How can the Cauchy integral and Fourier integral produce the same result?

[meopemuk]

I think causality and measurment are not the reason for space-like commutation. We never measure eigenstates of the field operator. We measure only asymptotic particle states. The reason for space-like commutation is really lorentz invariance. When you introduce interactions, the S-matrix in perturbation theory involves time-ordered products of the fields, and time-ordering is only lorentz invariant when the two points being time ordered are not space-like separated.

This point of view is best expressed in S. Weinberg "The quantum theory of fields", vol. 1. Weinberg's idea is that the reason for introducing quantum fields with their specific properties (covariant transformation laws, (anti)commutativity at space-like separations, etc) is that when we construct interaction Lagrangians (or Hamiltonians) as polynomials of such fields we immediately obtain non-trivial generators of the Poincare group in the Fock space, so that the theory is relativistically invariant. Moreover, this construction trivially satisfies the requirement of cluster separability.

I agree with Weinberg that these are the most important reasons. This leads me to a heretical idea that maybe these are *the only* reasons for introducing fields. Perhaps quantum fields do not play any other role, except as some formal mathematical expressions, which are "building blocks" of relativistic interaction operators? Then there is no need to be concerned about difficult issues of the physical interpretation of fields and corresponding (KG and Dirac) equations.

 

[Demystifier]

This point of view is best expressed in S. Weinberg "The quantum theory of fields", vol. 1. Weinberg's idea is that the reason for introducing quantum fields with their specific properties (covariant transformation laws, (anti)commutativity at space-like separations, etc) is that when we construct interaction Lagrangians (or Hamiltonians) as polynomials of such fields we immediately obtain non-trivial generators of the Poincare group in the Fock space, so that the theory is relativistically invariant. Moreover, this construction trivially satisfies the requirement of cluster separability.

I agree with Weinberg that these are the most important reasons. This leads me to a heretical idea that maybe these are *the only* reasons for introducing fields. Perhaps quantum fields do not play any other role, except as some formal mathematical expressions, which are "building blocks" of relativistic interaction operators? Then there is no need to be concerned about difficult issues of the physical interpretation of fields and corresponding (KG and Dirac) equations.

I agree.

 

[Demystifier]

Except that I think you should have the square root there like in my previous post. With the square root the norm $\langle\psi|\psi\rangle$ becomes Lorentz's invariant. At least if we have $[a_p,a^{\dagger}_{p'}]=(2\pi\hbar)^3\delta^3(p-p')$. This is the convention P&S use. Do other sources put $2E_p$ in front of the delta function?

Or maybe not. I don't have the book right here, and instead just looked at my notes, but now I started doubting if I have them correctly... Srednicki seems to indeed put that energy factor in front of the delta function. Blaa...

You are right, sorry for the mistake.

 

[Demystifier]

I cannot agree when you arbitrarily replace factors. All factors must follow from some fundamental principles. In first five chapters of http://www.arxiv.org/physics/0504062 I describe principles that I consider fundamental and find that the measure $$d^3p$$ should be used in formulas for localized wave functions. If you wish to change the measure to $$d^3p/2E$$, there should be some justification more rigorous than the desire to keep measure "Lorentz invariant".

I am sure that you have a good reason to take the measure you choose. It, indeed, seems very natural. Still, I would not say that a requirement of Lorentz invariance is not a good argument. Instead, it seems that one cannot satisfy two (or more) natural requirements at the same time, so one must reject at least one of them, either Lorentz invariance or something else.

 

[meopemuk]

I am sure that you have a good reason to take the measure you choose. It, indeed, seems very natural. Still, I would not say that a requirement of Lorentz invariance is not a good argument. Instead, it seems that one cannot satisfy two (or more) natural requirements at the same time, so one must reject at least one of them, either Lorentz invariance or something else.

If we use a non-invariant measure, this doesn't mean that the entire theory is non-invariant. Integration measure is not an observable thing. We need to look at transformations of observable properties in order to decide whether the theory is Lorentz invariant or not.

Eugene.

 

[Demystifier]

If we use a non-invariant measure, this doesn't mean that the entire theory is non-invariant. Integration measure is not an observable thing. We need to look at transformations of observable properties in order to decide whether the theory is Lorentz invariant or not.

