How observation leads to wavefunction collapse?

[Fra]

The problem I have with the classical probability theory is not only the issue of a fixed prior - this is solved in the bayesian approach, the other thing is that the probability space itself is supposedly given - this I can not wrap my head around. I suggest that even this space is subject to dynamics.

It shoudl be noted that this approach will simplify to the standard approach when the probability space is sufficiently stable, and when the prior is fairly stable we get the very classical probability like we have in classical thermodynamics too.

/Fredrik

 

[Mike2]

"truly statistical" - what exactly is that? To me it's an idealisation that doesn't quite make sense. Apparently or expected statistical or random yes, but "truly"? This is really one of the critical focus points IMO. Unless there is a proper discrimination between truly and apparent, then apparent is all we've got, and i think this distinction really does make a difference.

Actually, "truly statistical" and "derived from the most general principles of probabilities themselves" are meant as synonomous statements. So all I've stated is a tautology. What seems odd to me is that probabilities should at all be involved in the dynamics of particles interacting. Classically, we have dynamics driven by continuous fields between particles, and the outcoume is only determined by the initial conditions. What might be the case in other interactions of the same particles with the same initial conditions is not a consideration classically. It only depends on the particle properties at one given spacetime point. But in quantum mechanics some outcomes are made statistically impossible because of the interference pattern of the wavefunction (think double slit experiment). The probable nature of other possibilities seems to be taking priority over definite properties at each continuous spacetime point. This makes me think that the properties themselves have their origin in probability theory - and thus "derived" from the sample space considerations of probability theory.

The only counter argument I can think of is that there are no continuous particle properties, and we can't know which descrete level of property a particle might have. This would mean our theories can only be statistical in nature. But does that mean that nature itself is statistical in nature? The fact that we never see interaction where our statistical theories says none should occur does argue for the true statistical nature of reality. QM does assume and imply that reality is purely statistical in its very nature, right? What does purely statistical mean? I mean that every physical entity and every property of every entity has its origin in and is derived from the probability considerations of a sample space.

 

[Fra]

I think I understand what you mean, but...

...how do you imagine to *derive/deduce* (as opposed to guess & gamble) this sample space? Or is this sample space somehow not subject of suspicion, and does it never change?

If you place dice, once you know your dice you can play, but where did you get the dice in the first place?

If we know that the particle is in one state of a list of possibilities. Then of course there is no discussion and you already accepted the axioms of probability. But how do you know the premise in the first place? I think that's really the key point.

/Fredrik

 

[Mike2]

I think I understand what you mean, but...

...how do you imagine to *derive/deduce* (as opposed to guess & gamble) this sample space? Or is this sample space somehow not subject of suspicion, and does it never change?

If you place dice, once you know your dice you can play, but where did you get the dice in the first place?

If we know that the particle is in one state of a list of possibilities. Then of course there is no discussion and you already accepted the axioms of probability. But how do you know the premise in the first place? I think that's really the key point.

/Fredrik

You are thinking of particular examples of a sample space based on already known physical situations such as dice, cards, quantum effects, etc. But if ALL elemental physical entities, properties, and interactions are derived from probability theory, then we cannot start with a sample space of any known physical situation. We can only start with the principles of probability theory that are completely general. That would be to give it a feel of being derived from first principle. But then again, general principles are general principles precisely because they handle ALL situations. And we are trying to find a theoy that does handle ALL situations. So I think we need to try to derive physics from complete generality. What's that called, a top-down theory or a bottom-up, I don't remember which it is.

 

[Fra]

Reading your last post... I wonder what are we talking about at this point? I agree with parts of what you write but now I wonder if we are arguing past each other? I got the impression that you questioned what you now seem to argue in favour of.

I am definitely looking for a general model. The dice was of course but an example to suggest that one might need to generalize the formalism of probability theory even. Also the mere notion of "particle", and "space" are other examples.

/Fredrik

 

[diegzumillo]

I totally lost track of the discussion.. lol
Just out of curiosity, is there an interpretation for the wavefunction alone? we keep talking about probability, but that's the wavefunction by it's complex conjugate. is there is a way to interprete a complex function physically?

 

[scarecrow]

Whether the wave function is real, and what it represents, are major questions in the interpretation of quantum mechanics. Many famous physicists have puzzled over this problem, such as Schrödinger. Some approaches regard it as merely representing information in the mind of the observer. Others argue that it must be objective:

"If we are to believe that anyone thing in the formalism is 'actually' real for a quantum system, then I think it has to be the wavefunction or state vector that describes quantum reality."

