrum. But he thought this was because of the matter.

>> No.5834799

>>5834772
So was it Einstein in 1905 that said to Planck,
"Hey I think it's the EM field itself that has quantized energy traveling through it" ???

>> No.5834808

>>5834799
Yeah.

>> No.5834847

>>5834799
Honestly it's a lot more complicated than that. Quantum physics has a very muddled history.

But Einstein did win the Nobel Prize for that, yes.

>> No.5834850
File: 31 KB, 340x354, 12345.jpg [View same] [iqdb] [saucenao] [google]
5834850

Well, Richard Feynman was right. No one actually understands quantum mechanics.

>> No.5834874

>>5834328
I think the issue here is that you both have slightly different conventions for what it means for something to be "deterministic".

One of you is saying that the difference between deterministic and in-deterministic is that in an in-deterministic system we can't know anything about its future states given it's past states and in a deterministic we can. The other one is splitting it differently, in-deterministic, probabilistic, deterministic. The first is essentially the same, but the difference between probabilistic is that we can only predict possible future states probabilistically while deterministically means exact.

I myself am partial to the second set of definitions.

>> No.5834877

>>5834874
No. One is saying that a system is deterministic iff we can gain complete knowledge of it, and the other is saying it is deterministic iff its future states are entirely determined by its history.

So if an event is properly random, it is not "determined" by anything, but if there is an underlying cause, even one we cannot directly observed, then it is determined, and thus deterministic.

>> No.5834892

>>5834877
I should caution for those reading this thread that an (elementary) "history" in CH means the full sequence of events which in Copenhagen would be called wavefunction collapses, whether they happen in the past or the future. Histories are the outcomes that make up the probability space. (There are also "compound histories" which are elementary histories added together, which are the events.) Some people might see the word "history" and think it refers to only things in the past. I'm not sure which way you're using the term in your post, but if you mean stuff in the past, which is what it sounds like, you should clarify.

>> No.5834902

/sci/ cannot into QM

>> No.5834920

If I'm understanding this decoherence thing correctly, then wouldn't it violate Bell's theorem?

If the collapse is deterministic even if it's so complex we can't calculate it, it should still violate the theorem shouldn't it?

>> No.5834924

>>5834920
all I remember about that stuff is

YOU MUST ABANDON ALL HOPE FOR ONE OF THESE:
a) locality
b) counterfactual definiteness

>> No.5834926

>>5834892
>Some people might see the word "history" and think it refers to only things in the past.
Actually, it doesn't matter. The Schrodinger equation is path-independent, so the state of the universal wavefunction at any point in time is sufficient to determine it at all points in time.

>>5834920
>If I'm understanding this decoherence thing correctly, then wouldn't it violate Bell's theorem?
No. Consistent Histories is nonlocal, because entanglement is inherently nonlocal.

>> No.5834928

>>5834920
Two issues:

First, decoherence doesn't mean that the outcomes of collapses are deterministic. Anyone who's saying that is simply wrong. If you believe in many-worlds, then there is no collapse and all outcomes happen, but that's a different story.

Second, Bell's theorem isn't really about determinism; it's about locality (no faster-than-light influences) and counterfactual definiteness (whether the arguments about "what would have happened" if one experimenter chose a different setting are valid). You don't need determinism to derive the inequality that contradicts quantum mechanics.

>> No.5834937

>>5834928
>First, decoherence doesn't mean that the outcomes of collapses are deterministic.
Not on its own, no. But decoherence can be used as a mechanistic explanation of collapse.

>Second, Bell's theorem isn't really about determinism; it's about locality and counterfactual definiteness
Counterfactual definiteness and determinism kind of go hand-in-hand, though you're right that there is a subtle distinction.

>> No.5834976

>>5834415
>>5834422
>>5834415
>>5834422
>>5834460
>>5834580
>>5834926
If you're going to use the term, you need to define what "universal wavefunction" means in terms of the language of Consistent Histories. Preferably in agreement with the conventions of something readily available online like
http://quantum.phys.cmu.edu/CQT/index.html
Otherwise you're just talking to yourself.