So, did you prove that, in your case, all observable properties are Lorentz invariant?

For example, if a particle is localized for one observer and not localized for another observer (recall the Newton-Wigner paper), then the property of being localized is not Lorentz invariant. On the other hand, I think that this property is an observable one.

 

[meopemuk]

So, did you prove that, in your case, all observable properties are Lorentz invariant?

Yes, there is a proof. This proof is based on the fact that operators performing transformations of observables from one inertial frame of reference to another, form a unitary representation of the Poincare group. This property guarantees the relativistic invariance and conservation of probabilities. No extra conditions are needed.

For example, if a particle is localized for one observer and not localized for another observer (recall the Newton-Wigner paper), then the property of being localized is not Lorentz invariant. On the other hand, I think that this property is an observable one.

Yes, localization is an observable property. But why do you think that localization should be observer-independent? For example, observed particle momentum depends on the velocity of the observer, and there is no conflict with relativity, because we know that momentum (like velocity, position, energy, and whole bunch of other properties) is a relative quantity. Then why you insist that particle should look localized to everyone?

Eugene.

Eugene.

 

[Demystifier]

because we know that momentum (like velocity, position, energy, and whole bunch of other properties) is a relative quantity. Then why you insist that particle should look localized to everyone?

It is a question of relativity, so let us, for simplicity, consider a classical (not quantum) particle. Such a particle can be viewed invariantly as a timelike curve in spacetime. Now take the intersection of this curve with a hypersurface of constant time. The intersection is a point in spacetime, which means that the particle is local. Of course, the hypersurface of constant time depends on the observer. Nevertheless, for ANY observer, the corresponding constant-time hypersurface intersects the curve at one (and only one) point. Therefore, the particle is a local object for ANY observer.

Now you tell me how you imagine that this would change in the quantum case?

 

[meopemuk]

It is a question of relativity, so let us, for simplicity, consider a classical (not quantum) particle. Such a particle can be viewed invariantly as a timelike curve in spacetime. Now take the intersection of this curve with a hypersurface of constant time. The intersection is a point in spacetime, which means that the particle is local. Of course, the hypersurface of constant time depends on the observer. Nevertheless, for ANY observer, the corresponding constant-time hypersurface intersects the curve at one (and only one) point. Therefore, the particle is a local object for ANY observer.

Now you tell me how you imagine that this would change in the quantum case?

What you said could be true for classical particles, but quantum particles cannot be described by trajectories and worldlines. Even if a quantum state is localized at one time instant, the next instant its wave function will spread out. It can be proven that a similar spreading-out should occur in moving frames of reference (event at time t=0) as well.

Eugene.

 

[Demystifier]

What you said could be true for classical particles, but quantum particles cannot be described by trajectories and worldlines. Even if a quantum state is localized at one time instant, the next instant its wave function will spread out. It can be proven that a similar spreading-out should occur in moving frames of reference (event at time t=0) as well.

That is fine. But in this case, all observers will agree that the particle is localized at one and only one instant of time. In this sense, the fact that the particle is localized at some time is observer independent.

 

[meopemuk]

That is fine. But in this case, all observers will agree that the particle is localized at one and only one instant of time. In this sense, the fact that the particle is localized at some time is observer independent.

Yes, this would be the case if particle wavefunctions transformed under boosts like $\psi (x) \to \psi (\Lambda x)$. We are discussing this question in another thread https://www.physicsforums.com/showpost.php?p=1380999&postcount=193 I disagree with your transformation law. So, we need to reach some conclusion on that point, before continuing this thread.

Eugene.

 

[Hans de Vries]

Yes, this would be the case if particle wavefunctions transformed under boosts like $\psi (x) \to \psi (\Lambda x)$. I disagree with your transformation law. So, we need to reach some conclusion on that point, before continuing this thread.

Eugene.