Penrose, R. Road To Reality, p508

maybe.

 

[Mike2]

I totally lost track of the discussion.. lol
Just out of curiosity, is there an interpretation for the wave function alone? we keep talking about probability, but that's the wave function by it's complex conjugate. is there is a way to interprete a complex function physically?

Sorry if the conversation has become obscure. If you're curious about my perspective, you can of course always check out my home page on my personnel profile of Physics Forums.

But you ask what is the physical interpretation of the wave function. I think it is how one state propagates to the next state. I do not think that the wave function changes/collapses upon measurement. For it seems arbitrary where and when to do a measurement. We simply combine that wave function for propagation in one direction of time with the wave function for propagation in the reverse direction to get a probability of propagation from start to finish. So we are calculating the probability of two facts in conjunction. We calculate the probability of the initial state in conjunction with the finial state of a measurement,

I simply take it as no coincidence that this is similar to conjunction of two facts being equivalent to one fact implying the second in conjunction with the second fact implying the first, in symbols, a*b=(a->b)*(b->a). For it is most intuitive for me to understand propagation as a form of implication.

 

[meopemuk]

I totally lost track of the discussion.. lol
Just out of curiosity, is there an interpretation for the wavefunction alone? we keep talking about probability, but that's the wavefunction by it's complex conjugate. is there is a way to interprete a complex function physically?

Real physical meaning can be assigned only to things that are measured directly: values of observables and probabilities. Wavefunctions and state vectors are not real things. They are parts of our mathematical model of reality. They live not in the real world, but in abstract Hilbert spaces. We use them, because they provide for us a convenient mathematical tool for calculations of measured values of observables and probabilities.

Eugene.

 

[jostpuur]

Yes, your derivation is correct. However, why did you stop at taking the first time derivative of the Schroedinger equation? Why didn't you take 2nd derivative, 3rd derivative, etc?

The Klein-Gordon equation is particularly useful, because we know it to be Lorentz invariant, and know that its solutions don't lead to causality paradoxes. Immediate consequence is, that the solutions of the relativistic Shrodinger's equation don't lead to causality paradoxes either.

This is how the proof goes:

Solutions of relativistic SE are also solutions of the KGE. Because solutions of KGE are know to be paradox free, we conclude that so are solutions of relativistic SE paradox free too.

Can you say what's wrong in this?

I can tell what's wrong in the most popular proofs for the claim, that the relativistic SE would lead to paradoxes. One proof uses Taylor series of the square root. It is wrong because the series don't converge. One proof uses propagator, and a conclusion that if a function doesn't approach zero, then it's integral (or behaviour as a distribution) doesn't approach zero either. It is wrong, because integral can approach zero without the integrand approaching it.

Here,

F. Strocchi, "Relativistic quantum mechanics and field theory", Found. Phys. 34 (2004), 501; http://www.arxiv.org/abs/hep-th/0401143

I didn't even understand how the proof was supposed to work.(I thought it would be funny if we kept discussing about this same topic in several different threads. You know. New threads, old debate. I intended the other thread "propagation speeds in literature" to be about the history of this stuff.)

 

[genneth]

Real physical meaning can be assigned only to things that are measured directly: values of observables and probabilities. Wavefunctions and state vectors are not real things. They are parts of our mathematical model of reality. They live not in the real world, but in abstract Hilbert spaces. We use them, because they provide for us a convenient mathematical tool for calculations of measured values of observables and probabilities.

Eugene.

I think this is the clearest sentence so far in this thread. People keep confusing the state of the system with its representation in a position basis. The latter presupposes some operator which measures the position of a particle. In non-relativistic situations, we can foliate spacetime into simultaneous surfaces, and the trajectory of a particle can only cross each surface once, and must cross each surface once; therefore, the wavefunction in position basis can be understood to be a probability measure over space. Clearly, this breaks down badly in GR, where global foliation may not be possible. In SR, the situation is still complicated.

jostpuur: The problems with the KGE are mostly coming from a historical view. Historically, people wanted a theory for particles that's SR-compatible. However, in KGE, there is the possibility that the "current" goes negative, which made people think that it's not a valid model for a particle. The modern view is that the KGE equation doesn't model a particle, but a charged spin-0 boson, and the current is actually the charge-current, as opposed to the particle-current. Nevertheless, the "failure" of KGE pushed research, until the Dirac equation was found. Now, we understand bosons and fermions better, and we're more interested in field equations rather than just particles.

