How observation leads to wavefunction collapse?

[meopemuk]

genneth, thanks for this great analysis!

Space and time are intertwined, as SR teaches us.

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

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

Okay, I stop here.

Eugene.

 

[jostpuur]

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

is not Lorentz invariant while the Klein-Gordon equation

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

is Lorentz invariant?

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

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

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

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

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

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

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

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

 

[genneth]

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

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

 

[jostpuur]

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

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

But we have

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

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

 

[meopemuk]

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

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

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

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

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

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

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

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

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

Eugene.

 

[jostpuur]

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

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

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

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

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

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

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

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

Eugene.

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

 

[meopemuk]

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

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

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

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

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

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

Eugene.

 

[jostpuur]

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

Eugene.

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

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

 

[meopemuk]

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

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

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

Eugene.

 

[jostpuur]

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

Eugene.

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

 

[Hans de Vries]

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

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

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

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

Which becomes the following operator in configuration space:

\Box^{-1}\ \ -\ \ m^2\Box^{-2}\ \ +\ \<br /> m^4\Box^{-3}\ \ -\ \ m^6\Box^{-4}\ \ +\ \ ...

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

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

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

 

[Hans de Vries]

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

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

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


Regards, Hans

 

[meopemuk]

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

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

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

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

Eugene.

 

[Haelfix]

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

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

 

[jostpuur]

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

So good point!

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

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

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

 

[Hans de Vries]

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

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

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


Regards, Hans.

 

[meopemuk]

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

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

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

Eugene.

 

[meopemuk]

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

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

Sorry, this is not a proof.

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

Eugene.

 

[Hans de Vries]

Sorry, this is not a proof.

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

Eugene.

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


Regards, Hans

 

[meopemuk]

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

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

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

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

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

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

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

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

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

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

Eugene.

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

 

[Hans de Vries]

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

Shouldn't we keep separated Wigner's and Mackey's work on the Poincaré group on one
hand, and Newton, Wigner's paper "Localized states for elementary particles" on the
other hand? The issues seem to be with the latter paper.

Scanning through Weinberg vol. I (page 68,69) I see that he uses the Wigner rotations
as representations of the little group for m>0 to show that relativistic moving masive
particles have the same transformation under rotations as in non-relativistic quantum
mechanics which is a fundamental result.

Not discussed by Weinberg is that these rotations of relativistic moving particles
correspond to the Pauli Lubanski (spin) vector, which is a four vector which
reduces to the ordinary 3 component rotation vector for a particle at rest.
(the time component becomes zero). This four vector transforms like a Lorentz
pseudo vector in a way which is described in Jackson's classical electrodynamics
chapter 11.11A. The spin four-vector and the Pauli Lubanski vector differ only
by a factor m. See also Ryder's "Quantum Field Theory chapter 2.7

Expressing the Poincaré group with the usual operators you get two basic equations
corresponding to the two Casimir invariants of the Poincaré group (m>0). The 1st Casimir invariant (mass) gives rise to the Klein Gordon equation:

<br /> \left\{<br /> \frac{\partial^2 }{\partial t^2}\ <br /> -\ \frac{\partial^2 } {\partial x^2}\<br /> -\ \frac{\partial^2 } {\partial y^2}\<br /> -\ \frac{\partial^2 } {\partial z^2}\ <br /> \right\}\ \psi<br /> \ = \ <br /> -m^2\ \psi <br />

The 2nd Casimir invariant (angular mom.) gives rise to the "Pauli Lubansky" equation:

<br /> \frac{\partial^2}{\partial t^2}\left\{ <br /> \left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y} \right)^2 +<br /> \left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z} \right)^2 +<br /> \left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x} \right)^2 <br /> \right\}\ \psi\ =\ m^2 \ell(\ell+1)\ \psi<br />So, in the end these Wigner rotations result in the "Pauli Lubanski" equation which
expresses angular momentum as a Casimir invariant of the Poincaré group Regards, Hans

 

[meopemuk]

Shouldn't we keep separated Wigner's and Mackey's work on the Poincaré group on one
hand, and Newton, Wigners paper "Localized states for elementary particles" on the
other hand? The issues seem to be with the latter paper.

