Does Collapsing a Wave Function Require Inputting Energy into a System?

[Kidphysics]

When we collapse a wave function, say a photon into it's single particular form, are we inputting energy into a system? Is it an overall endothermic reaction i.e. is the photon more or less energetic after the collapse-do we even know this? Are we not closing it's possible locations and are we not reversing the arrow of disordered states ? Great help much appreciated very confused kid here.

 

[prajor]

Collapsing of wavefunction is a mathematical concept. It is a term used for saying probability of object, in your case photon, was something in both the paths before detection- however now that it is detected probability is 1. So mathematically it is like collecting probabilities from different paths and putting into one place. I.e. collapsing to one location.

There is no physical movement of anything, no reaction ,no heat is involved.

 

[arkajad]

Yes, the process of the collapse is accompanied by an irreversible coupling with the detectors, the energy is not conserved. This can be seen by studying multiple collapses and calculating the change of the averaged energy with time. I do not know if a general calculations has been ever done, but in simple cases when you can do the explicit calculations or simulations the average energy of the system slowly grows with time).

 

[Kidphysics]

Collapsing of wavefunction is a mathematical concept. It is a term used for saying probability of object, in your case photon, was something in both the paths before detection- however now that it is detected probability is 1. So mathematically it is like collecting probabilities from different paths and putting into one place. I.e. collapsing to one location.

There is no physical movement of anything, no reaction ,no heat is involved.

I understand this concept, however, something must change for an interference pattern to conglomerate into a single location, is this assumption correct?

Yes, the process of the collapse is accompanied by an irreversible coupling with the detectors, the energy is not conserved. This can be seen by studying multiple collapses and calculating the change of the averaged energy with time. I do not know if a general calculations has been ever done, but in simple cases when you can do the explicit calculations or simulations the average energy of the system slowly grows with time).

Can we say that we are inputting some sort of energy into the system to collapse it into one location and it is an endothermic reaction to the system?

Thank you both for the comments thus far.

 

[arkajad]

Well, the collapsing wave gives us something - some bits of information. It needs to receive something back, there is no free lunch in the universe, so it receives an energy kick. Whether there is a fixed minimal price (in terms of energy) for one bit of information - I do not know. I would rather expect that the price is one quantum of action, and that translates into energy depending on the actual exchange rate and how on impatient we are (that is how fast we want to receive our bit with some acceptable reliability).

 

[eaglelake]

When we collapse a wave function, say a photon into it's single particular form, are we inputting energy into a system? Is it an overall endothermic reaction i.e. is the photon more or less energetic after the collapse-do we even know this? Are we not closing it's possible locations and are we not reversing the arrow of disordered states ? Great help much appreciated very confused kid here.

We can do all the necessary calculations in quantum mechanics without any reference to collapsing wave functions. As far as we know, the wave function is not an observable thing. It is just a mathematical construct that allows us to determine the probability that a given experiment will yield a particular result. The wave function does not propagate in 3-space and it does not interact with photon detectors that cause it to collapse. But, the collapsing wave function is part of our attempt to explain "what is really happening" and many learned people accept it as fact.
We must not confuse the photon, or any quantum particle, with the wave function. The photon is not the wave function. The photon does exist in 3-space and it does interact with detectors. The photon has energy, momentum, and other objective attributes associated with real particles. The wave function does not. For example, I know of no way to determine the energy of the wave function. Nor does the wave function describe the behavior of the photon. For example, it cannot tell us how the photon gets from the source to the detector. Rather, it only tells us the probability of the photon getting there. Further, the wave function is determined by the entire experiment, including the measuring device, and not by the photon alone.
We must also distinguish between detecting a single photon and detecting many photons. A single photon is always detected as a particle, i.e. it is localized in space and time; we only see a single dot on a detection screen. There is no wave behavior associated with the detection of a single photon. It is never spread out over the entire observation screen. It is wrong to say that a quantum particle is a wave. In some experiments, particles do exhibit wave behavior, but we must detect many particles to see it. It is the probability distribution of many particles that looks like an interference pattern. There are web sites where you can see individual particles hitting a screen one at a time, and after many hits the interference pattern becomes evident. But, a single particle is always seen as a dot. Interference fringes begin to emerge only after many individual particles have been detected.
I'm sorry I did not answer your original question, but IMHO collapsing wave functions only create a lot of confusion. It is a can of worms!

