How observation leads to wavefunction collapse?

[meopemuk]

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.

Yes, I agree with that. And, in my opinion, the Wigner-Weinberg approach is the valid one.

Eugene.

 

[Fra]

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.

I guess it depends on how you define the "wavefunction" but assuming it's devised from information in possesssion of the observer in question, I agree that this transformation law makes no sense.

What bothers me is that the discussion doesn't distinguish between information possessed by different observers.

For example, to me, $$\psi(x) \to \psi(\Lambda x)$$ represents the "information about x" relative to O, transforming only x as per transformations that disrespect relativity of information.

In my thinking transformation of _information_ is a physical process involving dynamics, not just a simple "change of variables". IMO, no one has yet to my understanding given a satisfactory explanation of the real transformation. In my thinking, which isn't done yet, this transformation is not a simple change of variables, it is a process, taking a certain amount of time. I haven't figured it out yet, but I figured out so much as to see that something is unsatisfactory with the current views.

I ask generally, what is the relation between $$\psi(x)$$ and $$\psi'(x')$$. In the classical world,$$x' = \Lambda x$$, but $$\psi \neq \psi'$$

IMO, to understand $$\psi(x|O)$$ and the relative probability $$\rho(x|O)$$, I believe in analyzing the measurement process and definition of $$\psi$$ from the point of view of the observer O, and I think this naturally leads also to a revision of space itself. Also what process really brings by a coherent line of reasoning $$\rho(x|O)$$ into $$\rho(x'|O')$$? It's no way as simples as $$\rho(x'|O') = \rho(\Lambda x|O)$$.

/Fredrik

 

[Hans de Vries]

Yes, I agree with that. And, in my opinion, the Wigner-Weinberg approach is the valid one.

Eugene.

But does Weinberg agree? Now remembering his openings statement:

If it turned out that some physical system did not obey the rules of
quantum mechanics and relativity, it would be a cataclysm.

I guess not Regards, Hans

 

[Demystifier]

The point is that the "Wigner-Weinberg" is not an appropriate name, because Weinberg does NOT propose to use the non-covariant Wigner position operator. In fact, Weinberg does not introduce any position operator, which makes his approach inconsistent with nonrelativistic QM. The question is: Is Weinberg aware of that?

 

[gptejms]

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.

It wasen't my assumption--it was Demystifier's.As far as evidence is concerned he said 'Because this provides that the equation of motion for psi (the Klein Gordon equation) will have the same form in all Lorentz frames.'I think what he said is true for the field(here too, I would say the condition should be slightly modified to $$\psi^\prime (x^\prime)=\psi (x)$$ where $$x^\prime = \Lambda x$$) not the wavefunction.


May be you can tell us what it should be for the wavefunction.

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.

Eugene.

For me the statement that the equation for the wavefunction is non-local is a very significant statement.

 

[meopemuk]

The point is that the "Wigner-Weinberg" is not an appropriate name, because Weinberg does NOT propose to use the non-covariant Wigner position operator. In fact, Weinberg does not introduce any position operator, which makes his approach inconsistent with nonrelativistic QM. The question is: Is Weinberg aware of that?

You are right, Weinberg does not introduce position operator and position-space wavefunctions in the first 13 chapters of his book. However, he could easily do that without contradicting anything else said in his book.

The situation changes in chapter 14, where he discusses bound states and the Lamb shift. In eqs. (14.1.4) and (14.1.5) he introduces electron's position-space wavefunctions, which are equivalent to your earlier definition

$$\psi(x) = \langle 0 | \phi(x) |1 \rangle$$

where $\psi(x)$ is the wavefunction and $\phi(x)$ is the quantum field. I think, Weinberg is not fully consistent in his book. His "wavefunctions" are vulnerable to the counterargument that I suggested in post #241.

It would be great to ask Weinberg himself what he thinks about that. But I suspect he is not reading Physics Forums.

Eugene.

 

[meopemuk]

It wasen't my assumption--it was Demystifier's.As far as evidence is concerned he said 'Because this provides that the equation of motion for psi (the Klein Gordon equation) will have the same form in all Lorentz frames.'I think what he said is true for the field(here too, I would say the condition should be slightly modified to $$\psi^\prime (x^\prime)=\psi (x)$$ where $$x^\prime = \Lambda x$$) not the wavefunction.

