How observation leads to wavefunction collapse?

[gptejms]

He is allowed to read what I suggested above as well. That would be my answer to his objection.

I liked your comments in the section 'Does QFT solve the problems of relativistic QM' in your article that you referred, and I am reproducing the section below:-

Does QFT solve the problems of relativistic QM?
After this brief overview of QFT, we are finally ready to cope with the validity of the title of this section. How QFT helps in solving the interpretational problems of relativistic QM? According to QFT, the fundamental objects in nature are not particles, but fields. Consequently, the fundamental wave function(al) that needs to have a well-defined probabilistic interpretation is not ψ(x, t), but $$\Psi [\phi; t)$$. Thus, the fact that, in the case of Klein-Gordon equation, ψ(x, t) cannot be interpreted probabilistically, is no longer a problem from this more fundamental point of view. However, does it really solve the problem? If QFT is really a more fundamental theory than the first-quantized quantum theory of particles, then it should be able to reproduce all good results of this less fundamental theory. In particular, from the fundamental axioms of QFT (such as the axiom that (89) represents the probability in the space of fields), one should be able to deduce that, at least in the nonrelativistic limit, ψ*ψ represents the probability in the space of particle positions. However, one cannot deduce it solely from the axioms of QFT. One possibility is to completely ignore, or even deny [47], the validity of the probabilistic interpretation of ψ, which indeed is in the spirit of QFT viewed as a fundamental theory, but then the problem is to reconcile it with the fact that such a probabilistic interpretation of ψ is in agreement with experiments. Another possibility is to supplement the axioms of QFT with an additional axiom that says that ψ in the nonrelativistic limit determines the probabilities of particle positions, but then such a set of axioms is not coherent, as it does not specify the meaning of ψ in the relativistic case. Thus, instead of saying that QFT solves the problems of relativistic QM, it is more honest to say that it merely sweeps them under the carpet.

I think my solution would be not to deny the validity of the probabilistic interpretation of ψ,but to say that ψ is still a field and since ψ satisfies a continuity equation,it may be given a probabilistic interpretation also.

 

[Demystifier]

I think my solution would be not to deny the validity of the probabilistic interpretation of ψ,but to say that ψ is still a field and since ψ satisfies a continuity equation,it may be given a probabilistic interpretation also.

But then again: Can you give a probabilistic interpretation in the relativistic case as well? If yes, can you do it in a relativistic covariant way? How?

 

[Fra]

Demystifier, have you considered the concept of relative probabilities of the bayesian formalism? What is your liking of that? This is a key ingredient in my thinking.

(Btw is Nikolic your first or last name?)

/Fredrik

 

[Demystifier]

Demystifier, have you considered the concept of relative probabilities of the bayesian formalism? What is your liking of that? This is a key ingredient in my thinking.

(Btw is Nikolic your first or last name?)

I do not see how bayesian way of thinking would help in relativistic QM. Can you explain it?

(Last.)

 

[gptejms]

But then again: Can you give a probabilistic interpretation in the relativistic case as well? If yes, can you do it in a relativistic covariant way? How?

In the relativistic case,you can't give the field ψ a probabilistic interpretation,but then that's the way it isarticles are getting created and destroyed,so the particle no. is not fixed.You can however give prob. interpretation to $$\Psi [\phi; t)$$--why should this be expected to reduce to ψ of Schrodinger equation in the non-rel. limit?It has an independent existence.

 

[Fra]

I don't claim to have all the answers, but I've got some pretty decent ideas that is enough for me to work on it, and formalising these is part of what I'm trying to do. I am not prepared to explain it in detail yet, but I was curious if you had considered it on your own. My ideas is clear enough for myself, but I think you'd math the consistent math, and that is still missing. And it's not because I don't know math, it's because I don't know exactly what it will be like yet.

The only reason I ask is because I have a feeling that at least part of your questions could be addressed by going back to the foundations of the probability and the axioms of QM.

I have found out thought that what I want to do, quickly gets complex, and I suspect it will take some times for it to mature, because I think one has to revise many things.

But the basic ideas is this:

- Consider tha A interacts with B.

- interaction ~ communication

- A's response towards B, doesn't depend on B, it rather depends on A and A's subjective information about B. The key here is that if the information A has about B is "incorrect" relative to C, this doesn't influence A's response. A responds only to local information.

In this context, it's like a learning model. A reaching equilibrium with an environment can be thought of as A learning about it's environment. Here comes more things, and that's information capacity. A single particle can obviously never grasp the truth of a overwhealming environment. There are several reasons to suspect an association with energy and information capacity. But learning also means time, processing. This suggest a deep relation with time and energy from a point of view that even has not assumed anything of GR. Once you start to think along these lines, the relational concepts that lead to GR in the first place seem to come in on their own.

I am not wasting time to think about an alternative explanation to the standard stuff, if I didn't think there was an additional benefit beyond interpretation. I think once I can figure out how to do this, I think the formalism is going to be powerful.