 

[jostpuur]

genneth, check out my post #287 on the page 20 of this thread, and some discussion that followed. Some cooI stuff about KGE.

 

[meopemuk]

The Klein-Gordon equation is particularly useful, because we know it to be Lorentz invariant, and know that its solutions don't lead to causality paradoxes. Immediate consequence is, that the solutions of the relativistic Shrodinger's equation don't lead to causality paradoxes either.

This is how the proof goes:

Solutions of relativistic SE are also solutions of the KGE. Because solutions of KGE are know to be paradox free, we conclude that so are solutions of relativistic SE paradox free too.

Can you say what's wrong in this?

What is your definition of "Lorentz invariance"? For example, why do you say that the Scroedinger equation


$$i \hbar \frac{\partial}{\partial t} \psi(x,t) = \sqrt{-\hbar^2 c^2 \frac{\partial}{\partial t} + m^2 c^4} \psi(x,t)$$... (1)

is not Lorentz invariant while the Klein-Gordon equation

$$-\hbar^2 \frac{\partial^2}{\partial t^2} \psi(x,t) = (-\hbar^2 c^2 \frac{\partial}{\partial t} + m^2 c^4) \psi(x,t)$$...(2)

is Lorentz invariant?


What do you mean by "solutions of KGE are know to be paradox free"? How exactly you are using this statement for proving that solutions of (1) cannot exhibit superluminal propagation?

Here,

F. Strocchi, "Relativistic quantum mechanics and field theory", Found. Phys. 34 (2004), 501; http://www.arxiv.org/abs/hep-th/0401143

I didn't even understand how the proof was supposed to work.

In his proof Strocci uses a (supposedly well-known) lemma which states that the Fourier transform of an analytical function has a compact support (i.e., it is non-zero only in a finite region of its argument). And inversely, the Fourier transform of a function with compact support is analytical. I don't know how these statements are proved. However, intuitively, they make sense. An analytical function is supposed to be differentiable infinite number of times. So, its Fourier spectrum should not contain infinite frequencies, because they usually correspond to discontinuities of the function.

Once we established this, the Strocci's proof becomes simple. Suppose that the wave function $\psi(x,0)$ has compact support (i.e., localized) at time t=0. If we assume that the spreading cannot be superluminal, we conclude that $\psi(x,t)$ for finite t > 0 also has a compact support (the support at t=0 can expand only by ct, so it remains compact). Then, the time derivative $\partial \psi(x,t)/ \partial t$ at t=0 also has a compact support. Now we can take the Fourier transform of both sides of the Schroedinger equation (1) at t=0

$$i \hbar \frac{\partial}{\partial t} \psi(p,t) = \sqrt{p^2c^2 + m^2 c^4} \psi(p,t)$$....(3)

where (according to the Lemma) $i \hbar \partial \psi(p,t) / \partial t$ and $\psi(p,t)$ are both analytical functions of p. However, $\sqrt{p^2c^2 + m^2 c^4}$ is not an analytical function of p. So, there cannot be equality between the left and right hand sides of (3). This controversy demonstrates that our assumption (that $\psi(x,t)$ propagates with a finite speed) was wrong.

 

[meopemuk]

People keep confusing the state of the system with its representation in a position basis. The latter presupposes some operator which measures the position of a particle.

This is exactly the point that I was trying to make several times. One cannot write wave function (Schroedinger, Klein-Gordon, Dirac, or whatever) $\psi(x,t)$ before the operator of position is defined. Because $\psi(x,t)$ is nothing but projections of the state vector on eigenvectors of the position operator.

Moreover, in order to know how the wave function $\psi(x,t)$ transforms with respect to boosts one needs to know boost transformations of the position operator, i.e., the commutators of this operator with the boost generators.

Unfortunately, most textbook discussions of relativistic quantum mechanics do not bother to define the position operator. Some even claim that this operator does not exist. They simply postulate boost transformations of Klein-Gordon or Dirac "wave functions" without checking whether these transformations are consistent with time translations (i.e., whether the Poincare group properties are satisfied) and the conservation of probabilities (unitarity). This is especially strange since the correct way of doing these things is well-known for many decades:

E. P. Wigner, "On unitary representations of the inhomogeneous Lorentz group", Ann. Math. 40 (1939), 149.

T. D. Newton and E. P. Wigner, "Localized states for elementary systems", Rev. Mod. Phys. 21 (1949), 400.

Eugene.

 

[genneth]

Unfortunately, most textbook discussions of relativistic quantum mechanics do not bother to define the position operator. Some even claim that this operator does not exist.