Sure, we can do that. But then we will not be able to discuss wave functions in the position representation (that is needed to address the "superluminal propagation" issue), because Wigner-Mackey construction works only in the momentum representation. Also, we will not be able to find out how position-space wave functions transform with respect to boosts.

Expressing the Poincaré group with the usual operators you get two basic equations
corresponding to the two Casimir invariants of the Poincaré group (m>0).


The 1st Casimir invariant (mass) gives rise to the Klein Gordon equation:

<br /> \left\{<br /> \frac{\partial^2 }{\partial t^2}\ <br /> -\ \frac{\partial^2 } {\partial x^2}\<br /> -\ \frac{\partial^2 } {\partial y^2}\<br /> -\ \frac{\partial^2 } {\partial z^2}\ <br /> \right\}\ \psi<br /> \ = \ <br /> -m^2\ \psi <br />

I am not against using the Klein-Gordon equation. Yes, position-space wavefunctions of free particles do satisfy this equation. However, this equation has a very important weakness. This equation involves 2nd time derivative. Therefore, by solving this equation one cannot predict the wavefunction at time t \psi(x,t) from the knowledge of the wavefunction at time t=0 \psi(x,0). We know that in quantum mechanics the wave function fully describes the state. Therefore, the Klein-Gordon equation does not provide a full description of the time evolution. To solve this equation one should know the time derivative \partial \psi(x,t)/ \partial t at t=0 in addition to the wave function at t=0. And this time derivative can be specified in a completely arbitrary way.

Fundamentally, the time derivative \partial \psi(x,t)/ \partial t at t=0 must be uniquely determined by the wavefunction (state) \psi(x,0). This fact is completely missed in the Klein-Gordon equation. So, in fact, this equation has many more solutions than are physically permissible, so it is not very useful.

The relativistic Schroedinger equation (which is first order in t) does not have this problem. It allows us to find the wave function (state) \psi(x,t) once the wave function (state) at time t=0 \psi(x,0) is known. So, this equation provides a unique and complete description of the time evolution. All solutions of the Schroedinger equation also obey the Klein-Gordon equation. However, the reverse is not true. The Klein-Gordon solutions, which do not obey the Schroedinger equation are useless. They are not physical.

Eugene

 

[lightarrow]

I am not against using the Klein-Gordon equation. Yes, position-space wavefunctions of free particles do satisfy this equation. However, this equation has a very important weakness. This equation involves 2nd time derivative. Therefore, by solving this equation one cannot predict the wavefunction at time t \psi(x,t) from the knowledge of the wavefunction at time t=0 \psi(x,0). We know that in quantum mechanics the wave function fully describes the state. Therefore, the Klein-Gordon equation does not provide a full description of the time evolution. To solve this equation one should know the time derivative \partial \psi(x,t)/ \partial t at t=0 in addition to the wave function at t=0. And this time derivative can be specified in a completely arbitrary way.

Fundamentally, the time derivative \partial \psi(x,t)/ \partial t at t=0 must be uniquely determined by the wavefunction (state) \psi(x,0). This fact is completely missed in the Klein-Gordon equation. So, in fact, this equation has many more solutions than are physically permissible, so it is not very useful.

The relativistic Schroedinger equation (which is first order in t) does not have this problem. It allows us to find the wave function (state) \psi(x,t) once the wave function (state) at time t=0 \psi(x,0) is known. So, this equation provides a unique and complete description of the time evolution. All solutions of the Schroedinger equation also obey the Klein-Gordon equation. However, the reverse is not true. The Klein-Gordon solutions, which do not obey the Schroedinger equation are useless. They are not physical.

Eugene

Sorry if I enter your discussion. Can you give me a simple example of such non-physical solutions of KG eq.?
Thanks.
Alberto.

 

[Haelfix]

I don't know why people get so caught up with this.

