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Non Life Sciences => Physics, Astronomy & Cosmology => Topic started by: geordief on 28/12/2023 17:53:46

Title: Are quantum systems dynamic?
Post by: geordief on 28/12/2023 17:53:46
A classical system is clearly dynamic.

Can the same be said about a quantum system?

Can quantum system be said to evolve (I think so)?

Do they ,then evolve in discrete ways and not in a continuous fashion except when viewed as part of a classic system?



Hope the question makes some kind of sense as I don't have a very deep acquaintance with QM.

Title: Re: Are quantum systems dynamic?
Post by: Origin on 28/12/2023 20:04:09
Can the same be said about a quantum system?
Yes
Do they ,then evolve in discrete ways and not in a continuous fashion except when viewed as part of a classic system?
Large objects are made up of quantum particles.  The individual particles act in a quantum fashion and the large object acts in a classical fashion.  By large object I am talking about 10^23 particles and larger.
Title: Re: Are quantum systems dynamic?
Post by: geordief on 28/12/2023 20:28:06
Large objects are made up of quantum particles.  The individual particles act in a quantum fashion and the large object acts in a classical fashion
Thanks. Do the quantum systems evolve in discrete  ways rather than continuously?
Title: Re: Are quantum systems dynamic?
Post by: Origin on 28/12/2023 20:39:16
Thanks. Do the quantum systems evolve in discrete  ways rather than continuously?
I thought I just answered that question.  If your asking just about quantum particles then the answer is quantized or discrete.
Title: Re: Are quantum systems dynamic?
Post by: geordief on 28/12/2023 21:33:51
Thanks. Do the quantum systems evolve in discrete  ways rather than continuously?
I thought I just answered that question.  If your asking just about quantum particles then the answer is quantized or discrete.
I think I was trying  to distinguish between quantum particles and quantum  systems.

Does a quantum system lie between a quantum  object and a classical  system?

Can that distinction  be made?
Title: Re: Are quantum systems dynamic?
Post by: evan_au on 28/12/2023 22:45:33
Quote from: OP
Do they evolve in discrete ways and not in a continuous fashion except when viewed as part of a classic system?
There may be a confusion here between:
- a quantum state, which is quantised when it is measured
- a quantum state which evolves in the absence of external influences

My simplistic understanding of quantum field theory says that a quantum object (eg an electron in a uniform magnetic field) can be described by the Schroedinger equation.
- This describes the state of the electron as a complex wave function (ie involving the "imaginary" number i), which evolves over time. This evolution is continuous rather than quantised.
- However, if you measure the spin of the electron, it will be either "spin up" or "spin down" in the magnetic field - ie it is quantised. In some interpretations of quantum theory, the measurement "collapses" the wave function, so that the imaginary part of the wave function disappears, and you are just left with the real values. The probability of finding the electron spin up or spin down can be derived from the wave function.

There are some other components of the electron's behaviour that are not quantised (as far as we can tell with current techniques): for example, the x, y & z position of the electron in an infinite vacuum.
- The probability of finding the electron at a particular position can also be derived from the wave function.


PS: I hope I haven't simplified/twisted quantum field theory beyond all recognition!
Edit: Correction by Eternal Student... see below
Title: Re: Are quantum systems dynamic?
Post by: Eternal Student on 30/12/2023 18:52:26
Hi.

I essentially agree with what  @evan_au  just said,  with just some minor differences.
So, that does mean that @Origin was also correct - in that if you kept measuring the state of the quantum system (let's say at every minute, on the minute), then you would see evolution in discrete steps.  Indeed if you were to be continuously making a measurment of the quantum system (so that is at all times and not just one minute apart with no measurement made in between) then it may be effectively frozen into one qunatum state and not show any evolution.
Some of this was discussed in another fairly recent forum post but it will be easier just to give a relevant reference here:

    The quantum Zeno effect (also known as the Turing paradox) is a feature of quantum-mechanical systems allowing a particle's time evolution to be slowed down by measuring it frequently enough with respect to some chosen measurement setting.
    Sometimes this effect is interpreted as "a system cannot change while you are watching it".  One can "freeze" the evolution of the system by measuring it frequently enough....

 
[quote taken from Wikipedia, "Qunatum Zeno Effect":      https://en.wikipedia.org/wiki/Quantum_Zeno_effect ]


     I appreciate that it is difficult to understand quantum mechanics, so I will repeat what @evan_au has said but in a slightly different way  -  in the hopes that another way of saying it may be useful.    Note that I am not really attempting to explain "why" quantum mechanics works,  I am just stating "how" it works.

This is what you do, or how quantum mechanics works:

   1.    You have a wave function.   This is something that effectively describes the system at a time t.

   2.   The wave function evolves as determined by the time dependent Schrodinger equation   (TDSE).   That is, it will evolve like this provided you do not measure some observable for the system.   Provided the potential in which the particle exists is sensible then this time evolution will be continuous and even perfectly "smooth".   (I put "smooth" in quotes because it will be as smooth as the potential allows - but we have a tendancy to look at simplified situations like square well potentials where the potential can make abrupt and infinite jumps).
     I would write down the formula to determine  Ψ(x, t+δt)   given   the initial  Ψ(x,t) and the TDSE but this forum doesn't currently support mathematical notation in LaTex script.
     The main thing is that the time evolution of the wave function is continuous (it has no sudden jumps or discontinuities),  this is simply because the wave function at a later time Ψ(x,t+δt) is determined as a definite integral  from  t to  t+δt   of the TDSE.

   3.  The wave function actually determines everything that can be known (or observed) about the system.  So it is a complete description of the system.    However, there are only certain things that can be known or observed.   Something is an observable if it can be formulated as a Hermitian operator acting on the space of wave functions.  We don't need to worry too much about the details - something like position is an observable,  something like whether the electron likes strawberry milkshake is not an observable.    As you may know, there are a few things that may seem like you should be able to know them but actually cannot be formulated as an observable:   For example, in a classical system you would expect to be able to measure the angular momentum vector  (ωx , ωy, ωz ) = the x, y and z components of the angular momentum  of some particle.   However, in a quantum mechanical system you can only measure the x, y or z components seperately and could not measure all three simultaneously.   The total angular momentum vector (with x, y, z components) is evidently a vector and so it would be represented as something like a vector-valued operator, specifically it would map the space of wave functions, H,  to  the cartesian product space  HxHxH.   To say that more clearly, the vector-valued angular momentum operator takes H  --> HxHxH.    As such there is no reason to assume the angular momentum vector would be "an observable", it cannot be described as an operator just on H , i.e.  mapping  H -> H.     It takes a little while to get used to but, in general, we just need to accept that in QM,  something is an observable   if and only if    it can be formulated as a Hermitian opertaor acting on the space of wave functions   (specifically taking H -> H  and being Hermitian with respect to the inner product defined on that wave function space H).
    In the case of angular momentum, we can find a perfectly satisfactory Hermitian operator acting on H for any one of the components   (say the z-axis component),   so that component is an observable.   Now you may think that all we need to do is go and measure the  x, y and z components seperately and then we can put all this information together and have the complete angular momentum vector BUT it's not quite that simple.  As we will discuss in a moment, once an observation has been made the wave function is said to collapse (ie. it changes),  as such any previous value for another component is no longer valid.   So, you can certainly measure the x component and get a definite value but then when you go on to measure the y component you will change the wave function and mess up the previous x component value....  it's no good trying to go back and measure the x component again because that will mess up the y-component...... etc. etc.

4.    As we have already hinted at in the discussion above,  when an observation is made then the evolution of the wave function changes its rules completely.  Under the Copenhagen interpretation of QM, a new set of rules applies:
     (i)  The evolution upto this time, t, which was determined as described in Step (2) above is not completely ignored.  Specifically the evolution as determiend by the TDSE is taken into some consideration.
     (ii)  Only certain values of the observable could be obtained.
     (iii)  Having obtained that measurement, the wave function immediately changes.  It is now entirely just the eigenfunction for the observable corresponding to that measurement.