 

[arkajad]

We can do all the necessary calculations in quantum mechanics without any reference to collapsing wave functions.

Then do it. Show me how with the "usual quantum mechanics" you will simulate particle track formation on your computer so that it will have the same essential characteristics as a real track. Or point me to a textbook or a paper where it is done...

 

[eaglelake]

Then do it. Show me how with the "usual quantum mechanics" you will simulate particle track formation on your computer so that it will have the same essential characteristics as a real track. Or point me to a textbook or a paper where it is done...

Quantum mechanics predicts the possible results of a measurement and the statistical distribution of those results. That's all there is! Wave functions are used to calculate probabilities, not trajectories. Quantum mechanics gives the probability that a particle will reach the detector, not how it gets there. There is no way to determine the quantum particle's trajectory. The uncertainty principle prevents it. As Wheeler has said, in discussing a photon experiment, "---------we have no right to say what the photon is doing in all its long course from point of entry to point of detection". The trajectory is a classical concept and you are asking me to make a classical calculation. In order to determine the trajectory, we must know the position and momentum at every instant, which is easily obtained in classical physics. But, in quantum mechanics, measuring the position causes an uncertainty in the momentum. As a result, the particle can be deflected away from its "original direction". Thus, the trajectory is a meaningless concept. Actually, it's worse than that. A quantum particle has no trajectory! Experiments that assume a trajectory are incompatible with quantum mechanics. All quantum predictions based on such experiments, as far as I know, are erroneous. Any experiment in which you can determine the trajectory is a classical experiment and has nothing to do with quantum mechanics. EPR and Bell-like experiments are examples of such erroneous thinking. But, we do keep trying to make quantum events look classical. We find it very difficult to accept quantum mechanics at face value. I think it is fair to say that we all struggle with this. There are many interpretations which attempt to bring classical reality, such as trajectories, back into quantum mechanics. But, unfortunately, nature refuses to cooperate. The fact remains that quantum particles do not behave like their classical counterparts. I suspect from your comments that you are familiar with Bohm's interpretation of quantum mechanics, which does allow for trajectories. But that interpretation has troubles of its own.
Best wishes

 

[arkajad]

Thus, the trajectory is a meaningless concept. Actually, it's worse than that. A quantum particle has no trajectory!

The result of a measurement is therefore, according to you, a meaningless concept. There are no results. There are not dots on the screen. There are no photographs.

But, if you will protest, and will say, yes there are tracks in cloud chamber - then, do please, tel me, using quantum mechanics, what is the mechanism of their formation. Try to si,ulate, using quantum mechanics, their formation in real time. And you will not be able. Because that is where quantum mechanism in the old-fashioned but still taught form fails and where you need more powerful tools, like those used by many physicists involved with quantum optics.

Formation of results, of events that are being recorded is a stochastic process with its laws. Elements of probability are being taught to physic's students. But not of stochastic processes, and especially not those that are needed in order to simulate events as they appear in real time.

 

[Kidphysics]

We can do all the necessary calculations in quantum mechanics without any reference to collapsing wave functions. As far as we know, the wave function is not an observable thing. It is just a mathematical construct that allows us to determine the probability that a given experiment will yield a particular result.

First of all thank you eagleake for the response. There is much for me to learn about the wave function and about the wave function collapse. It was pointed out to me earlier that the wave function is only but a tool used to help us determine the probability amplitude of finding something at a certain location. Your post helped me get a better feel of what a wave function is and how to really talk about it. What I should have said is not "if we collapse a wave function" which now sounds sort of silly, but instead "at the time of a wave function collapse" which, as far as I know the wave function collapse is arguably indeed something.

Well, the collapsing wave gives us something - some bits of information. It needs to receive something back, there is no free lunch in the universe, so it receives an energy kick. Whether there is a fixed minimal price (in terms of energy) for one bit of information - I do not know. I would rather expect that the price is one quantum of action, and that translates into energy depending on the actual exchange rate and how on impatient we are (that is how fast we want to receive our bit with some acceptable reliability).