Unsolved things are howto couple dynamics with representation, because in a relational learning model there aren't separated. I've tried to start out assuming a certain information capacity/mass but I immediately realize that the whole logic requires a real feedback, and at this point it's not consistent to take the information capacity to be constant. Also spacetime is emergent in my thinking (emergent from information; this is close to Ariels thinking), not something I start out with. QFT for example starts out as if it was obvious what the spacetime background. This is really unacceptable in my thinking. It's a approximation, a good one, but if you probe the logic from the poitn of view of relative information it is inconsistent. But the good part is that there is plenty of support for that the principles of relativity will infact emerge as consequences in this thinking. Ie. the are implications of more fundamental "first principle".

I'm sorry but I don't want to even give illusion that I try to explain it seriously at this point. These comments are probably silly enough. Once I have at least more of the founding formalism that I can convince myself about, I will be back. I've been trying to find published work along the lines and Ariel Catichas (http://www.albany.edu/physics/ariel_caticha.htm ) think is the closest, but it's not quite either. In either case his papers are plausible reading.

If you want to get in the ballpark of the philosophy, check our Ariels papers. For example http://arxiv.org/PS_cache/math-ph/pdf/0008/0008018v1.pdf. But this is only a small part of the total picture. He is working in the continuum approximation, I don't like that. So I am trying to start below that.

I think your questions are good, and relating to my own thinking, I think (of course I could be all wrong) that the solution is to take a few steps back. That's what I have forced myself todo.

/Fredrik

 

[Demystifier]

In the relativistic case,you can't give the field ψ a probabilistic interpretation,but then that's the way it isarticles are getting created and destroyed,so the particle no. is not fixed.

But free particles are neither created nor destroyed. On the other hand, the probabilistic interpretation of psi does not work even for free particles. How do you interpret that? (Don't tell me that there are no free particles, because, when you calculate the probabilities for particle creation/destruction in QFT, you assume that particles are asimptotically free. Without that assumption, you could not make a prediction at all.)

 

[Demystifier]

I think your questions are good, and relating to my own thinking, I think (of course I could be all wrong) that the solution is to take a few steps back. That's what I have forced myself todo.

I agree that the solution is to take a few steps back. The problem is that there is more than one direction to do that.

 

[Fra]

I agree that the solution is to take a few steps back. The problem is that there is more than one direction to do that.

I agree with that. And it's not only a couple of directions to go either. It's more directions to go, than I could exhaust. So I need to find a selection rule, or get lucky.

In fact the point you raise is one of the reasons for the learning approach. The idea is that I should distribute my resources in a direction that I can defend even in the case where my future information suggests that anothre direction would have been better. But since I can't allow future information affect the past, chosing the right direction would have been luck. but stil, the subjective probability of a failure was higher.

Still you are right that by my own logic, my best shot may consistently differ from yours. That's life. I have no illusions that I this will work, but my conviction is that I do the best I can, and that's about as "right" as it ever gets.

/Fredrik

 

[Fra]

I agree with that. And it's not only a couple of directions to go either. It's more directions to go, than I could exhaust. So I need to find a selection rule, or get lucky.

Of course I still need to get lucky, but I need less luck. So the principle is to make use of a minium amuont of "luck" :) So I estimate my odds to the best of my capabilities, aim and fire. That's subjective reasoning. That's what I would to if I was an electron too. Aim, fire. Receive feedback aim and fire.

/Fredrik

 

[Fra]

I think this sounds obvious? Too obvious to be useful maybe? But this principle can I think lead to non-trivial predictions. A challage is howto formalize the "best odds", and how to couple that to real interactions, into a evolutionary mechanism that leads to dynamics. This is where I have some ideas.

/Fredrik

 

[Demystifier]

I agree with that. And it's not only a couple of directions to go either. It's more directions to go, than I could exhaust. So I need to find a selection rule, or get lucky.

I would not count on pure luck. A selection rule is a better strategy, but how to find it? My strategy is to follow my nose and intuition. The main problem is then to make a distinction between intuition and prejudices.

 

[gptejms]

But free particles are neither created nor destroyed. On the other hand, the probabilistic interpretation of psi does not work even for free particles. How do you interpret that? (Don't tell me that there are no free particles, because, when you calculate the probabilities for particle creation/destruction in QFT, you assume that particles are asimptotically free. Without that assumption, you could not make a prediction at all.)

May be what we call as a free particle is not really free in some sense--but yes,this is quite unlikely.

 

[meopemuk]

It seems so obvious to me that I can hardly imagine what kind or argument you would like to see. But let me try: Because this provides that the equation of motion for psi (the Klein Gordon equation) will have the same form in all Lorentz frames.

