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

[meopemuk]

This is probably not what you are talking about, but I'll continue with "very simple QFT questions". What would

<br /> \int\frac{d^3p}{(2\pi\hbar)^3}\frac{1}{\sqrt{2E_{\boldsymbol{p}}}} e^{-i\boldsymbol{p}\cdot\boldsymbol{x}/\hbar} a^{\dagger}_{\boldsymbol{p}} |0\rangle<br />

be?

There are two different things in QFT that look similar. One thing is a one-particle eigenstate of the position operator. A spin-zero particle localized at point \mathbf{x} is described by the state vector

<br /> \int\frac{d^3p}{(2\pi\hbar)^{3/2}}e^{-i\boldsymbol{p}\cdot\boldsymbol{x}/\hbar} a^{\dagger}_{\boldsymbol{p}} |0\rangle<br />

(note that I corrected factors and signs in the formula I wrote previously). The absence of the factor \frac{1}{\sqrt{2E_{\boldsymbol{p}}}} in the integrand is not accidental. It follows from careful examination of the properties of the Newton-Wigner position operator and normalization. This analysis was performed in section 5.2 of http://www.arxiv.org/physics/0504062

A completely different thing is formula for the quantum field associated with spin-0 neutral particles

<br /> \psi (\mathbf{x},t) = \int\frac{d^3p}{(2\pi\hbar)^{3/2}}\frac{1}{\sqrt{2E_{\boldsymbol{p}}}} (e^{\frac{i}{\hbar}(\boldsymbol{p}\cdot\boldsymbol{x} - E_{\boldsymbol{p}}t)} a^{\dagger}_{\boldsymbol{p}} + e^{-\frac{i}{\hbar}(\boldsymbol{p}\cdot\boldsymbol{x} - E_{\boldsymbol{p}}t)} a_{\boldsymbol{p}} )

If you allow this operator to act on the vacuum vector at t=0, you obtain a state vector

<br /> \psi (\mathbf{x},0) |0 \rangle = \int\frac{d^3p}{(2\pi\hbar)^{3/2}}\frac{1}{\sqrt{2E_{\boldsymbol{p}}}} e^{\frac{i}{\hbar}\boldsymbol{p}\cdot\boldsymbol{x} } a^{\dagger}_{\boldsymbol{p}} |0 \rangle

which is basically what you wrote. What is the meaning of this state? I don't know. In my opinion, it doesn't have any good meaning. Certainly, it is not a state of a localized particle.

As I tried to argue elsewhere (https://www.physicsforums.com/showpost.php?p=1375933&postcount=6) the role of quantum fields is to provide "building blocks" for interaction operators in QFT. There is no good reason to interpret quantum fields or their action on vacuum as some kinds of "wave functions" or "state vectors".

Eugene.

 

[meopemuk]

No, I am Dr. Nikolic. But I do not blame you, for some reason everybody makes the same misprint in my name. There must be a reason for that, but I cannot figure it out.

I apologize for misspelling your name.

Of course, but QFT is not the same thing as relativistic QM.

I would like to disagree. In my opinion, QFT is not that different from a relativistic QM. The only significant difference is that QFT deals with systems in which the number of particles is not fixed, and creation/annihilation processes are allowed.

 

[meopemuk]

What is your interpretation of probability? Frequentists? Bayesian? Do you just stick to the Kolmogorovs axioms of the formalism? What I see clearly is that while there isn't anything obviously wrong with the axioms themselves from the point of view of mathematics. The problem is howto make a satisfactory connection to reality, which is supposedly the business here at least for me. I consider this to be one of the prime issues. I can't accept that way these parts are rushed over as if it's obvious that while the particle position isn't deterministic, it's probability is? It's very typical of physics, to grab a nice mathematical formalism and adapt it. Are everything you need to define the probability space itself, observables? It seems not. This bothers me. And I won't sleep until it's fixed.

I described my interpretation of probability in section 2.3 of http://www.arxiv.org/physics/0504062 . Probabilities are normally assigned to experimental "propositions", i.e., statements that can have two values: either "true" or "false". We prepare N identical copies of the same system and perform N measurements of the proposition. Then we count the number of instances (M) in which the value of the proposition was found "true". Then we take the limit N \to \infty and say that the probability of the proposition to be true is equal to the ratio M/N in this limit. So, in principle, probabilities can be measured to an arbitrary precision.

I find it easier to see what I think is wrong now when I've had a break from physics. I try to forget what it's "supposed" to be like, and instead try to think from scratch in a critical manner. I found that to be very hard to do when you are in the middle of something, a course for example, because they you are kind of fighting against yourself and it makes everything 10 times harder becuase you are trying to "learn something" and the critical to it at the same time. This is what I strongly disliked during the courses I took. I tried to be critical, which annoyed the teachers because it was "counterproductive" for the classwork. So clearly, you were encouraged to accept the ideas, trying to be critical only made it harder from everyone by arguing that what is taught doesn't quite make sense.