This was most along the lines of my curiosity so I thank you too arkajad. I am assuming the bits of information are correlated to the position. Now this may be above my head but why is it that we get the information before we pay? In my understanding I thought that the detectors input the necessary energy, which then causes the localization of the particle, giving us information by observation.

 

[arkajad]

Now this may be above my head but why is it that we get the information before we pay? In my understanding I thought that the detectors input the necessary energy, which then causes the localization of the particle, giving us information by observation.

The exchange process is one event. So, this not a problem. The problem is that it is a nonlocal event, which to many is hard to accept because we are used to think of "events" as of a something that is pretty much localized. But quantum theory, whether we want or not, has in itself some kind of nonlocality that is usually kept in the background so that we can sleep without nightmares. The collapse mechanism brings it to the front. But there is nothing to be scared about. The relations between registered events are usually, and on average, causal. The wave collapses, but our material universe is not collapsing because of it.

 

[Kidphysics]

The exchange process is one event. So, this not a problem. The problem is that it is a nonlocal event, which to many is hard to accept because we are used to think of "events" as of a something that is pretty much localized. But quantum theory, whether we want or not, has in itself some kind of nonlocality that is usually kept in the background so that we can sleep without nightmares. The collapse mechanism brings it to the front. But there is nothing to be scared about. The relations between registered events are usually, and on average, causal. The wave collapses, but our material universe is not collapsing because of it.

Can you be specific as to which part of this process contains the non locality? I can only guess that there are non local interactions between the different probabilities as they vanish, but this as you probably already know, is a pretty wild guess. Also, I am having a hard time finding a good place to read up about what exactly a "Registered Event" actually is. From there I should be able to study more about non locality and hopefully answer more of my own questions. Thanks again

 

[arkajad]

The calculation process necessary to determine probabilities of the space-time events is non-local. It involves integration.

Register even is, for intance a detector click that happens at a certain time and at a certain place. It involves irreversibility, arrow of time.

 

[eaglelake]

The result of a measurement is therefore, according to you, a meaningless concept. There are no results. There are not dots on the screen. There are no photographs.

But, if you will protest, and will say, yes there are tracks in cloud chamber - then, do please, tel me, using quantum mechanics, what is the mechanism of their formation. Try to si,ulate, using quantum mechanics, their formation in real time. And you will not be able. Because that is where quantum mechanism in the old-fashioned but still taught form fails and where you need more powerful tools, like those used by many physicists involved with quantum optics.

Formation of results, of events that are being recorded is a stochastic process with its laws. Elements of probability are being taught to physic's students. But not of stochastic processes, and especially not those that are needed in order to simulate events as they appear in real time.

Measurement results are the essence of quantum theory. I never would suggest that, "there are no results". The measurement result is a necessary part of any quantum experiment. According to Bohr, it gives closure to the quantum event. In quantum theory there is no "before" or "after" the measurement. Yet, that is what is needed to determine a trajectory. The trajectory is a classical process involving a sequence of measurements, the results of which are then combined. Quantum events, on the other hand cannot be broken into such a sequence. Complementarity (Bohr again) requires that the entire experiment, including the particle, the particle source, the preparation apparatus, the measuring device, and the measurement result be treated as a single entity. This is what distinguishes quantum mechanics from classical physics. Quantum mechanics does not describe how the particle travels through 3-space, as does classical mechanics. It is all about probabilities. You refuse to accept that. So be it! You are in good company. I am sure you know that Einstein also wanted quantum particles to have trajectories, but, as far as I know, there is no experimental evidence for the existence of quantum trajectories. I do not know why you are bothered by tracks in a cloud chamber. Atomic particles do not always require quantum mechanics and in the appropriate experiment they behave classically. Charged particles traversing a cathode ray tube or a cloud chamber do have trajectories in agreement with the classical laws of Newton, Einstein, and Maxwell. There is nothing quantum mechanical going on. In the freshman lab we actually see the trajectory of an electron beam using a Helmholtz coil. (But, quantum mechanics is required to explain the emission of light from excited atoms, which makes the trajectory visible.) Quantum mechanics is not required when the ionizing radiation has momentum much greater than its uncertainty in momentum. All experiments that generate a trajectory measure the position of the particle with some uncertainty $$\Delta x$$. In a cloud chamber this would be the same order of magnitude as the droplet size. The Heisenberg Uncertainty Principle then gives the approximate value of the uncertainty in momentum $$\Delta p = \hbar /2\Delta x$$. If the momentum $$p$$ of the ionizing radiation is much greater than $$\Delta p$$, $$p \gg \Delta p$$, then we can measure the trajectory. Or, in terms of the deBroglie wavelength, the particle behaves in a classical way when the uncertainty in position, e.g. the droplet size, is much greater than the deBroglie wavelength. The bottom line is this: Complementarity does not allow a quantum particle to have a trajectory. Any attempt to observe a quantum trajectory is doomed to failure because it requires a sequence of infinitesimal steps, which is impossible in a quantum experiment. I suggest that we get some experimentalist to insert a photon into a very narrow tube, with diameter the same order of magnitude as its wavelength, and then see if the photon ever comes out of the far end of the tube. If the photon has a trajectory then we should be able to observe it as it emerges from the far end. Let's stop the speculation and do the experiment. Isn't that what science is all about?
Best wishes