Your reference to the Klein-Gordon equation is not very convincing. Normally this equation is introduced (in the context of relativistic wavefunctions) by making formal substitutions $E \to i \hbar \frac{\partial}{\partial t}$ and $\mathbf{p} \to i \hbar \nabla$ in the relativistic energy-momentum-mass formula

[tex] E^2 = m^2c^4 + \mathbf{p}^2c^2 [/itex]

This is not a proof.

Second objection is this: Whatever you do, you must be consistent with Rules of Quantum Mechanics. These rules define wavefunctions as coefficients in the expansion of the state vector $| \psi(t) \rangle$
in some orthonormal basis in the Hilbert space. In the case of position-space wavefunction, the basis vectors are eigenvectors of the position operator $\psi(\mathbf{r},t) = \langle \mathbf{r} | \psi(t) \rangle$. The argument $\mathbf{r}$ has the meaning of eigenvalue of the position operator. In your construction of "relativistic wavefunctions" you don't mention the position operator and its eigenvalues. So, in my opinion, you are not even allowed to identify your "wavefunction arguments" $x = (\mathbf{r}, t)$ with particle positions measured in experiments.

Another thing. In quantum mechanics, wavefunction transfrormations to different (e.g., moving) reference frame should be accomplished by unitary operators. What is the form of unitary operator which performs the transformation $\psi(x) \to \psi(\Lambda x)$?

Eugene.

 

[meopemuk]

See, e.g., eqs. (87)-(89) in
http://xxx.lanl.gov/abs/quant-ph/0609163
Compare it also with eq. (80).

Hi Demystifier,

I enjoy reading your papers. You try to attack very important foundational questions, which are rarely mentioned in textbooks. You do not try to hide behind incomprehensible jargon, as many other authors do. Your reasoning is straightforward and honest.

I also tried to understand such issues as "what is the wavefunction?", "what are quantum fields?" for many years. I reached conclusions very different from yours. I hope to have a productive discussion with you, and get to the bottom of things, eventually.

Eugene.

 

[strangerep]

[Demystifier's] reference to the Klein-Gordon equation is not very convincing. Normally this equation is introduced (in the context of relativistic wavefunctions) by making formal substitutions $E \to i \hbar \frac{\partial}{\partial t}$ and $\mathbf{p} \to i \hbar \nabla$ in the relativistic energy-momentum-mass formula

[tex] E^2 = m^2c^4 + \mathbf{p}^2c^2 [/itex]

This is not a proof.

I see this as just a construction of a representation of the
Poincare group in a particular Hilbert space (i.e: the space of
square-integrable functions of spacetime position). The KG eqn is
just a statement of the $mass^2$ casimir invariant of the
Poincare group. To express that in this Hilbert space, we need
representations for the energy-momentum operators, and these are the
usual d/dx you mentioned. In this sense, I think it is indeed
a proof - although not many textbooks express it clearly.

 

[meopemuk]

I see this as just a construction of a representation of the
Poincare group in a particular Hilbert space (i.e: the space of
square-integrable functions of spacetime position). The KG eqn is
just a statement of the $mass^2$ casimir invariant of the
Poincare group. To express that in this Hilbert space, we need
representations for the energy-momentum operators, and these are the
usual d/dx you mentioned. In this sense, I think it is indeed
a proof - although not many textbooks express it clearly.

You are right that in order to qualify as a Schroedinger equation for relativistic particles, KG equation should be constructed on the basis of a representation of the Poincare group in some Hilbert space. However, there are more conditions that need to be satisfied:

1. The representation should be unitary.

2. The representation should be irreducible.

Wigner (1939) provided a classification of all such representations. His construction involved wave functions in the momentum representation. In order to switch to the position representation and to identify momentum operator with d/dx one also needs to

3. define the relativistic operator of position in this Hilbert space.

The most important condition is that

4. any relativistic generalization of the time-dependent Schroedinger equation must involve the first time derivative of the wave function.

KG equation is 2nd order in $t$. This means that the wavefunction at time $t + \Delta t$ is not completely determined by the wave function at time $t$. This means that the KG wavefunction $\psi (\mathbf{r}, t)$ does not provide a complete description of the quantum state. This is against the most fundamental rule of quantum mechanics.

So, you don't have a proof yet.

Eugene.

 

[Hans de Vries]

The most important condition is that

4. any relativistic generalization of the time-dependent Schroedinger equation must involve the first time derivative of the wave function.

KG equation is 2nd order in $t$. This means that the wavefunction at time $t + \Delta t$ is not completely determined by the wave function at time $t$. This means that the KG wavefunction $\psi (\mathbf{r}, t)$ does not provide a complete description of the quantum state. This is against the most fundamental rule of quantum mechanics.

You can determine the 1st order time derivative by taking the 1st order
space derivatives and use the continuity relation to get the derivative in t.
Probability conservation is now build in and you can use the 2nd order time
derivative of the Klein Gordon equation to further evaluate the progression
in time.