Now let's explain each of these rules in turn:

   (i)  The wavefunction evolution by the TDSE upto this time t is not completely ignored:    The wave function is actually going to be a combination of eigenfunctions for the observable that is just about to be measured.  This combination is important.  If there was 0 amount of eigenfuction F1  in the wave function then there is no opportunity for the wave function to collapse to that eigenfunction  F1.   If the wave function was a lttle bit of eigenunction F1, then there is a chance it will collapse to become entirely F1.    The chance or probability of collapse to the function   Fi  increases as the amount of  Fi   increases.
   I would like to say the probability of collapse to Fi  is proportional to the amount of Fi  in the wave function but that isn't quite right or would need some further defintion of what we consider as "the amount" of the eigenfunction.     The wavefunction  Ψ(x,t)  at the given time t is a linear combination of the eigenfunctions:

      Ψ(x,t)  =    a F1    +   b F2  + c F3 + ...... more...
where a,b, c, .....  are called the coefficients of the eigenfnctions and I have called the eigenfunctions  F1, F2, ...

The probability of collpase to the eigenfunction Fi isn't proportional to the coefficient for Fi,  it's proportional to the square of that coefficient. ** LATE EDITING:  Actually the square of the modulus of the coefficient.   We are always interested in the size squared and we can allow complex valued coefficients.   If you're an engineer then you are probably familiar with Fourier Analysis as applied to some waveform that is a combination of sinusoidal waves of different wavelengths.   The eigenfunctions are like the individual sinusoidal waves present in the total waveform and the square of the modulus of the coefficients is much like the amplitude of the power spectrum.  What we are saying then is that the combined wavefunction collapses to an elementary waveform Fi that was present in the power spectrum and it will do this with a probablity that is proportional to amplitude of the power spectrum for that elementary waveform.  To put it more casually, it collapses to the elementary waveform Fi with probability proportional to the amount of Fi that was there.**   

   (ii)  Only certain values for the measurement are possible:   The only values possible are the eigenvalues for the eigenfunctions.  The wavefunction is about to collapse to precisely one of the eigenfunctions present in its eigenfunction expansion shown just above.  When this collapse occurs, the eigenvalue corresponding to that eigenfunction is returned as the value for the measurement that was made.
    As outlined above - it is seen that the evolution of the wave function by the TDSE upto the time of measurment was important:  It determines which values for the measurement could be obtained and how likely each such value would be.   For example, if there was no amount of an eigenfunction Fi then there will be no chance to measure the value corresponding to that eigenstate.  (I'm going to assume the simple situation where the eigenfunction expansion is non-degenerate,  i.e. you will not have an identical eigenvalue for two different eigenfunctions).

   (iii) Having obtained that measurement, the wave function immediately changes.   This is the single most important thing that happens when a measurement is made (under the Copenhagen interpretation).   The wave function just before measurement was   
Ψ(x,t)  =    a F1    +   b F2  + c F3 + ...... more...
as stated under  subsection (i) just above.    After measurment we have:
   Ψ(x, t+)  =     Fi   =   1 . Fi
where I have written  t+  to denote an arbitrary value of time that is (even just a tiny bit) greater than t (the instant when the measurment was made).   It is no longer a combination of several different eigenfunctions for the observable, it is entirely just one part the eigenfunction corresponding to the value obtained from that measurment.

(iv)   OK, I know there wasn't a sub-section (iv) but you'll want to know what happens next....   We just go back to subsection (i),   the new  wavefunction   [which was just one eigenfuction at time  t(+a little bit) ]  now continues to evolve as directed by the TDSE  until a new time is reached when another measurement is made.

- - - - - - - - - - - -
  That's it,  that's how QM works  (under the Copenhagen interpretation).   I'm not saying it seems reasonable or suggesting we know "why" it should work like this.   Indeed "wave function collapse" and the "measurement problem" is something that is extremely weird.   Why is it that the wave function evolves one way (by the TDSE) until a measurement is made and then a new set of rules comes into effect?   However, let's just be in the camp of people who just do the calculation rather than worrying about why it works like this.
 
Can quantum system be said to evolve (I think so)?
    Yes.   Without measurement the wavefunction evolves by the TDSE.   Since the wavefunction determines everything that can be known or measured for the system,  we can reasonably say that the system evolves (changes with time) if the wavefunction changes with time.
    Note that there could be some situations where the wavefunction would actually not change with time.   This is comparable to a classical system like a ball that has no initial velocity and is placed at the bottom of a valley -  the ball just sits there and is the same for all time.  We could say the ball system was showing no evolution with time if you wanted to be annoying.   I would prefer to say that we can determine the evolution with time, it just so happens that there is no change.
    If a measurment is made on a QM system then there is also some evolution of the system.  It is much more immediate, abrupt or discontinuous compared to evolution by the TDSE  (usually).   
    Note that we can prepare a QM system so that it was initially in an eigenstate for a particular measurment and have a potential such that this eigenstate is undisturbed by evolution under the TDSE.  Then provided we only measure that observable for which the system was in an eigentstate then we will change nothing (the wavefunction collpases to percisely what it was anyway).    A concrete example would be "a particle in a box" system,  let's say an electron in an infinite square well.  We can prepare the system so that initially it had precisely the energy required to be in an energy eigenstate and then it will evolve under the TDSE to remain in that state.  We can also go and measure the energy again at some later time, the wavefunction is only that energy eigenstate and nothing else so it can only collapse to that same energy eigenstate.   So, leaving it to evolve un-measured or just measuring the energy does not change the wavefunction in any way.   However, if you measured something else, let's say the position of the electron then you would immediately change the wavefunction and hence disturb the evolution in a discontinuous way.
    Overall summary:   Yes, we can determine how a QM system evolves with time.  It has different rules depending on whether you have measured something or not.  When un-measured, the evolution of the system changes smoothly and continuously with time  (for sensible potentials).  When measured the change can be sudden or discontinuous.


Do they ,then evolve in discrete ways and not in a continuous fashion except when viewed as part of a classic system?
   This is where it gets a bit interesting.   The wavefunction can evolve smoothly BUT you cannot go out and see or observe the wavefunction.    The tea cup infront of me has a wavefunction but I cannot see it, hear it or smell it.  I have no equipment with which I could measure it and indeed no Physics laboratory has the equipment to measure and record the wavefunction of my tea cup.  It is generally thought that the wavefunction is not any sort of physical thing.
     As previously mentioned the only things I can see or measure are the values of observables.    Observables are Hermitian operators acting on the space of wave functions.    When I observe something, a value is reported to me.  An observable can only report or be measured as having a value that is a single real number  (it can only be an eigenavlue for that operator and it's Real valued because the operator was Hermitian).  So we know instantly that there is no valid observable that would report the entire function  Ψ(x,t=now)  because that is a function of  x and would need to be given as an infinite set of values.   (For example,  I need to know what  value  Ψ(at x=0, t=now) is  and also what value  Ψ(at x= 1, now) is and also what value  Ψ(x=√2, now) is supposed to be  etc.  etc...   I need to know an infinite set of values for Ψ because there is an infinte set of positions, x, that I need to know about).   The best I can hope for is that there are observables that may report the real and imaginary parts of Ψ  at just one given x position and time,  t = now).    Just as with the previous discussion about angular momentum and its components,  we have no reason to beleive that we could measure Ψ(at a given x, t=now) and then go on to measure  Ψ(at another x position, t=slightly later)  without messing up the previous value for Ψ at the other x position.
    So what does this mean?   You cannot see the evolution of the wavefunction happening as it unfolds according the TDSE.  In theory it is happening, you just cannot see it.   What you can see are only the valid QM observables.  These may be things like the position of my tea cup.   Depending on the situation (e.g. the potential in which my tea cup exists), the position of my tea cup may very well be qunatised.  So it may only ever be found at x=1 , 2, 3, 4, ... positions  and never at   x = √2 position.    This means that if the position of my cup does change with time, then it could only take discrete hops from one place to another.   For typical systems, the discretisation is so fine grain that you would not notice it is discrete rather than being continuous.
    That's OK and we could leave it there but I'm sure that someone (that'll be @Halc with his "counterfactual" comments about Qunatum Mechanics) will raise an argument.    It is very human to assume my QM tea cup which I used as an example must always have some definite position but that is not actually required or suggested by Quantum Mechanics.  Just to be clear, there is no need to assume the cup had to be occupying one given position when it is not observed.  When un-obsserved the wavefunction can be in a superposition of states,  so that if the position was measured then it could be found to be at x=1 or x =2 with weighted probability such that the average or expected position did turn out to be x = √2.   This does not mean that the cup was actually at x=√2 at this time,  as stated, we do not know where it is until the position is measured so it just does not have a definite position.   However, for classical systems where many (millions) of particles may be involved, the ensemble will behave according to the average or expected positions.   So that a classical system (let's say a model for the structural integrity of my table which is supporting the tea cup) will show stress in various parts of the table as if it was supporting a mass equal to my tea cup at position x=√2 at that time.     To say that another way:   Although the tea cup could never be found at x=√2  when the position is observed,  it is possible to have a classical system that responds as if the tea cup does slide smoothly and continuously along the table rather than making discrete "hops" to different positions.