This is true in the case of multicomponent (spinor, vector, etc.) quantum fields $$\phi_i(x)$$. In this case, the transformation law is

$$\phi_i(x) \to \sum_j D(\Lambda^{-1})_{ij} \phi_j (\Lambda x)$$

where $$D_{ij}$$ is a finite-dimensional (non-unitary) representation of the Lorentz group associated with the field $$\phi_i(x)$$. In the case of scalar field $D = 1$.

May be you can tell us what it should be for the wavefunction.

I don't have a closed formula, but the boost transformation of the wavefunction $$\psi(\mathbf{r}) \to U(\Lambda) \psi(\mathbf{r})$$can be obtained in the following three steps:

1. Change to the momentum representation

$$\psi(\mathbf{p}) = (2 \pi \hbar)^{-3/2}\int d^3r \psi(\mathbf{r}) \exp(-\frac{i}{\hbar} \mathbf{pr})$$

2. Apply boost transformation to the wavefunction momentum arguments

$$U(\Lambda) \psi(\mathbf{p}) = \psi(\Lambda \mathbf{p})$$

Transformations of momenta under boosts are given by simple formulas. For example, if $$\Lambda$$ is a boost along the z-axis with rapidity $$\theta$$, then

$$\Lambda (p_x, p_y, p_z) = (p_x, p_y, p_z \cosh \theta + \frac{E_p}{c} \sinh \theta)$$

where $$E_p = \sqrt{m^2c^4 + p^2c^2}$$

3. Perform inverse transformation back to the position space

$$U(\Lambda) \psi(\mathbf{r}) = (2 \pi \hbar)^{-3/2}\int d^3p U(\Lambda) \psi(\mathbf{p}) \exp(\frac{i}{\hbar} \mathbf{pr})$$

 

[gptejms]

3. Perform inverse transformation back to the position space

$$U(\Lambda) \psi(\mathbf{r}) = (2 \pi \hbar)^{-3/2}\int d^3p U(\Lambda) \psi(\mathbf{p}) \exp(\frac{i}{\hbar} \mathbf{pr})$$

If $$\psi(\mathbf{p}) =1$$, then you get back $$U(\Lambda)$$ acting on the wavefunction(which is a delta function in this case).Now what do you do?

 

[Fra]

How do you transform the boundaries of the integrals during the boost? This is a critical part as the set of integration is the effective "event space" of the observer. The question is how does sets of simultaneous sampling transform, this seems to be the basis for the entire probabilistic reasoning, these sets seems generally not to be invariants. The observers doesn't possesses the same set of information. So it seems we are comparing apples to pears.

To consider infinity as sample space is a bit amgious in my thinking.

/Fredrik

 

[meopemuk]

If $$\psi(\mathbf{p}) =1$$, then you get back $$U(\Lambda)$$ acting on the wavefunction(which is a delta function in this case).Now what do you do?

Oops! I am sorry. I missed an important part in the boost transformation of the momentum-space wave function. Please replace step 2. in my previous posty with

2. Apply boost transformation to the momentum space wavefunction

$$U(\Lambda) \psi(\mathbf{p}) = \sqrt{\frac{E_{\Lambda p}}{E_p}}\psi(\Lambda \mathbf{p})$$ (1)

The (previously missed) square root factor is important in order to guarantee the conservation of probabilities under boosts. For example, the wavefunction normalization is now preserved, because

$$\int d^3p |U(\Lambda) \psi(\mathbf{p}) |^2= \int d^3p\frac{E_{\Lambda p}}{E_p}|\psi(\Lambda \mathbf{p})|^2$$ (2)

where one can make the substitution $\mathbf{q} = \Lambda \mathbf{p}$ and use the well-known property

$$\frac{d^3p}{E_p} = \frac{ d^3(\Lambda p)}{E_{\Lambda p}}$$


to show that the normalization integral (2) is equal to

$$\int d^3q|\psi( \mathbf{q})|^2 = 1$$

With this (now corrected) boost transformation law (1) the position-space delta function is not transformed to another position-space delta function. There is a wavefunction spreading caused by boosts.

