Question on observer created reality

[vanesch]

I was under the impression that Bohmian mechanics does not explain well the collapse of wavefunctions, i.e. the sudden discontinuity in Bohm's "quantum potential" after measurement.

I don't see what you mean (not pejorative: please explain). I thought that the quantum potential was a function of the actual positions of the particles (which are well-defined in Bohm's theory), and the wavefunction (the same as the one in quantum theory, and which continues to evolve according to unitary QM: that's why it has some MWI aspects). Both evolve smoothly and are not influenced particularly by a measurement, no ? A measurement (in Bohm's theory, there is only one kind of measurement possible, namely a position measurement ; hence the preferred basis problem is solved this way as there is explicitly a preferred basis: the position) just gives us information about what trajectory (from a statistical ensemble which had an initial Born rule given distribution: the deux ex machina in Bohm :-) we had.

The only objection one can have against Bohmian mechanics is its blunt violation of SR (the expression of the quantum potential) - ok, and the deux ex machina that the initial uncertainty on particle positions has to be given by the Born rule... If one is willing to let go SR, I think it is one of the finest resolutions of the measurement problem that exist. However, I find a violation of SR too gross to take it seriously.

cheers,
Patrick.

 

[seratend]

First of all, yes, I like the good wine and the smelly cheese (and lots of other stuff around here :-) ; one of the reasons to stick to a certain ontology :-)

: ))).

I have indeed the impression that we are talking next to each other, and I'm affraid I'm not clear on your viewpoint as well.
I'm *for the moment* a kind of MWI fan, with the hope that I'm wrong

I think so (for the impression). I am a fan of what I call the minimalist interpretation: the consistent logical content of QM formalism, especially the collapse postulate (hence, I hope it should be included in any consistent interpretation). I am sure it is not far from the logical content of MWI.

I think the main difference in our views seem to be your search for a reality behing the descriptive view given by QM. Personnally, I do not care if there is a concrete "reality", I can continue to appreciate good wine, and the rest, without that ; ))).

I have difficulties with a purely epistemological view because *something* exists (a la Descartes, also from cheese and wine fame).

So you think we do not live in the matrix?
Ok, I understand what you mean (would like to have).

Where does it lead us ?
If we take the wavefunction to be a representation of reality, clearly the collapse gives us a problem: how can a meager observation by a humble human change the state of the entire universe ?
So this "collapse" stuff must be something EPISTEMOLOGICAL.

I agree with that (all what I say is around this). QM formalism is, formally, a descriptive tool (in space and time). It does not require or assume other properties to work logically.
I would like to get a clear separation between the descriptive part (what we currently have) and the search of ontological properties or anything else (further improvements). The different posts we find in this forum highlights the confusion resulting from this mixture (especially concerning the collapse postulate and the HUP).
Moreover, most of the different existing interpretations participate to this confusion in mixing (by a lack of a clear separation in the formulation) the equivalent mathematical reformulation of the formalism (e.g. bohmian meachanics) together with the definition of external ontological properties (e.g. the path of a bohmian particle). MWI suffers from this default (e.g. especially by the choice of "many worlds", I would have chosen instead the ensemble interpretation ; ).

But when you look at the "ontology" (if that's what the wavefunction represents), this measurement did in fact more to me than to the system, which is still with amplitude a in |u> and with amplitud b in |v> except that it's now entangled with my bodystates ; that's what I meant.

The main consistency problem with this interpretation relies on the , may be implicit, meaning "before" "after" and "still" words.
Collapse postulate just defines a property (description and not prediction of what a system is and not what it would be or may be, etc ...). Collapse postulate does not reject "ontology" nor it assumes it: (all properties may be assumed to be "real" as long as we keep the logical consistency of the description). The only interpretational problem comes from assuming the system had a different property "before" the property given by the measurement result (i.e. the state before the measurement and the state "after" the measurement).
Concerning this aspect, the consistent histories formalism allows one to understand better how the things work when we consider the measurement results on space and time: a measurement result is, formally, part of a collection of measurement results (which is also a "measurement result"). (A good introduction to general stochastic processes)

Seratend.

 

[vanesch]

So you think we do not live in the matrix?

Even that is ontological. It is in fact pretty close to the view I'm advocating.

Concerning this aspect, the consistent histories formalism allows one to understand better how the things work when we consider the measurement results on space and time: a measurement result is, formally, part of a collection of measurement results (which is also a "measurement result"). (A good introduction to general stochastic processes)

Yes, exactly. When we learned something about a system at time t1, say: that the system was in state a|1> + b|2> and we found, at t1, that the result was "1", we are allowed to work with the state |1> alone (projection postulate) because all FURTHER measurement results are conjunct with "and at t1, the result was equal to 1". So a second measurement, at t2, with result U, is not an independent result but is a CORRELATION with "and at t1 we had result = 1". The possibility doesn't exist anymore of performing the measurement at t2, INDEPENDENT of what was the result at t1: we irreversibly HAVE that result.
This is in fact what allows us to apply unitary evolution and only apply the Born rule "at the end of the day", but this time with the measurement result "at t1 we had result 1 and at t2 we had result U". It will implicitly lead to the projection, first, of |1> and then of |U>
This is indeed the same as considering the entire history of measurement results as one single measurement.

cheers,
Patrick.

 

[Nicky]

[...] I thought that the quantum potential was a function of the actual positions of the particles (which are well-defined in Bohm's theory), and the wavefunction (the same as the one in quantum theory, and which continues to evolve according to unitary QM: that's why it has some MWI aspects). Both evolve smoothly and are not influenced particularly by a measurement, no ? [...]

I am no expert in Bohmian mechanics, so maybe you are right. I read some criticism of Bohm's account of wavefunction collapse, but it was just informal commentary on the 'net, not a serious analysis.