I think this is due to the heuristic "levelling" of position and time, which basically goes something like this:

Space and time are intertwined, as SR teaches us. In quantum mechanics, we end up isolating the time variable, which is also not an operator like the positions. Therefore, many failures to make a consistent special relativistic quantum mechanics can be attributed to this differing treatment of position and time. We therefore have two options: lift time to be an operator, or "lower" position to be non-operators. To pursue the latter, we note that in the Heisenberg picture, observables are actually families of observables, indexed by t: not A, but A(t). So now, we use the positions in a similar way: A(x, t). This actually wrecks havoc with any attempt to make a relativistic *particle* (as opposed to field) theory, but that's okay, because clearly we need fields in the relativistic case (yes -- that's a bit circular, but I never said I was defending this traditional view...). Now, we can say that the value of fields at each spacetime point is a doubly (or rather quadruply, but who's really counting?) indexed operator family, A(x, t). The rest, as they say, is history.

Now, you might wonder what about the other option? What about promoting time to be a real operator? Well, at least one textbook, by Srednicki, says that although it's possible, it turns out to be harder, due to the possibility of multiplicity of times (in the parameter of motion sense), and in any case, it's equivalent to the other way for special relativistic theories. Now, I should actually say that Srednicki deserves props for even mentioning the two possibilities, even if I'd say that the discussion is a bit lacking. Essentially, the inability for QFT to generalise to a background-free setting can be traced directly to this decision, but that's going into speculative grounds (or at least not well-accepted grounds).

So, back to the original point: people sometimes say that position isn't an operator in QFT, and that's what they mean. I personally think that's just because QFT is broken, but it's hard to find that many people who'd agree (by the way, let's not let this thread degenerate into a my-pet-theory-is-better-than-your-pet-theory -- and focus on the flaws and attributes of the standard, accepted theory; I know that meopemuk has at least differing views to me on this, but I don't think now is the right time to discuss them).

 

[meopemuk]

genneth, thanks for this great analysis!

Space and time are intertwined, as SR teaches us.

I am questioning the wisdom of this conclusion. I suspect this is a "nice" assumption rather than a solid proven fact.

I know that meopemuk has at least differing views to me on this, but I don't think now is the right time to discuss them).

Okay, I stop here.

Eugene.

 

[jostpuur]

What is your definition of "Lorentz invariance"? For example, why do you say that the Scroedinger equation $$i \hbar \frac{\partial}{\partial t} \psi(x,t) = \sqrt{-\hbar^2 c^2 \frac{\partial^2}{\partial x^2} + m^2 c^4} \psi(x,t)$$... (1)

is not Lorentz invariant while the Klein-Gordon equation

$$-\hbar^2 \frac{\partial^2}{\partial t^2} \psi(x,t) = (-\hbar^2 c^2 \frac{\partial^2}{\partial x^2} + m^2 c^4) \psi(x,t)$$...(2)

is Lorentz invariant?

I meant that it is not obvious in the first place, that SE would be Lorentz invariant, but we know already that the KGE is Lorentz invariant. So we can use invariance of KGE to prove the invariance of the SE.

What do you mean by "solutions of KGE are know to be paradox free"?

A small typo. "know" should have been "known". They are known to be free of paradoxes. That means that the disturbances propagate with speeds not greater c and so on.

How exactly you are using this statement for proving that solutions of (1) cannot exhibit superluminal propagation?

Because if solutions of (1) exhibit superluminal propagation, then also solutions of (2) could exhibit superluminal propagation, which would be a contradiction.

In his proof Strocci uses a (supposedly well-known) lemma which states that the Fourier transform of an analytical function has a compact support (i.e., it is non-zero only in a finite region of its argument). And inversely, the Fourier transform of a function with compact support is analytical. I don't know how these statements are proved. However, intuitively, they make sense. An analytical function is supposed to be differentiable infinite number of times. So, its Fourier spectrum should not contain infinite frequencies, because they usually correspond to discontinuities of the function.

Gaussian peak doesn't have a compact support, is analytical, and its Fourier transform is another Gaussian peak. Isn't that a counter example, or did I understand the theorem wrong?

The expression $\sqrt{|p|^2+m^2}$ has no discontinuities in its domain $p\in\mathbb{R}^3$. I cannot see why discontinuities somewhere in the $\mathbb{C}^3$ would matter at all.