 

[gptejms]

With this (now corrected) boost transformation law (1) the position-space delta function is not transformed to another position-space delta function. There is a wavefunction spreading caused by boosts.

Good!
But as I asked earlier also,what happens to the time coordinate?

 

[meopemuk]

Good!
But as I asked earlier also,what happens to the time coordinate?

The time coordinate is not present at all. These formulas relate wavefunction $\psi(\mathbf{r})$ seen by the observer at rest at time t=0 (measured by his clock) to the wavefuinction $U(\Lambda)\psi(\mathbf{r})$ seen by the moving observer at time t=0 (measured by her clock).

I guess you would like to condemn this transformation on the basis of unequal treatment of space and time coordinates. Before you do that, consider the following:

1. You haven't proved yet that space/time symmetry and covariant transformation laws follow necessarily from the principle of relativity. (I know, that's what most textbooks say. But I haven't seen a rigorous proof.)

2. It actually makes sense that the transformed wavefunction $U(\Lambda)\psi(\mathbf{r})$ depends on values of the initial wavefunction $\psi(\mathbf{r})$ at time t=0 only. Indeed, the latter wavefunction provides a complete description of the particle's state for the observer at rest at t=0. By knowing the state for one observer, we should be able to find exactly how the state looks like for any other (e.g., moving) observer. Therefore, it seems quite logical, that the transformed wavefunction $U(\Lambda)\psi(\mathbf{r})$ should have a unique expression in terms of the original wavefunction $\psi(\mathbf{r})$ at time t=0.

Eugene.

 

[Demystifier]

It would be great to ask Weinberg himself what he thinks about that. But I suspect he is not reading Physics Forums.

Completely agree.

 

[gptejms]

2. It actually makes sense that the transformed wavefunction $U(\Lambda)\psi(\mathbf{r})$ depends on values of the initial wavefunction $\psi(\mathbf{r})$ at time t=0 only. Indeed, the latter wavefunction provides a complete description of the particle's state for the observer at rest at t=0. By knowing the state for one observer, we should be able to find exactly how the state looks like for any other (e.g., moving) observer. Therefore, it seems quite logical, that the transformed wavefunction $U(\Lambda)\psi(\mathbf{r})$ should have a unique expression in terms of the original wavefunction $\psi(\mathbf{r})$ at time t=0.

Eugene.

The problem here is that $U(\Lambda)\psi(\mathbf{r})$ that you have determined is not all at one $$t^{\prime}$$,the time in the transformed frame.Is this even meaningful then?

 

[meopemuk]

The problem here is that $U(\Lambda)\psi(\mathbf{r})$ that you have determined is not all at one $$t^{\prime}$$,the time in the transformed frame.Is this even meaningful then?

I am not sure I understand what you are saying. $U(\Lambda)\psi(\mathbf{r})$ is the wavefunction seen by the moving observer at one particular time instant t=0.

From your question it seems to me that you are still trying to interpret wavefunction transformations in terms of Minkowski space-time diagrams. I think that this is not a useful approach.

Eugene.

 

[Demystifier]

I am not sure I understand what you are saying. $U(\Lambda)\psi(\mathbf{r})$ is the wavefunction seen by the moving observer at one particular time instant t=0.

From your question it seems to me that you are still trying to interpret wavefunction transformations in terms of Minkowski space-time diagrams. I think that this is not a useful approach.

Eugene.

Are you saying that two observers have the same time t?

 

[gptejms]

I am not sure I understand what you are saying. $U(\Lambda)\psi(\mathbf{r})$ is the wavefunction seen by the moving observer at one particular time instant t=0.

From your question it seems to me that you are still trying to interpret wavefunction transformations in terms of Minkowski space-time diagrams. I think that this is not a useful approach.

Eugene.

If t=0 ,then from Lorentz transformation, t' is not zero everywhere--so your transformed wavefunction is not defined at one t'(but multiple times).

 

[meopemuk]

Are you saying that two observers have the same time t?