Regards, Hans

 

[meopemuk]

You can determine the 1st order time derivative by taking the 1st order
space derivatives and use the continuity relation to get the derivative in t.
Probability conservation is now build in and you can use the 2nd order time
derivative of the Klein Gordon equation to further evaluate the progression
in time.

What about another important condition mentioned by strangerep? Now, as you defined how KG wavefunctions transform with respect to boosts and time translations, can you prove that these transformations respect the multiplication law of the Poincare group? Inertial transformations of observers certainly respect this law. Their representatives in the Hilbert space of KG wavefunctions must do the same. Otherwise, the theory would be not self-consistent.

Eugene.

 

[Demystifier]

Your reference to the Klein-Gordon equation is not very convincing. Normally this equation is introduced (in the context of relativistic wavefunctions) by making formal substitutions $E \to i \hbar \frac{\partial}{\partial t}$ and $\mathbf{p} \to i \hbar \nabla$ in the relativistic energy-momentum-mass formula

[tex] E^2 = m^2c^4 + \mathbf{p}^2c^2 [/itex]

This is not a proof.

Of course, this is not a proof. But QFT based on quantization of the Klein-Gordon equation seems to be in agreement with experiments. OK, we do not really have a fundamental spinless particle, but we have a photon which is also described by an equation similar to the Klein-Gordon equation sharing the same interpretational problems. Therefore, with an appropriate interpretation, the Klein-Gordon equation should be correct. How do you comment on this? The key, of course, is to make a clear distinction between fields and wave functions. We agree on that. But how exactly to do that distinction?

 

[meopemuk]

Of course, this is not a proof. But QFT based on quantization of the Klein-Gordon equation seems to be in agreement with experiments. OK, we do not really have a fundamental spinless particle, but we have a photon which is also described by an equation similar to the Klein-Gordon equation sharing the same interpretational problems. Therefore, with an appropriate interpretation, the Klein-Gordon equation should be correct. How do you comment on this? The key, of course, is to make a clear distinction between fields and wave functions. We agree on that. But how exactly to do that distinction?

True. QFT is based on quantum fields that satisfy relativistic wave equations, like Klein-Gordon or Dirac equation. And QFT is a very successful theory. However, my point is that the success of QFT does not depend on the probabilistic interpretation of quantum fields that you are suggesting. This interpretation is not used anywhere in QFT calculations.

The central quantity calculated in QFT is the S-matrix. If one knows the S-matrix, one can calculate scattering cross-sections, decay rates, energies of bound states, i.e., almost all properties that can be directly compared with experiment. All phenomenal successes of QED are related to very accurate calculations of certain S-matrix elements, which can be generally written as

$$S_{in,out} = \langle in | S | out \rangle$$

where $| in \rangle$ is an asymptotic state of free particles in the remote past, $| out \rangle$ is an asymptotic state of free particles in the remote future, and $S$ is the S-operator. The "in" and "out" states contain fixed (but not necessarily equal) number of particles, and they are normally obtained by acting with particle creation operators on the vacuum vector $| 0\rangle$. For example,

$$| in \rangle = a_{\mathbf{p}, \sigma}^{\dag} b_{\mathbf{q}, \tau}^{\dag} \ldots | 0\rangle$$

In accordance with the most common experimental situation, one usually considers particle states with definite momenta $\mathbf{p}, \mathbf{q}$ and spin projections $\sigma, \tau$.

The S-operator can be calculated from the interaction Hamiltonian $V(t)$ by the Feynman-Dyson perturbation formula (There are other methods to compute the S-operator, but they are all equivalent.)

$$S = 1 + \frac{i}{\hbar} \int \limits_{-\infty}^{\infty} V(t) dt - \frac{1}{2 \hbar^2} \int \limits_{-\infty}^{\infty} dt \int \limits_{-\infty}^{\infty}dt' T[V(t) V(t')] + \ldots$$

where T is the time-ordering sign.

The interaction operator $V(t)$ is normally expressed as a product of quantum fields. For example, in QED

$$V(t) = e\int d^3x \overline{\psi}(\mathbf{x},t) \gamma_{\mu} \psi(\mathbf{x},t) A^{\mu} (\mathbf{x},t)$$

where $\psi(\mathbf{x},t)$ is the electron-positron Dirac field,
$\overline{\psi} = \psi^{\dag} \gamma_0$, $A^{\mu} (\mathbf{x},t)$ is the photon field, and $\gamma_{\mu}$ are $4 \times 4$ Dirac matrices.

It is clear, that in order to perform S-matrix calculations it is sufficient to know how quantum fields $\psi(\mathbf{x},t)$ and $A^{\mu} (\mathbf{x},t)$ are expressed through particle creation and annihilation operators, and what are (anti)commutators of these operators.
That's basically all one needs to know to perform QED calculations. There are also issues related to ultraviolet and infrared divergences, but they do not change substantially the basic formalism described above.