On the other hand, I would point out where Bohm invokes some more "deus ex machina" as you call it. In second of two papers where Bohm lays out his theory (Phys Rev 85, p. 180 (1952)), the talk of "violent and extremely complicated fluctuations" in the wavefunction is a bit vague to me. Also he seems to take the Born rule as given:

Because the probability density is equal to |psi|^2, we deduce that the apparatus variable [...] must finally enter one of the packets and remain with that packet thereafter [...] Thus, for all practical purposes, we can replace the complete wave function [...] by a new renormalized wave function

But maybe he explains this better elsewhere. I still have not understood the whole paper.

 

[vanesch]

I am no expert in Bohmian mechanics, so maybe you are right.

I found http://plato.stanford.edu/entries/qm-bohm/ a very useful and readable introduction. Don't mind Bohm's original papers, after all, the originator of an idea doesn't always, at that moment, have the clearest view and explication of it (it is still brand new and has to be digested: as well by himself as by others).

cheers,
Patrick.

 

[George Jones]

What about the Kochen-Specker theorem? This says, roughly, that quantum systems do not, in general, possesses properties.

Consider the measurement of S_z of a spin-1 particle. If the state is psi = |1> just before measurement, then, with certainty, the result of the measurement is S_z = 1, and it seems to make sense that S_z = 1 was a property of the particle before measurement. Now suppose that the state is psi = (sqrt(3)|-1> + |1>)/2 just before the measurement, and that the result of the measurement is S_z = 1. Is it possible, within the framework of standard quantum theory, to say that the particle had the property S_z = 1 before the measurent? In other words, is possible that to say that the observable S_z possessed the value 1 before the measurent?

Now generalize this. Let A be an observable for a quantum system. Is it possible that a value function V that takes as input an observable and a state of the system, and spits out, say V(A ,psi), for the value of the observable A when the system is in state psi?

No one has ever thought of a way of doing this. Is this because no one has been able to hit on the right idea, or is it because it is impossible? If it were possible to do this, many of the ontological "problems" associated with quantum theory would disappear, and realists, like Patrick (I think) and me, would be happier campers :-). The Kochen-Specker theorem says that for quantum systems having state spaces of dimension 3 or higher, no value function (with reasonable properties) exists. :-(

This one of the things that makes a "disturbance" interpretation of the uncertainty principle difficult. If a system doesn't possesses a particular property, how is it possible to disturb that property.

Nice discussions of the Kocken-Specker theorem can be found in Lectures on Quantum Theory: Mathematical and Structure Foundations by Chris Isham, and in The Structure and Interpretation of Quantum Mechanics by R.I.G.Hughes. Hughes is a philosopher, and some philosophy he espouses in his preface is particularly interesting: "Having thus outlined my program and declared my allegiances, I leave the reader to decide whether to proceed further, or to open another beer, or both."

The Kochen-Specker theorem makes life somewhat difficult for realists, but positivists probably have no problem with it. Isham says "... very effective professionally without needing to lose any sleep over the implications of the Kochen-Specker theorem or the Bell inequalities. However, in the last twenty years there has been a growing belief among physicists from many different specialisations that even if modern quantum theory works well at the pragmatic level, it simply cannot be the last word on the matter."

Patrick, this is essentially your hope, isn't it? Maybe a subtle 4).

Isham says further "... there are now any any physicists who feel that any significant advance will entail a radical revision in our understanding in the meaning and significance of the categories of space, time and substance."

To me, this seems to point towards quantum gravity, which is no surprise coming from Isham. Many physicists would disagree this, but I am inclined to believe that a theory of quantum gravity might have interesting things to say about these issues. Unfortunately, despite claims by various camps to the contrary, I see no viable candidate for a quantum theory of gravity. Also, such a theory could bring forth new interpretational issues just as thorny as any it alleviates.

Regards,
George

 

[gptejms]

In order to say so, you have to consider the measurement system "classical", in that you put a "classical" potential in the Schroedinger equation of the electron. In doing so, you've neglected the quantum nature of the microscope. Of course, in practice this works in many occasions (semi-classical models work very well in different fields) ; after all that is the correspondence principle.

The potential was introduced to give an idea of what might be happening.The wavefunction changes by the introduction of a potential(in this case it gets localized).This at least makes the measurement process less mysterious.In any case I also mentioned the interaction of the photon field with the electron field to be fully quantum--this is more difficult to calculate,but I think it is worth exploring(and my guess is that the 'potential' mimics this effect well).With my little knowledge of QFT,I am not in a position to do it---but you or other QFTists in the forum can certainly do a small calculation--may be localization comes out of such two interacting fields.

that's true for those who think that decoherence solves the problem (they use the density matrix, which itself is based upon the Born rule and the projection postulate), it is even true for Deutsch (who introduces some "reasonable assumptions" for his rational deciders which comes down to the Born rule).

I agree with this totally--when you sum up over the bath coordinates,you are in a way already using the Born rule.But what I am saying is different--you are introducing a quantum interaction(mimicked roughly by a potential)when you do a measurement and this localizes the wavefunction.I think sudden introduction of a potential has merit in the following sense---evolution is unitary in the absence of the potential as well as when you have the potential 'on',but in between these two phases is the evolution unitary?May be not,but I need to think about this.

 

[vanesch]

I think sudden introduction of a potential has merit in the following sense---evolution is unitary in the absence of the potential as well as when you have the potential 'on',but in between these two phases is the evolution unitary?May be not,but I need to think about this.