I don't understand the meaning of your question. One observer measures wavefunction at time t=0 by his clock. The other (moving) observer also measures (a different) wavefunction at time t=0 by her clock. There is a formula that connects these two wavefunctions. Is there anything wrong with that?

Eugene.

 

[Fra]

> One observer measures wavefunction at time t=0 by his clock

Now it get's interesting. How does this measurement look like - I picture it to be a procedure, experimental procedure, possibly including calculations etc? Or perhaps the wavefunction isn't measurable?

I think it's safe to assume that only the observers own past can influence this "measurement", or?

But the two observers doesn't share the same past anyway, that's how it gets interesting to see how you can get the obvious connection of two different histories.

/Fredrik

 

[meopemuk]

If t=0 ,then from Lorentz transformation, t' is not zero everywhere--so your transformed wavefunction is not defined at one t'(but multiple times).

Wavefunction $\psi(\mathbf{r})$ refers to time t=0 from the point of view of observer O. Wavefunction $U(\Lambda) \psi(\mathbf{r})$ refers to time t'=0 from the point of view of the moving observer O'. Both wavefunctions are defined at a single time point for the respective observers.

What is the meaning of your statement "t' is not zero everywhere"?

I think you are trying to apply Lorentz transformations (formulated for times and positions of events viewed from different frames of reference) to arguments $(\mathbf{r}, t)$ of wavefunctions. What makes you believe that Lorentz transformations are applicable in this case?

Eugene.

 

[meopemuk]

> One observer measures wavefunction at time t=0 by his clock

Now it get's interesting. How does this measurement look like - I picture it to be a procedure, experimental procedure, possibly including calculations etc? Or perhaps the wavefunction isn't measurable?

I think it's safe to assume that only the observers own past can influence this "measurement", or?

But the two observers doesn't share the same past anyway, that's how it gets interesting to see how you can get the obvious connection of two different histories.

/Fredrik

Could you please be more precise? It seems that you find a contradiction in my statements. But I am not sure what is the contradiction that you see?

Eugene.

 

[gptejms]

Wavefunction $\psi(\mathbf{r})$ refers to time t=0 from the point of view of observer O. Wavefunction $U(\Lambda) \psi(\mathbf{r})$ refers to time t'=0 from the point of view of the moving observer O'. Both wavefunctions are defined at a single time point for the respective observers.

What is the meaning of your statement "t' is not zero everywhere"?

If t=0,then t'=0 only at x=0.Your wavefunction $\psi(\mathbf{r^{\prime}})$ is not defined at one particular t'.

 

[meopemuk]

If t=0,then t'=0 only at x=0.Your wavefunction $\psi(\mathbf{r^{\prime}})$ is not defined at one particular t'.

I don't understand what you mean by "then t'=0 only at x=0". In my understanding, the time shown by a clock does not depend on x. Could you please explain?

What I am saying is this: We have two observers O and O'. Both of them have clocks. Both of them perform measurements of the wavefunction of the (same) particle when their clocks show zero time. Can they do that? Yes they can. Both of them will obtain certain results for the position space wavefunction. Their results will be different and they will be related by the transformation I've outlined.

Eugene.

 

[Fra]

Could you please be more precise? It seems that you find a contradiction in my statements. But I am not sure what is the contradiction that you see?

Eugene.

I guess it's difficult, it is not a direct contradiction since I'm not sure I understand how you define the wavefunction $$\psi(r,t)$$.

Do you consider $$|\psi(r,t)|^2$$ to be the probability density to find whatever we are looking for at position r, at time t? If so, do you tehcnically consider the observer to be delocalized? Ie. does the observer have an array or detectors throughout the universe? Is t, the time where the observer gets informed, or the time in the observers frame, when the event happens simultaneously at another location, hitting one of the infinite number of detectors in the universe? or are the observer instantaneously informed?

The question is, how does the normalisation look like? on what data does the normalization take place? If you picture surfaces in minkowskispace your suggestions seems strange, so what is your thinking? Or are you thinking that the ensemble construction is observer invariant?

I don't understand what you mean by "then t'=0 only at x=0". In my understanding, the time shown by a clock does not depend on x. Could you please explain?