     I know that was a lot of words and may not be simple.   Here's an alternative summary:    We all want to say "yes, the wavefunction evolves smoothly and continuously when un-observed"  and then quickly suggest that's all there is to it, the system can evolve smoothly and continuously when un-observed.   However, the wavefunction just isn't a valid observable,  the only things you can observe may actually be discrete and therefore could only change (evolve in time) to take on other values in that discrete set when you do measure them.   When you don't observe them, those observables don't even need to have a definite value and we can therefore say very little about how they would "appear" to change because you just aren't observing them, they do not "appear" at all.    However, a classical system coupled to it can behave as if the observable value exists at all times (taking a value ~ the expected value of that observable) and that would show a continuous or smooth evolution.

   I hope that makes some sense and explains some of what you were asking about.

Best Wishes.
Title: Re: Are quantum systems dynamic?
Post by: Eternal Student on 31/12/2023 02:22:11
Hi again.
   
PS: I hope I haven't simplified/twisted quantum field theory beyond all recognition!
    Most of it makes sense, you haven't lead many people astray and you didn't accidentally destroy Canada on the way - so that's all good.
   It's also a lot shorter than anything I can write, which is obviously good. 

.....  the measurement "collapses" the wave function, so that the imaginary part of the wave function disappears, and you are just left with the real values.
   That's probably the bit that might give some people the wrong idea.
The measurement collapses the wave function  --->   that's OK.
the imaginary part disappears... (and stuff)...  --->  that's not quite right.

  A longer discussion will be put under this spoiler - because how on earth you ( @evan_au ) manage to make your points in just few lines just completely escapes me.   Leave the spolier and this post will look short and concise.

Spoiler: show

   The values of observables (like the z-component of spin for an electron) are hidden in the wave function in some sense but it is not as simple as being some value in the real part [ like +1/2 ("up")   or  -1/2 ("down") ]  which is cunningly disguised by the imaginary part.

    It is hidden in the wave function in that you can exhibit it by applying the operator that represents that measurement to the wave function.    When you do this,  the value of the observable will be the eigenvalue for that operator.  The collapse of the wave function reduces it to just one eigenfunction instead of being a sum of (often described as a superposition of) many eigenfunctions.
    When you think about it, it makes sense that the wavefunction is changed to be precisely just one eigenfunction of that operator when the measurement is made because then you can only obtain that eigenvalue corresponding to that eigenfunction..... let's do the example here....
     Before measurement you had    Ψ(x)  = a F1(x)   +  b F2(x)  + c F3(x)         [Equation 1]
 = a superposition of three functions   F1 ,  F2 ,  F3
     Each function,  F1 , F2   and  F3   is an eigenfunction for the operator, O that we will be applying.    So   O (F1)   =   λ1  F1,      applying the operator O to the function F1  just returns a multiple of the function F1.    That multiple, λ1 is the eigenvalue corresponding to the eigenfunction F1.    Similarly, O(F2) = λ2 F2   and  O applied to F3 returns the eigenvalue λ3 multiplied by F3.

    So if we apply the operator O  to the wavefunction just before collapse what we will obtain is a bit of mess....
O (Ψ)  =  O ( a F1   +  b F2  + c F3
   =  a O (F1)  + b O(F2) + c O(F3)          [using the linearity of the Operator]
   =  a.λ1 F1  + b λ2 F2  + c λ3 F3          [each F is an eigenfunction of O so we replace it with the apppropriate multiple of itself].
  = a bit of a mess.
 
In particlular, it is not just a simple multiple of the wave function Ψ to which we applied the operator O.  If you are incredibly lucky then all the eigenvalues are the same  λ1 = λ2 = λ3  and we can just call all of them  λ and factor that out to the front of the expression.   We would have:
O (Ψ)  =   λ (a F1   +  b F2  + c F3 )    and we recognise that  Ψ = a F1   +  b F2  + c F3   from  [Equation 1] above.   So we are incredibly lucky and we have:
O (Ψ)  =   λ Ψ,    so that the operator did (luckily) return just a multiple of the original wave function to which it was applied.   BUT most of the time, the eigenvalues are not all the same value,  so  the operator O applied to the wave function just returns a bit of a mess.

   However, if the wavefunction has been changed so that it is just a single eigenfunction of the Operator O,   so we have just that Ψ = F1      or     Ψ = F2        or    Ψ= F3  then we certainly do have what we want,   O (Ψ) =  λ Ψ    and  λ  will be (and can only be) one of  the values   λ1   ,  λ2   or  λ3  according to whether it was changed to be  the eigenfunction  F1,  F2   or  F3.     

---------
    Anyway, it can be helpful to recognise that the values of observables are hidden in the wave function in this way  (they are exhibited only by applying the apppropriate operator to the wave function)   rather than thinking that the values are somehow hidden by an imaginary part.
    If you accept that the value of an observable is hidden in the wave function in this way then there is only one way in which Quantum Mechanics can work - you must have exactly the sort of wave function collapse that the Copenhagen interpretation stipulates.    We must obtain a value for the observable, so the wave function must be an eigenfunction for the operator you have just applied.  It seems that even if the wavefunction was in a superposition of different eigenstates initially, Nature must choose only one of those eigenstates to remain or persist once the measurement is made.
    Alternative interpretations of QM exist but you must start by backing yourself into a corner and recognising that wavefunction collapse is the only way QM could work and be consistent.   Only after doing that will the other interpretations be interesting.

     Finally, it occurrs to me that I have spent a lot of time discussing an error in your ( @evan_au ) post and it may seem that I am being critical and unpleasant.   That is not the case,  I quite like what you have written and I would not have spent this much time and gone into this much detail if I hadn't thought you would be able to understand it.   Moreover, I get a better understanding of QM when I attempt to explain it and why it had to be a certain way to someone else.   I certainly need that improvement.


Best Wishes.
Title: Re: Are quantum systems dynamic?
Post by: geordief on 31/12/2023 03:50:04
@Eternal Student
just to acknowledge  your lengthy reply.

It may take me a  good few days to read and reread it.

Hopefully I will have something to answer it with then(yes,I had heard  of that Zeno effect.Quite weird .Maybe you have provided a context for me to understand it in.I'll see)
Title: Re: Are quantum systems dynamic?
Post by: alancalverd on 31/12/2023 11:38:42
I worry about "interpretations of quantum theory".

The evidence is that a lot of microscopic events occur in discrete jumps, i.e. quantum mechanics is an established fact, not a theory.

