What if the Bohmian model turned out to be correct?

[Schrodinger's Dog]

Isn't a photon even in QM the most fundamental of particles. I always thought the reason everything would end up under an infinitely expanding universe as photons and then as the fundamental sea of the vacuum because, that form of energy is the lowest common denominator of all matter. A particle and antiparticle pair is as far as it goes, excluding string theory, which I'm not a fan of anyway atm.

 

[Demystifier]

Isn't a photon even in QM the most fundamental of particles. I always thought the reason everything would end up under an infinitely expanding universe as photons and then as the fundamental sea of the vacuum because, that form of energy is the lowest common denominator of all matter. A particle and antiparticle pair is as far as it goes, excluding string theory, which I'm not a fan of anyway atm.

Well, that is wrong. A photon is not more (and not less) fundamental than an electron. The standard model of elementary particle contains a LOT of equally elementary particles (photons, gluons, W bosons, Z bosons, electrons, muons, tau, e-neutrinos, mu-neutrinos, tau-neutrinos, 6 types of quarks, Higgs). In fact, one of the main motivations for string theory is to derive all these seemingly "elementary" particles from a SINGLE elementary object - the string.

 

[Schrodinger's Dog]

Well, that is wrong. A photon is not more (and not less) fundamental than an electron. The standard model of elementary particle contains a LOT of equally elementary particles (photons, gluons, W bosons, Z bosons, electrons, muons, tau, e-neutrinos, mu-neutrinos, tau-neutrinos, 6 types of quarks, Higgs). In fact, one of the main motivations for string theory is to derive all these seemingly "elementary" particles from a SINGLE elementary object - the string.

By that I meant ultimately (under the infinitely expanding model) all matter in the Universe will end up as photons and then vacuum energy, and without strings the idea that a photon can become an electron is meaningless. Rather than it was more elementary than an electron. If I'm getting this straight though your points only make sense if you believe string theory to be more than a hypothesis, if not then they are meaningless, yes? If so carry on, I won't interfere again unless the subject changes.

I'm well aware quarks are more fundamental, that wasn't my point.

 

[Demystifier]

1. By that I meant ultimately (under the infinitely expanding model) all matter in the Universe will end up as photons and then vacuum energy,

2. and without strings the idea that a photon can become an electron is meaningless.

1. Even if that was true (which wasn't), so what? I do not understand your point at all.

2. It is not meaningless if you allow quantum jumps (whatever that means). But if all changes are continuous and if particles are pointlike objects, which is what Bohmian mechanics claims, then yes, it becomes meaningless.

 

[Demystifier]

If I'm getting this straight though your points only make sense if you believe string theory to be more than a hypothesis, if not then they are meaningless, yes?

Let me put it this way. If string theory is wrong, then Bohmian mechanics has problems that cannot be solved in a simple way (but can in a complicated, ugly way). If string theory is correct, then these problems of Bohmian mechanics are resolved automatically, in a simple way.

 

[Schrodinger's Dog]

1. Even if that was true (which wasn't), so what? I do not understand your point at all.

2. It is not meaningless if you allow quantum jumps (whatever that means). But if all changes are continuous and if particles are pointlike objects, which is what Bohmian mechanics claims, then yes, it becomes meaningless.

I'm sorry but yes it is.

The Big Freeze is a scenario under which continued expansion results in a universe that is too cold to sustain life. It could, in the absence of dark energy, occur only under a flat or hyperbolic geometry, because such geometries then are a necessary condition for a universe that expands forever. With a positive cosmological constant, it could also occur in a closed universe. A related scenario is Heat Death, which states that the universe goes to a state of maximum entropy in which everything is evenly distributed, and there are no gradients — which are needed to sustain information processing, one form of which is life. The Heat Death scenario is compatible with any of the three spatial models, but requires that the universe reach an eventual temperature minimum.

http://en.wikipedia.org/wiki/Big_Freeze

The Dark Age - from 10100 years until 10150 years

All Black Holes now Disintegrated: 10150 years

The remaining black holes evaporate: first the small ones, and then the supermassive black holes. All matter that used to make up the stars and galaxies has now degenerated into photons and leptons.