The Klein Gordon equation is fundamentally a constraint in Dirac quantization, not the full equation for time evolution of states. We further require the existence of a suitable scalar product, and once this is done there is no problem with unphysical states or anything of that nature.

 

[genneth]

Indeed -- and the thread has certainly drifted off-course from the original purpose. I suggest all interested parties take it to a new thread, where the context is introduced again, so that people don't have to read nearly 30 pages to catch up.

 

[meopemuk]

Sorry if I enter your discussion. Can you give me a simple example of such non-physical solutions of KG eq.?
Thanks.
Alberto.

I can give you a simple toy example, which illustrates the idea. Discussion of the full KG equation would require more work.

Consider a simple time-dependent function \psi(t) which satisfies the "Schroedinger equation", where the constant a plays the role of the "Hamiltonian"

d \psi(t)/dt = a \psi(t)...(1)

The general solution of (1) is \psi(t) = A \exp(at), where A is an arbitrary constant. If we know the value of the function at t=0, then we can find the time evolution. For example, if \psi(0) = 1, then A = 1, and the solution is

\psi(t) = \exp(at)...(2)

Now, let us take the time derivative of the both sides of (1) and obtain a new equation

d^2 \psi(t)/d^2 = a^2 \psi(t)...(3)

which can be regarded as a toy analog of the Klein-Gordon equation. The general solution of (3) is

\psi(t) = B \exp(at) + C \exp(-at)...(4)

where B and C are arbitrary constants. In contrast to the "Schroedinger" eq (1), this "Klein-Gordon" equation does not allow us to find the unique time evolution if we know the "initial state" \psi(0). For example, if \psi(0) = 1, then both \psi(t) = \exp(at) and \psi(t) = \exp(-at) are valid solutions of (4). However, the latter solution is unphysical, because it does not agree with the "Schroedinger" equation (1).

Eugene.

 

[lightarrow]

In contrast to the "Schroedinger" eq (1), this "Klein-Gordon" equation does not allow us to find the unique time evolution if we know the "initial state" \psi(0). For example, if \psi(0) = 1, then both \psi(t) = \exp(at) and \psi(t) = \exp(-at) are valid solutions of (4). However, the latter solution is unphysical, because it does not agree with the "Schroedinger" equation (1).

Eugene.

Thank you for the answer. However I still don't understand. That solution is unphysical because would correspond to negative energies? Why that solution have to agree with the Schrodinger equation? SR gives us more informations, with respect to a non-relativistic descriptions, so maybe it's not so strange that it could generate additional solutions (that is, additional informations) for the wavefunction. For example, SR says that it's not possible to distinguish between "A is stationary - B moves" and "B is stationary - A moves", so this only should broaden the class of allowed wavefunctions solution of the problem. What do you think?

 

[meopemuk]

Thank you for the answer. However I still don't understand. That solution is unphysical because would correspond to negative energies? Why that solution have to agree with the Schrodinger equation? SR gives us more informations, with respect to a non-relativistic descriptions, so maybe it's not so strange that it could generate additional solutions (that is, additional informations) for the wavefunction. For example, SR says that it's not possible to distinguish between "A is stationary - B moves" and "B is stationary - A moves", so this only should broaden the class of allowed wavefunctions solution of the problem. What do you think?

The important point is that in quantum mechanics the wave function at a given time determines the state of the system uniquely and completely. Therefore, it also uniquely determines the time evolution of this state. So, the time derivative of the wave function must depend on the wave function itself in an unambiguous way. This is exactly what Schroedinger equation does. It says: "time derivative of the wave function is equal to the Hamiltonian times the wave function". This is why I believe that Schroedinger equation is fundamental.

If we assume that the Klein-Gordon equation is sufficient to get the time evolution then we are led to the conclusion that the wavefunction at a given time does not determine the state completely. Then we also need to provide the time derivative of the wave function in order to predict the time evolution unambiguously. In my view, this is very serious deviation from the foundations of quantum mechanics.

Of course, there is a possibility to keep the Klein-Gordon equation and filter out its unphysical solutions by adding some extra conditions, such as requiring only positive frequencies. In my opinion, this approach makes everything more cumbersome and doesn't add anything to the physically transparent Schroedinger equation.