Predictions of how a system will behave are limited by Heisenberg's statement  Δp.Δx ≥ h and its equivalent reformulations in terms of time, energy, or whatever, but within the constraints of fundamental indeterminacy, we can assign probabilities derived from Schrodinger wave functions.

My point is that a wave function is a model, not a property, of a system. "Wave function collapse" just adds an unnecessary complication. The starting odds in a horse race are the summation of expert assessments, i.e. a model,  of probability. Nobody suggests that photographing the winner has any effect on the odds, but it is a measure of actual outcome. Over a large number of races, there will be very few winners at 100:1 and quite a lot of odds-on favorites in your photographs. 

Wave function collapse actually adds a new element of mystery! There must be a finite time interval associated with the collapse of a wave function. It surely can't begin before the observation, so there must be some delay between the act of observation and the outcome, and we would expect this delay to vary between observations of similar and different phenomena because each wave function is individual to the entity it describes. Is there any evidence of this? 
 
This is all quite distinct from the "observer effect of measurement". It is obvious but irrelevant that bouncing a photon off an electron is going to alter the momentum of said electron (classical mechanics!) but a lot of elementary textbooks confuse the two.
Title: Re: Are quantum systems dynamic?
Post by: Eternal Student on 01/01/2024 08:59:28
Hi.
There must be a finite time interval associated with the collapse of a wave function.
   You almost need to decide which camp you are in first.

CAMP 1:
My point is that a wave function is a model, not a property, of a system.
    You may very well be right.    As you may be aware, since the early 1900's (when QM was a new idea), people have been asking "what is the wave function actually a wave in?"   It is quite possible to consider the wavefunction just as an entirely abstract mathematical description of the state of a system.
     If it's not a physical thing then we have no reason to think it must take some time to collapse, indeed there is no reason it should be bound by any of the usual physical constraints like speed, distance and time.

- - - - - - - - -
 CAMP 2:   Then again, maybe it is something physical.   If that's what you think, then although Quantum mechanics may be a model (just like basically everything in science), it is attempting to model something that does have some physical existance.    Just to be clear then, our wave function that we have in our models may not be exactly right but there is something, lets call it the "real wave function thing" that is physical and exists and it may have physical properties and characteristics that we can identify.   It may then also be bound by the usual physical constraints of our universe, such as the constraints of speed, distance and time.   More-over it ought to show proper respect for the principle of causality.   There must be some reasonable explanation for why a wave function collpase over here can somehow fix the spin state of an electron over there.
- - - - - - - - -

There must be a finite time interval associated with the collapse of a wave function.
    Since the wave function may actually be something physical, a few scientists have looked into this.
Here's one paper that seems to be available online without cost:
Experimental test of the collapse time of a delocalized photon state ;  Garissi, Massara  and a long list of other authors ;   Scientific Reports;   Nature,  2019
https://www.nature.com/articles/s41598-019-48387-8

   I wouldn't claim it's the best article but most others are pay-walled.   I have skimmed through this one, it's OK and relevant enough to the question you asked.   Their conclusion is:
...If a nonzero collapse time is actually present, its duration Δ has to be shorter than 60 ps, which is the time resolution of the experiment.     
    To paraphrase that, the collapse time is quite possibly 0  (instant).   They could place no lower limit on the wavefunction collapse time with the time resolution they have in the experiment,  only an upper bound.

   One of their experiments was quite interesting if you want to consider that a wavefunction may actually be some physical wave in something.  It involved splitting a single photon with a beam splitter.   I've really got to keep this shorter, so we'll just describe that one casually.
     We can consider the photon as an ordinary particle, it won't split at the beam splitter, it will go one way or the other.   You have detectors at the end of the two paths.   The wavefunction is a different thing, we should have a wavefront for the wavefunction spreading up both paths.   The "detection" is some process that first triggers wave function collapse, there may be some collapse time and then the detector will click and we declare the entire detection process complete and the photon was detected.   The other detector, of course, must not detect the photon, so we have only the collapse time to have the collapse propogate through the medium.  It must propagate at least as far as the other detector.   Have the light move along paths that are on arms which can be pivoted with the detectors on the end of these arms and see what you get.  Maybe the collpase time will vary depending on how long was needed to get the wavefunction collapsed, or atleast collapsed far enough away that the other detector was not going to click.
   Anyway, that's the scene phrased casually and we can just see what their conclusion was:
    We can therefore put a lower bound to the collapse speed c′ of 1550 times c, since any other less direct path followed by the wavefunction would result in a higher collapse speed.
    So, I'll paraphrase some of that again:   The collapse of the wavefunction would seem to propagate through whatever medium in which the wave exist fairly quickly, orders of magnitude faster than light.

--------
While that's only one article and they only looked at one type of particle,  I'm inclined to think that the wavefunction is not any physical wave in some medium.   If it is only an abstract mathematical model then it has no physical restrictions and can collapse instantly.
    If the wavefunction of a particle is a physically existant thing then we probably need some very special fields that extend throughout all of space and might support it.  Quantum Field Theory already exists.

Best Wishes.
Title: Re: Are quantum systems dynamic?
Post by: alancalverd on 01/01/2024 11:50:49
I'm firmly in Camp 1 (Barth, not Lodz or Colditz) because I think the arguments of Camp 2 are absurd (inanimate objects being able to predict their fate) or so far not demonstrated (60 ps is an eternity in modern experimental physics: what were Garissi et al doing with the taxpayer's money?).

The "detection" is some process that first triggers wave function collapse, there may be some collapse time and then the detector will click and we declare the entire detection process complete and the photon was detected.
but that involves time reversal. We know that photons travel at a finite speed, so this model implies that the detection event at a distance d behind the slits determines the wavefunction at the slits d/c seconds earlier. Never mind tiny laboratory experiments: it says that every time I look up at the sky, I am influencing events that occurred somewhere between 2 seconds and 14 billion years ago! 

Time to saddle up my hobbyhorse: wave-particle duality is a dangerously misleading concept.
Title: Re: Are quantum systems dynamic?
Post by: geordief on 01/01/2024 23:31:54
@Eternal Student
I have read and reread (once) your post.

I think I may have got most of he gist of it.

So  if an observable of a quantum object is measured the waveform  suddenly  changes to a new waveform  which has a probability  attached to the "amplitude " of the particular observable at the time.

I see you listed several components of the waveform with corresponding coefficients

What might be the maximum number of coefficients for a typical quantum object  and what  might they correspond to?

Alternatively what might be a typical quantum object with very few such coefficients?

Might there  be such an object with ,say only 4(to represent spacetime)or 5 or 6 (to represent additional properties)?

Title: Re: Are quantum systems dynamic?
Post by: evan_au on 02/01/2024 01:25:14
Quote from: alancalverd
there must be some delay between the act of observation and the outcome
There is also the uncertainty principle; depending on how you structure the experiment:
- There is an uncertainty of when you actually took the measurement
- There is an uncertainty around the position of the particle when the measurement was taken, which also translates into an uncertainty of when the particle was measured, since most of our prospective measurement techniques are limited by c (you can sometimes overcome this by analysing the results retrospectively, which is what I assume they were doing...)

It is perhaps the battle against known uncertainties like this that limited the test results of (potential) wave function collapse to 60ps resolution?

Quote from: Garissi et al
We can therefore put a lower bound to the collapse speed c′ of 1550 times c
This is what Einstein criticised as "spooky action at a distance".

I assume that attempts to test the Bell inequality are measuring something related?
https://en.wikipedia.org/wiki/Bell%27s_theorem

Quote from: alancalverd
I think the arguments of Camp 2 are absurd (inanimate objects being able to predict their fate)
I assume that you also dismiss the "many worlds" interpretation (held in high credence by Sean Carroll)?
- This assumes that in measuring the arrival of a photon at a detector, the entire universe branches into two versions, one in which the photon ends at one detector, and another where the photon ends at the other detector?
- The probability that we see it arrive at one detector or the other is determined by which branch we are in
- This probability of each is determined by the wavefunction of the particle, and the measurement method...
Title: Re: Are quantum systems dynamic?
Post by: Eternal Student on 02/01/2024 02:10:17
Hi.