[edit] The Photon Age - 10150 years and Beyond

The Universe Achieves Low-Energy State: 1010³ years and beyond

The Universe now reaches an extremely low-energy state. What happens after this is speculative. It's possible a Big Rip event may occur far off into the future, or the Universe may settle into this state forever, achieving true heat death. Extreme low-energy states imply that localized quantum events become major macroscopic phenomena rather than negligible microscopic events because the smallest perturbations make the biggest difference in this era, so there is no telling what may happen to space or time. It is perceived that the laws of "macro-physics" will break down, and the laws of "quantum-physics" will prevail.

In this scenario where the Universe continues to expand more and more rapidly.

I agree with your second point.

It is meaningless if you accept string theory also I don't see what relevance it has to BM, not least because its a theory without any evidence and thus it shouldn't really be used to extrapolate except in philosophical terms . Personally I'd rather go with whatever theory is the most robust atm. String theory is a bit of a dead end IMO, unless something turns up soon, I think I'm personally going to consign it to the waste bin of nice ideas with no evidence. It's getting quite full actually.

 

[Demystifier]

I'm sorry but yes it is.

Oh, now I see the argument. OK, it is, provided two assumptions:
1. That Hawking radiation really exists (there is no proof yet).
2. That there are no other free massless particles except photons. (If, which is very likely, gravitons also exist, then Hawking radiation will produce gravitons as well.)

 

[Schrodinger's Dog]

Oh, now I see the argument. OK, it is, provided two assumptions:
1. That Hawking radiation really exists (there is no proof yet).
2. That there are no other free massless particles except photons. (If, which is very likely, gravitons also exist, then Hawking radiation will produce gravitons as well.)

Granted but at least we have some inferred evidence of black holes, so I'm willing to speculate on that. And actually knock your self out with string theory, pardon me being grumpy. This thread isn't exactly the least speculatory of threads anyway.

 

[akhmeteli]

1. In my view, these are merely two ways to say the same thing. As long as there is only ONE CONTINUOUS CURVE in spacetime, it does not matter whether we call it one particle that can move backwards in time, or many particles that can be created and destructed.

It took me quite some time to decide whether I should agree with this or not:-). OK, let us assume that I agree.

2. Well, if they don't join, then you have a problem.
Assume first that they do. Then, even with your many-particle interpretation, the positions of all these particles are determined by ONLY ONE initial particle position. This differs from the n-particle state in the usual sense, which (in the Bohmian interpretation) requires n initial particle positions.

Sorry, Demystifier, you just cannot have it both ways. You have just said that in this case "As long as there is only ONE CONTINUOUS CURVE in spacetime, it does not matter whether we call it one particle that can move backwards in time, or many particles that can be created and destructed."

Now assume that they don't. If these two parts of the curve are not joined, then how do you know that they actually belong to the same curve? If you simply say that they do not belong to the same curve, which indeed is in the spirit of your many-particle interpretation, then how do you know that one particle must be accompanied with another one?

I don't know (that they actually belong to the same curve), and I don't care. But it seems important to me that along with the first particle there is the second and the third ones, which are also real and can affect the first particle (as they are very close to it), so I 'm not over-enthusiastic when they are announced "unphysical" without bullet-proof arguments. How do I "know that one particle must be accompanied with another one"? I just take solutions of wave equations very seriously, and you are considering one of such solutions.

I hope you see that the arguments in 2. show that a single-particle interpretation has certain advantages (even if you are still not completely convinced).