>> No.5835062

>stop spewing nonsense
That is deliciously ironic.

>> No.5835064

>>5835062
faggot

>> No.5835071

>>5835062
>deliciously
Liberace-gay

>> No.5835076

>>5835071
even Liberace wasn't that gay

>> No.5835079

>>5834976
I'm using "universal wavefunction" according to the STANDARD Hartle-Hawking definition, given here: http://prd.aps.org/abstract/PRD/v28/i12/p2960_1

>> No.5835080
File: 25 KB, 300x300, christopher-hitchens[1].jpg [View same] [iqdb] [saucenao] [google]
5835080

>>5834311
>Consistent Histories
That which is asserted without evidence, can be dismissed without evidence.

>> No.5835086
File: 45 KB, 550x550, wonka.jpg [View same] [iqdb] [saucenao] [google]
5835086

>>5835080
>Thinks CH is asserted without evidence
>can't stop sucking cocks

CH is an interpretation of experimental data. Learn about it here:

http://www.springer.com/physics/particle+and+nuclear+physics/journal/11447

>> No.5835087

>>5835064
>>5835064
>>5835071
You are embarrassing yourself.

>> No.5835112

>>5834877
This looks correct. I'm partial to the second definition. Being able to experimentally obtain complete knowledge of a system has nothing to do with determinism. It's probably not even possible in classical mechanics.

In the second sense, quantum mechanics is certainly deterministic. The future evolution of any system is given by the Schrödinger equation.

>> No.5835122

I remember a professor last semester in one of my classes saying something like that QM is deterministic, but we just can't "see" the entirety of it. The measurements of our outcomes are probabilistic, but the systems still evolve deterministically, it's just what we can measure that is probabilistic.

>> No.5835127

>>5835112
this.

>>5835122
This. That's what OP was getting at, but he did a fucktarded job at explaining himself. He should have stuck to the universal wavefunction, without going into decoherence so much.

I suppose in that sense, he's right. It's a pretty widespread misconception that the universe evolves nondeterministically.

>> No.5835131

>>5835079
Nobody is asking about which quantum state you're using. I want to know precisely how one would translate what you have been saying into the language of consistent histories, which emphasizes histories, not states.

Now if I have this right, in terms of Griffiths' notation, an initial state <span class="math">|\psi_0\rangle[/spoiler] for the universe, which is the kind of object Hawking and Hartle were talking about, would be spoken of by choosing a family of histories like this:

One of the histories is
<span class="math">(I - |\psi_0\rangle \langle\psi_0|) \odot I[/spoiler]
and the other histories are things like
<span class="math">|\psi_0\rangle \langle\psi_0| \odot F_1 \odot F_2 \odot \ldots \odot F_n[/spoiler]
where each <span class="math">F_i[/spoiler] is a projection operator.

Having <span class="math">|\psi_0\rangle[/spoiler] as the initial state only excludes the first history. If I measure whether some observable A has value <span class="math">\lambda[/spoiler], there will in general be histories in the second group where the statement "A is found to be <span class="math">\lambda[/spoiler]" is true (the history contains the projector corresponding to this statement), and others where it's false. Consistent histories can tell me the probability that A will be <span class="math">\lambda[/spoiler], but it doesn't say whether it actually happens.

>>5834422
Yet there have been statements like
>CH doesn't determine what your measurement will yield.
>But if you knew the universal wavefunction, you could obviously calculate this.

Here I know the state you are calling the "universal wavefunction," and I only know the probability which which my measurement will yield any given value.

>> No.5835137

>>5835131
> Nobody is asking about which quantum state you're using.
If you don't think that the definition of the term "universal wavefunction" is relavent, then you're an idiot.