What might be the maximum number of coefficients for a typical quantum object  and what  might they correspond to?
   Sadly, there could be an infinite set of eigenfunctions, so that means you could have an infinite set of coefficients.
When you begin studying Quantum mechanics, you won't consider any situations with an infinite set of eigenfunctions in the basis.   
You wanted a typical example:  When the wave function is written in position basis, it is quite likely you would need an infinite set of eiegenfunctions to describe the state.    What this means is that there may be an infinite set of positions where the object could be located.
   
     
Alternatively what might be a typical quantum object with very few such coefficients?
      Probably something like the state of a system expressed in an eigenbasis of spin states.   If you were only interested in measuring the z-component of spin of your system,  then you can just express the wave function in a basis where you will only have as many eigenfunctions as the system has spin states  (so that'll be just two for an electron -> one for spin up and one for spin down).

- - - - - -
    As you may have gathered from the above comments,   a wavefuncton can be written in many ways.    Let's take an example:   
Suppose  Ψ(x) =   1 + x + x2/2    +  x3/ 6   + ....... more terms....
You would need an infinite set of  "monomials"  to represent that function   (a monomial is just a power of x).
However, I have chosen it because it just so happens to be precisely the same as    ex,   the usual exponential function.   So you need only one function to represent it if you use exponential functions.
     The wave function is the wave function and it doesn't matter which set of simple functions you allow yourself to describe it with.   However, it will matter when you start making observations on the system.   It will greatly simplify your calculations if you choose to write it as a sum of functions that are eigenunctions for the operator you are about to apply to the wave function.    Different observables will be represented by different operators and hence you will want to write the wavefunction in different ways depending on exactly which observable you were interested in.
    Let's go back to the wavefunction we have above.   Let's assume we are about to observe something,  I would like to say "position" but in all honesty monomials are not eigenfunctions of the usual position operator so instead I just need to make up an observable,  let's say the "Naked Scientists Monomial position"  or just "position" for short.    It may be that monomials are the eigenfunctions of this "position" operator.   In which case the wave function we had could collapse to any one of the infinite set of these monomials when you do measure the position of the particle.
     On the other hand momentum is represented by an operator that is just differentiation w.r.t. x  (well, with a constant but we'll choose units of measurment so that we can set "h bar" the reduced Planck constant to whatever we want).   So, well, wouldn't you know... it's almost as if I set up the example to do this deliberately....     we have   d/dx    (ex)  = ex     which means that   ex    is an eigenfunction  of the momentum operator  even though it wasn't an eigenfunction of the position operator.    Anyway, this means if we measured the momentum of the particle described by the wavefunction Ψ(x) given above, then it can only return one value and it will collapse to itself (i.e. not change in any way when we make that measurement).   The wave function wasn't a long sum (or superposition) of momentum eiegenfunctions, it was entirely just one eiegenfunction for momentum.   The particle can have only one momentum when it is measured.   

    I hope that clears up some ideas or misconceptions you might have had.   The number of eigenfunctions that are in the wavefunction actually depends on the observable (on the observation you are about to make).   What was an eigenfunction for one observable is rarely an eigenfunction for a different observable.
     So a simple electron system may be in a state that I can write  as     1/√2   |Spin Up>   +  1/√2  |Spin Down>,  i.e.   a superpopsition of just two states.   That's fine and very useful if all I want to measure was the z-component of its spin.
However, if I wanted to measure the position of the electron then although the wavefunction is still going to be the same wavefunction,  the way it has been expressed is not useful to me.

Might there  be such an object with ,say only 4(to represent spacetime)or 5 or 6 (to represent additional properties)?
   See above.   It's important to recognise that the eigenfunctions and what we call the "basis" or "eigenbasis" we use to describe the space of wavefunctions will usually change depending on the observation(s) we may be making.
    So, if  James tells you the coefficients  of the wavefunction are just  (1,2,3)  and  Harriet tells you the coefficients are actually complex numbers and there seems to be one-hundred of them,  they could both be right.    James may be considering the wavefunction with an eigenbasis for position  and   Harriet may be more interested in making measurements only on momentum.   The coefficients are tellling you something only about the amount of each eigenstate in the wavefunction for that specific observable you are interested in.   So they will tell you about the probability of finding the particle in a given position if you do choose to take a basis of position eigenfunctions.

Best Wishes.       
Title: Re: Are quantum systems dynamic?
Post by: Eternal Student on 02/01/2024 06:08:47
Hi.
   Let's try and take the latest @alancalverd and @evan_au replies together.
ES said:   "detection" is some process that first triggers wave function collapse.
Alancalverd replied:   ...but that involves time reversal....  Never mind tiny laboratory experiments: it says that every time I look up at the sky, I am influencing events that occurred somewhere between 2 seconds and 14 billion years ago!
   Not necessarily.   We could try to make a quick reply to Alancalverd's point (maybe a paragraph) or else we can set a firmer background to make everything clearer and then hit the comment much more squarely.   We're going to go for a half-and-half approach.   One page to set the background and one page to answer Alancalverd's comment.

    Probability, you know what it is.   When you throw a coin it has a probability of one-half to come up heads.   However, the process of throwing a coin is probably deterministic rather than completely random.   If you could identify precisely how it was thrown and knew all the releant parameters like mass, radius and shape of the coin etc.   and also had a fast computer to do all the Newtonian calculations... then you might actually be able to determine if the head or tail was going to be face up when it hit the table before it had actually hit the table.
   So what we really need to do is hypothesise that there is some physical process whch is genuinely random, let's assume a coin toss is actually like that.

   Now if you toss a coin the outcome is genuinely random (before it lands).  Let's say you toss the coin and it comes up heads.   Now what is the probability that the result was a head?  It's 100% isn't it?  I mean it's happened, it's a fact now not a random event.
    So, just to be clear, these are the "rules" of our coin toss:
   Before it lands on the table,  the probability of getting a head or tail is  1/2  and 1/2 and there is no way you can determine what it will be, for example there are no physical parameters in the universe you could measure or thing that determines the outcome above and beyond randomness and those probabilities which you can assign.   If you don't make that coin toss today but wait until tomorrow, it doesn't really change anything.   The probability of getting a head is still going to be 1/2 when you do finally make the coin toss.
   After it lands on the table the probability changes.  If it was a head, the probability of it being a head is now 100%,  similarly if it was a tail then there is a 100% probability of that coin toss giving you a tail.   Tomorrow, it will be the same, the result of that coin toss is just locked in,  whatever it was, that's what it was.

   Anyway, that's how probability should work and we tend to take it forgranted.   Probability changes in sudden and often discontinuous ways.   The probability of a coin toss coming up heads is not 50%,  it is only 50% before the coin lands.   Additionally, once the coin has landed no-one seems horrified and exclaims "...but that's just changed the past... the probability of it coming up heads while it was in the air wasn't 1/2."   We accept that the probability was 1/2 before it landed and so it is and forever more will be 1/2 at time t=5 minutes ago,   we have not changed the past,  we just have a different probability for the coin toss result at time t = now.

   Now, if you had a time machine (which we don't but we are hypothecating) and went back in time 5 minutes, then our hypothecated coin toss behaves in an interesting way.   You may have seen the coin come up heads before you got in the time machine but that doesn't matter.   The rules of our hypothecated coin toss were set out very clearly,  there are no deterministic rules for it,  it is genuinely random and that is all.   At the time t=5 minutes the result of that coin toss is genuinely random and it might come up tails with a probability of 1/2.   If it was any other way, then we have in fact actually done what the person exclaimed   "we just changed the past when the coin landed and came up heads".   Our hypothecated coin toss follows a simple rule - it is not deterministic before the coin lands, it is genuinely random.  If the time traveler has a coin toss that behaves differently and will come up heads with 100% probaility then they are not in our universe because there is atleast one object in it that is not following the same rules.