But your wavefunctions contains to datapoints, relating to position. And how does information about remote points propagate consistently to the observer and arrive as to allow simultaneous "sampling"? If you mention probability to see something at a certain time. One asks, at what time? Does the various events originate from a simulatenous events in the fram, or from different times? Either way may be fine, as long as it's consistent. In the first frame for example, did $$|\psi(r)|^2$$ refer to t=const in minkowski space? If not, how is if defined? $$\psi$$ is just a symbol, I'd like to see a logic how it's induced from something that is at leat in principle measureable. I'm not saying it can't be done, I'm just not sure I understand your view. If you have some interesting ideas I'd like to understnad them.

/Fredrik

 

[meopemuk]

Do you consider $$|\psi(r,t)|^2$$ to be the probability density to find whatever we are looking for at position r, at time t?

Yes, this is the standard definition of the wavefunction at time t.

If so, do you tehcnically consider the observer to be delocalized? Ie. does the observer have an array or detectors throughout the universe? Is t, the time where the observer gets informed, or the time in the observers frame, when the event happens simultaneously at another location, hitting one of the infinite number of detectors in the universe? or are the observer instantaneously informed?

I understand your concerns. Technically, it is difficult to collect information about extended wavefunctions at one time instant. However, in principle, it is not impossible.

Consider an array of detectors covering entire universe, and one processing unit in the middle. Suppose that each detector is connected to the processing unit by a cable of the same length L. For far away detectors the cables will be stretched. For detectors close to the processing unit we will need to fold their cables (in rings, or whatever). Assuming that signals propagate through all cables at the same speed c, it would take exactly the same time L/c to reach the central unit for signals from all detectors in the array. This arrangement will guarantee that our measurements will be done at the same time instant (or, if you like on the same "slice" of the Minkowski space-time) throughout entire universe.

This is for the observer at rest. The moving observer will have a similar array of detectors co-moving with her. Both observer at rest and moving observer can use their detector arrays to measure instantaneous (in their opinion) wavefunctions and then compare their notes. This comparison should agree with the 3-step wavefunction transformation that I outlined in previous posts. Does it make sense?


Eugene.

 

[gptejms]

I don't understand what you mean by "then t'=0 only at x=0". In my understanding, the time shown by a clock does not depend on x. Could you please explain?

Use $$t'=\frac{t-xv^2/c^2}{\sqrt{1-v^2/c^2}}$$.Your (transformed)wavefunction is defined at one particular t(not one t' as it should be).

In your post #252,you did Fourier transform over x--you need to take the time coordinate t also (and Fourier transf. over time as well) to complete the calculation.

 

[Fra]

I understand your concerns. Technically, it is difficult to collect information about extended wavefunctions at one time instant. However, in principle, it is not impossible.

Consider an array of detectors covering entire universe, and one processing unit in the middle. Suppose that each detector is connected to the processing unit by a cable of the same length L. For far away detectors the cables will be stretched. For detectors close to the processing unit we will need to fold their cables (in rings, or whatever). Assuming that signals propagate through all cables at the same speed c, it would take exactly the same time L/c to reach the central unit for signals from all detectors in the array. This arrangement will guarantee that our measurements will be done at the same time instant (or, if you like on the same "slice" of the Minkowski space-time) throughout entire universe.

This is for the observer at rest. The moving observer will have a similar array of detectors co-moving with her. Both observer at rest and moving observer can use their detector arrays to measure instantaneous (in their opinion) wavefunctions and then compare their notes. This comparison should agree with the 3-step wavefunction transformation that I outlined in previous posts. Does it make sense?

I suspect this boils down to our views of probability concepts and what the point of the wavefunction is, or should be.

Your prescription of filling the universe with detectors and then adjusting cable length to maintain simultanous sampling means that the "simultaneous sampling" will be delayed an incredible amount of time if the "universe" is large. Which would mean that the simultanous sampling at time t1, is delayed and not given to use until t2.

So, you can not in principle, in your thinking, "measure" $$\psi(x,t_1)$$until t2, if we agree on that, I think the construction is useless.