> an initial state <span class="math">|\psi_0>[/spoiler] for the universe
Sure.

> If I measure whether some observable A has value <span class="math">\lambda[/spoiler]
But you won't. The wave function of the universe never collapses.

>> No.5835138

>>5835137
>> If I measure whether some observable A has value \lambda
>But you won't.
Yes I will. People make measurements all the time. This is not something you can argue against.

>> No.5835146

>>5835127
>It's a pretty widespread misconception that the universe evolves nondeterministically.
In objective collapse theories, the quantum state of the universe does evolve nondeterministically.

>> No.5835145

>>5835138
No you won't. Measurements of a quantum system are made from outside the system.

My measuring anything doesn't collapse the wave function of the universe. Within the universe, my making a measurement is just part of the continuous evolution of the universe's wave function, which includes the wave functions of every particle in my measurement apparatus, as well as the microscopic subsystem I'm measuring.

>> No.5835151

Top fucking lel on physicists notation.

>> No.5835156

>>5835145
>No you won't.
Yes, I will. And so will you.

>My measuring anything doesn't collapse the wave function of the universe.
There you go. You measure things too. We all measure things. Now I'm not arguing about whether that collapses the wavefunction of the universe. The question is whether the result of my measurement can be predicted. The answer is no. And that's why we say quantum mechanics is nondeterministic.

(Unless you take the many-worlds view or modify it with something like the hidden variables of Bohmian mechanics.)

>> No.5835161

>>5834316
The triangle autists hide behind.

>> No.5835170

It's cool that you autistic anons have got this nice philosophy thread going, but vid related.

http://www.youtube.com/watch?v=ntxKRj9DRnc

>> No.5835176
File: 220 KB, 286x119, cant deal with it.gif [View same] [iqdb] [saucenao] [google]
5835176

>>5835156
No, you will never, ever measure an observable of the universal wave function. Jesus christ, how are you this retarded?

> Now I'm not arguing about whether that collapses the wavefunction of the universe.
You just stated that it did, when you asserted that projection operators were applied to the universal wave function.

> The question is whether the result of my measurement can be predicted.
No it's not, and it never has been. The question is whether, given an initial state of the universal wave function, it will evolve deterministically according to the Schrödinger equation.

You will never, ever be able to predict the result of your measurement, because you don't have the necessary information, and it's impossible to obtain it. So QM is deterministic, because *given this information,* the wave function of the universe, you could (at immense computational expense) predict the result.

It's the exact same situation in Bohmian mechanics. You will never, ever be able to predict the result of your measurement, because you don't have the necessary information, and it's impossible to obtain it. Yet Bohmian mechanics is deterministic, because *given this information,* the Bohmian coordinates in configuration space, you could (at immense computational expense) predict the result.

>> No.5835178
File: 27 KB, 367x451, ahahaha.jpg [View same] [iqdb] [saucenao] [google]
5835178

>>5835170
>calls others autistic anon
>is autistic anon

>> No.5835179

>>5834300
You lost me at "F"

>> No.5835181

>>5835179
lost me at "learn"

>> No.5835182

>>5835161
>responds to 12 hour old irrelevant post
>autist

>> No.5835185

>>5835151
?

>> No.5835190

>>5835176
>an observable of the universal wave function
Define that.

>> No.5835193

>>5835176
>>The question is whether the result of my measurement can be predicted.
>No it's not, and it never has been.
Yes it is, see:
>because *given this information,* the wave function of the universe, you could (at immense computational expense) predict the result.
That is wrong. By predict I mean whether it is predictable *in principle*. That is how you distinguish nondeterministic theories (like quantum mechanics) from deterministic theories like classical mechanics.

>> No.5835195

>>5835176
>>5835190
And if you mean "a Hermitian operator on the Hilbert space in which the state vector of the universe resides" then
>No, you will never, ever measure an observable of the universal wave function.
is wrong. Any observable on a subsystem of the universe can be thought of as an observable on the Hilbert space describing the whole universe simply by taking a tensor product with the identity operator.