 

[arkajad]

Quantum mechanics is not required when the ionizing radiation has momentum much greater than its uncertainty in momentum.

Quantum mechanics therefore does not cover all our reality. That is essentially your conclusion. A surrender. Then - what does cover? Are the limits sharp or fuzzy? How fuzzy?

 

[Phrak]

Yes, the process of the collapse is accompanied by an irreversible coupling with the detectors, the energy is not conserved.

Surely you did not mean energy but entropy. If not, I would like to see how you arrive at this conclusion.

 

[arkajad]

Surely you did not mean energy but entropy. If not, I would like to see how you arrive at this conclusion.

http://arxiv.org/abs/quant-ph/0506083"
Authors: Angelo Bassi, Emiliano Ippoliti, Bassano Vacchini
Journal-ref: J. Phys. A: Math. Gen. 38 (2005) 8017-8038

Abstract: A typical feature of spontaneous collapse models which aim at localizing wavefunctions in space is the violation of the principle of energy conservation. In the models proposed in the literature the stochastic field which is responsible for the localization mechanism causes the momentum to behave like a Brownian motion, whose larger and larger fluctuations show up as a steady increase of the energy of the system. In spite of the fact that, in all situations, such an increase is small and practically undetectable, it is an undesirable feature that the energy of physical systems is not conserved but increases constantly in time, diverging for $t \to \infty$. In this paper we show that this property of collapse models can be modified: we propose a model of spontaneous wavefunction collapse sharing all most important features of usual models but such that the energy of isolated systems reaches an asymptotic finite value instead of increasing with a steady rate.

I also did a computer simulation of a cloud chamber model - similar results. But the rate of energy increase is indeed model-specific, it depends on the details of the coupling.

 

[Phrak]

No, energy is not conserved in a class of quantum theories having so-called stochastic fields rather than an undecorated quantum theory. The intent of the authors is to identify the cause of energy non-conservation in these stochastic theories and propose a modified and stochastic Schrödinger equation that will still conserve energy.

 

[zonde]

The bottom line is this: Complementarity does not allow a quantum particle to have a trajectory.

You can't have trajectory with arbitrary precision. But how this is different from classical mechanics? There we will run into the problems too if we will try to determine trajectory with precision that exceeds precision with what we can determine boundaries of object.

Any attempt to observe a quantum trajectory is doomed to failure because it requires a sequence of infinitesimal steps, which is impossible in a quantum experiment.

We do not have to observe trajectory to speak about it and use it as an useful concept.

I suggest that we get some experimentalist to insert a photon into a very narrow tube, with diameter the same order of magnitude as its wavelength, and then see if the photon ever comes out of the far end of the tube. If the photon has a trajectory then we should be able to observe it as it emerges from the far end. Let's stop the speculation and do the experiment. Isn't that what science is all about?

You can't pass elephant through an eye of a needle. So elephants can't have trajectory?