My point is that nowhere in these calculation one needs to use the "probabilistic interpretation" of quantum fields. As I said, quantum fields do satisfy wave equations, but their knowledge is not crucial for S-matrix calculations. In his book, Weinberg mentions wave equations as some "side effects". They are useful to simplify certain calculation steps, but they are not critical.

Eugene.

 

[Demystifier]

Meopemuk, that is all correct, we agree on that. Can we now turn to the question of correct transformation law of relativistic wave function? What would that be in your approach?

 

[masudr]

Meopemuk, that is all correct, we agree on that. Can we now turn to the question of correct transformation law of relativistic wave function? What would that be in your approach?

Surely if the field variable satisfies a covariant equation, then we are sure the field variable transforms covariantly? That is one of the main motivations for tensor notation.

 

[meopemuk]

Meopemuk, that is all correct, we agree on that. Can we now turn to the question of correct transformation law of relativistic wave function? What would that be in your approach?

My point of view is best expressed in S. Weinberg, "The quantum theory of fields", vol. 1.: Quantum fields satisfy relativistic wave equations and covariant transformations rules, like $\phi(x) \to \phi(\Lambda x)$. The only role of quantum fields in QFT is to serve as "building blocks" for relativistic interactions.

One-particle wavefunctions and their transformations should be determined by the Wigner's method of unitary irreducible representations of the Poincare group. Wigner's approach is usually formulated in the momentum space. For a spinless particle, its wavefunction depends on 3 projections of momentum $\psi(\mathbf{p})$ and transforms under boosts like


$$\psi(\mathbf{p}) \to \psi(L\mathbf{p})$$ (1)


where $L\mathbf{p}$ is the usual action of boosts on particle momentum.

In order to find wave functions in the position representation one needs to define the operator of position. (It is not covered in Weinberg's book, but it doesn't contradict anything there either) This is the famous Newton-Wigner operator. Knowing the relationship between momentum and position operators one can rewrite Wigner's wavefunction in the position representation $\psi(\mathbf{r})$ and rewrite the boost transformation law (1) in the position representation as well. This transformation law is fundamentally different from $\psi(x) \to \psi(\Lambda x)$, because it doesn't involve the time coordinate.

In order to find the time evolution of wave functions, we should use the standard QM prescription

$$-i \hbar \frac{\partial}{\partial t}| \psi \rangle = H |\psi \rangle$$ (2)

where $H = \sqrt{\mathbf{P}^2c^2 + m^2c^4}$ is the free-particle Hamiltonian. Taking the product of (2) with position eigenvectors $\langle \mathbf{r} |$ from the left, we obtain a relativistic Schroedinger equation.

There are two most important advantages of this approach. First, probabilities are explicitly conserved for all kinds of inertial transformations, including time translations and boosts, because they are represented by unitary operators

$$|\psi (t) \rangle \to \exp(\frac{i}{\hbar} Ht) |\psi (0) \rangle$$

$$|\psi (\theta) \rangle \to \exp(\frac{i}{\hbar} K_x c \theta) |\psi (0) \rangle$$

(where $|\psi (\theta) \rangle$ is the state vector seen by observer moving with rapidity $\theta$ or speed $c \cdot tanh \theta$ along the x-axis.

Second, the Poincare group properties are explicitly preserved, because commutators of generators of inertial transformations (like $H$, and $K_x$) satisfy the Poincare Lie algebra.

The next question is to understand how these one-particle states and wavefunctions are related to quantum fields. This is well covered in Weinberg's book as well. Knowing one-particle state we can define creation and annihilation operators. Then quantum fields are simply linear combinations of these operators with coefficients adjusted in such a way that

1. Transformation laws are covariant, e.g., $\phi(x) \to \phi(\Lambda x)$.

2. Fields are (anti)commuting with themselves at spacelike separations.

These properties make it possible that interaction operators constructed from quantum fields satisfy all relativistic requirements.

Eugene.

 

[gptejms]

In order to find wave functions in the position representation one needs to define the operator of position. (It is not covered in Weinberg's book, but it doesn't contradict anything there either) This is the famous Newton-Wigner operator. Knowing the relationship between momentum and position operators one can rewrite Wigner's wavefunction in the position representation $\psi(\mathbf{r})$ and rewrite the boost transformation law (1) in the position representation as well. This transformation law is fundamentally different from $\psi(x) \to \psi(\Lambda x)$, because it doesn't involve the time coordinate.

Is it not incomplete then?What happens to the time coordinate?

In order to find the time evolution of wave functions, we should use the standard QM prescription

$$-i \hbar \frac{\partial}{\partial t}| \psi \rangle = H |\psi \rangle$$ (2)

where $H = \sqrt{\mathbf{P}^2c^2 + m^2c^4}$ is the free-particle Hamiltonian.