 

[genneth]

Remember that the problem with KGE is that it doesn't give unitary time evolution: probability isn't conserved. In other words:

$$\frac{d}{dt} \int \psi^*(x,t) \psi(x,t) \,dx \ne 0$$

 

[jostpuur]

Remember that the problem with KGE is that it doesn't give unitary time evolution: probability isn't conserved. In other words:

$$\frac{d}{dt} \int \psi^*(x,t) \psi(x,t) \,dx \ne 0$$

But we have

$$\frac{d}{dt} \int \psi^*(x,t) \psi(x,t) \,dx = 0$$

if we take only positive frequency solutions. In other words, if we take the solutions of the relativistic SE, which is the same thing as positive frequency solution of KGE.

 

[meopemuk]

I meant that it is not obvious in the first place, that SE would be Lorentz invariant, but we know already that the KGE is Lorentz invariant. So we can use invariance of KGE to prove the invariance of the SE.

My question was about your definition of "Lorentz invariance". I think we can agree that without first giving the definition you cannot say whether KGE is Lorentz invariant or not. It is important to be very precise, because otherwise we don't have a chance to solve the "superluminal" paradox, in my opinion.

[Solutions of KG equation] are known to be free of paradoxes. That means that the disturbances propagate with speeds not greater c and so on.

Because if solutions of (1) exhibit superluminal propagation, then also solutions of (2) could exhibit superluminal propagation, which would be a contradiction.

Can you prove that solutions of (2) never exhibit superluminal propagation?

Gaussian peak doesn't have a compact support, is analytical, and its Fourier transform is another Gaussian peak. Isn't that a counter example, or did I understand the theorem wrong?

That's a good point. I think you are right that a Fourier transform of an analytical function may not have a compact support. Perhaps, the Lemma only says that the Fourier transform of a function with a compact support is analytical (only this part is needed to prove Strocci's proposition 2.1). So, your Gaussian peak example doesn't satisfy the condition of the Lemma.

The expression $\sqrt{|p|^2+m^2}$ has no discontinuities in its domain $p\in\mathbb{R}^3$. I cannot see why discontinuities somewhere in the $\mathbb{C}^3$ would matter at all.

The proof is based on the requirement that both sides of eq. (3) must have the same analyticity properties on the complex plane.

Eugene.

 

[jostpuur]

My question was about your definition of "Lorentz invariance". I think we can agree that without first giving the definition you cannot say whether KGE is Lorentz invariant or not. It is important to be very precise, because otherwise we don't have a chance to solve the "superluminal" paradox, in my opinion.

The usual definition. When a solution wave function is transformed, it remains as a solution of the same invariant partial differential equation.

Can you prove that solutions of (2) never exhibit superluminal propagation?

I was afraid you were going to ask this. I know a very rigor proof for the special case m=0, which I have read from a book https://www.amazon.com/dp/0821807722/?tag=pfamazon01-20 For m>0 I don't know equally rigor proof, although I've been looking for it. For example here https://www.physicsforums.com/showthread.php?t=181383 But I'have good reasons to believe, that superluminal propagation will not appear as result of m=0 being replaced by m>0.

In general, the mass term has an effect of slowing down wave packets. The KGE is still Lorentz invariant, and superluminal propagation would mean that solutions actually propagate backwards in time, which would be very strange.

I have an idea how I could prove this rigorously through analysis of some oscillations, but I haven't seen the effort for it yet. Hans' has proved this also, but I haven't gone through his explanations very carefully.

That's a good point. I think you are right that a Fourier transform of an analytical function may not have a compact support. Perhaps, the Lemma only says that the Fourier transform of a function with a compact support is analytical (only this part is needed to prove Strocci's proposition 2.1). So, your Gaussian peak example doesn't satisfy the condition of the Lemma.

The proof is based on the requirement that both sides of eq. (3) must have the same analyticity properties on the complex plane.

Eugene.

For the special case m=0 this proof is in contradiction with well established mathematics about behaviour of the wave equation. That's why I remain reluctant to believe this.

 

[meopemuk]

The usual definition. When a solution wave function is transformed, it remains as a solution of the same invariant partial differential equation.

What is the boost transformation of the wave function? And why? I suspect that you are applying usual Lorentz transformations to the wave function arguments x and t. Aren't you? Can you explain why this can be done? Is there a proof? Note that we are talking about transformations of some purely quantum object - the probability amplitude. It is not at all obvious that one can apply the same Lorentz transformation formulas that were derived in SR for coordinates of macroscopic particles or light pulses.

