How observation leads to wavefunction collapse?

[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:

$$\left\{ \frac{\partial^2 }{\partial t^2}\ -\ \frac{\partial^2 } {\partial x^2}\ -\ \frac{\partial^2 } {\partial y^2}\ -\ \frac{\partial^2 } {\partial z^2}\ \right\}\ \psi \ = \ -m^2\ \psi$$

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

$$\frac{\partial^2}{\partial t^2}\left\{ \left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y} \right)^2 + \left(z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z} \right)^2 + \left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x} \right)^2 \right\}\ \psi\ =\ m^2 \ell(\ell+1)\ \psi$$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:

$$\left\{ \frac{\partial^2 }{\partial t^2}\ -\ \frac{\partial^2 } {\partial x^2}\ -\ \frac{\partial^2 } {\partial y^2}\ -\ \frac{\partial^2 } {\partial z^2}\ \right\}\ \psi \ = \ -m^2\ \psi$$

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

$$i\hbar\partial_t\Psi=\sqrt{-c^2\hbar^2\nabla^2 + (mc^2)^2}\Psi$$

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

$$i\hbar\partial_t\Psi=\sqrt{-\hbar^2\nabla^2 + (mc^2)^2}\Psi$$

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

$$\sqrt{-\nabla^2 + m^2}\psi(x) = \int\frac{d^3x'\;d^3p}{(2\pi)^3}\psi(x')\sqrt{|p|^2+m^2}e^{ip\cdot(x-x')}$$

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

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

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

$$\sqrt{-\nabla^2 + m^2}\psi(x) = \int\frac{d^3x'\;d^3p}{(2\pi)^3}\psi(x')\sqrt{|p|^2+m^2}e^{ip\cdot(x-x')}$$

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

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

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

$$\psi(x)\mapsto \int\frac{d^3x'\;d^3p}{(2\pi)^3} \psi(x')\big(|p|^2+m^2\big) e^{ip\cdot(x-x')}$$

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