/Fredrik

This is a very good point. This is one of the reasons I distanced myself from "mainstream" academia physics. I've noticed that when I was a part of the system I felt a pressure (to publish, to say certain things, to be "smarter" than my colleagues, etc.). This is not a strong pressure, and many would say that they can ignore this pressure and be independent thinkers. I am not so sure. I'm afraid that this subtle pressure is enough to align thoughts of many people in just one direction, which is not good. I feel much better when I am free of external influences and obligations and can study questions that interest me personally at my own pace. It is called freedom, and it is invaluable.

I wish you a good luck in exploring your intersting ideas.

 

[meopemuk]

Hegerfeldt wrote:

Let
us suppose that at time t = 0 one could prepare an ensemble of strictly localized
(non-interacting) particles by laboratory means, e.g. photons in an oven. Then
one could open a window and would observe some of them at time t = " later
on the moon. Or to better proceed by repetition, suppose one could successively
prepare strictly localized individual particles in the laboratory. Preferably this
should be done with different, distinguishable, particles in order to be sure when
a detected particle was originally released. Such a signaling procedure would
have very low efficiency but still could be used for synchronization of clocks or,
for instance, for betting purposes.

Yes, one can release localized particles at t=0 and find them almost instantaneously on the Moon. If these were classical particles, then the causality would be violated in a moving reference frame, and one could theoretically arrange a scheme to predict stock market or kill her grandfather before she was born. However, "unfortunately", we are dealing with quantum particles whose behavior is inherently probabilistic. When a moving observer looks at a localized quantum particle he has a chance to see it on the Moon already at time t=0. So, it is not possible for him to say what is the direction of particle propagation or what is the cause and what is the effect. Probabilities make everything fuzzy and blurred.

I tried to make this point also in
https://www.physicsforums.com/showpost.php?p=1374456&postcount=13

Eugene.

 

[meopemuk]

Yes but this still begs the question, this one-particle state is still only physically realized on the order of the Compton wavelength, beyond that extra degrees of freedom arise.

No, this state is localized in a single point. Realistically, of course, the volume of localization is not a point but a finite region in space. But this region can be much smaller than the Compton wavelength. No problem. It is not clear to me what "extra degrees of freedom" you have in mind.

Moreover this is also in the strict confines of the free particle, turn on interactions (the physical regime) and you can't even define this.

I don't think that interactions have any effect on localization either. The role of interaction is to modify the total Hamiltonian of the system. So, if the interaction is present, the time evolution of the localized state prepared at t=0 would be different from the non-interacting time evolution. But this shouldn't change our ability to prepare the localized state in the first place.


We could probably talk about semi localization and what not see
http://arxiv.org/PS_cache/quant-ph/pdf/0112/0112149v1.pdf
(see discussion from p33 onwards) [/QUOTE]

Thanks for the reference. I am familiar with Wallace's works, but I disagree with quite a few things that he wrote. My disagreements begin from the top of his page 2.

 

[meopemuk]

The problem is that the measure
d^3p
is not Lorentz invariant.
If you replace it with a Lorentz invariant measure
d^3p/2E
then the state is no longer localized, but "semilocalized" as Haelfix said.

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".

 

[meopemuk]

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&#039;}]=(2\pi\hbar)^3\delta^3(p-p&#039;). 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...

When you are thinking about this, please keep in mind the fundamental difference between wave functions and quantum fields. For wave functions the normalization condition is fairly straightforward. It follows from the fact that the total probability to find a particle somewhere in (position or momentum) space is 1. The norm of the state vector must be 1 in all frames of reference

I suspect that above you were talking about quantum fields |\psi\rangle, not about wave functions. As I said earlier, quantum fields are needed in QFT only as building blocks for interaction Hamiltonians. (This point of view is most clearly expressed in S. Weinberg "The quantum theory of fields", vol. 1). So, there is no any physical condition that would fix the normalization of fields. If you decided to change the field normalization, you could also change some factors in formulas expressing interaction Hamiltonians through the fields. And you would be OK.

Different authors choose different normalization conventions for quantum fields. This is OK as long as they stick to the same convention from the beginning to the end. But this is really frustrating when you want to compare formulas from different books.

 

[Fra]

Thanks Meopemuk for your comments.

I described my interpretation of probability in section 2.3 of http://www.arxiv.org/physics/0504062 . Probabilities are normally assigned to experimental "propositions", i.e., statements that can have two values: either "true" or "false". We prepare N identical copies of the same system and perform N measurements of the proposition. Then we count the number of instances (M) in which the value of the proposition was found "true". Then we take the limit N \to \infty and say that the probability of the proposition to be true is equal to the ratio M/N in this limit. So, in principle, probabilities can be measured to an arbitrary precision.

Ok, thanks. That's what I'd call the frequentist interpretation. I have several serious issues with this view and how N \to \infty is trivialized into "in principle", but it's along the lines we discussed before and perhaps we can get back to details another time. And I guess this is a common issue with that standard interpretations.