It is the Klein Gordon equation which actually PRODUCES the relativistic
transformation $\psi (x) \to \psi (\Lambda x)$. One should say that SR is the RESULT
of the relativistic equations like those of Maxwell, Klein Gordon, Dirac
et-cetera, rather than saying that these equations are "Lorentz invariant"
Lorentz contraction from the (free) Klein Gordon equation $$\mbox{Klein Gordon: }\ \ \ \hbar^2 \frac{\partial^2 \psi}{\partial t^2}\ -\ c^2\hbar^2 \frac{\partial^2 \Psi} {\partial x^2} \ + \ (mc^2)^2 \psi \ =\ 0$$We separate $\psi$ as below where $Q$ is a localized Quantum wave
packet Q times a planewave with energy E momentum p.

$$\psi \ \ \ \equiv \ \ \ Q_{xt} \ (e^{i2\pi \mathbf{x}/\lambda}) \ (e^{-i2\pi f\mathbf{t}}) \ \ \ \equiv \ \ \ Q_{xt} \ e^{ip\mathbf{x}/\hbar-iE\mathbf{t}/\hbar}$$

The second order derivative in time becomes written out:

$$\ \hbar^2 \frac{\partial^2}{\partial t^2} \psi \ \ = -\left\{\ E^2\ + \ \frac{2i\hbar E}{Q}\ \frac{\partial Q}{\partial t} \ - \ \frac{\hbar^2}{Q}\ \frac{\partial^2 Q}{\partial t^2}\right\} \psi$$Since we want Q to be a constant localized function which shifts
along with physical speed v we can express the derivatives in time
as derivatives in space:

$$Q\ = \ Q(\gamma(x-vt))\ \ \ \ \mbox{therefor:}\ \ \ \ \frac{\partial Q}{\partial t} = -v \frac{\partial Q}{\partial x},\ \ \ \ \frac{\partial^2 Q}{\partial t^2} = v^2\frac{\partial^2 Q}{\partial x^2}$$Which is valid for any non-changing wave packet Q moving at a
constant velocity v. We will use these identities to make our equation
time independent and write for the partial derivatives:$$\hbar^2 \frac{\partial^2 \psi}{\partial t^2} \ \ \ \ = -\left\{\ E^2\ \ \ - \ \frac{2i\hbar E v}{Q}\ \frac{\partial Q}{\partial x} \ - \ \frac{\hbar^2 v^2}{Q}\ \frac{\partial^2 Q}{\partial x^2}\right\} \psi$$$$\ \ \frac{\partial^2}{\partial x^2} \psi \ \ = -\frac{1}{\hbar^2}\left\{\ p_x^2 c^2\ - \ \frac{2i\hbar p_x c^2}{Q}\ \frac{\partial Q}{\partial x} \ - \ \frac{\hbar^2 c^2}{Q}\ \frac{\partial^2 Q}{\partial x^2}\right\} \psi$$$$\ \ \frac{\partial^2}{\partial y^2} \psi \ \ = -\frac{1}{\hbar^2}\left\{ - \ \frac{\hbar^2 c^2}{Q}\ \frac{\partial^2 Q}{\partial y^2}\right\} \psi, \qquad \mbox{idem for z}$$

We then insert these terms in the Klein Gordon equation. The first
order derivative terms cancel each other since $Ev = pc^2 = mc^2v$. The
remaining terms become:$$E^2 - c^2p_x^2 \ = \ m^2 c^4\ -\ \frac{\hbar^2 c^2}{Q} \left[ \left(1-\frac{v^2}{c^2}\right)\ \frac{\partial^2 Q}{\partial x^2} +\frac{\partial^2 Q}{\partial y^2} +\frac{\partial^2 Q}{\partial z^2}\right]$$That is, the moving packet Q is compressed by a factor $\gamma$ in the x
direction. The second order derivatives are higher by a factor $\gamma^2$
which is canceled by the factor in final formula.

The crucial step is replacing the derivatives in time by the derivatives
in space for a stable solution shifting along with speed v. This is in
essence what causes Lorentz contraction.


