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

[meopemuk]

No, I am saying the transformation from one observer to another must be covariant.

What does the "covariance" mean exactly, when applied to transformations of particle wavefunctions?

 

[Demystifier]

I still cannot see what could go wrong with choosing the Newton-Wigner operator as a relativistic generalization of position? I think this operator provides a perfect description of position in relativistic quantum theory.

From another point of view, there must be *some* position operator in relativistic quantum theory. You cannot just say: there is no operator, so I am not going to consider the position representation. Position is the most basic observable in physics, and it remains measurable in both non-relativistic and relativistic physics. If position is an observable, then in quantum theory there must be an operator corresponding to it. It is just inevitable.

I fully agree with you that position is the most basic observable. Still, I agree with majority that the Newton-Wigner operator is not satisfying as it is not covariant. To solve the puzzle, I suggest to modify the axiom that an observable must be defined by an operator. For a concrete proposal, see
http://xxx.lanl.gov/abs/0705.3542

 

[meopemuk]

Exactly! Relativistic QM does not satisfy this axiom [unitary Hamiltonian time evolution law]. This is also closely related to the fact that relativistic QM cannot be interpreted probabilistically.

This is too radical for me. I want to believe that laws of quantum mechanics remain valid in all regimes, including the relativistic one. If you say that these laws become invalid, then you need to substitute quantum mechanics with a more general (non-probabilistic?) theory. What is this theory?

 

[Demystifier]

What does the "covariance" mean exactly, when applied to transformations of particle wavefunctions?

No, the correct question is what does it mean when applied to transformations of particle position operator. The answer is: The position operator must be the space component of a 4-vector. Then the transformation to another observer is just a Lorentz transformation of this 4-vector, followed by taking the space part of new coordinates.

 

[Demystifier]

This is too radical for me. I want to believe that laws of quantum mechanics remain valid in all regimes, including the relativistic one. If you say that these laws become invalid, then you need to substitute quantum mechanics with a more general (non-probabilistic?) theory. What is this theory?

Yes, that is radical. But most physicists agree that you must be ready to do something radical in order to combine QM and relativity consistently.
To see what that theory might be, see the link in my recent post above.

 

[Hans de Vries]

Apparently, particles localized in the reference frame at rest don't look like localized in the moving frame. So what? Special relativity has taught us that many things previously considered absolute are, in fact, observer-dependent. Localization is just one of these relative things. That's how I see it.

A particle localized in one frame is also localized in another frame. You are
using an example where the particle is a Dirac function. So, in 4D it's a line,
(the t-axis). After a boost it will be another line, the t'-axis.

Rotations and Boost are exactly the same in moment space as they are in
configuration space, so doing a Fourier transform followed by a boost
followed by an inverse Fourier transform gives you the same result.

Now, the moving particle is of course anywhere at x at some time t, The
math in 10.1.2 doesn't consider time so that's where it might go wrong.

A real wave function will spread (with a maximum speed of c). In any other
reference frame it does spread as well, again with a maximum of c. If you
look at t=t'=0 where the particle is a point, then it's a point in any
reference frame with v<c.


Regards, Hans.

 

[meopemuk]

I fully agree with you that position is the most basic observable. Still, I agree with majority that the Newton-Wigner operator is not satisfying as it is not covariant. To solve the puzzle, I suggest to modify the axiom that an observable must be defined by an operator. For a concrete proposal, see
http://xxx.lanl.gov/abs/0705.3542

Thanks for the reference. I enjoyed reading this well-written paper. (I have a suspicion that you are Dr. Nicolic. Is it true?) However I cannot accept its basic premise. I am possibly too conservative to accept the idea that fundamental Rules of Quantum Mechanics should be abandoned. Moreover, I don't see anything wrong with using the Newton-Wigner position operator in quantum field theory.

I have strongest disagreements with your attempts to connect wavefunctions with quantum fields (e.g., in eq. (1)). I think that particle wavefunctions and quantum fields are two completely different beasts that have nothing in common, except, probably, the usual misleading practice to denote them by the same Greek letter $\psi$.