I have saved your large 600+ page paper and I hope to find in it your core ideas some day. I haven't gotten around to it in detail yet.

/Fredrik

 

[meopemuk]

Thanks Meopemuk for your comments.

I have saved your large 600+ page paper and I hope to find in it your core ideas some day. I haven't gotten around to it in detail yet.

/Fredrik

Talk to you later, then. I would be glad to answer your questions.

Eugene.

 

[kharranger]

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.

 

[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&#039;}]=(2\pi\hbar)^3\delta^3(p-p&#039;). 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 <br /> \mbox{Klein Gordon: }\ \ \ \hbar^2 \frac{\partial^2 \psi}{\partial <br /> t^2}\ -\ c^2\hbar^2 \frac{\partial^2 \Psi} {\partial x^2} \ + \ <br /> (mc^2)^2 \psi \ =\ 0<br />We separate \psi as below where Q is a localized Quantum wave
packet Q times a planewave with energy E momentum p.

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

The second order derivative in time becomes written out:

<br /> \ \hbar^2 \frac{\partial^2}{\partial t^2} \psi \ \ = -\left\{\ E^2\ <br /> + \ \frac{2i\hbar E}{Q}\ \frac{\partial Q}{\partial t} \ - \ <br /> \frac{\hbar^2}{Q}\ \frac{\partial^2 Q}{\partial t^2}\right\} \psi<br />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:

<br /> Q\ = \ Q(\gamma(x-vt))\ \ \ \ \mbox{therefor:}\ \ \ \ \frac{\partial <br /> Q}{\partial t} = -v \frac{\partial Q}{\partial x},\ \ \ \ <br /> \frac{\partial^2 Q}{\partial t^2} = v^2\frac{\partial^2 <br /> Q}{\partial x^2}<br />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:<br /> \hbar^2 \frac{\partial^2 \psi}{\partial t^2} \ \ \ \ = -\left\{\ <br /> E^2\ \ \ - \ \frac{2i\hbar E v}{Q}\ \frac{\partial Q}{\partial x} \ <br /> - \ \frac{\hbar^2 v^2}{Q}\ \frac{\partial^2 Q}{\partial x^2}\right\} <br /> \psi<br /><br /> \ \ \frac{\partial^2}{\partial x^2} \psi \ \ = <br /> -\frac{1}{\hbar^2}\left\{\ p_x^2 c^2\ - \ \frac{2i\hbar p_x c^2}{Q}\ <br /> \frac{\partial Q}{\partial x} \ - \ \frac{\hbar^2 c^2}{Q}\ <br /> \frac{\partial^2 Q}{\partial x^2}\right\} \psi<br /><br /> \ \ \frac{\partial^2}{\partial y^2} \psi \ \ = <br /> -\frac{1}{\hbar^2}\left\{ - \ \frac{\hbar^2 c^2}{Q}\ <br /> \frac{\partial^2 Q}{\partial y^2}\right\} \psi, \qquad<br /> \mbox{idem for z}<br />

We then insert these terms in the Klein Gordon equation. The first
order derivative terms cancel each other since Ev = pc^2 <br /> = mc^2v. The
remaining terms become:<br /> E^2 - c^2p_x^2 \ = \ m^2 c^4\ -\ <br /> \frac{\hbar^2 c^2}{Q} \left[ \left(1-\frac{v^2}{c^2}\right)\ <br /> \frac{\partial^2 Q}{\partial x^2} +\frac{\partial^2 Q}{\partial y^2} <br /> +\frac{\partial^2 Q}{\partial z^2}\right]<br />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

-i \hbar \frac{\partial}{\partial t}\psi(\mathbf{r}, t) = H \psi(\mathbf{r}, t) [/itex]<br /> <br /> 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&#039;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

-i \hbar \frac{\partial}{\partial t}\psi(\mathbf{r}, t) = H \psi(\mathbf{r}, t) [/itex]<br /> <br /> 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&#039;t it a controversy?

<br /> <br /> You are just repeating a 70 year old discussion. In 1934, Pauli and Weisskopf<br /> revived the Klein Gordon equation. Read for instance Sakurai, chapter 3.1<br /> &quot;Probability Conservation in Relativistic Quantum Mechanics&quot; where he <br /> shows the conservation of the 4-vector probability density current of the<br /> Klein Gordon equation via the continuity equation.<br /> <br /> That Pauli and Weisskopf were 100% right about their interpretation of <br /> the probability current as the charge current density has been proved, <br /> tens of thousands of times, by the huge and diverse industry, which uses<br /> the Pauli Weisskopf interpretation as the basis for chemical, medical, <br /> semiconductor and nano technology calculations.<br /> <br /> Googling for &quot;Density Functional Theory&quot; gives 1.7 million pages which<br /> should give you some idea about the industrial activity. It dwarfs the <br /> number of pages for instance containing &quot;Copenhagen Interpretation&quot; <br /> which is &quot;only&quot; 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.