# Re-Emergence of Indeterminacy

1. Nov 30, 2006

### Lelan Thara

Hello, folks -

The descriptions I read for laymen say that once a particle is measured or observed, it's wave function collapses and it becomes a well defined, deterministic object. Then if that same particle remains unobserved after the intitial measurement, the wave function re-emerges and the particle becomes indeterminate again.

My questions are:

What kind of time scale doe this return to indeterminacy happen on? How long does the particle remain in a well defined state before we need to describe it with a wave function again?

Is there some period of time after a particle is measured and its wave function collapses when we can accurately measure both it's position and momentum?

Have there ever been any experiments where we have seen the wave function of a measured particle return? For instance, imagine a double version of the 2 slit experiment. We detect the particle at the slits of a first screen, making it's wave function collapse - but then it speeds on not to a detector, but to a second screen with two slits, with the detector behind the second screen. If the particle regains its wave function after measurement at the first screen, we would see interference patterns on the detector behind the second screen. Has anything like this been tried?

Thanks!

2. Nov 30, 2006

### ZapperZ

Staff Emeritus
Can you please cite the exact source where you read this? It is difficult to analyze such a thing without a clear citation. We can't tell if you're reading the wrong thing, or simply that you interpreted it wrong.

This request applies to everyone else who intend to post similar "I read somewhere..." or "I heard somewhere..." type of question.

This is rather vague to answer. However, let's be clear on one fundamental issue that you may have missed.

There is something called non-commuting observables. This is the source of QM phenomenon such as the Heisenberg Uncertainty Principle (HUP). Let's say I have two non-commuting observables A and B, represented by operators. If I measure the value of observable A, I have collapsed the wavefunction to give me a definitely value for that observable (let's call that value "a" corresponding to observable A). However, I have NOT collapsed the wavefuction for observable B. It could still be in a superposition of values for observable B.

This means that your apparent statement that an act of measurement collapsed ALL the observables associated with that wavefunction is wrong. There is STILL something "indeterminate" in whatever remains after a measurement.

So already the starting premise of this question is rather shaky, because one already does not have the scenario that you described.

Zz.

3. Nov 30, 2006

### vanesch

Staff Emeritus
This is the Copenhagen interpretation of quantum theory. It is because this is how one does practical calculations in QM, but, as you point out with your questions, there are very vague points to it (which don't matter for most applications of quantum theory in practice, but do matter on the level of principle). There are also serious formal difficulties with it (mostly non-lorentz invariance). This is bothering people already for 80 years now, and is called the "Measurement Problem" in QM.

Well, given that this "projection thing" makes the quantum object become a classical object by an undefined operation called "observation", we have no theoretical framework to describe the transition with (we're in between two frameworks: classical and quantum). Moreover, it is not clear what/when exactly an observation occurs. The only thing you can practically say, is that when there is a clear macroscopic manifestation, you can safely consider that the observation "already occured" and you won't make a numerical error.

No, because the "collapse" just puts the state into *another* wavefunction, which is an eigenfunction of the measurement operator corresponding to the observation.

Things of the kind have been done. The conclusion is then simply that no "genuine observation" has occured (you don't have the result on a paper printout) and hence no projection occurred.

I would like to point out that there are other views on quantum theory, apart this "Copenhagen interpretation". I myself am rather in favor of a "many worlds view", and other people have other views. However, all of these views also have something uncomfortable to them.

4. Nov 30, 2006

### Staff: Mentor

And we all know how accurate descriptions for laymen usually are...

When you measure a particular quantity for a particle (position, momentum, energy, angular momentum...) its wave function collapses into an eigenstate of the operator corresponding to that quantity. It is not necessarily also in an eigenstate of any other quantity. In fact, because of the Heisenberg uncertainty principle, an eigenstate of position cannot simultaneously be an eigenstate of momentum, and this will be true for any pair of quantities whose operators do not commute.

After the measurement, you still have a wave function, and it propagates in the same fashion as the original wave function, but from a new "initial state", so to speak.