 

[meopemuk]

The position operator must be the space component of a 4-vector.

Why? I think this is a too narrow reading of the principle of relativity. According to Wigner, relativistic quantum mechanics requires that there should exist an unitary representation $U_g$ of the Poincare group in the Hilbert space of the system. Then transformation of any observable (operator F) to a different (e.g., moving) reference frame should be given by formula

$$F \to U_g F U_g^{-1}$$

In some cases (e.g., for the total energy-momentum ($E, \mathbf{P}$)) this formula, indeed, leads to the 4-vector transformation law. In other cases (e.g., for the Newton-Wigner position operator ($\mathbf{R}$)) the transformation law is different. I don't think there is anything wrong with it. Relativistic invariance of the theory is guaranteed by the fact that operators $U_g$ satisfy multiplication laws of the Poincare group. Covariant transformations of observables is an additional assumption, which cannot be rigorously justified, in my opinion.

You can possibly argue that my departure from canons of special relativity is too radical, but I consider it less radical than your departure from laws of quantum mechanics. It is true that Einstein's special relativity cannot peacefully coexist with quantum mechanics. Something's got to give. I think we both agree about that. We disagree about ways forward.

 

[meopemuk]

A particle localized in one frame is also localized in another frame. You are
using an example where the particle is a Dirac function. So, in 4D it's a line,
(the t-axis). After a boost it will be another line, the t'-axis.

Rotations and Boost are exactly the same in moment space as they are in
configuration space, so doing a Fourier transform followed by a boost
followed by an inverse Fourier transform gives you the same result.

Now, the moving particle is of course anywhere at x at some time t, The
math in 10.1.2 doesn't consider time so that's where it might go wrong.

In subsection 10.1.2 I considered two reference frame in relative motion. Both of them perform position measurements at the same time instant t=0. This is the reason why time is not present in these formulas. The presence of the boost-induced spreading was analyzed in the (previously cited) paper of Newton and Wigner. Another argument was presented in
https://www.physicsforums.com/showpost.php?p=1376402&postcount=29
I haven't seen a proof that (as you say) particle remains localized in all frames. If you have such a proof, it would great to compare our notes.

Regards.
Eugene.

 

[Hans de Vries]

I haven't seen a proof that (as you say) particle remains localized in all frames. If you have such a proof, it would great to compare our notes.

Regards.
Eugene.

The easiest thing is probably to visualize this with a Minkowsky type image
like the one here:

http://www-users.york.ac.uk/~mijp1/transaction/figs/fig02_h.gif

If you confine the wave function to a point at (t,x) = (0,0) and it is allowed
to spread at any speed up to c, then it will still be confined in the future
light cone.

Any x'-axis belonging to another reference frame is rotated in the point (0,0)
over an angle between +45 and -45 degrees. This means that it will never
cut through the future (or past) light cones. It will only cut through the
particle's wave function in the point (0,0). This means any observer with a
speed v<c will see it as a point a t=t'=0.


Regards, Hans

 

[meopemuk]

The easiest thing is probably to visualize this with a Minkowsky type image
like the one here:

http://www-users.york.ac.uk/~mijp1/transaction/figs/fig02_h.gif

If you confine the wave function to a point at (t,x) = (0,0) and it is allowed
to spread at any speed up to c, then it will still be confined in the future
light cone.

Any x'-axis belonging to another reference frame is rotated in the point (0,0)
over an angle between +45 and -45 degrees. This means that it will never
cut through the future (or past) light cones. It will only cut through the
particle's wave function in the point (0,0). This means any observer with a
speed v<c will see it as a point a t=t'=0.


Regards, Hans

In your proof you made one important tacit assumption. You assumed that boost transformations of wave functions can be represented by simple x-t "rotation" in the Minkowski space-time diagram.

Now, if you take a different route and decide to derive boost transformations of position-space wave functions from Wigner's theory of irreducible unitary representations of the Poincare group, you will obtain a different result. (see subsection 10.1.2 in my book).