>> No.5835199

>>5835193
Predictable in principle by an omniscient observer is different from predictable in principle by any classical measurement system, including humans. Which one are you talking about?

>> No.5835201

>>5835176
>It's the exact same situation in Bohmian mechanics. You will never, ever be able to predict the result of your measurement, because you don't have the necessary information, and it's impossible to obtain it. Yet Bohmian mechanics is deterministic, because *given this information,* the Bohmian coordinates in configuration space, you could (at immense computational expense) predict the result.
I think this is the most critical point. For Bohmian mechanics anyone can look up the method for calculating the result of a measurement given that hidden information. It's easy to find references that do this for toy examples. In fact the first four image hits on Google for "Bohmian mechanics" show the trajectories of particles through a double slit apparatus; you can see the path from initial condition to measured result. How does one do the same thing given the wavefunction of the universe? The answer is of course that you can't, as anybody remotely familiar with QM knows. You claim otherwise? Show me.

>> No.5835204

>>5835199
Hey, I'm not the one claiming it's predictable -- it isn't. It is neither predictable in principle by an omniscient observer or by a classical measurement system.

>> No.5835207

>>5835204
>doesn't answer question regarding definition of determinism

Answer the question, faggot. And for the record, it is predictable by an omniscient observer.

>> No.5835208

>>5835204
I should note that I take "predictable by an omniscient observer" to mean predictable given only the initial state vector of the universe and its dynamical laws.

>> No.5835213

>>5835208

Answer: how are you defining determinism, "Predictable in principle by an omniscient observer" or "predictable in principle by any classical measurement system"

>> No.5835214

>>5835213
I mean the first.

>> No.5835215

>>5835208
Then that is no different than a formal description of everything, which there may never be one. The actual principles founding the action are very solid, but the interaction of bodies "measuring" a superposition cause it to collapse. Precision concerns are negligible at the classical scale.

>> No.5835221

>>5835201
>How does one do the same thing given the wavefunction of the universe

Google "solving the time-dependent Schrödinger equation". 9/10 images depict plots from numerical methods.

Say you have the initial wavefunction, <span class="math">|\psi_0>[/spoiler] of a measurement device which contains a hydrogen atom whose electron's spin is <span class="math">s = \frac{1}{ \sqrt{2} } (\left|\uparrow\right\rangle + \left|\downarrow\right\rangle)[/spoiler].

Assume that in your initial condition, the measurement device is about to measure the electron's spin.

Time-evolve <span class="math">|\psi_0>[/spoiler] using the Schrödinger equation. Notice how the wave functions of billions of billions of atoms in the device interfere with the hydrogen atom's. Somewhere in <span class="math">|\psi_0>[/spoiler] is the wavefunction of the atoms of the measurement device's LCD. Learn which one illuminate. Prediction complete.

>> No.5835242

>>5835221
This thought experiment can be applied to arbitrarily large systems (like the universe), with the same results.

This is determinism, by your definition. (>>5835214)

>> No.5835246

sage

>> No.5835250

>>5835242
Ergo quantum mechanics <span class="math">is[/spoiler] deterministic after all.

/thread

>> No.5835255

>>5835221
I note that you haven't shown how to actually calculate the result. Your conclusion is wrong. You cannot predict this result by time-evolving the state as you say.

So the initial state is
<span class="math">|\psi_0> = \frac{1}{ \sqrt{2} } (\left|\uparrow\right\rangle + \left|\downarrow\right\rangle) \otimes |other~stuff\rangle[/spoiler].

Let's actually be explicit and create a toy model of measurement. We'll assume that there's only the particle and the measuring device, and everything else in the universe doesn't interact. If you want, we can follow all the states we write with <span class="math">\otimes |rest~of~universe\rangle[/spoiler] but that would be tedious. Our toy model of a measuring device is a simple two state system with states |0> and |1>, prepared in state |0>. Our initial state is

<span class="math">|\psi_0> = \frac{1}{ \sqrt{2} } (\left|\uparrow\right\rangle + \left|\downarrow\right\rangle) \otimes |0\rangle[/spoiler].