You see, come to think of it, I don't really care whether the single-particle or multi-particle interpretation is correct. I can even tolerate the Wheeler's idea that there is just one electron in the Universe. I just cannot see any advantages in deeming that part of the trajectory unphysical. You may say that the advantage is that difference then arises between predictions of BI and CI. I think this can be an advantage only if experiments confirm that you are right. If, however, that does not happen, BI will suffer through no fault of its own. Ultimately, your arguments cannot convince me just because they contradict my physical intuition. I understand, however, that your arguments can eventually prove right, and my intuition - wrong. But that would surprise me immensely.

Another thing. In thirties, Landau and Peierls wrote that the achievable accuracy of coordinate measurement is even further limited in the relativistic case, as a hard enough photon used for such measurement can generate pairs, so you'll never know the coordinate of which particle you are measuring (by the way, maybe this is the ultimate source of the uncertainty relation even in the non-relativistic case). This is actually the case that we are considering.

 

[akhmeteli]

Akhmetely, one additional remark. At the beginning, I was also hoping, using similar arguments that you do, that motions backwards in time could be related to genuine particle creation. Nevertheless, by using arguments that I presented above, I have concluded that it was not possible (or at least very unlikely).

However, a truly amazing result was when I recently realized that the Bohmian motions backwards in time ARE related to genuine particle creation - in string theory.
See my preprints arXiv:hep-th/0702060 and arXiv:0705.3542 (I am not giving the direct links because ZZ does not allow to do that for papers that are not yet published). See in particular Fig. 1 in the first paper that summarizes various views of particle creation.

Again, no offence, but, as you admit in your article, the experimental status of string theory is problematic, to put it mildly, so no arguments based on string theory can convince me. Sorry.

 

[Demystifier]

Again, no offence, but, as you admit in your article, the experimental status of string theory is problematic, to put it mildly, so no arguments based on string theory can convince me. Sorry.

But the experimental status of Bohmian mechanics is also problematic. So how arguments based on BM can convince you?

 

[Demystifier]

1. Sorry, Demystifier, you just cannot have it both ways. You have just said that in this case "As long as there is only ONE CONTINUOUS CURVE in spacetime, it does not matter whether we call it one particle that can move backwards in time, or many particles that can be created and destructed."

2. How do I "know that one particle must be accompanied with another one"? I just take solutions of wave equations very seriously, and you are considering one of such solutions.

3. I just cannot see any advantages in deeming that part of the trajectory unphysical.

4. Another thing. In thirties, Landau and Peierls wrote that the achievable accuracy of coordinate measurement is even further limited in the relativistic case, as a hard enough photon used for such measurement can generate pairs, so you'll never know the coordinate of which particle you are measuring (by the way, maybe this is the ultimate source of the uncertainty relation even in the non-relativistic case). This is actually the case that we are considering.

1. We cannot discuss it without drawings. Therefore, please see the picture in my
http://xxx.lanl.gov/abs/quant-ph/0208185 [Found.Phys.Lett. 17 (2004) 363].
Is it 1 particle or 3 particles?

2. The particle trajectories are not solutions of the wave equations. Perhaps you meant something else?

3. You missed the point. The dotted part is unphysical because this part is actually NOT a solution of the particle-trajectory equation of motion. This is because the interaction with the measuring apparatus changes the wave function. (Sorry, but I must suspect again that you are not familiar with the measurement theory in Bohmian mechanics. Please inform me, by PM if you want, if this is the case. It is essential for the efficiency of further discussion.) Now the dashed part is unphysical because it is no longer joined with the solid part, so it is NOT a part of the same trajectory.

4. Can you give me an appropriate reference in which this was shown? I've seen books with similar statements, but such claims were based on hand waving, not on serious calculations based on principles of quantum field theory.

 

[Schrodinger's Dog]

But the experimental status of Bohmian mechanics is also problematic. So how arguments based on BM can convince you?

