Decoherence

yuggib

May 11, 2016 09:59

I'm bored...I have to finish a paper but I'm lazy today

user116211

May 11, 2016 10:17

okay, I discovered that the old guy who formulated decoherence is....

user116211

can anyone guess?

user116211

No one ;_;

user116211

okay that old guy is from ACM's university! ACM must be acquainted with him ;/

yuggib

a google search reveals him immediately

user116211

@yuggib yeh!

yuggib

May 11, 2016 10:30

but I never heard of him before

user116211

@yuggib not even I. At least no undergrad QM books cover decoherence, let alone the man behind it.

yuggib

yes, but decoherence is something quite natural

even if poorly understood from a mathematical standpoint (apart very few examples)

nevertheless some of the ideas are interesting

user116211

@yuggib CuriousOne understands it VERY well.

yuggib

@MAFIA36790 he may understand the physics, surely not the mathematics behind it

;-P

user116211

I still do think decoherence doesn't explain collapse.

yuggib

May 11, 2016 10:34

well, it actually does

but that has been proved by Ozawa, following von Neumann's ideas

user116211

@yuggib ooh... then i've to wait for more studies.

yuggib

the point is that it is possible to define a measurement process (intended as a big system that couples the measurement apparatus to the observed system) that satisfies the required features of a measurement

at least it is possible, in QM systems, for every observable

and the apparatus has a scale on which you read the outcome of the measurement, and the system is suitably perturbed by the measurement, giving the apparent "collapse" of it if considered isolated

user116211

So, why do people still argue on that if it is well settled that collapse is explainable already by Decoherence?

yuggib

that is mathematically pretty well understood (even if very few people know about that I think)

because people do not know that result...also it is a mathematical result

that describes theoretically the existence of such apparati and measurement processes for any given observable

for me it's crystal clear, but I don't know what other people may think

user116211

At least one statement, I remember, from Moretti:

user116211

May 11, 2016 10:41

> The nature of the second type of evolution is still source of an animated debate
in the scientic community of physicists and philosophers of Science. There are
many attempts to reduce the collapse of the state to the standard time evolution
referring to the quantum evolution of the whole physical system, also including the
measurement apparatus and the environment (de-coherence processes). None
of these approaches seem to be completely satisfactory up to now.

user116211

Why that debate ;((

yuggib

scitation.aip.org/content/aip/journal/jmp/25/1/10.1063/1.526000

I find it satisfactory; but that is my opinion ;-)

user116211

People should stop spreading ambiguity then if the debate issue is solved ;(

user116211

Sigh

yuggib

I'm saying that I find that answer satisfactory

but I am not the repository of ultimate truth

so I may very well be wrong

user116211

May 11, 2016 10:45

@yuggib Did I sound otherwise :(

yuggib

I'm just saying I do not assert that I'm always right, and what I find satisfactory may not be such for other people

user116211

@yuggib ohh.

Secret

My current interpretation of quantum basically treat the maths as real entities. So far for most of the simple systems encountered in textbooks it worked pretty well because quantum states obey some kind of continuity equation, thus in some sense they behave like stuff in that they have some kind of conserved quantity. Under this interpretation, I found the interaction picture most natural of all the approaches

Slereah

oh no

He's going Max Tegmark

Secret

I can comprehend any concepts that are chunky, which means, can be in some way represented as elements in a block diagram

Slereah

May 11, 2016 10:52

He cannot understand love

Secret

Art is NOT one of them, though it seems somehow I do have emotions about them, but these cannot be described other than it is based on experience

Actually, I am not even sure if I understood romance, given that the minimal requirement I seek in all my friendships turns out to have qualities matches what other people called romance

Slereah

I believe in a thing called love

user116211

May 11, 2016 11:29

@CuriousOne aren't you slightly exaggerating the insights from de-coherence? It can explain why a density matrix is diagonalized, but it still cannot tell us which one of the diagonals will be selected by a measurement and that's still a non-unitary process. It doesn't solve the measurement problem. — innisfree Jul 3 '15 at 10:55

yuggib

@MAFIA36790 apart from the physics jargon, the dynamics of the full system apparatus+measured sys is unitary, as it is the isolated dynamics of the measured system

user116211

thanks for commenting @yuggib :)