This is definitely a contradiction. The question is, which one of these two methods is more consistent with the principle of relativity and quantum mechanics? I would like to argue that the latter method is more consistent. I tried to justify this conclusion axiomatically in my book.

 

[Haelfix]

Incidentally I object to the use of the Klein Gordon equation and the Dirac equation to make some sort of argument against localization as Hegerfield (sp?) seems to argue for.

Localizing any physical particle in that framework, beyond its compton wavelength will automatically enter a regime (QFT) where pair creation becomes important and where multiple fields must be considered.

 

[meopemuk]

Incidentally I object to the use of the Klein Gordon equation and the Dirac equation to make some sort of argument against localization as Hegerfield (sp?) seems to argue for.

Localizing any physical particle in that framework, beyond its compton wavelength will automatically enter a regime (QFT) where pair creation becomes important and where multiple fields must be considered.

This argument is often invoked in such discussions, but it doesn't apply to particles prepared in eigenstates of the Newton-Wigner position operator. This operator explicitly commutes with particle number operators, so it is possible to have a single particle localized at one point.

 

[Haelfix]

Ok. Define the particle number operator in the context you are thinking off that commutes with the Newton-Wigner operator.

I have a feeling what I'm thinking off, and what you are thinking off are going to be different. Regardless I suspect this operator is going to run into issues with its commutation relations with other observables like say the hamiltonian and hence disallowed on physical grounds.

 

[meopemuk]

Ok. Define the particle number operator in the context you are thinking off that commutes with the Newton-Wigner operator.

I have a feeling what I'm thinking off, and what you are thinking off are going to be different. Regardless I suspect this operator is going to run into issues with its commutation relations with other observables like say the hamiltonian and hence disallowed on physical grounds.

To keep things simple, I will write you expression for a one-particle state localized at point $\mathbf{x}$

$$|\Psi \rangle = \int d^3p e^{\frac{i}{\hbar} \mathbf{p} \cdot \mathbf{x}} a^{\dag}_{\mathbf{p}} |0 \rangle$$

Here $a^{\dag}_{\mathbf{p}}$ are particle creation operators and $| 0 \rangle$ is the vacuum vector. Of course, this state has infinite uncertainties of momentum and energy, but it is still an one-particle state. It is an eigenstate (with eigenvalue 1) of the particle number operator

$$N = \int d^3p a^{\dag}_{\mathbf{p}} a_{\mathbf{p}}$$

 

[jostpuur]

It is often said in textbooks that the Klein-Gordon equation is a relativistic analog of the Schroedinger equation, and therefore it can be used for the description of time evolution. Then, as you correctly pointed out, initial conditions should include the time derivative of the wave function in addition to the wave function itself. In other word, the state of the particle at time $t + \Delta t$ is determined not only by its state at time $t$, but also by the "state derivative" at time $t$.

This statement is in contradiction with the fundamental equation for time evolution in quantum mechanics

$$|\Psi(t) \rangle = \exp(\frac{i}{\hbar}Ht) |\Psi(0) \rangle$$

where H is the Hamiltonian. This equation show, in particular, that the state at time $t + \Delta t$ is determined by the state at time $t$, the Hamiltonian, and nothing else.

Exactly! Relativistic QM does not satisfy this axiom. This is also closely related to the fact that relativistic QM cannot be interpreted probabilistically.

But isn't the time evolution in QFT defined precisly with this equation?

 

[jostpuur]

To keep things simple, I will write you expression for a one-particle state localized at point $\mathbf{x}$

$$|\Psi \rangle = \int d^3p e^{\frac{i}{\hbar} \mathbf{p} \cdot \mathbf{x}} a^{\dag}_{\mathbf{p}} |0 \rangle$$

Here $a^{\dag}_{\mathbf{p}}$ are particle creation operators and $| 0 \rangle$ is the vacuum vector. Of course, this state has infinite uncertainties of momentum and energy, but it is still an one-particle state. It is an eigenstate (with eigenvalue 1) of the particle number operator

$$N = \int d^3p a^{\dag}_{\mathbf{p}} a_{\mathbf{p}}$$

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

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

be?