Eugene.

 

[jostpuur]

I'm now talking about m=0 case with the Klein-Gordon equation.

Okey, this is more tricky than I assumed.

If we have an initial configuration \phi,\partial_0\phi so that it vanishes outside some bounded subset of \mathbb{R}^3, then I am very confident that the field will not start spreading out of this bounded area in superluminal fashion according to the KGE. I mean... not only does the time evolution vanish in the linear approximation of time, but the spreading occurs only in the light cone.

However, if we only have an initial \phi defined on some bounded subset, and then take the positive frequency solution of the KGE, how can we know that also \partial_0\phi vanishes outside the bounded subset at time t=0?

If it is a mathematical provable fact, that \partial_0\phi will never be zero outside the bounded initial subset, I would not immediately draw the conclusion of superluminal propagation. It just means that the completely localized solutions do not exist. And in this case, the assumed local initial configuration wasn't so local as it might have seemed.

Also notice, that the positive frequency solutions don't exist for real solutions of the wave equation. The behaviour of the solutions of the relativistic SE can be expected to differ from the solutions of the traditional wave equation.

 

[Demystifier]

KG equation is a second-order differential equation, so the initial condition must specify both \phi and \partial_0\phi, period.
It is a mathematical fact, and any sound physical interpretation must respect it. If it confronts with some physical interpretation of \phi, them either this interpretation or the KG equation should be abandoned. But don't try to retain both!

 

[jostpuur]

KG equation is a second-order differential equation, so the initial condition must specify both \phi and \partial_0\phi, period.

Or we can specify only \phi, and take the positive frequency solution, which is unique. The demand of the solution being positive frequency also defines the initial \partial_0\phi uniquely, so it would be redundant to give it separately.

Or equivalently take the solution of the relativistic SE, which is a first order differential equation (in respect to time).

 

[Demystifier]

Or we can specify only \phi, and take the positive frequency solution, which is unique.

Or do not specify any initial condition, but simply pick up one particular solution ...
The point is that the issue of causality makes sense only if the picking of solutions is formulated merely in terms of initial conditions, NOT ANY OTHER CONDITIONS (like restriction to positive frequency solutions).

 

[jostpuur]

Or do not specify any initial condition, but simply pick up one particular solution ...
The point is that the issue of causality makes sense only if the picking of solutions is formulated merely in terms of initial conditions, NOT ANY OTHER CONDITIONS (like restriction to positive frequency solutions).

I am believing in the

<br /> i\hbar\partial_t\Psi=\sqrt{-c^2\hbar^2\nabla^2 + (mc^2)^2}\Psi<br />

now. But defining the time evolution of some initial wave function with this, is equivalent to defining it with the KGE and demanding the solution to be the positive frequency. So the positive frequenciness is not some ad hoc condition. But if it is still disturbing, then take the relativistic SE without any other conditions. It is the same thing.

 

[Demystifier]

I am believing in the

<br /> i\hbar\partial_t\Psi=\sqrt{-\hbar^2\nabla^2 + (mc^2)^2}\Psi<br />

now.

Fine. Now, if \Psi vanishes outside of some bounded subset of R^3, then so does \partial_t\Psi, due to the equation above. Again, there is no problem with causality.

 

[jostpuur]

Fine. Now, if \Psi vanishes outside of some bounded subset of R^3, then so does \partial_t\Psi, due to the equation above. Again, there is no problem with causality.

The behaviour of the operator \sqrt{-\nabla^2+m^2} is non-trivial. It is given by the integrals

<br /> \sqrt{-\nabla^2 + m^2}\psi(x) = \int\frac{d^3x&#039;\;d^3p}{(2\pi)^3}\psi(x&#039;)\sqrt{|p|^2+m^2}e^{ip\cdot(x-x&#039;)}<br />

where in particular also the variable x' is integrated over the all space. I don't see how

<br /> \psi(x)=0\quad\textrm{for}\quad x\in B(x_0,R)\quad\implies\quad \sqrt{-\nabla^2+m^2}\psi(x)=0\quad\textrm{for}\quad x\in B(x_0,R)<br />

could be trivial for some x_0 and R>0. But I haven't thought about this yet. Perhaps that is true, and it can be proven. I have a wild guess, that that is not true, however. The operator seems to be truly non-local, as concluded (IMO incorrectly) out of the Taylor series too.