 

[arkajad]

Are there any limits to the degree at which precision of a trajectory can be determined? I do not think so. At least no one claimed that there are such limits, though some people will talk about Planck length and such things. Or are we talking about practical limits today, and not of theoretical limits?

Yes, every trajectory determination has a back effect on the "particle" - but that happens also in classical physics, when we determine trajectories of celestial bodies, except there these effects are somewhat less bothering us than in the case of, say electrons.

It is necessary to distinguish between a) trajectory observation and b) back reaction.

In Bohmian mechanics, at least in the version I am familiar with, there is no back reaction, but there is no trajectory determination either. Trajectories in Bohmian mechanics "exist" even when no one tries to determine them. Why they appear in a cloud chamber the way they appear is left there as a mystery.

 

[zonde]

Are there any limits to the degree at which precision of a trajectory can be determined? I do not think so. At least no one claimed that there are such limits, though some people will talk about Planck length and such things. Or are we talking about practical limits today, and not of theoretical limits?

There are limits to any theoretical concept. To construct some theoretical concept you disregard some physical aspects in favor of simplicity. But if some of those ignored physical aspects become significant this theoretical concept will fail.

In case of trajectory you represent object as a point. Clearly this simplification will become inadequate as you go down the scale.

Yes, every trajectory determination has a back effect on the "particle" - but that happens also in classical physics, when we determine trajectories of celestial bodies, except there these effects are somewhat less bothering us than in the case of, say electrons.

It is necessary to distinguish between a) trajectory observation and b) back reaction.

I don't think that this back reaction is major issue. You can get around it if you can prepare many fairly similar situation and then make measurements at different places. But of course you will be limited to extent that is determined by similarity of those situations.
But I think that in QM one of the problems is interference effect of measuring particles.

 

[arkajad]

I don't think that this back reaction is major issue. ...
But I think that in QM one of the problems is interference effect of measuring particles.

You call `interference', I call it `back reaction'.

 

[zonde]

You call `interference', I call it `back reaction'.

I am not sure that we talk about the same thing.
I understand "back reaction" as a deflection of measured particle. Is it right?

 

[arkajad]

Back reaction is, for me, the extra term in the evolution equation of the wave function that we associate with a `particle'. This extra term is due to to the coupling with the apparatus that is described by an experimentalist in classical terms like 'cloud chamber, photographic plate, particle counter, retina etc. The apparatus is a physical `object', not a wave function, not a Hilbert space operator and not a Lagrangian or Hamiltonian. It takes some space and it acts according to some prescriptions that engineers can understand in order to be able to construct (or extracts from what already exists) a similar device.

 

[zonde]

Back reaction is, for me, the extra term in the evolution equation of the wave function that we associate with a `particle'. This extra term is due to to the coupling with the apparatus that is described by an experimentalist in classical terms like 'cloud chamber, photographic plate, particle counter, retina etc. The apparatus is a physical `object', not a wave function, not a Hilbert space operator and not a Lagrangian or Hamiltonian. It takes some space and it acts according to some prescriptions that engineers can understand in order to be able to construct (or extracts from what already exists) a similar device.

Well, I guess I understand better now what you mean with "back reaction". With this extra term you mean interference term, right?

But I am not sure if you can talk meaningfully about trajectory in context of wavefunction as trajectory is not really defined for wavefunction. Simply wavefunction is not the same as particle.

 

[arkajad]

Trajectory is what is seen on the photographic plate. In order to explain what is beeing seen in experiments we use wavefunctions. We do not see wavefunctions, that's fo sure, but we see tracks. These tracks are hard to explain in classical terms alone, by, say "traveling particles", so we use wavefunctions in order to explain what is being seen in experiments. There are two parts of the whole: one part is what we see, the other part is what we use in order to explain what we see. There is a complex mechanism of the relation between the two, a mechanism that can be organized into mathematically well defined relationships.

 

[zonde]

These tracks are hard to explain in classical terms alone, by, say "traveling particles"

Why do you think so?

 

[arkajad]

They are hard to explain for instance because classical particles are usually not created in pairs and they usually do not annihilate. Also classical particles when they come from the source to the detection device coming through a two-slit screen behave differently. You should always take into account all that is known about a given phenomenon. Not just a convenient part of it.