You repeatedly give this (ugly)square root equation for the wavefunction.You may like to read the following about this equation(taken from a book):-

'Unfortunately, this equation presents us with a number of difficulties. One
is that it apparently treats space and time on a different footing: the time
derivative appears only on the left, outside the square root, and the space
derivatives appear only on the right, under the square root. This asymmetry
between space and time is not what we would expect of a relativistic
theory. Furthermore, if we expand the square root in powers of $$\nabla^2$$, we get an infinite number of spatial derivatives acting on ψ(x, t); this implies that the equation is not local in space.'

 

[meopemuk]

[The wavefunction] transformation law is fundamentally different from $\psi(x) \to \psi(\Lambda x)$, because it doesn't involve the time coordinate.

Is it not incomplete then?What happens to the time coordinate?

In order to find the time evolution of wave functions, we should use the standard QM prescription

$$-i \hbar \frac{\partial}{\partial t}| \psi \rangle = H |\psi \rangle$$ (2)

where $H = \sqrt{\mathbf{P}^2c^2 + m^2c^4}$ is the free-particle Hamiltonian.

You repeatedly give this (ugly)square root equation for the wavefunction.You may like to read the following about this equation(taken from a book):-

'Unfortunately, this equation presents us with a number of difficulties. One
is that it apparently treats space and time on a different footing: the time
derivative appears only on the left, outside the square root, and the space
derivatives appear only on the right, under the square root. This asymmetry
between space and time is not what we would expect of a relativistic
theory.

I understand where you are coming from, and why you think my approach is "ugly". You believe (like many other people, including the author of your quote) that the essence of relativity is in the symmetry between space and time. In your opinion, all physical properties must be Lorentz scalars, or 4-vectors, or tensors. I know that most people consider these statements "self-evident". Can you prove them?


In my opinion, there are two fundamental requirements that any relativistic theory must obey:

1. All inertial frames of reference must be exactly equivalent. There should be no preferred position, time, velocity, or orientation.

This principle of relativity was formulated a long time ago by Galileo. It is the second postulate, which distinguishes modern (Einstein's) relativity from the old (Galilei) relativity

2. Transformations between inertial frames of reference must form the Poincare group.

I think we can agree that 1. and 2. are absolutely necessary, and without them a theory cannot be called relativistic. In relativistic quantum mechanics or QFT we should combine these postulates with the Rules of Quantum Mechanics (Hilbert space, state vectors, Hermitian operators, wavefunctions, etc.). Such a combined theory was rigorously constructed in

E. P. Wigner "On unitary representations of the inhomogeneous Lorentz group" Ann. Math. 40 (1939), 149

and Weinberg in his book builds entire apparatus of quantum field theory on these ideas. I follow the same path.

It appears that in this Wigner-Weinberg approach there is no symmetry between space and time coordinates. Is this "ugly" or wrong? I don't think so. At closer inspection it appears that the space-time symmetry and covariant transformation laws (which you would like to have) do not follow directly from the two above postulates. In order to prove these properties one needs to make some additional assumptions, which may or may not be justified.

Basically, our discussion can be boiled down to the following dilemma: There are two ways to reconcile quantum mechanics with relativity:

1. Accept "minimal relativity" as given in the two above postulates, and accept all basic Rules of Quantum Mechanics. This leads us to the Wigner-Weinberg path.

2. In addition to two relativity postulates, assume the symmetry between space and time and the manifest covariance. This requires some important changes in the formalism of quantum mechanics. For example, one must reject the idea of the position operator.

Interestingly, for S-matrix calculations in QFT one doesn't need to worry about these foundational issues. Formulas for the S-matrix can be written in a manifestly covariant form. So, it may appear that the approach 2. is working. However, at a closer inspection, it seems that perfect space-time symmetry is never present in quantum mechanical problems. Scattering is about system's evolution in time. So, time is playing a distinguished role there.

Furthermore, if we expand the square root in powers of $$\nabla^2$$, we get an infinite number of spatial derivatives acting on ψ(x, t); this implies that the equation is not local in space.'

I don't know why you think this is bad. Why the fouth (or higher) spatial derivative is worse that the second spatial derivative (as in the non-relativistic Schroedinger equation)?

 

[gptejms]

.

However, at a closer inspection, it seems that perfect space-time symmetry is never present in quantum mechanical problems. Scattering is about system's evolution in time. So, time is playing a distinguished role there.

All physics is about evolution in time.Time gets a distinguished role because of our initial conditions(and an arrow because of 2nd law of thermodynamics).You may like to read the following from 'Bjorken & Drell'

'The only vestige of a preferred role for the time coordinate lies in the statement of initial conditions and of commutation relations at say t=0.The surface t=0 is a non-covariant element in the theory.Even this, however, may be removed by the covariant notion of a space-like surface on which to specify initial cnditions and commutators.'

You say 'space-time symmetry is never present in q.m. problems'--on the contrary space-time symmetry has to be there in relativistic (q.m.) equations.