According to QM, of course it is unitary ! There is nothing else that can give you a time evolution. A time-dependent hamiltonian still gives you a unitary time evolution.
I think the problem in your approach is that you refuse to see the "thing that applies the potential" as a quantum object. I agree with you that FOR ALL PRACTICAL PURPOSES you can do that, by restricting your attention to the "system", and find a quite good approximation to the quantum evolution of the system alone ; also you wouldn't find much of a difference for a "full" quantum treatment IF AT ANY POINT YOU APPLY THE BORN RULE. But *applying the Born rule* means that you switch to a classical world, and unitary evolution doesn't apply there. That's the essential contents of the Copenhagen view: that there is a macroscopic, classical world, and that there is a microscopic quantum world and that both "communicate" by the Born rule (micro -> macro) and the projection postulate (macro -> micro). But this means that there is a barrier separating the two. So the macroscopic world is then NOT subject to quantum theory, which would be, in itself, an amazing statement: we don't have the same laws, and it is not exactly clear what objects obey what set of rules.
If you refuse that, and you say that EVERYTHING is ruled by quantum theory however, then you are stuck with your unitary evolution all the way up. This is the case I was considering.

cheers,
Patrick.

 

[gptejms]

According to QM, of course it is unitary ! There is nothing else that can give you a time evolution. A time-dependent hamiltonian still gives you a unitary time evolution.

When I wrote that I had a step function kind of a potential in mind--a potential that's introduced all of a sudden.Say the particle is free to begin with and has a wavefunction \exp[\iota(k_1 x - \omega_1 t)].Now a potential is introduced say at time t_1 which to begin with we take to be constant.The wavefunction now readjusts(over some time) to \exp[\iota(k_2 x - \omega_2 t)],where the new variables take care of the change in energy.What happens at t=t_1:-do the two wavefunctions merge?No they don't---there is some funny wavefunction during the readjustment phase which is a mixture of the two.Can you derive it from the Schrodinger equation?Can you give me the readjustment/response time?

Another problem:-Say the potential introduced is that of a harmonic oscillator,so the wavefunction not only goes from a plane wave to a Hermite Gauss function(if I remember correctly),but also the particle now falls into the nearest eigenstate i.e. it gains or loses some energy--can you describe this process thoroughly through unitary evolution/Schrodinger equation?May be you can,I need to think about these things.

I think the problem in your approach is that you refuse to see the "thing that applies the potential" as a quantum object. I agree with you that FOR ALL PRACTICAL PURPOSES you can do that, by restricting your attention to the "system", and find a quite good approximation to the quantum evolution of the system alone ; also you wouldn't find much of a difference for a "full" quantum treatment IF AT ANY POINT YOU APPLY THE BORN RULE. But *applying the Born rule* means that you switch to a classical world, and unitary evolution doesn't apply there. That's the essential contents of the Copenhagen view: that there is a macroscopic, classical world, and that there is a microscopic quantum world and that both "communicate" by the Born rule (micro -> macro) and the projection postulate (macro -> micro). But this means that there is a barrier separating the two. So the macroscopic world is then NOT subject to quantum theory, which would be, in itself, an amazing statement: we don't have the same laws, and it is not exactly clear what objects obey what set of rules.
If you refuse that, and you say that EVERYTHING is ruled by quantum theory however, then you are stuck with your unitary evolution all the way up. This is the case I was considering.

On the contrary I am taking everything to be quantum.What I am stressing is this:--what a measurement does is not to yield a particle with a definite x and a definite p.When you have made a measurement say by Heisenberg's microscope at one of the slits,you still observe a particle with some \Delta x and a \Delta p i.e. it is still a wavefunction--the orignal wavefunction corresponding to the two slits has now shrunk,but it's still a wavefunction or a quantum particle(without a definite x or p) that you observed when you made a measurement.I think the mistake one does is to think that 'since the particle has been observed,it must have a definite x and a definite p' and then one starts questioning how this mysterious collapse(of the wavefunction) comes out of unitary evolution.But if you keep in mind that all you do is to localize the wavefunction when you make a measurement,you can argue that the measurement process is akin to the introduction of a potential.

 

[vanesch]

Another problem:-Say the potential introduced is that of a harmonic oscillator,so the wavefunction not only goes from a plane wave to a Hermite Gauss function(if I remember correctly),but also the particle now falls into the nearest eigenstate i.e. it gains or loses some energy--can you describe this process thoroughly through unitary evolution/Schrodinger equation?May be you can,I need to think about these things.

First of all, I'd like to re-iterate my remark that there is no *interaction* between two systems that introduces a "potential" in the Schroedinger equation. You only introduce INTERACTION TERMS in the product hilbert space, and if the initial (product) state of both systems is not an eigenstate of that interaction term (which it most probably isn't), this will give rise to a time evolution of the total state which is an entangled state of both systems.

But now to come to your specific problem of the sudden introduction of a harmonic potential: clearly the particle will NOT evolve to an energy eigenstate, for the following simple reason:
at t0- (just before the potential is introduced), we assume that the particle is, say, in a plane wave state. Of course, we require continuity of the wavefunction (otherwise the schroedinger equation doesn't make sense: it is a second order differential equation), so at t0+ the particle is now still in this plane wave state. We now rewrite that plane wave state as a superposition of eigenstates of the harmonic potential (the Hermite functions). Clearly we do not have just one term !
Given that each individual term is a stationary state, the time evolution from t0+ onward is given by the hamiltonian of the harmonic potential, so each term (stationary state) just gets a phase factor exp(-i E_n t).
No term will "decay". So we remain in this superposition for ever.

cheers,
Patrick

 

[gptejms]

First of all, I'd like to re-iterate my remark that there is no *interaction* between two systems that introduces a "potential" in the Schroedinger equation. You only introduce INTERACTION TERMS in the product hilbert space, and if the initial (product) state of both systems is not an eigenstate of that interaction term (which it most probably isn't), this will give rise to a time evolution of the total state which is an entangled state of both systems.

You introduce a 'nucleus' and say you your system of interest is an electron--don't the two things interact?You mean the nucleus just introduces a potential?!