 

[Demystifier]

The behaviour of the operator \sqrt{-\nabla^2+m^2} is non-trivial. It is given by the integrals

<br /> \sqrt{-\nabla^2 + m^2}\psi(x) = \int\frac{d^3x&#039;\;d^3p}{(2\pi)^3}\psi(x&#039;)\sqrt{|p|^2+m^2}e^{ip\cdot(x-x&#039;)}<br />

where in particular the variable x' is integrated over the all space. I don't see how

<br /> \psi(x)=0\quad\textrm{for}\quad x\in B(x_0,R)\quad\implies\quad \sqrt{-\nabla^2+m^2}\psi(x)=0\quad\textrm{for}\quad x\in B(x_0,R)<br />

could be trivial for some x_0 and R>0. But I haven't thought about this yet. Perhaps that is true, and it can be proven.

The square-root operator above is either local or nonlocal. In my argument, I have tacitly assumed that it is a local operator. Your notes above suggest that it may actually be nonlocal. To determine whether it is local or nonlocal certainly requires a more careful mathematical analysis. However, if it turns out that it is nonlocal, then the relativistic Schrodinger equation violates the principle of locality. With this, one should not be surprised to violate causality as well. But the equation that violates locality does not seem to be physically acceptable. Which brings us back to the local KG equation.

Now, let me present a simple argument that it is a local operator. One can expand the square root as an infinite sum of powers of the local differential operator nabla^2. Any finite truncation of this infinite sum is a local operator. This suggests that the whole infinite sum also seems to be local, provided that the limit is defined in some appropriate sense. Of course, this is not a proof, just a heuristic argument which could actually be wrong.

Note also that an integral representation of a differential operator as above makes simple things very complicated. Consider, for example, the simple operator nabla^2.

 

[jostpuur]

Your notes above suggest that it may actually be nonlocal. To determine whether it is local or nonlocal certainly requires a more careful mathematical analysis.

One might think that

<br /> \psi(x)\mapsto \int\frac{d^3x&#039;\;d^3p}{(2\pi)^3} \psi(x&#039;)\big(|p|^2+m^2\big) e^{ip\cdot(x-x&#039;)}<br />

is non-local by the appearance of the integrals, but now this turns out to be local. So yeah, that is non-trivial. The locality or non-locality should be proven somehow. You cannot simply see it.

However, if it turns out that it is nonlocal, then the relativistic Schrodinger equation violates the principle of locality. With this, one should not be surprised to violate causality as well. But the equation that violates locality does not seem to be physically acceptable. Which brings us back to the local KG equation.

I cannot understand how Lorentz invariant time evolution could violate causality. This problem with the relativistic SE is merely about the lack of completely local solutions. If all initial wave functions are non-local (in the sense that even if \phi=0, still \partial_0\phi\neq 0), and then their time evolutions remain non-local correspondingly, there is no obvious violation of causality.

By "completely local" I mean local in such strong sense that the wave function remains zero outside some bounded area. Gaussian wave packet is of course local too in some sense, but not completely.

Now, let me present a simple argument that it is a local operator. One can expand the square root as an infinite sum of powers of the local differential operator nabla^2. Any finite truncation of this infinite sum is a local operator. This suggests that the whole infinite sum also seems to be local, provided that the limit is defined in some appropriate sense. Of course, this is not a proof, just a heuristic argument which could actually be wrong.

The Taylor series of \sqrt{1+x} don't converge for x>1, which becomes the real problem with this. The integral of p is supposed to be carried out over the whole momentum space, so local approximations of the integrand are not acceptable.