 

[Demystifier]

Thanks for the reference. I enjoyed reading this well-written paper. (I have a suspicion that you are Dr. Nicolic. Is it true?)

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.

 

[Demystifier]

But isn't the time evolution in QFT defined precisly with this equation?

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

 

[Demystifier]

You can possibly argue that my departure from canons of special relativity is too radical, but I consider it less radical than your departure from laws of quantum mechanics. It is true that Einstein's special relativity cannot peacefully coexist with quantum mechanics. Something's got to give. I think we both agree about that. We disagree about ways forward.

I agree that we agree on what you say that we agree and disagree on what you say that we disagree.
I also agree that my approach is more radical than yours. But let me quote one famous physicist: Your theory is crazy, but not crazy enough to be true.

 

[Fra]

I am possibly too conservative to accept the idea that fundamental Rules of Quantum Mechanics should be abandoned.

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.

But I don't think this change need to flip all of QM over, like QM didn't flip classical mechanics over. The main problem I see is that when you are taught something, and trained to see how well it works, it's basic psychology that it's easy that the thinking may get restricted. Just like Newtonian ideals are very rooted in us. It's not easy to think outside the box.

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

 

[Hans de Vries]

Hegerfeldt's discussion is even more general as he doesn't specify the position operator explicitly. You can find his article on the web http://www.arxiv.org/quant-ph/9809030

Hegerfeldt is apparently on a quest to find instantaneous and superluminal
communication. Don't be surprised to find opposition...

Could instantaneous spreading be used for the transmission of signals if it
occurred in the framework of relativistic one-particle quantum mechanics? 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.

Bad or naive math gives plenty of opportunities to find non-causal and non-
local phenomena: What about using a harp to predict bomb-attacks?
An explosion is basically a delta function, which Fourier spectrum contains a
wide spectrum of harmonics going from t = minus to plus infinity. Surely
these frequency components should start resonating the harp's strings well
before the bomb explodes...

Unfortunately, even intelligent people can fool them self. Very similar to
the above example is this paper based on Hegersfeldt ideas from Nobel
laureate for Chemistry I. Prigogine:

"NONLOCALITY AND SUPERLUMINOSITY"
http://www1.jinr.ru/Archive/Pepan/v-31-7a/E_otkr_08_p.pdf

The Fourier decomposition of about any localized function contains both
positive and negative frequencies, on shell and (mostly) off shell. Leave
anything out and you end up with something non-local. This doesn't imply
that we should abandon abandon Einstein locality because we see localized
wave functions.Regards, Hans

 

[Haelfix]

To keep things simple, I will write you expression for a one-particle state localized at point $\mathbf{x}$

$$|\Psi \rangle = \int d^3p e^{\frac{i}{\hbar} \mathbf{p} \cdot \mathbf{x}} a^{\dag}_{\mathbf{p}} |0 \rangle$$

Here $a^{\dag}_{\mathbf{p}}$ are particle creation operators and $| 0 \rangle$ is the vacuum vector. Of course, this state has infinite uncertainties of momentum and energy, but it is still an one-particle state. It is an eigenstate (with eigenvalue 1) of the particle number operator

$$N = \int d^3p a^{\dag}_{\mathbf{p}} a_{\mathbf{p}}$$

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. 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.

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)

 

[Demystifier]

To keep things simple, I will write you expression for a one-particle state localized at point $\mathbf{x}$

$$|\Psi \rangle = \int d^3p e^{\frac{i}{\hbar} \mathbf{p} \cdot \mathbf{x}} a^{\dag}_{\mathbf{p}} |0 \rangle$$

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.

 

[jostpuur]

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.

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...

 

[meopemuk]

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

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

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

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

(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

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

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

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

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 $[a_p,a^{\dagger}_{p'}]=(2\pi\hbar)^3\delta^3(p-p')$. 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...

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.