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

[Demystifier]

In scattering theory, the fundamental quantity is the S-operator, which is basis-independent. Matrix elements of this operator on momentum eigenvectors form the momentum space S-matrix, but you can also choose to find matrix elements of the S-operator in any other representation, including the position representation. This is not convenient, but possible.

In the relativistic case, there is no position representation because there is no position basis because there is no position operator. Fourier transform is not a transition to the position representation.

 

[meopemuk]

Now let the time evolution of this wave function be given by the Klein-Gordon equation... The point is that the Klein-Gordon equation is a SECOND order differential equation in time derivatives. Consequently, the initial condition is given not only by the initial wave function, but also by the initial time derivative of the wave function.

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.

 

[Demystifier]

Are you then saying that a particle that looks localized for observer ar rest must also look localized for any moving observer?

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

 

[Demystifier]

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.

 

[meopemuk]

In the relativistic case, there is no position representation because there is no position basis because there is no position operator. Fourier transform is not a transition to the position representation.

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.

 

[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

<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?

 

[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&#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...