Regards, Hans

 

[meopemuk]

It is the Klein Gordon equation which actually PRODUCES the relativistic
transformation $\psi (x) \to \psi (\Lambda x)$. One should say that SR is the RESULT
of the relativistic equations like those of Maxwell, Klein Gordon, Dirac
et-cetera, rather than saying that these equations are "Lorentz invariant"

Could you please tell the reasons why you think that Klein-Gordon is a good equation for relativistic wavefunctions? In my opinion, this equation does not comply with very fundamental Rules of Quantum Mechanics. In QM the time evolution of a single particle must be described by equation of the type

[tex] -i \hbar \frac{\partial}{\partial t}\psi(\mathbf{r}, t) = H \psi(\mathbf{r}, t) [/itex]

where H is the Hamiltonian. This form can be rigorously derived from such fundamental requirements as relativistic invariance and conservation of probabilities. Klein-Gordon equation does not have this form. Isn't it a controversy?

 

[Hans de Vries]

Could you please tell the reasons why you think that Klein-Gordon is a good equation for relativistic wavefunctions? In my opinion, this equation does not comply with very fundamental Rules of Quantum Mechanics. In QM the time evolution of a single particle must be described by equation of the type

[tex] -i \hbar \frac{\partial}{\partial t}\psi(\mathbf{r}, t) = H \psi(\mathbf{r}, t) [/itex]

where H is the Hamiltonian. This form can be rigorously derived from such fundamental requirements as relativistic invariance and conservation of probabilities. Klein-Gordon equation does not have this form. Isn't it a controversy?

You are just repeating a 70 year old discussion. In 1934, Pauli and Weisskopf
revived the Klein Gordon equation. Read for instance Sakurai, chapter 3.1
"Probability Conservation in Relativistic Quantum Mechanics" where he
shows the conservation of the 4-vector probability density current of the
Klein Gordon equation via the continuity equation.

That Pauli and Weisskopf were 100% right about their interpretation of
the probability current as the charge current density has been proved,
tens of thousands of times, by the huge and diverse industry, which uses
the Pauli Weisskopf interpretation as the basis for chemical, medical,
semiconductor and nano technology calculations.

Googling for "Density Functional Theory" gives 1.7 million pages which
should give you some idea about the industrial activity. It dwarfs the
number of pages for instance containing "Copenhagen Interpretation"
which is "only" 130,000.Regards, Hans

 

[Demystifier]

That Pauli and Weisskopf were 100% right about their interpretation of
the probability current as the charge current density has been proved,

If it has nothing to do with probability, but only with charge, then why the probabilistic interpretation is consistent with experiments, at least in the nonrelativistic limit?

 

[Fra]

Correct me if I got you wrong but I think what disturbs Demystifier and what is discussed here is that we would like to find a coherent line of reasoning of current understanding, that connects it to the past one. At least that is what I _thought_ we were discussing, and thus explains my actions ;) (Another illustration of my point)

If we don't care about that, and just does the magic and conclude that the current model now is consistent with experiment, then there is no problem. The question of wether the past is consistent with the future doesn't isn't asked.

Demystifier, to related to my odd thinking, I think of the consistency of reasoning, in that if we relate the probability appropriately to our "ignorance" or incompletness of initial conditions it might be consistent.

I think the inconsistencies is related to our thinking that we can't ever be wrong, or that we can never into a question to which the answer was't be predicted in the past. Ie. if we think that the ordinary QM and the one particle interpretation of non-rel stuff is CORRECT, and can't be wrong. It seems hard to understand the logic of evolving something, into a forbidden state. I think we need to understand how forbidden can also be relative, and that was seems "impossible" is in fact only impossible because our own limitations doesn't allow us to see the possibilities until they are right under our nose.

/Fredrik

 

[Fra]

I think it can't be understated that the standard probabilistic interpretations doesn't quite make sense, if we think they are "law". They are clearly abstractions only that _we think_ are the ultimate tool of science. Yet we preach about things must be measurable and relate to real experiments.

I dare you folks to make a foolproof from-scratch-deduction of the consistency of the probability space formalism when it comes to reality AND make sure that this "deduction" is at the same time observer invariant.

/Fredrik

 

[Hans de Vries]

If it has nothing to do with probability, but only with charge, then why the probabilistic interpretation is consistent with experiments, at least in the nonrelativistic limit?