So the wavefunction the observer writes down at t2, that is supposedly $$\psi(x,t2)$$,is really based on $$\rho(x,t1)$$. Clearly the information from nearby detector may have impacted the observer between t1 and t2. The whole point is that the observer needs his wavefunction for guidance. If this is not determined until the end of time, I see the whole objective beeing disrespected.

Anyway, I assumed that what the detectors would give you anyway, is particle position, so what we really get is $$\rho(x,t)$$, where t is way back in history once we receive it.

So suppose we define $$\psi(x,t) = (\rho(x,t))^{1/2}$$, then we determine \psi up to the complex phase only.

I think the probabilistic constructions should be made on truly simultaneously available information. The relevant question I ask, is what is my _currently_ available best information, and how does that induce my "probabilistic" estimates? But then, my view of probability in this context is the bayesian one. Supposedly the wavefunction represents this information, right? If not, the wavefunction is of little use. The information that is in my hand tomorrow, does not help me today.

You seem to suggest(?) that your best bets at t2 are based on old simultanous information originating from t1? I don't understand the point in such construction?

I am still working on my thinking, but if I introduce a wavefunction it represents the simultanous information I have, at hand. I will from then one define related probability spaces. The relations will mutually induce priors, and there will also be a relation to mass. You can not accumulated arbitrary amounts of information without getting huge mass. This doesn't contain the information I get tomorrow. Neither can I ignore information at hand, just to synchronize with collection of data sent from remote locations. This is a bit difficult though, and in progress. But in my thinking the proper transformation is not as simple as a rotation or translation and so on in some space. It would ultimatley be an equilibration process, where equilibrium is where the two observers are in agreement.

I think the most fundamental difference to my thinking here is the treatise of probability theory, and what it is used for? IMO, the purpose of probability in reality is for any subject to formulate it's best educated guess, accuonting to all available information. This is entangled with the definition of wavefunction. The simple postulates of QM are far from satisfactory. They don't match the complexity I see in reality. In my thinking this analysis demands accounting for several things, the notion of space time, "entropy" concepts and relative probabilities and the concept of information vs mass vs inertia.

I find the standard formulations simplistic and unsatisfactory from the point of view of a scientific method. The motivation is that "looks it works". That may be true, but the real question is if we are just lucky, and still await a proper explanation?

/Fredrik

 

[meopemuk]

Use $$t'=\frac{t-xv/c^2}{\sqrt{1-v^2/c^2}}$$
Your (transformed)wavefunction is defined at one particular t(not one t' as it should be).

I think you should be careful when extending Lorentz transformations to transformations of wavefunction arguments. You may have extended them beyond their range of validity.

Recall the physical meaning of Lorentz transformations. They relate times and positions of actual events measured in different reference frames. If from the point of view of observer O there was a physical event (for example, a collision of two particles) at point x at time t, then from the point of view of the moving observer O' the same event occurs at space-time point (x',t'), where

$$t'=\frac{t-xv/c^2}{\sqrt{1-v^2/c^2}}$$ (1)

$$x'=\frac{x-vt}{\sqrt{1-v^2/c^2}}$$ (2)

These transformations apply to definite classical events measured with 100% probability.

The physical interpretation of wavefunction $$\psi(x,t)$$ tells that if I place a particle detector at point x, then the probability (density) of registering a particle at this point at time t is $$|\psi(x,t)|^2$$. Your prescription is equivalent to the following statement. If the moving observer O' places a particle detector at point x', then the probability (density) of finding a particle there at time t' should be equal to $$|\psi(x,t)|^2$$ This statement doesn't necessarily follow from the original (classical) meaning of Lorentz transformations. At least, you haven't provided yet a convincing proof that one implies the other. If you want to sound convincing, I think you need to prove your statement from fundamerntal axioms of quantum mechanics and relativity.

On the other hand, I am ready to provide a complete proof of my wavefunction transformation law based on such axioms. The full proof would be quite lengthy, so I will outline just few important steps:

1. Particle states are described by unit vectors in a Hilbert space H.

2. Transformations between different inertial frames of reference (or observers) form the Poincare group (here I am considering not just boosts, but the full set of inertial transformations, which include also space translations, time translations, and rotations; all of them are treated on equal footing).

