How observation leads to wavefunction collapse?

[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 \exp(-\frac{i}{\hbar}P_x a) [/itex] and \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.<br /> <br /> Now let us form the following product of unitary transformations<br /> <br /> \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)<br /> <br /> By applying Poincare commutators it is not difficult to show that this product is equal to <br /> <br /> \exp(\frac{i}{\hbar}H a \sinh \theta) (3)<br /> <br /> 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). <br />

<br /> <br /> How can you compare something with transformation (only)under the boost(1) with (2) which involves the boost as well as translation?<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Is there anything wrong with non-locality? What exactly? </div> </div> </blockquote><br /> 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.

 

[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.