Let's write down a Hamiltonian for the toy system.

<span class="math">H = |\uparrow\rangle \langle\uparrow|
\otimes \frac12 (|0\rangle+|1\rangle)(\langle 0|+\langle 1|)[/spoiler] should do the trick.

>> No.5835257

>>5835250
Ergo quantum mechanics is deterministic after all when meassured in hindsight.

Ftfy

>> No.5835258

Let the system evolve under this Hamiltonian for time <span class="math">\pi[/spoiler].
<span class="math">e^{-iH \Delta t} = e^{-i\pi H}[/spoiler]
<span class="math">= - \left|\uparrow\right\rangle \left\langle\uparrow\right|
\otimes \frac12 (|0\rangle+|1\rangle)(\langle 0|+\langle 1|)
+ \left|\uparrow\right\rangle \left\langle\uparrow\right|
\otimes \frac12 (|0\rangle-|1\rangle)(\langle 0|-\langle 1|)
+ \left|\downarrow\right\rangle \left\langle\downarrow\right| \otimes I[/spoiler]
<span class="math">=-\left|\uparrow\right\rangle \left\langle\uparrow\right|
\otimes (|0\rangle \langle 1| + |1\rangle \langle 0|)
+ \left|\downarrow\right\rangle \left\langle\downarrow\right| \otimes I[/spoiler]
and the state of the system after the measurement is
<span class="math">\frac{1}{ \sqrt{2} } (-\left|\uparrow\right\rangle \otimes |1\rangle
+ \left|\downarrow\right\rangle \otimes |0\rangle)[/spoiler].

It's entangled with the measuring device now, but we still can't predict which measurement is made. For systems more complicated than this toy model, it will be a bit messier, but you still won't be able to see which measurement outcome you get by evolving the state vector under the Hamiltonian.

>> No.5835266

>>5835258
Now let's talk about what QM predicts for the probabilities of the measurement outcomes. Assume that it's valid to use the Born rule to figure the probability that the measuring device is in state |0> or |1>. (This will not always be true, and finding the exact conditions under which it is true means solving the measurement problem and picking a precise interpretation to work with. That's not our concern for the moment, though; we want to demonstrate how to "shut up and calculate.")

Simply, the probability of the measuring device being in state |1> is

<span class="math">\langle\psi_0| e^{iH \Delta t}
(I \otimes \left|1\right\rangle \left\langle1\right|)
e^{-iH \Delta t} |\psi_0\rangle[/spoiler]

which works out to 1/2.

>> No.5835276

>>5835255
>>5835258
>>5835266

Except that the device isn't a binary system, and its Hamiltonian isn't that simple. There are octillions of components in its wave function, and these are tightly coupled with each other and to the electron's wave function.

When the electron & device decohere, the device's initial condition will determine (with sensitive dependence) the electron's final state.

>> No.5835278

>>5835276
Prove it.

>> No.5835280

>>5835276
No Hamiltonian, no matter how complicated, violates linearity.

If |up, ready> evolves to |up, measured up>
and |down, ready> evolves to |down, measured down>
then |up, ready> + |down, ready> must evolve to |up, measured up> + |down, measured down>.

>> No.5835281

>>5835278
I can't provide calculations or numerical plots, faggot, because I'm pretty sure this has never been attempted before.

I'm pretty sure this has never been attempted because my old quantum professor is currently trying to do this, to prove that QM subsumes classical measurement and that the measurement process is chaotic (it displays sensitive dependence to initial conditions).

>> No.5835299

riddle me this.

How do we know that the operator for momentum is
-1h(d/dx) ???

Is it just a theory?? A geuss?