Both probabilistic and charge-current density behavior are extensively proven.
(and both must be explained by any theory describing the underlaying physics)

The wavefunction of a single free electron can extend beyond one micron:

Two split single electron interference:
http://www.hqrd.hitachi.co.jp/em/doubleslit.cfm
http://www.hqrd.hitachi.co.jp/em/movie.cfm

While electrons can also be used to visualize the dumbbells of the atomic
electronic structures with sub Ångström resolution:

http://fei.com/Portals/_default/PDFs/content/2006_06_MicroscopyToday_.pdf
http://fei.com/Portals/_default/PDFs/content/2006_06_LithiumImagingOkeefe_wp.pdfRegards, Hans.

 

[Demystifier]

Demystifier, to related to my odd thinking, I think of the consistency of reasoning, in that if we relate the probability appropriately to our "ignorance" or incompletness of initial conditions it might be consistent.

Initial conditions of what? Wave functions? Particle positions? Something else?
In my Bohmian proposal, probabilities emerge from our ignorance of initial particle positions, which are the quantities that are actually measured. As the probabilities are not fundamental in the Bohmian approach, the fact that an a priori relativistic probability density of particle positions is not well defined is not really a problem.

 

[Demystifier]

Both probabilistic and charge-current density behavior are extensively proven.
(and both must be explained by any theory describing the underlaying physics)

That is exactly my point too. Both are proven experimentally. And both should be explained by a single coherent self-consistent theory. The problem is that we do not seem to have such a theory, or at least not a widely accepted one. Instead, we have TWO widely accepted theories (nonrelativistic QM and QFT) that we frequently mix in an incoherent manner.

 

[Fra]

Here are some attempts to explain more... but keep in mind that I'm still working on it..so the comments serve to propagate ideas only.

Initial conditions of what? Wave functions? Particle positions? Something else?

Eventually this will be given a exact definition, but I mean Initial conditions of our information - prior estimates. The same set of data can in the general case be given several interpretations. We can invent concepts, like space and particles, charge... but that's just "lables".. what is a particle? I'd that that any support for that concept is in the data we have. Ie. I may be exposed to a stream of photons, propagating as nerve signals to my brain... now in principle, it's clear that the "picture" I created of the outside world is not ambigously derived from input, however in the context of live a fitness etc, one can probably consider some interpretations more or less useful, or to get back on track, to have various estimated probabilities to be successful. Against this is a subjective estimate.

The initial conditions is all our knowledge, and record of history - but constrained to the fact that our memories are limited, ultimately relating to our limited size and mass and so on. We would probably grow be be black holes if we could store every piece of information without reduction. This leads to the concept of data compression adn storage effiency, and here hte interpretation comes in. Which in turn is related to our interaction properties.

In my thinking, space, particles and other "abstractions" are emergent structures in my view. They are _selected_ as the (in context) expectedly most constructive/fit interpretations. In this sense interpretation can be thought of loosely as as choice of data reduction. The task is to reduce data storage, but loose a minimum amount of _significant_ information.

This can be linked to several interesting interpretations, black body radiation beeing a way to dispose of energy, and the distribution of the energ is what the emitter considers (subjectively) to be the least useful. But not necessarily lacking information comlpetely, this is not the same thing.

In my Bohmian proposal, probabilities emerge from our ignorance of initial particle positions, which are the quantities that are actually measured. As the probabilities are not fundamental in the Bohmian approach, the fact that an a priori relativistic probability density of particle positions is not well defined is not really a problem.

You think differently, so it's hard to see exactly, but in a certain sense perhaps your missing information of bohmian particle may be given some interpretation in my view. But I'm not sure I like the word particle though, or to assume a "shape" of what's missing. Anyway, I aim to start off at a lower level... I'm first of all trying to operatively define space in terms of correlations in a random walk... and the result of the dimensionality is nothing that can predicted from the formalism, it must come from real data... but the formalism should define the relation betwene input and prediction - the best induction. But at each stage there is uncertainty. And the result is also observer dependent. For example. I am not so sure that the simplest possible elementa we know, can GRASP the full dimensionality. How does an electron really percept reality? Of course we could never know, but that thinking is interesting can I think even in lack of a perfect answer, provide us to insight.