But now to come to your specific problem of the sudden introduction of a harmonic potential: clearly the particle will NOT evolve to an energy eigenstate, for the following simple reason:
at t0- (just before the potential is introduced), we assume that the particle is, say, in a plane wave state. Of course, we require continuity of the wavefunction (otherwise the schroedinger equation doesn't make sense: it is a second order differential equation), so at t0+ the particle is now still in this plane wave state. We now rewrite that plane wave state as a superposition of eigenstates of the harmonic potential (the Hermite functions). Clearly we do not have just one term !

Again I take the example of an electron approaching a nucleus.Take for the moment the electron to be a classical particle.Depending on its speed,it should either fall into a circular orbit(rather spiral into the nucleus) or be scattered off.In a quantum scenario,if it is scattered off, it doesen't remain the same plane wave--its wavevector changes.If,as you say, the plane wave remains as it is,the electron should merrily pass on away from the nucleus with the same momentum and in the same direction as if nothing has happened--obviously this is not correct.In the case the electron gets trapped by the nucleus,the wavefunction again changes.

Let's however remain focused on the central issue.I stated that when you 'observe' a particle you are just observing a shrunk form of the wavefunction--the particle is still quantum and has a delta x and a delta p.The only thing a measurement does is to give you this shrunk form which comes out very well from quantum mechanics itself.Need to see some comments on this.

 

[seratend]

Again I take the example of an electron approaching a nucleus.Take for the moment the electron to be a classical particle.Depending on its speed,it should either fall into a circular orbit(rather spiral into the nucleus) or be scattered off.In a quantum scenario,if it is scattered off, it doesen't remain the same plane wave--its wavevector changes.

Attention, you are attaching a reality to the wave function (in the same way you may speak wrongly about the path of particle in a double slit experiment). Try to think with the Heisenberg view where the wave function is a constant in time in order to avoid false deductions.
Note that in the quantum as well as in the classical scenario, a particle may be scattered of or captured by the scatterer (it depends on the interactions).

Seratend.

 

[vanesch]

Again I take the example of an electron approaching a nucleus.

Ok, it is an interesting example.

Take for the moment the electron to be a classical particle.Depending on its speed,it should either fall into a circular orbit(rather spiral into the nucleus) or be scattered off.

Let us more say that the speed is fixed, and that we work with the impact parameter b (the distance between the straight line on which the incoming electron moves, and the nucleus) in the classical case.

Indeed, for a large impact parameter, the particle almost goes straight through, while for smaller and smaller impact parameters, first the particle track bends lightly, a bit and finally, for small enough values, the particle scatters under large angles. Let us not consider the case of capture. The nucleus will undergo the appropriate recoil for momentum to be conserved.

In a quantum scenario,if it is scattered off, it doesen't remain the same plane wave--its wavevector changes.If,as you say, the plane wave remains as it is,the electron should merrily pass on away from the nucleus with the same momentum and in the same direction as if nothing has happened--obviously this is not correct.In the case the electron gets trapped by the nucleus,the wavefunction again changes.

In a semi-quantum scenario (incoming particle quantum, nucleus classical), if the particle's momentum is well known, it is an incoming plane wave, and hence it "contains" all impact parameters at once. The nucleus being considered classical, we can replace it by a potential in the electron's schroedinger equation. We now solve that Schroedinger equation, and out will come a rather complicated looking wavefront. If we apply the Born rule to it (meaning, we make a transition to classical), it gives us the probability of the particle to scatter under a certain angle. From that *classical* result, we can then find the corresponding probability of the nucleus to recoil with a certain momentum.

In the full quantum scenario, the initial state of the nucleus-electron system is a product state |nucleus> |electron> in which the nucleus corresponds to some localized lumpy wavefunction, and the electron an incoming plane wave. Now, there is no external potential, but an interaction term in the hamiltonian over the hilbert space of the two particles, which describes their Coulomb interaction. If we apply this hamiltonian's time evolution to
|nucleus> |electron>, this product state will evolve in an entangled state, which I intuitively think will take on something like:

|psi_final> = Integral db f(b) |nucleus-plane-wave-p(b)> |electron-plane-wave-p(b)>

Here, f(b) is some complex number as a function of a variable which I call b, because it more or less corresponds to some impact parameter, and the two states correspond to a product state of a plane wave with momentum p(b) for the nucleus and a plane wave with momentum - p(b) for the electron.

I'm not sure about this, I just guess it: it is my guess for the form of the solution of the schroedinger equation (in Schmidt decomposition).

So we end up with a global wavefunction for the nucleus and the electron, which is not a product state. If we now apply the Born rule to this system, measuring both momenta of nucleus and electron, we notice several things:

1) there is a very strong correlation between the momentum of the electron and the momentum of the nucleus (they are exactly opposite). That's what you obtained classically by imposing momentum conservation, but it will simply come out of the Schroedinger equation.

2) the probability distribution of the momenta of the electron-nucleus system will be VERY CLOSE to the probability distribution of the semiclassically obtained potential-scattering solution, where the nucleus was replaced by a potential.

I didn't claim that the wave function of the electron had to remain "as if nothing happened" ; after all there is an interaction with a nucleus ; and this entangles both wavefunctions, and transforms them unitarily. But you see again that no SINGLE scattering angle comes out of this: all angles are realized (that is: there is a term in the final wavefunction corresponding to each possible scattering angle for the electron) and to each angle for the electron, corresponds the matched angle for the nucleus.

Let's however remain focused on the central issue.I stated that when you 'observe' a particle you are just observing a shrunk form of the wavefunction--the particle is still quantum and has a delta x and a delta p.The only thing a measurement does is to give you this shrunk form which comes out very well from quantum mechanics itself.Need to see some comments on this.