For a relatively simple example in 1-D QM, construct a Gaussian wave packet for a free particle, This wave function has a width of $\Delta x$ and is a superposition of plane waves with wavelengths corresponding to a range of momenta $\Delta p$. Each one of these plane waves propagates unaffected by the others. As time passes, the details of the interference between these waves changes, such that the width $\Delta x$ increases while $\Delta p$ remains constant. The wave packet "spreads out" as it travels. (This is discussed and derived in any number of QM textbooks.)

Now suppose that at some point you measure the position of the particle so as to reduce $\Delta x$. This collapses the wave function into a new one that contains a different superposition of plane waves. Each of these plane waves propagates as before, from their new "initial state", and the wave packet starts to spread out again. The time evolution of the wave packet is a smooth continuous process, except at the moment where you make the measurement.

5. Nov 30, 2006

### Lelan Thara

jtbell said:

"And we all know how accurate descriptions for laymen usually are..."

That's exactly why I am here, sir.

Zapper - when I post questions here, I phrase things in my own words for a very good reason - if I can't summarize what I read in my own words, I don't understand it. I am looking for a critique of my own understanding, not a critique of John Gribben, Paul Davies, Michio Kaku, Wikipedia or the other sources I rely on.

I recognize this can create a level of frustration for the people who try to answer me - all I can say is I appreciate your efforts to explain things to me.

Despite the imprecision of my question, I believe you folks have answered it, for the most part. The fundamental idea I needed to grasp was that the collapse of the wave function for one property of a particle doesn't mean the collapse of the wave function for all properties. I actually knew enough already to grasp this, and I apologize if the question was naive.

However, let me rephrase my question in terms of a two-slit experiment described by John Gribben in his book, Schrodinger's Kittens.

Gribben says that we can set up a two slit experiment so that the particles are detected at the slits - either one or both slits. Gribben does not describe the methodology, so neither can I.

Gribben says that if we do this - detect the particles at the slits - that when these particle pass on to the detector behind the screen, we will not see wave interference patterns. Instead, the particles will distribute themselves as we would expect for classical particles like bullets, despite the fact that we have two open slits, which would result in interference patterns had we not detected the particles at the slits.

So, you folks are telling me that even after the detection of the particle at the slit, and corresponding collapse of the wave function that describes it position, we must still describe the particle's other properties - momentum, for instance - with a wave function.

Yet what we see is a distribution pattern beyond the slits, on the detector, that resembles classical macroscopic particles. Macroscopic particles do not need to be described with wave functions. So we seem to have a contradictory situation - a distribution pattern that seems deterministic, but a mathematical description of the particle which is not deterministic.

Can anyone explain this apparent contradiction?

Or am I just rephrasing what everyone has been scratching their heads over for close to a century?

Vanesh - I am aware there are other interpretations than the Copenhagen Interpretation - like you, I favor hidden variables models over observer created reality. Not that it means much coming from me.

Last edited: Nov 30, 2006
6. Nov 30, 2006

### ZapperZ

Staff Emeritus
But what if there is a disconnect between all those sources and your understanding? We could be haggling over something moot if you interpreted it wrong. This could have easily been solved by pointing out where the misunderstanding occured and that those sources didn't actually say that. I'm not telling you to actually quote those sources verbatim. I'm asking you to at least give a citation of the sources, which is something you have to learn how to do anyway in paying attention to physics news.

Let's just start with ONE slit, shall we?

When you have photon, electrons, neutrons, buckyballs, etc. pass through a single slit, you have essentially made a POSITION meansurement in a particular direction. If the slit has a width in the x direction, then you have essentially determined that the particle was at that x-position to be able to pass through it, and it has an uncertain in position equal to the width of the slit.

Now, what happened after it passed though the slit? For one, you actually do not know the x-component of the momentum. In fact, the smaller you make the width of the slit, the LARGER is the possible spread in the momentum that this particle can acquire. This is what I meant before when I said that the superposition of the non-commuting observable still remains.

But here's the kicker... what if you forget about momentum, but instead, try to again predict where the particle is AFTER is passed through the slit? In other words, make another position measurement like before. Can you make a very accurate prediction of its position simply because you already know where it was using the slit?