Anway, I hope to get back to comparing the QM equations once I've found out how to treat space and time better. My previous attempts did resemeble the bohmian formalism, but not the bohmian interpretation (of particles). I tried to consider relations connecting different probability spaces... and the phase seems to receive a special interpretation, as a way to bundle the ignorance, but this was too shaky and I stepped back again to revise the notion of space and time. Because I was uncomfortable talking about functions of space and time, before the whole issue of space and time is clearly defined from my first principles.

/Fredrik

 

[Fra]

Initial conditions of what? Wave functions? Particle positions? Something else?

Another clarification: In my (personal) rethinking here, there is at fundamental level, nothing I call wave functions. I start with "labels", and the concept of distinguishability. From there on I consider that an observer in order "to make comparasions" and perceive change, must at minimum, at least transiently somehow be able to store and compare the present with the nearest past. This gives rise to the notion of "change". Again, correlations in the observers memory (particle or system state if you like) can define things like distinguishable structure, but the distinguishability soon ends up beeing fuzzy... so you run into the concept of relative frequencies and estimated probabilites. So far there is nothing I call wave function. That is further up in the abstractions, or I think it's probably just an alternative mechanism... further on "changes" of patterns is observed, which indirectly makes dynamics observable. But not observable in the same meaning as in QM.

Eventually the system of changing but somewhat stable patterns, can be given names. And in this way I hope to understand the information theoretic connection in the hierarchy of things... from the most basic boolean observation, through emergent structures of space and dimensionality, and further structures that are really relations between the more primary patterns.

This way I hope to see the choherent line of reasoning I want. But it's a long way to go.

/Fredrik

 

[Anonym]

Both probabilistic and charge-current density behavior are extensively proven(and both must be explained by any theory describing the underlaying physics).

Please look at A.Tonomura et al, “Double –biprism electron interferometry”, Applied Physics Letters, 84(17), 3229 (2004); Fig. 3(b),(c),(d) and (a).

I see E. Schrödinger, Zs. Phys., 14,664 (1926) coherent wave packet. Absence of the relevant set-up parameters prevent the geometrical optics calculations to be sure. Please, provide your comment/explanations.

If I am right, it is impossible with M.Born statistical approach.

Regards, Dany.

 

[meopemuk]

That is exactly my point too. Both are proven experimentally. And both should be explained by a single coherent self-consistent theory. The problem is that we do not seem to have such a theory, or at least not a widely accepted one. Instead, we have TWO widely accepted theories (nonrelativistic QM and QFT) that we frequently mix in an incoherent manner.

What about Weinberg's approach in his volume 1? I think it is an ideal way to formulate QFT. The limit to non-relativistic QM and the probabilistic interpretation of wavefunctions are readily available. Quantum fields and wavefunctions are well-separated.

Eugene.

 

[Micha]

I

I did extensive numerical simulations of Klein Gordon propagation
(in many different spatial dimensions) and one never sees any
propagation outside the light cone. Also analytically one doesn't see
anything outside the light cone.

Hi Hans,
I also checked this and I think, Feynman and the textbooks are right.
The propagator 1/(p^2-m^2 + i*epsilon) (written in Fourier space)
is not strictly zero outside the lightcone, when written
in real space, although dropping of fast.

I checked this in 1+1 dimensional spacetime.
First I did the integral over energy by integration along contours.
Then I put t=0 (i.e. +0, because of the teta function in the result you have to decide).
Now I am left with a one dimensional integral and I can easily check its
value as a function of x numerically.

Of course, my analysis is very primitive, but therefore I can not see, where the error
should lie.

For example, I think, you can do better and solve the second integral analytically by integrating along the branch cut. This way you get a factor exp(-kx) in the remaining integral pointing to an exponential drop off as well.

I can only guess, but maybe the problem with your analysis is, that you work with the massless propagator, and this
gets singular.

 

[Hans de Vries]

Hi Hans,
I also checked this and I think, Feynman and the textbooks are right.
The propagator 1/(p^2-m^2 + i*epsilon) (written in Fourier space)
is not strictly zero outside the lightcone, when written
in real space, although dropping of fast.