I was afraid you were going to ask this. I know a very rigor proof for the special case m=0, which I have read from a book https://www.amazon.com/dp/0821807722/?tag=pfamazon01-20 For m>0 I don't know equally rigor proof, although I've been looking for it. For example here https://www.physicsforums.com/showthread.php?t=181383 But I'have good reasons to believe, that superluminal propagation will not appear as result of m=0 being replaced by m>0.

In general, the mass term has an effect of slowing down wave packets. The KGE is still Lorentz invariant, and superluminal propagation would mean that solutions actually propagate backwards in time, which would be very strange.

I have an idea how I could prove this rigorously through analysis of some oscillations, but I haven't seen the effort for it yet. Hans' has proved this also, but I haven't gone through his explanations very carefully.

I think you must provide a rigorous proof rather than claim "good reasons to believe". What you are saying (the absence of superluminal solutions of the relativistic Schroedinger equation) directly contradicts a vast literature on this subject (Strocci, Hegerfeldt, and many others). So, you better be very precise in your statements.

Eugene.

 

[jostpuur]

I think you must provide a rigorous proof rather than claim "good reasons to believe". What you are saying (the absence of superluminal solutions of the relativistic Schroedinger equation) directly contradicts a vast literature on this subject (Strocci, Hegerfeldt, and many others). So, you better be very precise in your statements.

Eugene.

When somebody says that in solutions of KGE in general the disturbances can propagate with speeds greater than c, he is also saying that in solutions of the wave equation disturbances can propagate with these speeds, because the wave equation is a special case of the KGE. This contradicts mainstream mathematics, and mainstream mathematics is stronger than mainstream physics, when it comes to mathematical issues. So this isn't really my personal claim.

There still seems to be a small chance, that perhaps superluminal propagation doesn't occur with m=0, but with m>0 it does. But I haven't seen even physicists claiming so. They always show that superluminal propagation occurs with arbitrary m, and that's when they make a mistake. Apparently these physicists don't know they are fighting against mainstream mathematics.

 

[meopemuk]

When somebody says that in solutions of KGE in general the disturbances can propagate with speeds greater than c, he is also saying that in solutions of the wave equation disturbances can propagate with these speeds, because the wave equation is a special case of the KGE. This contradicts mainstream mathematics, and mainstream mathematics is stronger than mainstream physics, when it comes to mathematical issues. So this isn't really my personal claim.

There still seems to be a small chance, that perhaps superluminal propagation doesn't occur with m=0, but with m>0 it does. But I haven't seen even physicists claiming so. They always show that superluminal propagation occurs with arbitrary m, and that's when they make a mistake. Apparently these physicists don't know they are fighting against mainstream mathematics.

Can we then conclude that neither you nor I can point to the rigorous proof of this statement (the (im)possibility of superluminal solutions for KG)? However, we have different feelings regarding its plausibility.

Eugene.

 

[jostpuur]

Can we then conclude that neither you nor I can point to the rigorous proof of this statement (the (im)possibility of superluminal solutions for KG)? However, we have different feelings regarding its plausibility.

Eugene.

Temporarily yes. I'm at the moment busy with some courses, and cannot put much time into my own stuff. But I'll try to return to this.

 

[Hans de Vries]

I was afraid you were going to ask this. I know a very rigor proof for the special case m=0, which I have read from a book https://www.amazon.com/dp/0821807722/?tag=pfamazon01-20 For m>0 I don't know equally rigor proof, although I've been looking for it. For example here https://www.physicsforums.com/showthread.php?t=181383 But I'have good reasons to believe, that superluminal propagation will not appear as result of m=0 being replaced by m>0.

Indeed, Of course for m=0 the Klein Gordon equation becomes the wave
equation for the propagation of the free electromagnetic potentials, and
light doesn't propagate faster than c...

The point I'm trying to get across is that you can write the case m>0 as
a sequence of the m=0 case. So if you prove the m=0 case then you also
proof the m>0 case.

$$\frac{1}{p^2-m^2}\ =\ \frac{1}{p^2}+\frac{m^2}{p^4}+\frac{m^4}{p^6}+\frac{m^6}{p^8}+...$$

Which becomes the following operator in configuration space:

$$\Box^{-1}\ \ -\ \ m^2\Box^{-2}\ \ +\ \ m^4\Box^{-3}\ \ -\ \ m^6\Box^{-4}\ \ +\ \ ...$$

Where $$\Box^{-1}$$ is the inverse d'Alembertian, which spreads the wave function
out on the lightcone as if it was a massless field. The second term then
retransmits it, opposing the original effect, again purely on the light cone.
The third term is the second retransmission, et-cetera, ad-infinitum.