Because the particle has a spread in $p_x$, it also automatically implies that it will have an x position as undetermined as the momentum itself. All this happens AFTER an initial position measurement. This is what we notice as diffraction from a single slit.

Now, how does this jive with what Gribbin said? Gribbin was trying to illustrate that the pattern we see from one 2-slit experiment is different from the pattern we see from two 1-slit experiment! Remember, even when we do know that an electron passed through one OR the other (and not one AND the other as in the 2-slit experiment), it is STILL going through a slit that is small enough to generate diffraction pattern. What is being loosely called as "classical distribution" is such pattern, i.e. from two 1-slit experiments where the superposition of paths from the 2 different slits have been removed.

Conclusion? When you know which slit the particle passed through, then you do not get the 2-slit interference pattern, but rather simply patterns from 2 1-slit diffraction. But you STILL have superposition in terms of observables for that particular system. Everything that has been explained to you here are consistent with that.

Zz.

7. Nov 30, 2006

### Lelan Thara

Point taken. I'd just say that if we're haggling over something moot because I've interpreted it wrong - that's part of what I aim to find out.

Your overall point, if I understand it, is that even classical, macroscopic objects can be described with probabilities - that uncertainties exist even in the Newtonian world. I can see how that is true, but...

What Gribben was trying to illustrate in this example is that quantum phenomena are counter-intuitive in ways classical phenomena aren't. He uses this example of detecting particle at the slits, with both slits open, to make this point - if a single particle's wave state is able to pass through two slits and interfere with itself, common sense would tell us that this should happen regardless of whether an observer is watching at the slits or not.

But this is not the case, according to Gribben. If an observer measures the particle as it passes through the slit, the experimental results change, and the two-slit setup acts like two one-slit setups.

Gribben raised this to illustrate the concept of observer created reality. So now let me give an exact quote from the book. Gribben quotes Heinz Pagels, the president of the New York Academy of Sciences in 1981, as saying, "There is no meaning to the objective existence of an electron at some point in space, for example at one of the two holes, independent of actual observation. The electron seems to spring into existence as a real object only when we observe it!"

I don't believe any physicist would make this claim about bucky balls, or bullets, or any other classical object. Yet, you're telling me the uncertainties of particles passing through slits are fundamentally the same as the uncertainties of classical objects passing through slits.

Really, do you blame laymen who read about "electrons springing into existence", "particles that 'know' whether two slits are open", "photons that 'smell' which paths to take" and so on for being confused? For thinking of modern physics as some form of scientific mysticism?

My question is, if quantum uncertainties are analagous to classical uncertainties, is this notion sold by the popularizers - that the observer creates reality, that there is no meaningful objective reality on the quantum level - a bunch of hype designed to sell books?

Last edited: Nov 30, 2006
8. Nov 30, 2006

### ZapperZ

Staff Emeritus
I am not sure where you got from my post that these are CLASSICAL particles. Buckyballs, neutrons, protons, electrons, photons are QUANTUM particles. All of them have been show to exhibit interference patterns via analogous 2-slit experiments. These are NOT "classical uncertainties". Classical uncertainties do not show "diffraction pattern" that gets broader as the slit gets smaller.

Zz.

9. Nov 30, 2006

### Lelan Thara

So sorry - I don't know what a bucky ball is and assumed it was something macroscopic.

Nevertheless - if I fire macroscopic particles like bullets at a slit, I will get a Bell curve distribution behind the slit. It's still a distribution - it still describes the probability of finding a particle in a given area. Right?

And if my only detection of the bullet is a measurement of its position at the slit - I still have uncertainty about its momentum, and a limit of precision to predict its position in a later measurement - right?

So the analogy between the classical and quantum holds to that extent, right?

My original question, if I phrased it in Pagel's terms, was - if the particle "springs into existence" and becomes "real" when measured, how long does it take until the particle becomes "unreal" again? JTBell's answer was that this process starts immediately and the particle's position becomes increasing uncertain, in a smooth progression, over time.