    This was a simple example of what "random" and "probability" means.  In the real world a coin toss, as I stated, probably isn't genuinely random, it may very well be deterministic.   There are incredibly few things in nature that are genuinely random, a whole lot of things are likely to be determinstic but with so many variables that it only appears to be random.   Indeed it is very likely that Quantum Mechanics provides us the main example of something in Nature that is genuinely random and probabilistic rather than deterministic.

   With this understanding, a lot of alancalverds comments should evaporate.   When you observe light from the night sky you have not changed the past.   There just has been a sudden and discontinuous change in the probability for certain things.   The coin has landed.    The probability for things to have been some way at time t=1 minute ago has not been changed.   1 minute ago (or maybe longer) when that photon was passing through some slits it could have gone some other way.   Indeed if you could get in your time machine and just wind back the time,  then when you look up in the night sky you might not actually see the photon this time around.

   I don't know, as a human being we may assume that everything was deterministic and possibly as an engineer you would want to assume that even more so.   If your bridge collapses, there ought to have been some reason and some way you can trace the changes in the stress back through time to identify exactly which bit failed and why.   Quantum Mechanics suggests that maybe the universe isn't governed by the hard rules that we would like it to be.  On a microsocopic scale Nature is genuinely random or probabilistic (and if it is, then a lot of Science Fiction ideas based on time travel aren't really going to work, knowing how a coin toss landed first time around may not help you).

- - - - - - - - -


ES said:    We can therefore put a lower bound to the collapse speed c′ of 1550 times c
Evan-au replied:     This is what Einstein criticised as "spooky action at a distance".
I assume that attempts to test the Bell inequality are measuring something related?
https://en.wikipedia.org/wiki/Bell%27s_theorem

Yes.
There are some ways around the problem of "spooky action at a distance" and causality violations that seem to arise from Quantum Mechanics.   This is one thing (there are others):
    We need to be carefull about what we identify as the "cause" of something else.   The Copenhagen interpretation doesn't have to be phrased as "Measurement causes wave function collapse",  all we need is that there is wave function collapse when a measurement is made,  exactly what caused all of this is a separate issue.
   For example, you can often get around problems by just considering the quantum system to be slightly more than you thought it was.   The observer or device that did the detection is itself a quantum mechanical object and there should be one (large) wavefunction that describes the entirity of this larger system.  By extension there is one wave function for the universe.

I assume that you also dismiss the "many worlds" interpretation (held in high credence by Sean Carroll)?
   I don't know about alancalverd but I quite like the many worlds interpretation.
- This assumes that in measuring the arrival of a photon at a detector, the entire universe branches into two versions, one in which the photon ends at one detector, and another where the photon ends at the other detector?
  I don't know why there's a question mark at the end of this.  Yes.   Although, there's really no reason why the branching should only occurr when there was a set of different outcomes.   The branching may just happen all the time, say every second (possibly using co-moving co-ordinate time).   It's just some nominal attempt to be as economic as you can be with the creation of whole new branched universes, you do only need them when there was a different set of outcomes possible.   I'm really not sure that Nature has the same view of things as a human being and sees the creation of new parallel universes as some work and therefore wishes to make as few as possible.

Best Wishes.
   
Title: Re: Are quantum systems dynamic?
Post by: alancalverd on 02/01/2024 12:08:58
I assume that you also dismiss the "many worlds" interpretation (held in high credence by Sean Carroll)?
Again, it's a mathematical model and doesn't imply that there are many worlds! Feynman used to talk about multiple alternative paths, and that nature followed the path of least action, but never said why!

I can distinguish between probability, our assessment of the distribution of outcomes of many future trials, and fact, the actual outcome(s). Fact is that the photon that hits the screen always has the same energy as the source, so you need an "indivisible particle" model to describe each interaction event, and the distribution of those events in space is best modelled by a wave propagation equation that gives you a probability function. One world is enough for me and the bookmakers.
Title: Re: Are quantum systems dynamic?
Post by: Zer0 on 03/01/2024 18:37:57
@Eternal

I donno Why you did it, but Luckily for me, a Coin Toss Analogy is an Opportunity not to be missed!

I have a few Basic doubts, shall number them out, even if You respond with Yes/No/Maybees it's Totally Fine.

1) If a Billion people Toss, someone might get Heads a hundred times in a row.
Sheer chance or Probability?

2) While Coin is in air, is it Heads & Tails both at the Same time.
Superposition?

3) Say the Coin is tumbling down Stairs(t1 t2 t3 t4 t5)
If i exist inside of Only t3, i see Heads, but Others existing inside t1245 can see anything at Random(headsORtails)
Simultaneity?

4) A Coin has more than 3 sides.
What if it lands on the 3rd side.
What i am tryin to ask is, can/has the Wave Function ever predicted an Inaccurate result, or is it 100 % Correct every single time.
Uncertainty?

5) Has it Ever happened that a Coin was tossed & it Disappeared into thin air & never landed.
(lol sorry)

ps - i sincerely appreciate all the time n effort you put into your Posts.
Surely Everyone here notices it.
On behalf of All of Us, Thank You Very Much!
Title: Re: Are quantum systems dynamic?
Post by: alancalverd on 03/01/2024 19:52:24
5) Has it Ever happened that a Coin was tossed & it Disappeared into thin air & never landed.
Eddington said that the student of physics must become accustomed  to having his common sense violated five times before breakfast, and if he disappeared through the floor and rematerialised in the cellar, he would not consider it mysterious but merely a lucky observation of a very rare phenomenon.
Title: Re: Are quantum systems dynamic?
Post by: alancalverd on 03/01/2024 20:02:31
1) If a Billion people Toss, someone might get Heads a hundred times in a row.
Sheer chance or Probability?
The probability of scoring 100 consecutive heads is obviously 1/2100, which is a lot less than a billion. You would have to do the trial a billion billion times to be reasonably sure that it happened.

Your other examples underline the difference between propagation and interaction (2 and 3)

(4) doesn't make sense. A quantum wave function extends throughout the whole of space but just gets very small at long distances.
Title: Re: Are quantum systems dynamic?
Post by: geordief on 03/01/2024 20:23:16
If we observe 2 stars at a very great distance  and the light arriving from them only amounts to a very low number of pixels , are there any quantum effects involved  in the observation?

Let's say it a is binary system.

How many pixels would you need to form any kind of a judgement  as to position and relative moment between them?

Would quantum effects come into play if the amount of light reaching the receptor was small enough?
Title: Re: Are quantum systems dynamic?
Post by: evan_au on 04/01/2024 08:17:47
Quote from: ??
Let's say it a is binary system.
If we observe 2 stars at a very great distance  and the light arriving from them only amounts to a very low number of pixels , are there any quantum effects involved  in the observation?
Let's take the star Algol as an example (an Arabic name meaning "the ghoul"; sometimes called the demon star)
- It is 90 light-years away, so the diameter of the star's image on your retina would be less than the wavelength of visible light.
- Every 2.86 days, the brightness visibly halves (violating assumptions at the time about the unchangeable nature of the heavens)
- This is due to eclipsing binary stars

I don't see too many quantum effects here (apart from the fusion powering the star)
- Although the image is very small on your retina, it is very large at its source
- The finite size of your iris means that the image will be blurred to a larger size
- Passing through the atmosphere blurs the image considerably
- It doesn't matter how small the image is on your retina. if you can measure the brightness, you don't need the resolution needed to separate the stars in a telescope.

https://en.wikipedia.org/wiki/Algol
Title: Re: Are quantum systems dynamic?
Post by: alancalverd on 04/01/2024 12:40:52
Quantum effects were very important in film-based astronomy and radiology.