3. The action of inertial transformations on state vectors are such that probabilities are preserved.

4. It follows from 3. that inertial transformations of observers are represented by unitary operators in the Hilbert space H, and that there exists an unitary representation $U_g$ of the Poincare group in H. This means that if $|\Psi \rangle$ is a state vector of the particle from the point of view of observer O, then observer O' will describe the same state by the vector $U_g|\Psi \rangle$, where g is the boost transformation that connects O to O'. So, we found our desired transformation in the language of abstract state vectors. All we need to do is to translate this transformation in the language of position-space wavefunctions.

5. In order to do that we need to define the position operator $\mathbf{R}$ and its eigenvectors $|\mathbf{r} \rangle$ in the Hilbert space H. Then the wavefunction from the point of view of observer O will be $\langle \mathbf{r}| \Psi \rangle$, and the wavefunction from the point of view of O' will be $\langle \mathbf{r}| U_g |\Psi \rangle$

6. The position operator $\mathbf{R}$ satisfying all relativistic requirements was constructed by Newton and Wigner in

T. D. Newton, E. P. Wigner, "Localized states for elementary systems" Rev. Mod. Phys. 21 (1949), 400.

If you put all these ideas together you'll obtain exactly the wavefunction transformation law given in post #252 (with correction in post #255).

You may wonder whether this transformation law agrees with classical Lorentz transformations? It appears that yes, they do agree with each other. It is possible to show (I will not do it here, but I am ready to reproduce calculations, if you are interested) that if we

1) take the classical limit of the above theory and consider trajectories of two non-interacting particles that intersect at the space-time point (x,t) from the point of view of observer O and
2) transform these trajectories according to commutators between the Newton-Wigner operator and generators of boost transformations $U_g$,

then

3) we will find that from the point of view of O', the intersection of trajectories occurs at point (x',t') which is related to (x,t) by usual Lorentz formulas (1) and (2).

Eugene.

 

[meopemuk]

Hi Fredrik,

Yes, in my detector design the signal generated by the particle today and here will reach the central processing unit only after very long time L/c. So, I'll need to wait forever to analyze these results. But we are not talking about practical matters here, we are talking about matters of principle. So, in principle, it is possible to measure the wavefunction in entire space at a fixed time instant.

You are right that this will give us only the square of the position-space wavefunction, which doesn't determine the state uniquely. To overcome this problem, one can also build detectors measuring probability distributions of particle momentum, angular momentum, and other observables not commuting with position. So that the full set of data would allow us to reconstruct the particle state in an unique fashion.

I don't want to speculate how exactly these detectors should be built. Personally, I believe that nothing useful can be gained from considering exact designs of such detectors. I think that for theoretical purposes it is sufficient to postulate that each observer can somehow precisely determine the wavefunction of any state in any given basis at each fixed time instant t. This postulate is, of course, an idealization. But I believe that this is a reasonable price for getting a simple and transparent theory.

Eugene.

 

[Demystifier]

I don't understand the meaning of your question. One observer measures wavefunction at time t=0 by his clock. The other (moving) observer also measures (a different) wavefunction at time t=0 by her clock. There is a formula that connects these two wavefunctions. Is there anything wrong with that?

Not necessarily. But can you write down a general mathematical relation (not only for the case when both are zero) between times of the two observers? If I understand you correctly, you suggest that it might NOT be the Lorentz transformation.

 

[Fra]

I don't want to speculate how exactly these detectors should be built. Personally, I believe that nothing useful can be gained from considering exact designs of such detectors.

I agree the exact detector design may not be the biggest issues at this point, but still details like how information from detectors are transformed into possession of the observer is i think important. If we are careless here I think we are compromising our own endavours.

I think that for theoretical purposes it is sufficient to postulate that each observer can somehow precisely determine the wavefunction of any state in any given basis at each fixed time instant t. This postulate is, of course, an idealization. But I believe that this is a reasonable price for getting a simple and transparent theory.

I think we disagree about the validity of some of the idealizations that we admittedly all do at some level. But the exact concept of information, what it means, and what it's for, is I think so paramount that I disagree with your foundations here. I want to find more solid ground.