Mark Mitchison

@MAFIA36790 I would agree fully with the comment of innisfree

Well, almost

yuggib

simply, to describe the measurement process happening in an interval $[0,T]$ of time you need to consider both: the resulting global evolution would be time-dependent, and the full Hamiltonian would contain an interaction term that is non-zero only in the time interval $[0,T]$

Mark Mitchison

The selection of the diagonal doesn't have to be non-unitary, since it can be a process that happens in your mind, i.e. just Bayesian updating of a probability distribution. But that implies a radical denial of realism to the point of solipsism

yuggib

May 11, 2016 11:33

(interaction between system and measurement apparatus)

@MarkMitchison still that is not necessary, since the process can be explained in satisfactory and unitary terms considering a more complete system+apparatus picture

Mark Mitchison

@yuggib That's not true, but I assume you didn't understand what I meant. The interaction between quantum systems only explains the reduction of marginal states from coherent superpositions to incoherent mixtures. My point is that it remains unexplained why only a single element of the incoherent mixture is ever observed.

yuggib

May 11, 2016 11:48

@MarkMitchison Either I do not understand the language, or that is an oversimplification. It is not true, apart from special observables with discrete spectrum, that "a single element of the mixture is ever observed".

In a measurement you observe a value on your scale that is included in an interval of the real numbers (due to the limited precision of your scale or whatever; possibly the interval may consist of just a single point)

that induces accordingly a change on the reduced state of the system (partial trace of the whole state), that it is possible to understand theoretically

and that represents the fact that the measurement yielded the aforementioned outcome lying in an interval of real values

that is how I would explain the "collapse" (actually, von Neumann and Ozawa provided the explanation)

Mark Mitchison

@yuggib Of course you are right (except that I disagree with you about continuous spectra: all actual observables have discrete spectra). What I am talking about is the process whereby one of these values or intervals of values is actually "selected", which cannot be described simply by invoking interaction with a larger system.

yuggib

@MarkMitchison well, it is hard (actually impossible) to explain how values are "selected" in a probability theory; and quantum theories are (non-commutative) probability theories

Mark Mitchison

@yuggib Yes, but in classical probability theory you can always assume that there was only one "real" underlying value, which was obscured by uncertainty over initial conditions. And therefore there is no philosophical problem with the selection process. In QM you cannot do this (unless you want to use non-local hidden variables). This is the crux of the problem. If QM admitted an acceptable realistic interpretation then of course everything would be fine.

yuggib

@MarkMitchison I don't care less about the philosophical interpretation, sorry ;-P For me probability theory is still perfectly acceptable without that assumption you make.

Mark Mitchison

@yuggib Sure, you don't have to care. But what I was talking about is the philosophical problem you don't care about, i.e. the measurement problem. So it's not true that extending the Hilbert space to include the measurement device somehow solves anything regarding this problem.

yuggib

May 11, 2016 12:03

@MarkMitchison if the measurement problem is a metaphysical problem, then it has nothing scientifical.

and nothing could help solve a metaphysical problem apart from belief and hope

Mark Mitchison

@yuggib Yep. This is of course well known.

Although perhaps it deserves to be better known.

yuggib

Maybe, but that applies also to the supposed existence of a god... ;-P

(maybe the two questions are related somehow)

Mark Mitchison

But the metaphysics is nevertheless important for science, because it frames the language in which we speak. For example, if there is nothing real or physically meaningful about "observables" such as position and momentum, perhaps we are missing out on a much more efficient description.

yuggib

For you of course they're not physical, since you said before that "actual" observables have only discrete spectrum

Secret

isn't quantum is already being experimentally proved to be a non realist theory?

user116211

May 11, 2016 12:07

Okay, if the measurement problem is metaphysical, then why not discard it? There would be no debate then.

Mark Mitchison

@yuggib I'm just saying that position and momentum cannot be measured to arbitrary precision.

@Secret No, it's been proven to be incompatible with local realism. Non-local realism is still "fine".

yuggib

@MarkMitchison Yes, so that qualifies them as not observables?

Mark Mitchison

@MAFIA36790 See my comment above, scientific interpretation is based on metaphysics. I'm not saying you should waste your time debating the measurement problem, though it is fun when drunk

@yuggib I'm just saying operators with continuous spectra are a mathematically convenient model for position and momentum observables, but there is no fundamental need for them to be continuous.

yuggib

@MarkMitchison apart that, roughly speaking, no non-continuous observables can satisfy the canonical commutation relations?

user116211

@MarkMitchison There is no waste of time as it is a physical problem and not just based on trivial metaphysical arguments.