I don't know why you think this is bad. Why the fouth (or higher) spatial derivative is worse that the second spatial derivative (as in the non-relativistic Schroedinger equation)?

Higher spatial derivatives make it more and more non-local as said in the original statement.BTW how do you propose to solve this equation with infinite no. of spatial derivatives?

 

[meopemuk]

All physics is about evolution in time.Time gets a distinguished role because of our initial conditions(and an arrow because of 2nd law of thermodynamics).You may like to read the following from 'Bjorken & Drell'

'The only vestige of a preferred role for the time coordinate lies in the statement of initial conditions and of commutation relations at say t=0.The surface t=0 is a non-covariant element in the theory.Even this, however, may be removed by the covariant notion of a space-like surface on which to specify initial cnditions and commutators.'

You say 'space-time symmetry is never present in q.m. problems'--on the contrary space-time symmetry has to be there in relativistic (q.m.) equations.

Let us agree to disagree. There are two valid approaches. One of them assumes perfect space-time symmetry, but violates some quantum-mechanical rules. Let us call it the Bjorken-Drell approach. The other (Wigner-Weinberg) approach keeps all Rules of Quantum Mechanics, but violates the symmetry between space and time.

Both approaches lead to the same formulas for the S-matrix in QFT, so they cannot be distinguished by modern experimental methods. Can we agree about that?

Higher spatial derivatives make it more and more non-local as said in the original statement.BTW how do you propose to solve this equation with infinite no. of spatial derivatives?

Yes, it is difficult to solve this equation in the position space. However, one can switch to the momentum space (Fourier transform) where momentum operator acts simply by multiplication, and then return back to the position space (another Fourier transform).

Eugene.

 

[gptejms]

One of them assumes perfect space-time symmetry, but violates some quantum-mechanical rules. Let us call it the Bjorken-Drell approach.

What quantum mechanical rules does it violate?

The other (Wigner-Weinberg) approach keeps all Rules of Quantum Mechanics, but violates the symmetry between space and time.

I can't comment on this approach because I haven't read it,but I guess it can't be much different from the standard one.

Yes, it is difficult to solve this equation in the position space. However, one can switch to the momentum space (Fourier transform) where momentum operator acts simply by multiplication, and then return back to the position space (another Fourier transform).

Yes, you can do that for solving--but it doesen't take away the non-locality(in position space).

 

[meopemuk]

What quantum mechanical rules does it violate?

As we discussed already, this method does not allow introduction of the position operator. It suggests using "position-space wavefunctions" $\psi(\mathbf{r}, t)$ without providing an interpretation of $\mathbf{r}$ as eigenvalues of the position operator. I think this is a contradiction.

There is another contradiction related to transformations of wavefunctions under boosts:

$$\phi (x) \to \phi(\Lambda x)$$ (1)

This formula implies that if the wavefunction is localized in the origin for the observer at rest, it remains localized for any moving oberver. I would like to show that this leads to a contradiction.

Rules of quantum mechanics dictate that boost transformations should be representable as actions of unitary operators on the wavefunction, i.e., in the case of boost along the x-axis

$$\phi (x) \to \exp(-\frac{i}{\hbar}cK_x \theta) \phi (x)$$

where $K_x$ is Hermitian generator of boosts. Similarly, space and time translations can be represented by unitary operators [tex] \exp(-\frac{i}{\hbar}P_x a) [/itex] and [tex] \exp(\frac{i}{\hbar}Ht) [/itex], where $P_x$ and $H$ are operators of momentum and energy, respectively. All these operators must satisfy commutation relations from the Poincare Lie algebra.

Now let us form the following product of unitary transformations

$$\exp(\frac{i}{\hbar}cK_x \theta) \exp(-\frac{i}{\hbar}P_x a) \exp(-\frac{i}{\hbar}cK_x \theta) \exp(\frac{i}{\hbar}P_x a \cosh \theta)$$ (2)

By applying Poincare commutators it is not difficult to show that this product is equal to

$$\exp(\frac{i}{\hbar}H a \sinh \theta)$$ (3)

Now, all factors in (2) preserve particle localization, so the product must do the same as well. However operator (3) does not preserve localization (this is known as the wave-packet spreading). This means that we made an error somewhere along the way. In my opinion, the error is in the assumption (1).

I can't comment on this approach because I haven't read it,but I guess it can't be much different from the standard one.

We are discussing here very esoteric things. They can be very important for theoretical foundations, but they have no connection to experiment. For example, with modern equipment it is impossible to verify or disprove eq. (1) to any reasonable level of accuracy.

As I said earlier, in experiments we can measure things related to the S-matrix. Sure enough, both Wigner-Weinberg and Bjorken-Drell approaches lead to the same S-matrix, so, currently they cannot be distinguished by experiment.

Yes, you can do that for solving--but it doesen't take away the non-locality(in position space).

Is there anything wrong with non-locality? What exactly?