Though you are completely right that what I "want", will not be as simple, and I'm still fighting with "basics" that you already idealized away. We will see if I succeed in resolving it to a way that satisfies me. Meanwhile m view of the standard models is as a kind of effective theories, but whose rigorous formulation is yet to be seen. And with rigour I mean not just mathematical rigour of formalism, I mean rigour of reasoning.

/Fredrik

 

[meopemuk]

Not necessarily. But can you write down a general mathematical relation (not only for the case when both are zero) between times of the two observers? If I understand you correctly, you suggest that it might NOT be the Lorentz transformation.

Lorentz transformations are understood as transformations between times and positions of events in different reference frames. When you have a single quantum-mechanical particle with its wavefunction, there is no specific event for which you could exactly pinpoint its space-time location. The wavefunction is diffuse, and measurements are not certain. So, Lorentz transformations need not to apply in this case.

The situation is different in the classical limit. Then particle's behavior can be described by a trajectory x(t) in the reference frame O and by a trajectory x'(t') in the reference frame O'. These trajectories can be regarded as series of events (x,t) and (x',t'). Within my theory x and x' can be represented as expectation values of the Newton-Wigner position operator in different reference frames. By applying boost transformations to the Newton-Wigner operator it can be shown that events (x',t') are related to (x,t) by usual Lorentz formulas. So, this theory does produce Lorentz transformations, but only in the classical (h -> 0) limit.

Eugene.

P.S. It is important to note that Lorentz formulas can be obtained in this fashion only if the particle is non-interacting, i.e., it has a straight-line trajectory. It appears that for interacting particles (with events defined as, for example, intersections of their trajectories) this approach leads to different transformations (x,t) -> (x',t'), which deviate from Lorentz formulas if the interaction is strong. So, in fact, predictions of my theory are different from those of ordinary special relativity in some cases. This is another "can of worms", and I am not sure if we are ready to open it at this moment.

 

[gptejms]

These transformations apply to definite classical events measured with 100% probability.

If the probability is not 100%,I don't see why it should follow a different law/transformation.

At least, you haven't provided yet a convincing proof...

I asked you to complete your calculation in post #252 with the time coordinate included.Anyway---if the wavefunction in one frame is $$\exp{\iota p_\mu x^\mu}$$(where $$\hbar=1$$),then you can see that the transformed wavefunction is going to be not much different.

It is possible to show (I will not do it here, but I am ready to reproduce calculations, if you are interested)

Go ahead!

 

[Fra]

If the probability is not 100%,I don't see why it should follow a different law/transformation.

I disagree with some of Eugene's thinking, but I agree on the fact that probabilities can't just be transformed like a lorentz scalar. I think Eugene's reason for thinking so is different than mine, but by reason is that probability is not an objective thing, it's not an object that everyone agress upong. It's intrinsically relative to the observer. What is probably for me, need not be probably to you for the simple reason that we devise our ensembles differently.

It does not make sense to think that there is someone, like a God, that knows everything and thus therefore can give objective status to probabilities.

This debate is not specific to quantum mechanics, it's a bit philosophical and has to do as much with your view of probability theory.

The frequentists and the bayesian thinking can probably be partly united with some abstractions, if you consider that the frequency of information is relative anyway. Two observers by definition doesn't see the same things. They live in the same universe, but it's a universe we all are fighting to understand, and no one has an omnipotent view of things. Thus anything claiming to find a simple connection between the views, by means of a mathematical transformation must incorporate assumptions about information they do not have. I suggest it's formulate in a bayesian sense, as degrees of information in different things, relative to what we know. This is essence, IMO, one of the few honest implementations of the scientific method, which I think of as finding the BEST guess, given the current evidence.

I think most people would agree that, just because we don't know, this doesn't mean we might as well guess randomly. The fact that we still know _something_ can induce a bayesian probability, as to guide us forward. This can be formalised. The proof is in the succcess and survival implying a selection, I think there are no formal proofs.

/Fredrik

 

[Fra]

You can ponder and reflect over this in many ways, and many have done so in the past.

Here in one link with some elaborations.
http://math.ucr.edu/home/baez/bayes.html

/Fredrik