All propagators in this series are on the lightcone. The wave function does
spread within the light cone because of the retransmission, but it does
never spread outside the light cone, with superluminal speed.

In the Standard Model the particles are in principle massless (m=0) but
they acquire mass due to interactions with the hypothetical Higgs field.
The above just describes a peturbative expansion of the interaction with
the Higgs field just like in QED the Feynman diagrams are a peturbative
expansion of the interaction with the electromagnetic field. Regards, Hans.

 

[Hans de Vries]

What is the boost transformation of the wave function? And why? I suspect that you are applying usual Lorentz transformations to the wave function arguments x and t. Aren't you? Can you explain why this can be done? Is there a proof? Note that we are talking about transformations of some purely quantum object - the probability amplitude. It is not at all obvious that one can apply the same Lorentz transformation formulas that were derived in SR for coordinates of macroscopic particles or light pulses.

It's shown that Special Relativity holds at length scales down to 10^-19 meter.
This is what's done by high energy collider experiments.

So, if the electron's wave-function in an atomic orbital would be the size of
the Earth then the scale at which SR is proven to hold would be about 1 centimeter...


Regards, Hans

 

[meopemuk]

It's shown that Special Relativity holds at length scales down to 10^-19 meter.
This is what's done by high energy collider experiments.

So, if the electron's wave-function in an atomic orbital would be the size of
the Earth then the scale at which SR is proven to hold would be about 1 centimeter...

Thank you for reminding that. It is true that some results of special relativity hold to very high precision in experiments. Kinematical relationships between momenta and energies of scattering particles are a good example of such a success. However, it is also true that boost transformations of particle wavefunctions in the position space have not been measured directly, even approximately. Of course, you can believe that kinematical Lorentz formulas remain valid in this case as well. I was asking if you can offer something in addition to this belief? Some kind of proof...

I believe that superluminal propagation of relativistic wavefunctions is a real and interesting paradox. There are numerous papers that support this point of view. I think that the only way to resolve the paradox is to carefully examine all assumptions and beliefs that went into its formulation.

Eugene.

 

[Haelfix]

"the (im)possibility of superluminal solutions for KG"

Stated another way, commutators of gauge invariant observables in field theory vanish outside the light cone. This can be proven (rigorously) to be a requirement for the analyticity of the SMatrix. Turned around it implies micro locality.

 

[jostpuur]

Indeed, Of course for m=0 the Klein Gordon equation becomes the wave
equation for the propagation of the free electromagnetic potentials, and
light doesn't propagate faster than c...

So good point!

I think you must provide a rigorous proof rather than claim "good reasons to believe". What you are saying (the absence of superluminal solutions of the relativistic Schroedinger equation) directly contradicts a vast literature on this subject (Strocci, Hegerfeldt, and many others). So, you better be very precise in your statements.

Isn't it paradoxical, that according to these many authors, how can be considered mainstream scientists, the positive frequency light waves, being solutions the Maxwell equations and the massless KGE (in proper gauge), and also solutions of the relativistic SE, propagate faster than light?

Edit:
hmhmh... I think I'm not removing this, because it was funny enough. But now I just recalled, that the solutions of the Maxwell equations cannot be purely positive frequency, because they are real valued. Anyway, if somebody believes that the locality is lost when the real wave function is replaced with a complex one, I would like to hear explanations.

 

[Hans de Vries]

Kinematical relationships between momenta and energies of scattering particles are a good example of such a success. However, it is also true that boost transformations of particle wavefunctions in the position space have not been measured directly, even approximately. Of course, you can believe that kinematical Lorentz formulas remain valid in this case as well. I was asking if you can offer something in addition to this belief? Some kind of proof...

The kinematic behavior is a proof that SR holds at these length scales. For instance,
matter waves do acquire a wavelength only due to the non-simultaneity of the Lorentz
transform. A particle in it's rest frame has equal phase throughout it's wave function,
while in a frame where it is boosted it's phase becomes different at different locations,
and does so purely because of the non-simultaneity in the direction of motion.

There is also the fact that the fundamental mechanism governing the propagation
of matter-waves is the four dimensional version of Huygens principle, connecting
geometry with kinematics. SR is the foundation of wave behavior and visa versa.


Regards, Hans.