I'm willing to buy into the idea that you can have a particle representation in non relativistic QFT. Although the maths is somewhat above me, I can follow it well enough to see what they are driving at. But I think you've hit the nail on the head there: how can arguments on BI (Bohmian Interpretation) convince you if the experimental status of BM (Bohmian Mechanics) is problematic. It's kind of the same with CI (Copenhagen Interpretation) isn't it? Why doesn't CI convince some people, doesn't this again boil down to a measurement problem? Isn't the same problem with CI present in BI.

I think the best you can hope for is to convince people that both a particle and wave model are possible in QFT (and that so far CI has a better track record) since that is already self evident given the nature of photons, that's not exactly controversial. I don't think you can say much more or indeed expect people to be convinced much more than that, after all people aren't that convinced by CI either.

Pardon me if I'm way off the beaten track here, just trying to follow your discussion before it disappears off my radar.

 

[Demystifier]

Schrodinger's Dog, but what IF I have a theory that naturally unifies BM and string theory? Not much if one does not find BM or strings (or both) convincing.

But what IF, in addition, my theory predicts something new that can be tested by experiments, but cannot be predicted without BM or without strings? Then someone who finds this theory "elegant" (whatever that means) could be motivated to make the experimental test in a laboratory.

Finally, what IF the experiment turn out to agree with this prediction? Then we have an experimental proof that both BM and string theory are correct.

Indeed, this is how science (a dialog between theories and and experiments) works.

Just for the record, the first two "IF's" are actually "yes". What remains is to do an experimental test.

 

[Schrodinger's Dog]

Schrodinger's Dog, but what IF I have a theory that naturally unifies BM and string theory? Not much if one does not find BM or strings (or both) convincing.

But what IF, in addition, my theory predicts something new that can be tested by experiments, but cannot be predicted without BM or without strings? Then someone who finds this theory "elegant" (whatever that means) could be motivated to make the experimental test in a laboratory.

Finally, what IF the experiment turn out to agree with this prediction? Then we have an experimental proof that both BM and string theory are correct.

Indeed, this is how science (a dialog between theories and and experiments) works.

Just for the record, the first two "IF's" are actually "yes". What remains is to do an experimental test.

Well that's a no brainer, in that case, that is science. If you can validate it experimentally then it becomes a matter of determining if your underlying principles are valid, just like in CI. If you managed that through experimentation - although it might be inferred not directly proved - your hypothesis is more than, well, a mere hypothesis, then you'd be talking. I'm sure if you could do that, then funding would follow.

I see what you are saying.

 

[akhmeteli]

1. We cannot discuss it without drawings. Therefore, please see the picture in my
http://xxx.lanl.gov/abs/quant-ph/0208185 [Found.Phys.Lett. 17 (2004) 363].
Is it 1 particle or 3 particles?

I would say there are 3 particles, although you can say that it is 1 particle (along the lines of the Wheeler's idea), and I could agree. As you said, it's pretty much the same. However, I tend to believe that even if the first particle disappears, the other two will be real, although they may annihilate later.

2. The particle trajectories are not solutions of the wave equations. Perhaps you meant something else?

No, that's what I meant. You see, the Bohmian trajectory can be built on the basis of a wavefunction (which is a solution of a wave equation), as a current line. The current for such solutions is conserved, so there are no sources and no sinks. I see no reasons for instantaneous disappearance of the two remaining particles.

3. You missed the point. The dotted part is unphysical because this part is actually NOT a solution of the particle-trajectory equation of motion. This is because the interaction with the measuring apparatus changes the wave function. (Sorry, but I must suspect again that you are not familiar with the measurement theory in Bohmian mechanics. Please inform me, by PM if you want, if this is the case. It is essential for the efficiency of further discussion.)

I already admitted that I had mixed up the dotted part of the trajectory (related to the future) with the dashed one (related to the past), and I apologized. I could agree that the dotted part is unphysical.

I still don't know how this is relevant, but I am familiar with the measurement theory in Bohmian mechanics, though I am not sure my knowledge is very profound.

Now the dashed part is unphysical because it is no longer joined with the solid part, so it is NOT a part of the same trajectory.