To form an image in a silver halide crystal, you need  at least two photons to arrive within a fairly short time. In "normal" photography this isn't a problem as you have zillions of photons arriving in a fraction of a second, so the film blackening is proportional to the number of photons hitting each grain. But at very low intensities or very short exposures, you get "reciprocity failure" where the effect of the first photon decays before the second one hits the grain,or the exposure is too short for the second one to arrive at all. The astronomer's trick was to "pre-fog" the photographic plate with a hint of visible light, thus reducing the theoretical detection threshold to a single photon. The radiographer's trick was to convert the few high-energy x-ray photons to lots of visible photons and capture those, and at very low doses with an efficient fluoroscopic converter, you could see "quantum mottle" where each incoming photon had generated an isolated spot on the film.

In theory you can detect a single visible photon with a cooled photomultiplier or semiconductor - the latter being more common with x-ray photons.
Title: Re: Are quantum systems dynamic?
Post by: Eternal Student on 05/01/2024 03:48:56
Hi.
   I think many of your ( @geordief  )  questions have been reasonably answered by others already.  Here is just a quick reply to each.

1) If a Billion people Toss, someone might get Heads a hundred times in a row.
Sheer chance or Probability?
   I don't understand what you see as the difference between the words "chance" and "probability" as used in your sentence,  sorry.

2) While Coin is in air, is it Heads & Tails both at the Same time.
Superposition?
    If you treat the coin as a QM object then you could consider it as being in a superposition of states while in the air.    Most people wouldn't treat a real world coin as a QM object, it's a macrosocopic object.

3) Say the Coin is tumbling down Stairs(t1 t2 t3 t4 t5)
If i exist inside of Only t3, i see Heads, but Others existing inside t1245 can see anything at Random(headsORtails)
Simultaneity?
   Is this a metaphor for people who exist in different universes as described in the many worlds interpretation of QM?   If it was then yes,  they can see different results and they see them simultaneously.   The branched worlds were just the one world until branching occurred.  So whatever event or outcome lead to the branching, that should have happened when the same past applied to all the worlds created in that branching.   Since the different worlds should not influence one another after branching, it's a bit arbitrary to ask if their time co-ordinates or flow of time would continue to be synchronous afterwards.   However, the same laws of physics should apply in all branches.

What i am tryin to ask is, can/has the Wave Function ever predicted an Inaccurate result, or is it 100 % Correct every single time.
Uncertainty?
    For certain, physicists sometimes make inaccurate predictions.   Furthermore, Quantum Mechanics is almost certainly not the final or ultimate theory that models and explains everything corrrectly.
    There are many inherent problems with Quantum Mechanics.   Problems concerning measurement and wave function collapse have been discussed elsewhere.   Other problems include assuming a particle does have a mass - which you will need to do when constructing the Hamiltonian that appears in the Shrodinger wave equation.   QFT is one extension of (basic) Quantum Mechanics where "mass" doesn't have to be an inherent property of particles (it can be just an interaction from a field).   QM, even in the more developed form of QFT, still doesn't reproduce results consistent with all of General Relativity (GR).

5) Has it Ever happened that a Coin was tossed & it Disappeared into thin air & never landed.
(lol sorry)
   Maybe.  QM predicts some incredibly strange things.

If we observe 2 stars at a very great distance  and the light arriving from them only amounts to a very low number of pixels , are there any quantum effects involved  in the observation?
   I think most of us are interpreting  "low number of pixels" as indicating only a low number of photons were also received.
   Yes...   for a small number of microsopic objects like photons, all the usual quantum mechanical behaviours could or should apply.   However, they are stated as having been detected by the receptor.   So measurement has been made, the wave function has collapsed, the photon is definitely at the detector and should behave much as you expect an ordinary particle to behave.

How many pixels would you need to form any kind of a judgement  as to position and relative moment between them?
    This sounds very specific to the equipment being used and what was being attempted.   The light probably went through some imperfect lens before it was brought to focus on this light recpetor,  it almost certainly went through a lot of air in our atmosphere.   QM effects that may happen involving detection at a pixel aren't the most limiting effect.
    Lots of information already exists on the internet about telescope and detector resolution limits. The wave-like behaviour of light (inlcuding X-ray, radio or any wavelength) generally puts the main constraint on "angular resolving power" or resolution but there will also be particle-like behaviour and QM effects.  @alancalverd has already said quite a lot about this.

Best Wishes.
Title: Re: Are quantum systems dynamic?
Post by: alancalverd on 05/01/2024 09:21:10
If we observe 2 stars at a very great distance  and the light arriving from them only amounts to a very low number of pixels
Don't confuse between photons (what the source emits) and pixels (how the receiver is constructed).
In a direct-conversion system any pixel can be activated in principle by a single photon. The art of receiver engineering involves optimising the ratio of Pixel width to Boundary width within the constraints of the manufacturing process. If P>B you will lose resolution because each pixel will respond to photons from a wider area than necessary, and if P<B you will lose sensitivity because some photons will hit the boundary rather than the pixel.

How many pixels would you need to form any kind of a judgement  as to position and relative moment between them?
Two. If we initially detect intensity I1 at pixel A and 0 at A + 1, then subsequently detect I A < I1 and I A+1 > 0, we can infer that the binary is rotating.
Title: Re: Are quantum systems dynamic?
Post by: Eternal Student on 05/01/2024 16:49:53
Hi.

any pixel can be activated in principle by a single photon.
    I'm not an engineer and I honestly don't know what sort of detection devices exist so I went shopping on the internet.   Does "in principle"  mean  ?4, 000   (four thousand GBP) per pixel?
 [ Invalid Attachment ]

I also thought these SPD  (single photon detectors) were considerably less than 100% effective.   Sometimes only about 10% effective.   The one in this advert claimed 60% peak detection efficiency but there is a "dead time" of at least 50 ns after each detection during which the SPD will not click on impact with a photon. 

If we initially detect intensity I1 at pixel A and 0 at A + 1, then subsequently detect I A < I1 and I A+1 > 0, we can infer.......
    ..... that the pixels just don't click with 100% effectiveness?   that the air changed its density locally?   or maybe that the source image was rotating.

Best Wishes.
Title: Re: Are quantum systems dynamic?
Post by: Zer0 on 05/01/2024 19:34:27
Hi.
   I think many of your ( @geordief  )  questions have been reasonably answered by others already.  Here is just a quick reply to each.

i had Observed this before, in another Thread, you perhaps were referring to me, but you ended up typing Origin's nickname.
I thought it must be a typo or system error.
But here We are, it seems to have happened again.
Instead of my nickname, you typed Geordief's.
But i still Wish to confirm this, was this a typo, or is there a Glitch in the System?

Coz in another Thread, i had typed Varsigma's nickname expecting a response from them, but instead Neilep ended up responding.

Is This phenomenon taking place with Anybody Else?
(to clarify a bit, i type username @abcde...i see it as @abcde.
But the user @abcde sees it as @1234567.
& eventually user @1234567 ends up responding to me, on a query, which i had asked to user @abcde)



1) If a Billion people Toss, someone might get Heads a hundred times in a row.
Sheer chance or Probability?
   I don't understand what you see as the difference between the words "chance" and "probability" as used in your sentence,  sorry.

There is No difference.
Sorry i confused you.
Rephrase - if a billion people toss the Coin, one of em gets heads 10 times in a row.
This is simply Probability Right?
Not a Miracle!



2) While Coin is in air, is it Heads & Tails both at the Same time.
Superposition?
    If you treat the coin as a QM object then you could consider it as being in a superposition of states while in the air.    Most people wouldn't treat a real world coin as a QM object, it's a macrosocopic object.

Most people are Sane enough to Not do that.
Agreed!
((2) - got it)



3) Say the Coin is tumbling down Stairs(t1 t2 t3 t4 t5)
If i exist inside of Only t3, i see Heads, but Others existing inside t1245 can see anything at Random(headsORtails)
Simultaneity?
   Is this a metaphor for people who exist in different universes as described in the many worlds interpretation of QM?   If it was then yes,  they can see different results and they see them simultaneously.   The branched worlds were just the one world until branching occurred.  So whatever event or outcome lead to the branching, that should have happened when the same past applied to all the worlds created in that branching.   Since the different worlds should not influence one another after branching, it's a bit arbitrary to ask if their time co-ordinates or flow of time would continue to be synchronous afterwards.   However, the same laws of physics should apply in all branches.