If you mean that you got now entangled with that "shrunk part" (which is just a term in the overall wavefunction), then yes, I agree of course!
If you say that quantum theory by itself unitarily shrunk the wavefunction, I'm telling you that that is mathematically impossible for a unitary operator.

cheers,
Patrick.

 

[vanesch]

I want to lift a potential confusion:

I said:

We now rewrite that plane wave state as a superposition of eigenstates of the harmonic potential (the Hermite functions). Clearly we do not have just one term !
Given that each individual term is a stationary state, the time evolution from t0+ onward is given by the hamiltonian of the harmonic potential, so each term (stationary state) just gets a phase factor exp(-i E_n t).
No term will "decay". So we remain in this superposition for ever.

However, this doesn't mean that, after t0+, the particle remains in a PLANE wave! This wavefunction will evolve now according to the harmonic potential hamiltonian, and this will "dephase" the different terms in the initial plane wave decomposition in Hermite functions, so that after a finite time, their recomposition will give you something else than a plane wave. I only showed you that whatever that complicated function is, it cannot decay into the ground state of the harmonic potential.

 

[seratend]

If you say that quantum theory by itself unitarily shrunk the wavefunction, I'm telling you that that is mathematically impossible for a unitary operator.

Only for finite times. : ))

Seratend.

 

[vanesch]

Only for finite times. : ))

I don't know exactly what you mean ? Do you allude to the result of DeWitt (I think) that all branches in which there is a serious deviation from the Born rule have a total (Hilbert) norm 0 for t-> Infinity ?
Or do you allude to the slightly imaginary slope given to the time axis to derive the LZH result (but that, to me, is just a sloppy trick) ?
Or still something else ?

cheers,
Patrick.

 

[seratend]

I don't know exactly what you mean ? Do you allude to the result of DeWitt (I think) that all branches in which there is a serious deviation from the Born rule have a total (Hilbert) norm 0 for t-> Infinity ?
Or do you allude to the slightly imaginary slope given to the time axis to derive the LZH result (but that, to me, is just a sloppy trick) ?
Or still something else ?

cheers,
Patrick.

: ).

We just need to take any translation generator. We have (with the good units) <x|exp(-ip.a)|psi> = psi(x+a) for all finite a and all states |psi> but not necessarily for infinite a (a group is closed under finite operations and not necessarily for infinite operations) where everything is possible (without external properties on this limit, e.g. sigma additivity).
The rest is just application of this property (e.g. classical limits, decoherence etc ...).

Some formal applications of this result for the time (i.e. the unitary evolution) are used in the formal study of the boundary limit between classical world and quantum world especially in the decoherence area . If you are interested on this aspect I recommend the excellent paper of of N.P. Landsman "Between Classical and quantum" quant-ph/0506082 (especially section 6.6 and 7 that deals with the results of infinite time on measurements).
(Note: it requires some knowledge on the C*-algebra formalism results, but it has a lot of pointers).

(I hope you will appreciate this paper as it may also help you in your search for an ontological reality ).

Seratend.

P.S. My comment does not invalidate what you say, just that we need to take some care with the unitary evolution for the sake of consistency.

P.P.S. I hope you may have a new view on the scattering matrix and more precisely the “limit at t --> +oO”.

 

[gptejms]

Attention, you are attaching a reality to the wave function (in the same way you may speak wrongly about the path of particle in a double slit experiment).

When you have shrunk the wavefunction to one slit,obviously the particle's path is certain to the the extent that the particle passes through this slit.

Try to think with the Heisenberg view where the wave function is a constant in time in order to avoid false deductions.

Good suggestion.If the wavefunction is a constant,obviously no operator can cause the particle to pass through one slit only.However when you suddenly introduce a potential,you can no longer continue to work with the old wavefunction.

Note that in the quantum as well as in the classical scenario, a particle may be scattered of or captured by the scatterer (it depends on the interactions).

Of course, and I have mentioned the latter case also.

 

[gptejms]

If we apply this hamiltonian's time evolution to |nucleus> |electron>, this product state will evolve in an entangled state, which I intuitively think will take on something like:

|psi_final> = Integral db f(b) |nucleus-plane-wave-p(b)> |electron-plane-wave-p(b)>

Here, f(b) is some complex number as a function of a variable which I call b, because it more or less corresponds to some impact parameter, and the two states correspond to a product state of a plane wave with momentum p(b) for the nucleus and a plane wave with momentum - p(b) for the electron.

I'm not sure about this, I just guess it: it is my guess for the form of the solution of the schroedinger equation (in Schmidt decomposition).

Ok,the nucleus is now entangled with the electron.You will use this to argue that the measuring apparatus also gets entangled with the system being measured etc. etc. and finally give only a 'conscious' observer the authority to break the superposition(by MW splitting) and that too to your own consciousness only and not that of a cat or for that matter not even any other human--this is a kind of wishful thinking that I don't subscribe to.

If you mean that you got now entangled with that "shrunk part" (which is just a term in the overall wavefunction), then yes, I agree of course!
If you say that quantum theory by itself unitarily shrunk the wavefunction, I'm telling you that that is mathematically impossible for a unitary operator.

When you are entangled with the shrunk part,at least the electron passed through a particular slit--you are entangled with one slit only.Regarding the unitarity,I like seratend's suggestion of the Heisenberg picture.Read my response to him.

 

[gptejms]

I want to lift a potential confusion:

I said:



However, this doesn't mean that, after t0+, the particle remains in a PLANE wave! This wavefunction will evolve now according to the harmonic potential hamiltonian, and this will "dephase" the different terms in the initial plane wave decomposition in Hermite functions, so that after a finite time, their recomposition will give you something else than a plane wave. I only showed you that whatever that complicated function is, it cannot decay into the ground state of the harmonic potential.