It may not be a part of the same trajectory, but the particles that the dashed part describes are real, if you ask me, not unphysical. You state that this part is unphysical. I fail to see convincing arguments in favor of such statement.

4. Can you give me an appropriate reference in which this was shown? I've seen books with similar statements, but such claims were based on hand waving, not on serious calculations based on principles of quantum field theory.

There may be different opinions on whether it was appropriately "shown", but the article is Zs. Phys., 69, 56, 1931 (I guess there is an English translation somewhere, allegedly, in Wheeler, J. A., and W. H. Zurek, eds., 1983, Quantum Theory and Measurement, (Princeton University Press, Princeton, NJ). ) They don't mention pair production (for obvious reasons:-), it is mentioned in Berestetskii, V. B., Lif****z, E. M., & Pitaevskii, L. P. 1982, Quantum Electrodynamics (Oxford:
Pergamon).

 

[akhmeteli]

But the experimental status of Bohmian mechanics is also problematic. So how arguments based on BM can convince you?

First, I would say the experimental status of the Bohmian mechanics is better than that of the string theory, as its predictions coincide with those of quantum mechanics, at least in the nonrelativistic case (one may say that the same is true for the string theory with certain values of parameters, but there are many more additional radical assumptions in the string theory than in the Bohmian mechanics.). Second, we both use arguments based on QFT, and they seem much more reliable to me, as QFT is properly tested. So I don't think I just refuse to accept anything as a starting point, but I cannot accept everything in this capacity.

 

[Demystifier]

It may not be a part of the same trajectory, but the particles that the dashed part describes are real, if you ask me, not unphysical. You state that this part is unphysical. I fail to see convincing arguments in favor of such statement.

I think I see now where is the source of misunderstanding between you and me. Not really disagreement, but misunderstanding of each other (at least I hope so). So let me explain the things more carefully.

Consider first a nonrelativistic 1-coordinate wave function. All Bohmian trajectories are potentially physical. However, in a single physical case, only one is actually physical. A priori you cannot know which one, but you know that there is one of them that is actually realized. That's what nonrelativistic Bohmian interpretation says. Do you agree?

Now the generalization to a relativistic 1-coordinate wave function. Again, all Bohmian trajectories are potentially physical, including the dashed ones. However, only one connected trajectory is actually physical. For example, if a dashed trajectory is actually physical, then no solid trajectory is actually physical. Conversely, if a solid trajectory is actually physical, then no dashed trajectory is actually physical. Do you still agree?

Now the main point. IF you know with certainty that particle were present somewhere at Sigma_0, THEN you know with certainty that the actually physical trajectory is one of the solid ones. Therefore, you know that the actually physical trajectory is not a dashed one.

Do you agree with such a sharpened explanation?

 

[akhmeteli]

Finally, what IF the experiment turn out to agree with this prediction? Then we have an experimental proof that both BM and string theory are correct.

What bothers me is: what IF the results of the experiment disagree with this prediction? Whose fault would that be? BM's? String theory's? Or maybe both will be falsified? I am afraid you are pushing both the one-particle quantum mechanics and the one-particle Bohmian mechanics to the limit where they are known to fail or will probably fail due to pair production. When you try to compare them in the domain they are not supposed to describe correctly, even potential disagreement with experimental results may be not very meaningful, because we know the natural factor that can cause it.

 

[Demystifier]

First, I would say the experimental status of the Bohmian mechanics is better than that of the string theory, as its predictions coincide with those of quantum mechanics, at least in the nonrelativistic case (one may say that the same is true for the string theory with certain values of parameters, but there are many more additional radical assumptions in the string theory than in the Bohmian mechanics.). Second, we both use arguments based on QFT, and they seem much more reliable to me, as QFT is properly tested. So I don't think I just refuse to accept anything as a starting point, but I cannot accept everything in this capacity.