Mark Mitchison

May 11, 2016 12:14

@yuggib Of course, but that's hardly a problem. It's clear that no deviations from the CCR would be observed assuming a sufficiently dense discrete spectrum for $q$ and $p$

Really the point I'm making is that it's a convenient, yet entirely metaphysical, choice to assume that $x$ and $p$ are really continuous. There is no experiment that could ever disprove that hypothesis.

So again it's not science. But you nevertheless find it important enough to discuss with me ;) So I think we all like a bit of metaphysics somewhere.

yuggib

Well, science is about making falsifiable statements, and the statement $x$ and $p$ are continuous observable is a falsifiable statement, at least in theory

the fact that no experimentalist is (supposedly) able to falsify it, does not make it less scientifical

Mark Mitchison

@yuggib Actually yes, that's a fair point. An experiment demonstrating violation of the CCR on some scale is conceivable. So in that case, I guess what I'm saying is that I am sceptical about the unproved hypothesis that $x$ and $p$ are continuous.

yuggib

I say supposedly because I am absolutely not convinced that you could "approximate" CCR to any given precision by means of sufficiently dense discrete spectra

@MarkMitchison Ok, I see your point. Nevertheless you have to admit that such an assumptions, i.e. the assumptions of CCR gives you a lot of very accurate and tested predictions about quantum systems

would you be able to reproduce all the aforementioned confirmed predictions by means of an alternative theory? Is it worth it?

Mark Mitchison

@yuggib In the end the CCR are irrelevant. What matters is that observable properties are the same either way. In my experience, lattice models always correctly reduce to the continuum limit when taken.

Slereah

Lattice models are done via path integrals, though

They reduce to a path integral in the end which has Ways to derive the CCR

via the class of paths that have a non-0 measure

Mark Mitchison

May 11, 2016 12:22

@yuggib No, it's of course not worth it because calculus is easy and discrete summations and difference equations are hard.

Secret

If x and p has a continous spectrum, must it also necessary mean that spacetime is also continous (as otherwise I cannot see how x and p states can be quantised at all, what kind of mechanism can induce them to be quantise like in a bound state scenario, if any?

Mark Mitchison

@Slereah Um, path integrals are defined on a lattice.

Slereah

Depends

You can define them on a lattice

Or not

Lattice is certainly easier

Mark Mitchison

If you want to actually solve them, you define on a lattice

Otherwise it's just a notation

(i;.e. shorthand for the lattice definition)

Slereah

Nah, it can work out alright if you use like

a wiener measure

(heheheh wiener)

For a somewhat restricted class of potentials, but still

Mark Mitchison

May 11, 2016 12:25

Yeah, but now you're going to undue effort just to restore the continuum description!

From the point of view of actually calculating anything in modern QFT, space and time are discretised.

Slereah

well gee sorry I didn't know doing physics was undue effort

Mark Mitchison

Of course, one does assume that the fields are continuous.

John Duffield

@HariPrasad : I know exactly how magnets work.

Mark Mitchison

@Slereah Ha, no I was more referring to the idea that it's always easier to assume continuous variables.

Formally speaking

Slereah

Though to be fair, yeah most computations are done on a lattice

It is way easier

Though you lose some in rigor

Mark Mitchison

May 11, 2016 12:28

Actually, I would say that the choice of $x$ and $p$ being discrete or continuous is arbitrary and philosophical, until such time as one could experimentally falsify one choice or the other.

And given the lack of any direct experimental evidence for continuous spectra, I choose to be sceptical.

Slereah

How do you check for that, though

You can always say that the discrete scale is arbitrarily small

Mark Mitchison

You can falsify the hypothesis that they are continuous

You can't strictly speaking falsify the converse

yuggib

@MarkMitchison and I would say that the predominance of a theory over another is given by the predictive power; and the extent of predictive power given by usual quantum mechanics, i.e. the non-commutative probability given by the algebra of CCR over finite dimensional symplectic spaces, is unsurpassed by any other alternative theory

Secret

on what scale will we expect to see p and x became discrtised?