Say you have a plane wave \exp[i(kx-\omega t)] to start with.You Fourier analyze the x part of this to Hermite functions---the time dependence continues to be \exp(-i \omega t) !If you insist that you'll Fourier analyze the t part also into different frequencies,even then that grand sum will sum up to act like \exp(-i \omega t) only.Now you may say,you meant that the t part would change upon interaction--ok fine,but then you see that the energy has changed.In your model the x part has remained the same i.e. the momentum has not changed,whereas the energy has--I see no point in preserving the momentum.

 

[vanesch]

If you are interested on this aspect I recommend the excellent paper of of N.P. Landsman "Between Classical and quantum" quant-ph/0506082 (especially section 6.6 and 7 that deals with the results of infinite time on measurements).
(Note: it requires some knowledge on the C*-algebra formalism results, but it has a lot of pointers).

(I hope you will appreciate this paper as it may also help you in your search for an ontological reality ).

Thanks for the reference ! I started (p 20) reading it (but it's LONG ! 100 p).

cheers,
Patrick.

 

[vanesch]

Good suggestion.If the wavefunction is a constant,obviously no operator can cause the particle to pass through one slit only.However when you suddenly introduce a potential,you can no longer continue to work with the old wavefunction.

Of course you can ! The potential now just changes the evolution of the observables. "Suddenly introducing a potential" V(x) at time t0 just comes down to having a hamiltonian:

H(t) = H0 + h(t-t0) V(x)

(h(t) is the Heavyside step function)

U(t) is the solution to the equation:

hbar dU/dt = - i H(t)

and is a unitary operator

(the solution is not anymore exp(-i H/hbar t) because that only applies for time invariant hamiltonians).

Given your (time independent) waverfunction psi (which, say, coincides with the Schroedinger view at t = ta), we have now:
x(t) = U(t)U_dagger(ta) x U_(ta) U_dagger(t)
and the same for p(t),
with x and p the Schroedinger position and momentum operators, and x(t) and p(t) the Heisenberg position and momentum operators, evolving according to your hamiltonian with sudden potential.

psi remains always psi(ta).

cheers,
Patrick.

 

[seratend]

Good suggestion.If the wavefunction is a constant,obviously no operator can cause the particle to pass through one slit only.However when you suddenly introduce a potential,you can no longer continue to work with the old wavefunction.

?? What do you mean?
If the potential is bounded, you still have your constant state.

Seratend.

EDIT:

P.S. Damned too slow : ))

 

[vanesch]

Say you have a plane wave \exp[i(kx-\omega t)] to start with.You Fourier analyze the x part of this to Hermite functions---the time dependence continues to be \exp(-i \omega t) !

No, it doesn't of course ! This time dependence is what you get for the free hamiltonian ; this time dependence was correct as long as the potential was not switched on (for t < t0, say). But when the hamiltonian changes (it is a time-dependent hamiltonian, because some potential is switching on at t = t0), all these different hermite functions get DIFFERENT omegas (namely E_n / hbar).

cheers,
Patrick.

 

[vanesch]

Ok,the nucleus is now entangled with the electron.You will use this to argue that the measuring apparatus also gets entangled with the system being measured etc. etc. and finally give only a 'conscious' observer the authority to break the superposition(by MW splitting) and that too to your own consciousness only and not that of a cat or for that matter not even any other human--this is a kind of wishful thinking that I don't subscribe to.

Apart from your last statement, yes, you captured my thought :-) HOW do you "break the superposition" ? My hope is that gravity will, somehow, do so, because I'd also rather not subscribe to this view, but I simply don't know how to avoid it if unitary QM is correct, and we're talking here about how to interpret unitary QM.

However, you missed a point: I'm not saying that *my* consciousness somehow breaks the superposition, as some objective thing it does to the state of the universe, as if I were somehow a special physical or devine construction. The superposition is still there ; my consciousness just only observes one term of it. Now, of course, TO ME, that comes down to exactly the same thing (and that's why the projection postulate WORKS FAPP) - epistemologically. The problem only exists if you say that there is a real world out there (ontologically) and that its objective state is given by the wavefunction ; if QM is correct on all levels, this wavefunction cannot do anything else but evolve according to a Schroedinger equation, and if it does, there's no way to break the superposition.
(except if one is going to introduce infinities and so on, but if we take it that the universe has only a finite, but very large, number of degrees of freedom, I don't see how this is going to solve the issue).
As not one single physical interaction in the universe (possessing a hamiltonian) can break the superposition, clearly something unphysical has to do it when our subjective observations are to be explained.
Again, IF unitary QM is correct.

 

[gptejms]

?? What do you mean?
If the potential is bounded, you still have your constant state.

Seratend.

If what you say is right,no atom should radiate!The wavefunction remains a constant and no amount of interaction causes it to change--think about it.

 

[vanesch]

If what you say is right,no atom should radiate!The wavefunction remains a constant and no amount of interaction causes it to change--think about it.

The wavefunction of the atom alone ? But then it is not interacting with something ! Or do you mean, the wavefunction of the atom + the EM field ?

But if you are talking about the Heisenberg picture, the wavefunction there is constant by construction, so I do not see how it ever could do something else but "remain constant": it is not a function of time !

 

[gptejms]

U(t) is the solution to the equation:

hbar dU/dt = - i H(t)

and is a unitary operator

Given your (time independent) waverfunction psi (which, say, coincides with the Schroedinger view at t = ta), we have now:
x(t) = U(t)U_dagger(ta) x U_(ta) U_dagger(t) ...(1)
and the same for p(t),
with x and p the Schroedinger position and momentum operators, and x(t) and p(t) the Heisenberg position and momentum operators, evolving according to your hamiltonian with sudden potential.

psi remains always psi(ta).