There is a lot of confusion and disagreement between different physicist and
different textbook. Pauli's paper "Spin and Statistics" disagrees with Feynman
in the analytical result. Pauli then uses the commutation argument as well to
claim there is no propagation outside the lightcone. That is he literally, quote:
"postulates" this to be the case without giving the math...


Regards, Hans

 

[Micha]

There is a lot of confusion and disagreement between different physicist and
different textbook. Pauli's paper "Spin and Statistics" disagrees with Feynman
in the analytical result. Pauli then uses the commutation argument as well to
claim there is no propagation outside the lightcone. That is he literally, quote:
"postulates" this to be the case without giving the math...Regards, Hans

Solving the integral is a well defined and rather simple mathematical question.
We should be able to settle on the right answer without referring to
either history or commutation relations. With this for the moment I just mean the question,
whether the propagator is zero outside the light cone or not.
I claim, this question can be settled in 1+1 dimensions.
Of course a full analytical answer in 4d is even better, if it is correct.

 

[Hans de Vries]

I can only guess, but maybe the problem with your analysis is, that you work with the massless propagator, and this
gets singular.

Both the massive and massless case have poles in momentum space and they need
careful attention. In position space you do not get infinities for any t smaller than
infinity. Why? The reason for the infinities is the plane wave representation which
stretches from x = +infinity to x = -infinity. Contributions from farther and farther
away regions keep coming in and, at the pole frequency, they all add up. The result
is infinite at t=infinte.

Any physical process doesn't continue until t=infinite nor is infinite in size.

One can obtain the analytical Green's function with the series development:

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

The analytical results then coincide with the numerical simulations and there is
no propagation outside the light cone. There is no need to use commutation
arguments to preserve causality. Regards, Hans

 

[Hans de Vries]

One can obtain the analytical Green's function with the series development:

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

The analytical results then coincide with the numerical simulations

The procedure to solve this analytically is as follows:
1) Solve the 1+1d case for 1/p²

2) Extend this to the full series given above for the 1+1d case. The series
becomes a Bessel J function of order zero.

3) Extend this to any dimensional space using the "inter-dimensional operator"
The Bessel J function of order zero becomes one of first order in 3+1d space
and the total result is:$$\Theta(t) \left(\ \frac{1}{2\pi}\delta(s^2)\ + \frac{m}{4\pi s} \Theta(s^2)\ \mbox{\huge J}_1(ms)\ \right), \qquad \mbox{with:}\ \ \ s^2=t^2-x^2$$
Steps 1) and 3) can be found in my paper here:
http://chip-architect.com/physics/Higher_dimensional_EM_radiation.pdf

Step 1: See page 3: "IV Derivation of the propagators" (${\cal H}\equiv\Theta$ is Heaviside step-function)
Step 3: The ïnter dimensional operator is proved at page 5: section V.Now for step 2 I'll write up another post.Regards, Hans

 

[Hans de Vries]

Now for step 2 I'll write up another post.

First we need the series expansion for the Bessel function which you can find here:
http://functions.wolfram.com/BesselAiryStruveFunctions/BesselJ/02/

In our case we need:

$$J_0(s)\ =\ \sum_{k=0}^\infty\ \frac{(-1)^k}{(k!)^2} \left( \frac{s}{2} \right)^{2k}$$

Written out:

$$J_0(s)\ =\ 1\ -\ \left( \frac{s}{2} \right)^2\ +\ \frac{1}{4}\left( \frac{s}{2} \right)^4\ -\ \frac{1}{36}\left( \frac{s}{2} \right)^6 ...$$Now we need to go back to page 3 of my paper "IV derivation of the propagators"
to get the higher order terms of the series.