Slereah

Planck scale, hopefully

Secret

May 11, 2016 12:31

so pretty much that question can be expt tested under quantum gravity conditions, I guess

Mark Mitchison

@yuggib The predominant formalism in modern quantum mechanics is the path integral, which to all practical purposes is based on a discrete lattice of spacetime points not a continuous manifold. And is also known to reproduce the predicitons of other formulations of QM. So I don't agree with that statement at all.

yuggib

@MarkMitchison path integral is not even wrong, since it is a non-rigorous pseudo-mathematical theory (at least at the present level of knowledge)

so statements given using path integrals are, in my opinion, moot

Slereah

Well there are rigorous path integrals

yuggib

apart from the very few cases where everything has a rigorous meaning

Slereah

The hard part is making them rigorous for interacting fields

Mark Mitchison

May 11, 2016 12:33

@yuggib Well that is what I really call metaphysics (for mathematicians). Path integrals have survived every experimental test, genuine science doesn't care about (mathematical) rigour. Or rather, science has a different kind of rigour (experimental rigour).

Secret

QFT without path integrals? I am not aware of that

Mark Mitchison

@yuggib But I can see this is another one of those "agree to disagree" situations between physics and mathematics approaches.

Slereah

there's plenty of QFT without path integrals

yuggib

@MarkMitchison I don't think so; it is an example of people saying "who cares" until they find something for which it is important to care

Secret

in physics, we cared more about whether the model fits the observations and predict new phenomenon, more than the mathematical rigor. Personally, I prefer the maths to be rigorous, as it provide a soild frame work to extend a model to fit new observations, but ultimiately it is the experiments that has the final verdict on whether a model survives

Mark Mitchison

May 11, 2016 12:36

@yuggib How do you mean?

Do you mean you expect that one day we will find a physical situation that cannot be modelled properly with non-rigorous path integrals?

yuggib

@MarkMitchison Yes exactly

Mark Mitchison

OK. But until such time, this remains metaphysics ;)

yuggib

well, you may say that physicists at the beginning modelled the volume of space we live in with a lattice of integers (and you still believe it to be true); however the discovery of the circumference of a circle put that belief in jeopardy

Mark Mitchison

@yuggib And for what it's worth I would claim the opposite: that one day some illustrious mathematician will finally make the formalism rigorous. Of course, physicists will not be that interested.

yuggib

since there is no experiment that can fix the integer value of the circumference of a circle of radius one

Slereah

May 11, 2016 12:40

But @yuggib, how will you know

With a small enough lattice it would drop below experimental error

yuggib

I am saying that no known experiment could prove that the circumference has an integer value

no matter how much precise the experiments became, there is no way of proving that

Slereah

Tell that to Leopold Kronecker

yuggib

you may still believe that the circumference has an integer value

Mark Mitchison

@yuggib Or you could just choose not to really pick a strong side, accepting that you'll never prove things one way or another.

yuggib

But I am perfectly aware of that, as is every mathematician; I am not sure physicists are so willing to accept that, and stick to beliefs and interpretations

in fact, if I produce a mathematical result, I do it in a certain theory (i.e. "choosing one side"), but I am perfectly aware it is possible to prove different results in a different theory (the other side)

Mark Mitchison

May 11, 2016 12:45

Well, I agree with the sentiment, but it's also just a practical choice. Path integrals are a wonderful tool. It's a shame not to use them just because we don't understand them fully. As soon as you prove that they aren't self-consistent, of course they'll be dropped immediately. But until then, it's crazy not to use them.

yuggib

the results have equal value for a mathematician, i.e. they are true in the given theory

Mark Mitchison

@yuggib And of course you are right that physicists like to believe and interpret a lot.

But as long as it's remembered that all interpretations must be ultimately limited by scientific rigour (i.e. agreement with empirical results), then it's ok I think

yuggib

@MarkMitchison I agree about that, nevertheless there is the risk of saying something false when you use a theory that is not well understood. My personal belief is that everything that is done using path integrals may be justified one day mathematically

Slereah

QFT is kind of weird really

yuggib

but there is always the risk of some unexpected thing popping out

Slereah

May 11, 2016 12:48

So many things in it have big theorems saying THIS SHOULD NOT WORK

Mark Mitchison

@yuggib Yes, but of course the correct approach is to use the tool until it breaks (or breaks something else), not to leave it out of fear of the unknown.

yuggib

@Slereah well everything that is used in practice would work in some sense, simply you have to be careful

@MarkMitchison Well that is your job, mine is to justify why that tool should work (and possibly understand something deeper in doing that)

Mark Mitchison

@yuggib Indeed

The h Bar

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