I see. But let me explain my way of thinking (for which I do not claim that it is better then other ways of thinking). The main reason I like BM is NOT because its predictions are consistent with observations (although they are, but so are the predictions of other interpretations), but because I find BM very elegant. Many scientists will say that "elegance" is irrelevant, but for me (and certainly not only me) it is very relevant.

Now, if you want to make BM consistent with particle creation and quantum field theory (QFT), it is possible to do that, but NOT in an elegant way. But I insist on elegance, that's my guiding principle in search for a more fundamental theory. What I find is that, if QFT is replaced by strings, or more precisely, if QFT is viewed as only an effective theory that emerges from fundamental strings, then the elegance of BM is recovered. For me, it is a good hint that the string version of BM could be the right theory.

Even if this is not your way of thinking (which is understandable), I hope you understand now why I think the way I do.

 

[Demystifier]

What bothers me is: what IF the results of the experiment disagree with this prediction? Whose fault would that be? BM's? String theory's? Or maybe both will be falsified?

What would be falsified is a single elegant theory that unifies BM and string theory. Neither BM nor string theory per se would be falsified because different (for me less elegant) versions of BM and string theory are also possible.

 

[akhmeteli]

Consider first a nonrelativistic 1-coordinate wave function. All Bohmian trajectories are potentially physical. However, in a single physical case, only one is actually physical. A priori you cannot know which one, but you know that there is one of them that is actually realized. That's what nonrelativistic Bohmian interpretation says. Do you agree?

I agree that this is what nonrelativistic Bohmian interpretation says.

Now the generalization to a relativistic 1-coordinate wave function. Again, all Bohmian trajectories are potentially physical, including the dashed ones. However, only one connected trajectory is actually physical. For example, if a dashed trajectory is actually physical, then no solid trajectory is actually physical. Conversely, if a solid trajectory is actually physical, then no dashed trajectory is actually physical. Do you still agree?

No. I believe both the solid and the dashed trajectories are physical at Sigma.

No the main point. IF you know with certainty that particle were present somewhere at Sigma_0, THEN you know with certainty that the actually physical trajectory is one of the solid ones. Therefore, you know that the actually physical trajectory is not a dashed one.

Do you agree with such a sharpened explanation?

No. For example, if there is no measurement, and the solid trajectory is physical, then the dashed trajectory is also physical, otherwise there is a sink. If there is a measurement, that does not make the dashed trajectory unphysical. Yes, it means that there are more than one particle. Yes, it means that one-particle theories are not entirely sufficient in this case. This is how I see it. Apparently, you disagree, and maybe you're right, and I am wrong. But so far I don't see convincing arguments that could make me change my position.

 

[Demystifier]

No. I believe both the solid and the dashed trajectories are physical at Sigma.

But for a given solid trajectory, how will you determine WHICH dashed trajectory is the physical one? Since the dotted part is no longer physical (we agree on that), the solid and dashed curves are not two parts of one connected curve, so for a given solid trajectory, the dashed trajectory is not unique. Do you agree?

Now, if you allow that more than one trajectories are physical, how you will determine how many of them are physical? For example, can TWO almost parallel dashed trajectories by physical? If not, why?

 

[Demystifier]

Here I attach I picture that should make more clear what I am talking about.

 

[akhmeteli]

I see. But let me explain my way of thinking (for which I do not claim that it is better then other ways of thinking). The main reason I like BM is NOT because its predictions are consistent with observations (although they are, but so are the predictions of other interpretations), but because I find BM very elegant. Many scientists will say that "elegance" is irrelevant, but for me (and certainly not only me) it is very relevant.

Now, if you want to make BM consistent with particle creation and quantum field theory (QFT), it is possible to do that, but NOT in an elegant way. But I insist on elegance, that's my guiding principle in search for a more fundamental theory. What I find is that, if QFT is replaced by strings, or more precisely, if QFT is viewed as only an effective theory that emerges from fundamental strings, then the elegance of BM is recovered. For me, it is a good hint that the string version of BM could be the right theory.