Your expression (1) above is not right.There is another term i \hbar U^\dagger \dot U which you have missed out.Let's start afresh.Say your hamiltonian is
H(t) = H_0 + \Theta(t-t_0)V(x)

U(t) = \exp\frac{-i \int_0^t [{H_0 + \Theta(t^\prime-t_0)V(x)}]dt^\prime}{\hbar} = \exp(-iH_0 t/ \hbar) \exp[-i V(x)(t-t_0)/ \hbar]

Now x(t) is something like this:-

x(t) = \exp[-i/\hbar(...)] x(0) \exp[i/\hbar(...)])]+i\hbar U^\dagger(t).\dot U (t),

where \dot U(t) has terms like:-
H_0(..) + V(x)(..) + V^\prime(x) \dot x(t)(...)

Work out what this leads to---I am already tired of writing in tex.

 

[gptejms]

No, it doesn't of course ! This time dependence is what you get for the free hamiltonian ; this time dependence was correct as long as the potential was not switched on (for t < t0, say). But when the hamiltonian changes (it is a time-dependent hamiltonian, because some potential is switching on at t = t0), all these different hermite functions get DIFFERENT omegas (namely E_n / hbar).

cheers,
Patrick.

Vanesch,again you have not chosen to comment on the x part---do you accept that it won't remain as such,it would change?

In any case my point was that the energy of the particle changes when it is trapped by the nucleus and its wavefunction also changes to be that correponding to the introduced potential---this is a point you were not accepting earlier.

 

[vanesch]

Your expression (1) above is not right.There is another term i \hbar U^\dagger \dot U which you have missed out.Let's start afresh.Say your hamiltonian is
H(t) = H_0 + \Theta(t-t_0)V(x)

U(t) = \exp\frac{-i \int_0^t [{H_0 + \Theta(t^\prime-t_0)V(x)}]dt^\prime}{\hbar} = \exp(-iH_0 t/ \hbar) \exp[-i V(x)(t-t_0)/ \hbar]

?

If H0 and V don't commute, you can't do that ! What you write is not the general solution to the operator differential equation. Check out "Dyson series".

Now x(t) is something like this:-

x(t) = \exp[-i/\hbar(...)] x(0) \exp[i/\hbar(...)])]+i\hbar U^\dagger(t).\dot U (t),

I have no idea where this comes from.

If U(t) (let us take ta = 0) is the time evolution operator (solution of the Schroedinger equation), then the Heisenberg operator is nothing else but
O_H(t) = U(t)_dagger O_S U(t).

You can check this easily: O_H defined in this way satisfies the Heisenberg evolution equation
i hbar d O_H/dt = [O_H(t), H(t)]

So where does this extra term you give, come from ?
(I think I put the dagger on the wrong side in my previous post)

cheers,
Patrick.

 

[vanesch]

Vanesch,again you have not chosen to comment on the x part---do you accept that it won't remain as such,it would change?

Of course it will change, that was not the point. You wanted it to evolve into the ground state, something that won't happen.

In any case my point was that the energy of the particle changes when it is trapped by the nucleus and its wavefunction also changes to be that correponding to the introduced potential---this is a point you were not accepting earlier.

It is difficult to say what you mean by the "energy of the particle" in this situation, and we are making different points simultaneously, which may lead to confusion.

The point I tried to make was, that if you treat EVERYTHING quantum-mechanically, including the measurement apparatus, the environment, your body, etc... then a "measurement" is nothing else but an interaction of the measurement apparatus with the system, and simply leads to the overall wavefunction of the entire system to be a non-product state of the system, and the apparatus and all the rest, so there is no MEANING attached anymore to "the wavefunction of the apparatus" or "the wavefunction of the system": the overall wavefunction is not a product state anymore which would be necessary to talk about the individual wavefunctions of the subsystems. There is no escaping of this, when you consider all these interactions to be described by a hermitean hamiltonian, which automatically leads to a unitary evolution operator.

However, you can also work semi-classically, by considering the measurement apparatus and everything classically, and imposing a wavefunction onto the system (this is how people use quantum theory in practice, and corresponds to the Copenhagen view of things). Mind you, this means that part of the universe DOES NOT FOLLOW THE LAWS OF QUANTUM THEORY.
Now, as the measurement apparatus has a classical description (no hilbert space, but a phase space) and the quantum system has a quantum description (a hilbert space, no phase space), the interaction between both needs to go through a "bridge". This bridge is THE POTENTIAL you introduce in the hamiltonian of the quantum system, and THE BORN RULE and PROJECTION postulate, which then introduces an effective potential in the classical equations of motion of the measurement system.

I tried to point out that the introduction of the potential alone is not sufficient to make a system "fall to its ground state" in the cases you mentionned. However, there is some on-going work in looking how coupled classical-quantum systems, with added noise, CAN do this. Nevertheless, this approach NECESSARILY places you outside of quantum theory: part of the universe is to be considered NOT subject to quantum mechanics. So even if this work is successfull, it doesn't contradict my statement that WITHIN QM, as it stands, you can never obtain this.


It is the essence of the Copenhagen view, to which I do not subscribe AS SUCH. Of course, in practice, everybody who uses QM works with it this way. But I find it fundamentally disturbing that quantum mechanics somehow stops to be valid, and classical theory takes over, and that's it. I'd rather see then both theories as limiting cases of a more fundamental theory.

cheers,
Patrick.

 

[gptejms]

If H0 and V don't commute, you can't do that ! What you write is not the general solution to the operator differential equation. Check out "Dyson series".

That's right.I am being a bit sloppy here,but the idea was to show that x(t) is significantly altered by the introduction of a potential--thus the path of an electron through a double slit arrangement would be altered and may be reduced to be 'through one slit' by a suitable potential(even in the Heisenberg picture).I agree this is easier said than done,but you get the basic idea.

I have no idea where this comes from.