 

[Demystifier]

1. I cannot understand how Lorentz invariant time evolution could violate causality.

2. The Taylor series of \sqrt{1+x} don't converge for x>1, which becomes the real problem with this.

1. It is not at all clear that the relativistic Schrodinger equation is Lorentz invariant. We know that the Dirac equation is Lorentz invariant, but the relativistic Schrodinger equation does not seem to be so. For example, the notion of frequency requires the notion of time, so it is not at all obvious that the difference between positive and negative frequencies is relativistic invariant.

In addition, Lorentz invariant time evolution CAN violate causality. The best known example is a tachyon. Of course, the tachyon does not appear in our case, but now I am talking about general principles.

2. Perhaps the Borel resummation could solve this technical problem.

 

[jostpuur]

1. It is not at all clear that the relativistic Schrodinger equation is Lorentz invariant.

The solutions of relativistic SE are also solutions of KGE, so the Lorentz invariance comes from there. The positive and negative frequency solutions of KGE don't mix in Lorentz transformations, so the positive frequency solution remains positive frequency in all frames.

 

[Demystifier]

OK, let us attack the problem from another point of view. Take the positive frequency solution of the KG equation that is a spatial delta-function delta^3 at t=0. One can show that the time derivative of that solution does not vanish outside any bounded region at t=0. This is not a problem for the KG equation. However, this is also a solution of the relativistic Schrodinger equation. When interpreted as a solution of the Schrodinger equation, this shows that the square-root operator is nonlocal when acts on this particular solution. Indeed, the requirement that only positive frequency solutions are taken is not a local requirement, so it is clear where this nonlocality comes from. Without locality, one should not be surprised to obtain violation of causality.

Now, how violation of causality may be compatible with Lorentz invariance? Your argument for Lorentz invariance was based on the fact that it is a solution of the KG equation, which is a Lorentz invariant equation. This is true, but at the level of KG equation, this solution is not picked up by a choice of the initial condition, which automatically breaks causality. (As another example, consider the procedure of choosing the solution by fixing phi at two different times. Clearly, this also violates causality.)

Now you will argue that the argument above does not imply violation of causality when our solution is interpreted as a solution of the relativistic SE. But I have shown above that this equation is not local. Therefore, the violation of causality is a consequence of the violation of locality. But you have presented a good argument that this equation is relativistic invariant. Then how relativistic invariant equation may violate locality? Well, it can. For example, consider action of the form
\int d^4x \int d^4y L(x,y)
If L transforms as a biscalar, then this action is Lorentz invariant. Nevertheless, it is nonlocal. This simple example demonstrates the general rule that relativistic invariance is compatible with nonlocality.

To conclude, relativistic SE is relativistic invariant but not local. Violation of locality induces violation of causality.

 

[jostpuur]

I'm not sure I (or we) know what causality means anymore in this context.

If some guy who doesn't know relativity asks why something cannot move faster than light, then we could explain him about Lorentz transformations, and how faster than light travel would be movement backwards in time in some other frame.

Now we have a Lorentz invariant differential equations. The solutions are non-local, although on the other hand wave packets still travel with speeds that don't exceed the speed of light.

What is causality here? If wave packets don't travel faster than light, then we cannot, according to this theory, create an experiment where causality would be violated in sense that we could it observe.

Now theoretician says that the theory would still in principle imply violation of causality. But what is this "violation of causality in principle"? It is Lorentz invariant theory! How could there be something wrong with it?

 

[Demystifier]

I'm not sure I (or we) know what causality means anymore in this context.

Now we have a Lorentz invariant differential equations. The solutions are non-local, although on the other hand wave packets still travel with speeds that don't exceed the speed of light.

The key is to understand in what sense the theory is nonlocal. Assume that Psi is a classical physical field (not a quantum wave function with a possibly problematic interpretation). The fact that the differential equation is a first-order one in time derivatives implies that it violates the usual dynamical principle (discovered by Newton) that initial velocities should be arbitrary. It is this constraint on the initial velocities that is nonlocal. By choosing initial Psi, you instantaneously send information at other locations what initial velocities must be. However, once the initial Psi and the velocities (the time derivative of phi) are fixed, then further evolution of Phi is causal.