& what if t12345 was Time, and i checked heads/tails 5 times.
Will i see a Pattern emerge?
t1h  t2t t3h t4t t5h
Or
t1t t2h t3t t4h t5t
Or
t1h t2h t3h t4h t5h
Or
t1t t2t t3t t4t t5t
Will i be able to Predict future readings of t6 t7 t8 t9 t10 based on emergent pattern or Do the Readings remain Absolutely Random without predictable Patterns?
(t/1/h --- time/1/heads
&
t/1/t --- time/1/tails)



What i am tryin to ask is, can/has the Wave Function ever predicted an Inaccurate result, or is it 100 % Correct every single time.
Uncertainty?
    For certain, physicists sometimes make inaccurate predictions.   Furthermore, Quantum Mechanics is almost certainly not the final or ultimate theory that models and explains everything corrrectly.
    There are many inherent problems with Quantum Mechanics.   Problems concerning measurement and wave function collapse have been discussed elsewhere.   Other problems include assuming a particle does have a mass - which you will need to do when constructing the Hamiltonian that appears in the Shrodinger wave equation.   QFT is one extension of (basic) Quantum Mechanics where "mass" doesn't have to be an inherent property of particles (it can be just an interaction from a field).   QM, even in the more developed form of QFT, still doesn't reproduce results consistent with all of General Relativity (GR).

Cordially agreed!
QM is Incomplete.
((4) - got it)



5) Has it Ever happened that a Coin was tossed & it Disappeared into thin air & never landed.
(lol sorry)
   Maybe.  QM predicts some incredibly strange things.

Strange things like Virtual Particles?
That pop into existence from nothing & go back into nothing.
If VPs are truly physical, Why then does the Universe not glow with a Blinding light when matter/antimatter VP pairs collide?

Or VPs just exist in 2d sheets of mathematical equations & the Energy released from anti/matter collisions remains trapped behind thick book covers, busy illuminating the bookmarker Within.



Best Wishes.
Much Appreciated & Thanks!

Title: Re: Are quantum systems dynamic?
Post by: evan_au on 05/01/2024 20:45:11
Quote from: Eternal Student
I also thought these SPD  (single photon detectors) were considerably less than 100% effective
A cryogenically-cooled semiconductor sensor as used by astronomers has something like 70% photon-capture efficiency, at chip level.
- You have to treat any lens or window with anti-reflective coatings, or you lose another 10% of photons due to reflections at the air-glass-air boundary
- The mirrors on the telescope lose about 10% of the photons before they hit the detector (which is why they recoat the mirrors every year).
- More is lost due to the prime focus mirror, and its support structures

These semiconductor sensors have the advantage that they can form an image of a region of the sky, rather than recording the intensity at just one point
- One amazing observatory that should start operations in about a year has a 3 Gigapixel camera, formed from a matrix of semiconductor sensors.
https://en.wikipedia.org/wiki/Vera_C._Rubin_Observatory#Camera

The example you found appears to be a photomultiplier tube, which (in the past) tended to be a single pixel, ie not an image-forming device? (I am sure alancalverd will know!)
https://en.wikipedia.org/wiki/Photomultiplier_tube

Quote from: Zer0
Will i be able to Predict future readings of t6 t7 t8 t9 t10 based on emergent pattern
As far as we can tell, quantum phenomena are truly random.
- There are some interpretations of Quantum Theory that involve "hidden variables" - if you could peek at these, you may be able to predict the outcome - but peeking at these quantum variables changes their value
- It's a bit like peeking at the unseen Wave Function - it collapses
Title: Re: Are quantum systems dynamic?
Post by: Eternal Student on 05/01/2024 22:34:54
Hi.
Instead of my nickname, you typed Geordief's.
   I.T. support would say it was a PEBKAC issue (problem exists between keyboard and chair).   
I'm going to claim the forum software didn't make it easy.    When the post gets long it splits replies into separate pages.  I could only see the name on the immediately preceeding post when I was writing the reply.   I can only apologise @Zer0.
   
Title: Re: Are quantum systems dynamic?
Post by: alancalverd on 05/01/2024 23:27:27
Quote from: alancalverd on Today at 09:21:10
If we initially detect intensity I1 at pixel A and 0 at A + 1, then subsequently detect I A < I1 and I A+1 > 0, we can infer.......
    ..... that the pixels just don't click with 100% effectiveness?   that the air changed its density locally?   or maybe that the source image was rotating.
Detection efficiency isn't relevant - you have to integrate over a reasonable period to determine intensity.
Air density and turbulence may have been problems for neanderthal astronomers, but any respectable Waitrose customer buys his data from space telescopes nowadays.   
Title: Re: Are quantum systems dynamic?
Post by: alancalverd on 05/01/2024 23:35:26
The example you found appears to be a photomultiplier tube, which (in the past) tended to be a single pixel, ie not an image-forming device? (I am sure alancalverd will know!)
Not really even a pixel, though we have used arrays of photomultipliers to determine the location of an event in a scintillator crystal or swimming pool full of carbon tetrachloride*, and display the resulting vector as a pixel on a flat screen.

*"micellar water" be damned! a million gallons of CCl4 will really shift a layer of makeup. Wax on, wax off.
Title: Re: Are quantum systems dynamic?
Post by: evan_au on 06/01/2024 22:38:26
Quote from: alancalverd
Air density and turbulence may have been problems for neanderthal astronomers, but any respectable Waitrose customer buys his data from space telescopes nowadays.   
If you want high resolution images of really faint objects, the best option is still to use really large telescopes on top of really high mountains (a number of them in the Atacama desert of South America).
https://en.wikipedia.org/wiki/List_of_largest_optical_reflecting_telescopes

These telescopes use adaptive optics and high-powered computing to decode the atmospheric distortions in real time, and adjust the shape of a compensating mirror every millisecond or so to cancel the atmospheric distortion.
- This is most useful on telescopes of 8m diameter and above (a 40m diameter telescope is under construction in Chile, intended to be operational in 2028)
- I am assuming that the calculations involved treating light from the star as a classical light wave passing through a classical medium with time-varying refractive index, so no quantum effects need to be calculated.
https://en.wikipedia.org/wiki/Adaptive_optics#Wavefront_sensing_and_correction
Title: Re: Are quantum systems dynamic?
Post by: alancalverd on 06/01/2024 23:13:08
- I am assuming that the calculations involved treating light from the star as a classical light wave passing through a classical medium with time-varying refractive index, so no quantum effects need to be calculated.
At some point in the system, you need to generate an electrical signal representative of the wavefront distortion. This involves a photon-electron converter (usually a CCD), which, like all photon detectors, employs quantum phenomena!
Title: Re: Are quantum systems dynamic?
Post by: varsigma on 07/01/2024 10:45:51
There's this astronomical effect called the Hanbury Brown-Twiss effect. Something about interferometry from optically distant light sources that you could look at, that might be connected to the question.

It's both classical (it's observed at radio frequencies I.O.W), and quantum (it involves "photon bunching"),
Title: Re: Are quantum systems dynamic?
Post by: Zer0 on 08/01/2024 17:21:44
@Evan

Hmm...so no playing peek-a-boo! in the Quantum world.
 : /
booring!


@Eternal

haha@PEBKAC
But loong hours at the Table is not nice, wrist bones n elbows might ache.

Why don't you get one of those automatik massager chairs?
hehe@TickleTickle
: )
(sending Apology back to Sender as refused to be collected by the Recipient)


@Var
Thnx for your Valuable input.

https://en.m.wikipedia.org/wiki/Hanbury_Brown_and_Twiss_effect


ps - @G
i wish you'd speak your mind a bit more.
cya!