>> No.5835302

>>5835299
that ofcourse should say
-ih(d/dx)

but hte question remains the same.
How do we KNOW that that is the operator for momentum (in the x direction)?

>> No.5835311

>>5835299
>>5835302
Get out.

>>5835280
Idk. I've seen his formulation, it looks absolutely nothing like yours. You're not modeling the detector or the interaction correctly. I've emailed him for it, we'll see if he responds.

I suggest you troll the literature for discussions of the influence of decoherence on the measurement problem, and on interaction. I remember reading a good, non-Zurek review which discusses exactly this, and the implications for modern interpretations of QM.

It reaches precisely my conclusions, in which decoherence reformulates collapse as a dephasing process characterized by the order of the environment.

>> No.5835313 [DELETED] 

>>5835302
All separable Hilbert spaces over a complete field are isomorphic. This means the predictions of QM will be the exact same no matter which of those Hilbert spaces you use to do your calculations with.
So you obviously choose the space such that the operators have such a simple form.

>> No.5835320

>>5835313
so why not use
-hd/dx as teh operator for momentum?

>> No.5835324

>>5835320
Because you touch yourself at night.

>> No.5835326

>>5835299
>>5835302
Every symmetry gives rise to a conserved quantity via Noether's theorem, and momentum is the one coming from translation. You want a momentum eigenstate to be shifted by a phase when you translate the wavefunction, so they should look like e^(ipx). (I'm setting hbar=1.) And when you differentiate the sum of a bunch of e^(ipx) terms, you multiply each one by its respective ip.

Also it's worth working out how the expectation value of position changes with time; you should find its derivative to be proportional to momentum.

>> No.5835334

>>5835326
>You want a momentum eigenstate to be shifted by a phase when you translate the wavefunction, so they should look like e^(ipx).
you haven't justified this line.

How did people almost a hundred years ago KNOW that this was the case

>> No.5835336

Random question.

Do people come on this board and expect people to say "bazinga"?

>> No.5835354 [DELETED] 

>>5835326
>smart enough to find missing step in my argument and force me to resort to an appeal to authority, pending sleep
>feeds troll
from reluctant respect to disappointment. anon was almost not a fag today. almost

>> No.5835355

>>5835326
>smart enough to find missing step in my argument and force me to resort to an appeal to authority, pending sleep
>feeds troll
from reluctant respect to disappointment. anon was almost not a fag today. almost

>> No.5835370

Why can people never agree on what quantum mechanics is?

>> No.5835378

what's the wave function for a quark in a box?

>> No.5835386

>>5834307
Could you give us some cliff notes on what decoherence is? Math appreciated.

>> No.5835443

>>5834300
Don't ask me why, but I just read that entire post to the hoedown tune from whose line is it anyway

>> No.5835497

>>5835178
>>calls others autistic anon
>>is autistic anon

>> No.5835738

>>5835334
>How did people almost a hundred years ago KNOW that this was the case
Because that's what we mean by "momentum." It's just a word. Specifically, this definition of momentum is one that applies to quantum systems but corresponds to the classical definition of momentum at the classical limit.

There is no God defining things like "momentum" for us to discover, we make our own terms for certain characteristics.

>> No.5835741

>>5835378
>what's the wave function for a quark in a box?
No such thing.

>> No.5836830 [DELETED] 

>>5834300

>> No.5837053

>>5834300
Was a vectur....why are you eigenvectoring in this thread? What happened to my pop /sci/ discussion....the fuck is this?
>fucking engineers...

>> No.5837101

>>5837053
Excellent bump.

>> No.5838248

>>5834300
yeah but that's all just fucking saying that I've started drawing on this black box because I'm board, and fucking math just does that all the time and wonders why they can fucking see that that bs just doesnt hold up when you try to look at whats underneath

>> No.5838257

>>5838248
Excellent bump.

>> No.5838863

>>5838248
>inconceivable retard