 

[meopemuk]

"the (im)possibility of superluminal solutions for KG"

Stated another way, commutators of gauge invariant observables in field theory vanish outside the light cone. This can be proven (rigorously) to be a requirement for the analyticity of the SMatrix. Turned around it implies micro locality.

This is a completely different issue. We were not talking about any commutators. We were discussing time evolution of initially localized 1-particle wave functions.

Eugene.

 

[meopemuk]

The kinematic behavior is a proof that SR holds at these length scales. For instance,
matter waves do acquire a wavelength only due to the non-simultaneity of the Lorentz
transform. A particle in it's rest frame has equal phase throughout it's wave function,
while in a frame where it is boosted it's phase becomes different at different locations,
and does so purely because of the non-simultaneity in the direction of motion.

There is also the fact that the fundamental mechanism governing the propagation
of matter-waves is the four dimensional version of Huygens principle, connecting
geometry with kinematics. SR is the foundation of wave behavior and visa versa.

Sorry, this is not a proof.

For a rigorous derivation of relativistic wave functions and their transformation properties I can recommend (again) papers by Wigner and Newton & Wigner. However, these derivations do not lead to the KG equation and to Lorentz transformations of wave functions under boosts. I feel that from this discrepancy between rigorous (Wigner) and commonly accepted (KG, Dirac) approaches one can learn some important lessons about relativistic quantum theory.

Eugene.

 

[Hans de Vries]

Sorry, this is not a proof.

For a rigorous derivation of relativistic wave functions and their transformation properties I can recommend (again) papers by Wigner and Newton & Wigner. However, these derivations do not lead to the KG equation and to Lorentz transformations of wave functions under boosts. I feel that from this discrepancy between rigorous (Wigner) and commonly accepted (KG, Dirac) approaches one can learn some important lessons about relativistic quantum theory.

Eugene.

It's the KG equation which leads straightforwardly to relativistic matter waves and
Lorentz transform. If you want to claim otherwise then you have to discuss the math
itself rather then just make these claims and refer to physicist which, according to your
claims, would support your opinions...


Regards, Hans

 

[meopemuk]

It's the KG equation which leads straightforwardly to relativistic matter waves and
Lorentz transform. If you want to claim otherwise then you have to discuss the math
itself rather then just make these claims and refer to physicist which, according to your
claims, would support your opinions...

I would be happy to discuss the math of Wigner's approach and its difference from the textbook approach (KG and Dirac equations are basically postulated; boost transformations of wave functions are postulated as well). Here is a short list of steps involved in this theory.

1. This approach is based on the principle of relativity: physical laws (e.g., probabilities of measurements) are invariant with respect to the Poincare group of transformations between inertail observers.

2. It is based on postulates of quantum mechanics: Hilbert space, state vectors, Hermitean operators, etc.

3. From combination of these two principles it follows that there is an unitary representation of the Poincare group in the Hilbert space of each isolated physical system.

4. It is postulated that simplest physical systems - elementary particles - are described by simplest - irreducible - representations of the Poincare group.

5. The theory of unitary irreducible representations of the Poincare group was developed by Wigner. A more modern formulation was given by Mackey in terms of the theory of "induced representations". Here we should be interested in representations associated with massive (m>0) particles of arbitrary spin (s=0, 1/2, 1, ...).

6. Wigner-Mackey theory is constructed in the momentum representation and wave function transformations with respect to all Poincare group elements are explicitly written. Note that for s>0 transformations with respect to boosts involve non-trivial "Wigner rotations".

7. In order to find wave functions and their transformations in the position space, one need to define the position operator and its commutators with generators of the Poincare group. This has been done by Newton and Wigner in 1949. They formulated a few postulates that must be satisfied by any sensible relativistic position operator, and they showed that there is only one choice that obeys these postulates.

8. As a result we obtain position-space wave functions whose time evolution is described by the Schroedinger equation (it involves only 1st time derivative, unlike the KG equation). Transformations of the wave functions to moving reference frame are not given by usual Lorentz formulas. So, this approach is profoundly different from what textbooks say about KG and Dirac "wave functions". However, the advantage is that Wigner's approach is fully axiomatic. It follows rigorously from a set of postulates each of which is perfectly reasonable and generally accepted.

Eugene.

P.S. I know one textbook (vol. 1 of Weinberg's "The quantum theory of fields") in which the Wigner and Klein-Gordon-Dirac approaches are clearly and correctly described and the relationships between the two approaches are established. Bottomline: particle wave functions are described by the Wigner's approach. KG-Dirac equations and manifestly covariant transformation laws are properties of quantum fields, rather than wave functions.