 

[gptejms]

As we discussed already, this method does not allow introduction of the position operator. It suggests using "position-space wavefunctions" $\psi(\mathbf{r}, t)$ without providing an interpretation of $\mathbf{r}$ as eigenvalues of the position operator. I think this is a contradiction.

There is another contradiction related to transformations of wavefunctions under boosts:

$$\phi (x) \to \phi(\Lambda x)$$ (1)

This formula implies that if the wavefunction is localized in the origin for the observer at rest, it remains localized for any moving oberver. I would like to show that this leads to a contradiction.

Rules of quantum mechanics dictate that boost transformations should be representable as actions of unitary operators on the wavefunction, i.e., in the case of boost along the x-axis

$$\phi (x) \to \exp(-\frac{i}{\hbar}cK_x \theta) \phi (x)$$

where $K_x$ is Hermitian generator of boosts. Similarly, space and time translations can be represented by unitary operators [tex] \exp(-\frac{i}{\hbar}P_x a) [/itex] and [tex] \exp(\frac{i}{\hbar}Ht) [/itex], where $P_x$ and $H$ are operators of momentum and energy, respectively. All these operators must satisfy commutation relations from the Poincare Lie algebra.

Now let us form the following product of unitary transformations

$$\exp(\frac{i}{\hbar}cK_x \theta) \exp(-\frac{i}{\hbar}P_x a) \exp(-\frac{i}{\hbar}cK_x \theta) \exp(\frac{i}{\hbar}P_x a \cosh \theta)$$ (2)

By applying Poincare commutators it is not difficult to show that this product is equal to

$$\exp(\frac{i}{\hbar}H a \sinh \theta)$$ (3)

Now, all factors in (2) preserve particle localization, so the product must do the same as well. However operator (3) does not preserve localization (this is known as the wave-packet spreading). This means that we made an error somewhere along the way. In my opinion, the error is in the assumption (1).

How can you compare something with transformation (only)under the boost(1) with (2) which involves the boost as well as translation?

Is there anything wrong with non-locality? What exactly?

In post #235,I quoted an author against the use of the square root operator--this was when he discussed fields.Later when he talks of the wavefunction,he introduces back the square root thing(!)--as if it is the most natural thing to do, and without a trace of explanation.May be I should then conclude that the non-locality is, obviously, not allowed for the field,but allowed for the wavefunction.I'll appreciate if the mentors and the science advisors chip in with their comments.

 

[meopemuk]

How can you compare something with transformation (only)under the boost(1) with (2) which involves the boost as well as translation?

I am not comparing (1) and (2). I am comparing (2) and (3). My logic is the following: According to (1) boosts preserve localization. Space translations preserve localization as well. Therefore, operator (2) must also preserve localization, because it is composed of boosts and space translations only. From Poincare group properties, operator (2) is equal to operator (3). But (3) does not preserve localization, because time translations are responsible for the wave packet spreading effect. This is the contradiction.

Eugene.

 

[gptejms]

I am not comparing (1) and (2). I am comparing (2) and (3). My logic is the following: According to (1) boosts preserve localization. Space translations preserve localization as well. Therefore, operator (2) must also preserve localization, because it is composed of boosts and space translations only. From Poincare group properties, operator (2) is equal to operator (3). But (3) does not preserve localization, because time translations are responsible for the wave packet spreading effect. This is the contradiction.

Eugene.

Still not clear--if (2) leads to (3),why should (1) be blamed?Anyway,what is your solution?
You haven't commented on the 2nd part of the post(non-locality etc.).

 

[meopemuk]

Still not clear--if (2) leads to (3),why should (1) be blamed?Anyway,what is your solution?

I think that my derivation is completely rigorous. Still it leads to a contradiction. The only weak point is assumption (1) that wavefunctions transform covariantly. This was your assumption, not mine. I think that (1) is a wrong boost transformation law for wavefunctions in the position representation. At least, so far you haven't provided any good evidence for it.

You haven't commented on the 2nd part of the post(non-locality etc.).

I cannot speak for the author of your quote. My position is that the relativistic Schroedinger equation for particle wavefunctions involves the "ugly" square root. This equation may be non-local, but I don't see anything wrong with this non-locality.

Wave equations for quantum fields are covariant Klein-Gordon, Dirac, etc. equations (the choice depends on particle's spin and mass). Wavefunctions and quantum fields are completely unrelated objects that play very different roles in QFT. Therefore, there is no contradiction in the fact that they obey different equations.

Eugene.

 

[Demystifier]

There are two valid approaches. One of them assumes perfect space-time symmetry, but violates some quantum-mechanical rules. Let us call it the Bjorken-Drell approach. The other (Wigner-Weinberg) approach keeps all Rules of Quantum Mechanics, but violates the symmetry between space and time.

I think we could take this as the conclusion.
Except that I don't agree that they are both "valid". At least one should be invalid.