Even if this is not your way of thinking (which is understandable), I hope you understand now why I think the way I do.

I share some of your beliefs, but I wish I could be more enthusiastic about such an approach. I am not in a position to assess mathematical beauty of the string theory, but from the physical point of view it seems like it may need a touch of the Occam's razor. Anyway, even if such approach is absolutely correct, that is not very encouraging, as it means that the problem of interpretation of quantum mechanics will not be solved until the string theory is confirmed experimentally, and that may take a couple of centuries... So I very much hope this is not the way Nature works:-) . Furthermore, I feel that BM, however elegant, is not elegant enough. The fact that the electromagnetic potential can replace the quantum potential as a guiding field for the Klein-Gordon-Maxwell system suggests that BM can be made much more elegant.

 

[Schrodinger's Dog]

I share some of your beliefs, but I wish I could be more enthusiastic about such an approach. I am not in a position to assess mathematical beauty of the string theory, but from the physical point of view it seems like it may need a touch of the Occam's razor.

No one is except string theorists. It's a fortified position where peer review comes only from within, which is another reason I deeply distrust the "theory", apart from the obvious one - it has no evidence at all.

I agree, there's no problem with creating imaginative elegant theories, provided they actual have some relation to the real world at some point.

I genuinely hope that most physicists are not in a position to discard any interpretation out of hand or completely, that would be very arrogant, given they are all flawed to some extent or incomplete. There's something about string theory, like MWI that sets my spidey sense a tingling though. I have to admit I am prejudiced.

 

[akhmeteli]

What would be falsified is a single elegant theory that unifies BM and string theory. Neither BM nor string theory per se would be falsified because different (for me less elegant) versions of BM and string theory are also possible.

That's what I mean. In general, BM does not predict those experimental differences, that is a prediction of your version of BM. Again, maybe you're absolutely right, but the mechanism that you suggest to prove that such differences exist does not seem very natural to me.

 

[akhmeteli]

But for a given solid trajectory, how will you determine WHICH dashed trajectory is the physical one? Since the dotted part is no longer physical (we agree on that), the solid and dashed curves are not two parts of one connected curve, so for a given solid trajectory, the dashed trajectory is not unique. Do you agree?

I'm afaid not. The wavefunction is also real, so the physical dashed trajectory is the dashed trajectory that is a continuation of the solid trajectory for that wavefunction.

Now, if you allow that more than one trajectories are physical, how you will determine how many of them are physical? For example, can TWO almost parallel dashed trajectories by physical? If not, why?

I explained above that there is a way to define the unique physical trajectory, so I don't need to answer this question. That does not mean that there cannot be some additional physical trajectories that are just not described by our one-particle theory (so those additional trajectories relate to some other particles).

However, I should point out that the above was written for the standard BI. On the other hand, the following two possibilities seem also feasible to me: 1). ALL trajectories possible for the wavefunction are physical (so actually we have an infinite number of particles). 2). No trajectories are physical (so there are no particles, just the electromagnetic field that they are supposed to interact with). But I guess this is speculative.

 

[Demystifier]

I'm afaid not. The wavefunction is also real, so the physical dashed trajectory is the dashed trajectory that is a continuation of the solid trajectory for that wavefunction.

My point is that there is NO continuation, simply because the doted part (that should interpolate between them) does not exist.

 

[Demystifier]

Anyway, even if such approach is absolutely correct, that is not very encouraging, as it means that the problem of interpretation of quantum mechanics will not be solved until the string theory is confirmed experimentally, and that may take a couple of centuries...

One of the nice properties of my theory is that it is not true. Namely, my theory predicts an effect of strings (or more precisely of a specific form of boson-fermion unification emerging from strings) that can be tested at LOW energies. A type of experiment that is supposed to test it is not of a CERN (big accelerator) type, but of a Zeilinger (foundations of QM) type.