Let me ask another question. Over and over I read that uncertainty is not an experimental limitation. That it is fundamental - that Pagel's statement that "there is no meaning to the objective existence of an electron...independent of actual observation" is literally true.

Can anyone explain why this must be so?

10. Nov 30, 2006

### selfAdjoint

Staff Emeritus
If you haven't already learned this, a buckyball is a molecule made up of 60 Carbon atoms arranged in a spherical configuration that recalls both Buckminster Fuller's geodesic domes and the stiching on a soccer ball, hence the name.

11. Nov 30, 2006

### ZapperZ

Staff Emeritus
That classical distribution is due to the uncertainty in the EXPERIMENT, as in measurement uncertainty. If all the classical particles have definite momentum, then in principle, the classical scenario give zero distribution. Classical physics does not prohibit the knowing of a value to an arbitrary precision. So if you have no measurement uncertainty (i.e. if you measure something, and your instrument has zero error), then you have no distribution in the value.

This is not true in quantum measurement. Even if your instrument has zero error, you WILL still get a distribution in the value of your observable depending on the experiment. It is not an instrumentation uncertainty. It is inherent in the nature of the system.

Zz.

12. Dec 1, 2006

### Lelan Thara

Zapper, the way I have understood things up to now is that with a quantum particle, when I look at non-commuting observables, I can choose one of the observables and measure it with arbitrary precision, at the expense of decreasing accuracy of measurements of the other non-commuting observable.

For example, if I have your instrument with zero error, I thought it was possible to measure the position of a particle with zero error - but that the particle's momentum could then not be measured with any accuracy.

Am I wrong about this? Am I misunderstanding "arbitrary precision" when I assume it can mean "perfect precision"?

(I understand that instruments with zero error are hypothetical - I think you will understand my question.)

Selfadjoint - thanks for the info on bucky balls.

Last edited: Dec 1, 2006
13. Dec 1, 2006

### ZapperZ

Staff Emeritus
The HUP has nothing to do with a SINGLE measurement. If you look at the definition of the HUP, it requires a statistical measurement. I can determine the position of something, and then measure the momentum of that something later on with an accuracy that is only determined by my instrument. These are instrumentation/methodology accuracy and has nothing to do with the HUP. The HUP states that if you your ability to PREDICT the next set of values depends how well you know the first measurement. This is because the more you know about the position, the LESS are you able to predict the momentum it will have. How do we verify this? We repeat the experiment many, many times and look at the statistical spread of the values of x and p. You will see that the spread in p will be larger as the spread in x is made smaller.

This is very unlike classical experiments. Under identical conditions, as the value of the first measurement of x is made more accurately, our ability to predict the momentum it will have becomes more accurate. Again, this should not be confused with the instrumentation uncertainty, which is present no matter if you're doing a classical or quantum experiments. Such uncertainty has no correlation between different observables.

Zz.

14. Dec 1, 2006

### Lelan Thara

Thanks, Zapper.

How do you feel about Pagel's statement? Do you feel it's an exaggeration to talk about "electrons springing into existence" when some observation on them is made?

15. Dec 1, 2006

### ZapperZ

Staff Emeritus
Well, many times I don't pay that much attention to the "words", unless I'm considering how to most accurately convey something. This is because when one tries to describe in ordinary human language a physics that has an underlying mathematical description, one tends to use one's own style and interpretation. Such things can often be subjective, even when the physics is actually pretty clear.

So I try to limit my involvement and pick my battles when things are simply a matter of "tastes" than anything else.

Zz.

16. Dec 1, 2006

### Lelan Thara

Thanks again, Zapper. I must admit I'm very dependent on the "words", since my math knowledge doesn't extend beyond everyday algebra and Euclidean geometry. I'm always conscious when I read about physics that I am reading a "translation" from math to English, and that was my motivation for joining here.

I appreciate you and everyone else here who are willing to try to be my "translators".

17. Dec 1, 2006

### ZapperZ

Staff Emeritus
Ooooh.. that's a good one! Do you mind if I steal that phrase?

:)

Zz.

18. Dec 1, 2006

### Lelan Thara

Be my guest. Glad I could contribute something of value.

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