Note also that the relativistic SE has a fixed time, so it is fixed in which Lorentz frame the transmition of information on initial velocities is instantaneous.

 

[meopemuk]

I'm not sure I (or we) know what causality means anymore in this context.

If some guy who doesn't know relativity asks why something cannot move faster than light, then we could explain him about Lorentz transformations, and how faster than light travel would be movement backwards in time in some other frame.

Now we have a Lorentz invariant differential equations. The solutions are non-local, although on the other hand wave packets still travel with speeds that don't exceed the speed of light.

What is causality here? If wave packets don't travel faster than light, then we cannot, according to this theory, create an experiment where causality would be violated in sense that we could it observe.

Now theoretician says that the theory would still in principle imply violation of causality. But what is this "violation of causality in principle"? It is Lorentz invariant theory! How could there be something wrong with it?

Can we now agree that (initially localized) solutions of the Schroedinger equation propagate faster than light? I think, yes. However, does it really mean the violation of causality? I think that we need to be a bit more careful with the definition of causality here.

In my opinion the causality would be definitely violated if in some reference frames the effect appears earlier than the cause. The "effect" and the "cause" must be certain events connected by the cause-effect relationship. This also means that these events should belong to a system of interacting particles. However, Schroedinger equation describes only free particles. So, logically, it seems to be too early to conclude that causality is violated from the (proven) fact that SE solution propagate superluminally.

Eugene.

 

[jostpuur]

If we say that \psi(0,x) vanishes outside some bounded set, then the wave function was non-zero outside this set also for arbitrarily small times t<0 (I'm now assuming the non-locality, for which I have not encountered a convincing proof though). So that is not "local initial state" really. It was non-local before time t=0, and is non-local after it.

 

[meopemuk]

If we say that \psi(0,x) vanishes outside some bounded set, then the wave function was non-zero outside this set also for arbitrarily small times t<0 (I'm now assuming the non-locality, for which I have not encountered a convincing proof though). So that is not "local initial state" really. It was non-local before time t=0, and is non-local after it.

Yes, this is true. The free particle is localized only during infinitesimally short time interval around time t=0. It becomes delocalized shortly before and shortly after this time.

It is also interesting to note that from the moving reference frame the wave function does not look localized at any time. So in quantum mechanics the violation of causality is not so obvious.

In the classical case we could say that if in the reference frame at rest the particle propagated from A to B superluminally, then in the moving frame the particle propagates from B to A, which violates causality.

In the quantum case, in the rest frame we can localize particle at A at t=0 and have a non-zero probability of finding it at B shortly after. However, in the moving reference frame we'll see an uncertain picture in which the particle is spread over entire space at all time instants, so it is not possible to say where is the source and where is the destination; where is the cause and where is the effect.

Eugene.

 

[Haelfix]

The restriction to positive energy solutions of the KG equation is not some adhoc procedure, its absolutely fundamental to Dirac quantization (or said another way, the existence of a scalar product). Without it, its perfectly nonsensical and theoretically does not make sense. This shows up all the times in different situation:

For instance in the Ashtekar program they are completely unable to find a suitable scalar product that remains in their Hilbert space, despite having simple equations for all the other constraints. This is one of the main problem that people in LQG face.

Another thing that I insist on keeping in mind. Physical states are gauge invariant observables, and time evolution, with that choice of parameter time is a gauge transformation. Note that the KG equation is thus not an equation about the dynamics of time evolution of physical states, rather its simply a constraint. Instead to extract physical information one must look at relationships between the observables with an appropriate slicing of spacetime as well as to follow the rest of the Dirac quantization procedure. Its very easy to forget this point, b/c things are so simple in Minkowski space and often looks the same. Not so with say a different background, like something that's highly curved. Its something people in "field theory in curved space" worry about all the time.

This is not some mere technicality, its a monumental problem. Consider DeSitter space. No one knows what observables exist for that, so we cannot even study the real time evolution of quantum states