The series in position space is:

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

For each extra term the procedure is:

a) Integrate over the t=+r line.
b) Integrate over the t=-r line.
c) Multiply by -m²

We start with the first term which we know and which is a simple Heaviside step
function. The paper uses ${\cal H}$ instead of $\Theta$ for the Heaviside step function.
Now we get the series:

$$\Theta(s^2)J_0(ms)\ \ =\ \ \Theta(s^2) \left(\ 1\ -\ \left( \frac{ms}{2} \right)^2\ +\ \frac{1}{4}\left( \frac{ms}{2} \right)^4\ -\ \frac{1}{36}\left( \frac{ms}{2} \right)^6 ...\right)$$
Regards, Hans

 

[Hans de Vries]

For each extra term the procedure is:

a) Integrate over the t=+r line.
b) Integrate over the t=-r line.
c) Multiply by -m²

Use the fact that:

$$s^2\ =\ (t+r)(t-r)\ =\ u v$$

Keep the terms expressed in the u and v and:

a) Integrate over u.
b) Integrate over v.
c) Multiply by -m²

Where u and v are independent (orthogonal) coordinates.Regards, Hans

 

[Micha]

Your machinery is quite impressive.
Yet, I think, if it predicts no propagation outside the light cone,
there must be an error somewhere.

Let's focus just on the 1+1 dimensional case for that
to keep things simple, ok?

Forget my remark about the massless propagator being singular.
It is just that I focused on the case with finite rest mass so far,
although the massless propagator should be also finite and probably even
simpler to calculate.

Your geometrical series for the propagator as such is obviously right.
I didn't follow your calculation from there so far.
But did you ever do the calculation in the way suggested by Zee in his
Quantum field theory in a nutshell?
You mentioned this book yourself in your earlier post.
The integration along contours for the integral over energy
is rather straightforward and the result can be found in the book.
The calculation is done for the 3+1d case, but you can copy this
step for the 1+1d case.

And then there is only a single integral left, which you can
evaluate for t=0 and x finite numerically.

 

[Hans de Vries]

Your geometrical series for the propagator as such is obviously right.
I didn't follow your calculation from there so far.

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

Using the series is sufficient to guarantee causality and SR because
none of the terms of this series has propagation outside the light cone.
$\Box^{-1}$ is the photon propagator on the light cone. $\Box^{-2}$ is a re-emission,
which is again on the light-cone, the third time is the second re-emission
and so on, all on the light-cone...


Regards, Hans

 

[Hans de Vries]

But did you ever do the calculation in the way suggested by Zee in his
Quantum field theory in a nutshell?
You mentioned this book yourself in your earlier post.
The integration along contours for the integral over energy
is rather straightforward and the result can be found in the book.
The calculation is done for the 3+1d case, but you can copy this
step for the 1+1d case.

You can find the source of all this in chapter 17 and 18 of Feynman's:

"The theory of fundamental processes"

It's is a non-physical artifact of perturbation theory where there are
pairs of diagrams which have a physical meaning. Independently
they have no physical meaning because they can be converted into
each other via a Lorentz transform, and only together they respect
causality and special relativity.

Now the term "non-physical" here is mine because Feynman himself,
In the spirit of his mentor John Archibald Wheeler (The good man is still
with us) had no problem with particles going not only forward but also
backward in time, or going at any speed faster than c.

Feynman cuts the propagator in two with his pole prescription which
leads to these pair of diagram's which have to be considered together
to get both poles back and special relativity restored.


Regards, Hans

 

[OOO]

It is frightening (or should I say enlightening) to see what a considerable amount of dissent can be generated by seemingly "very simple QFT questions". I didn't understand neither P&S' argument nor any other I have found in textbooks. It's also quite remarkable how much all the reputable authors avoid making definite observable statements for which they may be held responsible.

I have also done numerical simulations of KG and it seems quite obvious to me that there is no propagation outside the light cone. If one discretizes the d'Alembertian by finite differences of second order (for stability analysis see e.g. Numerical Recipes), one gets something like (in 1+1 dimension)

$$\Psi(t+1,x)+\Psi(t-1,x)-2\Psi(t,x) - (\Psi(t,x+1)+\Psi(t,x-1)-2\Psi(t,x)) = -m^2 \Psi(t,x)$$

If one solves for the next timestep one gets

$$\Psi(t+1,x) = -\Psi(t-1,x)+ \Psi(t,x+1)+\Psi(t,x-1) -m^2 \Psi(t,x)$$

The maximum speed with which information can propagate through this lattice is 1 (the speed of light in lattice units). You see this immediately from the above equation: the next time step is only influenced by the two adjacent lattice sites (and the one in the middle of course).

Of course it is not a strict proof but I think very convincing and very graphic (if you've seen the sim).