If U(t) (let us take ta = 0) is the time evolution operator (solution of the Schroedinger equation), then the Heisenberg operator is nothing else but
O_H(t) = U(t)_dagger O_S U(t).

You can check this easily: O_H defined in this way satisfies the Heisenberg evolution equation
i hbar d O_H/dt = [O_H(t), H(t)]

So where does this extra term you give, come from ?
(I think I put the dagger on the wrong side in my previous post)

The formula I have given you is right.I don't have the books with me(I left physics 9 years back as soon as I submitted my Ph.D. thesis)--but I remember it's there in Merzbacher's QM.

 

[seratend]

That's right.I am being a bit sloppy here,but the idea was to show that x(t) is significantly altered by the introduction of a potential--thus the path of an electron through a double slit arrangement would be altered and may be reduced to be 'through one slit' by a suitable potential(even in the Heisenberg picture).I agree this is easier said than done,but you get the basic idea.

I think I do not understand you at all. Double slit experiment is the plain old toy model used to explain why we cannot have the two properties: "electron passes through one slit" and "interference pattern on the screen".
Therefore, I do not see how you can reduce the "path" of an electron to the one through one slit and still have the interference pattern.

Seratend.

 

[Igor_S]

I think you can, according to Bohm's interpretation. Interference pattern is created by introduction of some (really) strange quantum potential - which (I think) contains stohastic part. I know very little about it and personally don't like Bohm's interpretation. If I remember correctly he claims that 1s electron in hydrogen atom is not moving at all; he is just standing there (not sure how is missing el. dipole moment is then explained, but I don't want to go offtopic).

However if you dismiss existence of quantum potential, then you surely cannot have electron passing through one particular hole and have interference (as you said).

 

[gptejms]

I think I do not understand you at all. Double slit experiment is the plain old toy model used to explain why we cannot have the two properties: "electron passes through one slit" and "interference pattern on the screen".
Therefore, I do not see how you can reduce the "path" of an electron to the one through one slit and still have the interference pattern.

Seratend.

My model is this:-a measurement process is like the introduction of a potential.In absence of the potential(i.e. when we do not make a measurement) of course we see an interference pattern and the wavefunction of the electron(or whatever) spans both the slits.This broad wavefunction gets affected when we introduce the potential--what I am proposing here is that the potential is such that the wavefunction shrinks to span only one slit(of course the interference pattern is destroyed).The question being discussed is if this model is right or not.

 

[vanesch]

The formula I have given you is right.I don't have the books with me(I left physics 9 years back as soon as I submitted my Ph.D. thesis)--but I remember it's there in Merzbacher's QM.

I'm sorry but it just doesn't make sense.
i hbar U-dagger dU/dt = U-dagger (i hbar dU/dt) = U-dagger H U.
so this means that you add to each observable in the Heisenberg picture, U-dagger H U. Now let us consider a trivial observable, namely I (the unit operator). It simply means that any state is an eigenvector of I with eigenvalue 1. As such, it is a good observable. Now, clearly, no matter what happens to the system, I will always measure "1" with this trivial observable.

But with your formula, its Heisenberg representation (which should also just be I) is given by:

I_H = U-dagger I U + U-dagger H U
= I + U-dagger H U

Clearly, for many states now, the expectation value of I_H, given by:

<psi_H | I_H | psi_H> = <psi_H | I | psi_H> + <psi_H | U-dagger H U | psi_H>
= 1 + <psi_S|H|psi_S>
= 1 + <H>

This cannot be: the expectation value in the Heisenberg picture cannot be different from 1.

cheers,
Patrick.

 

[vanesch]

My model is this:-a measurement process is like the introduction of a potential.In absence of the potential(i.e. when we do not make a measurement) of course we see an interference pattern and the wavefunction of the electron(or whatever) spans both the slits.This broad wavefunction gets affected when we introduce the potential--what I am proposing here is that the potential is such that the wavefunction shrinks to span only one slit(of course the interference pattern is destroyed).The question being discussed is if this model is right or not.

You can do this, but only if you allow for a stochastic potential. Some people work on this.

cheers,
Patrick.

 

[seratend]

I think you can, according to Bohm's interpretation. Interference pattern is created by introduction of some (really) strange quantum potential - which (I think) contains stohastic part. I know very little about it and personally don't like Bohm's interpretation. If I remember correctly he claims that 1s electron in hydrogen atom is not moving at all; he is just standing there (not sure how is missing el. dipole moment is then explained, but I don't want to go offtopic).

However if you dismiss existence of quantum potential, then you surely cannot have electron passing through one particular hole and have interference (as you said).

Currently, the only consistent bohmian interpretation I know may be reduced to "when you know result, e.g. a position at time t, you may infer the position at any time t<to if this position is not measured during the experiment. The rest is a play of words.
Yes, you can say that when the electron hits the screen it has crossed 10 times all the universe and passed through a given slit as long it is not measured or the electron is a human being before it is measured as long as these propositions are compatible with QM results.
If this discussion concerns physically non verifiable properties, I have nothing to say. However one who uses such description must first prove it is consistent with QM formalism.
Bohmian mechanics is a perfect example of this problem: as long as we do not require a reality of the bohmian particle, everything is ok. When we require some interpretation, well consistency is no more guaranteed: since its first publication in 1952, the interpretation has continuously evolved in order to keep the consistency with the results issued by the QM formalism.

Seratend.

 

[seratend]

My model is this:-a measurement process is like the introduction of a potential.In absence of the potential(i.e. when we do not make a measurement) of course we see an interference pattern and the wavefunction of the electron(or whatever) spans both the slits.This broad wavefunction gets affected when we introduce the potential--what I am proposing here is that the potential is such that the wavefunction shrinks to span only one slit(of course the interference pattern is destroyed).The question being discussed is if this model is right or not.

For the slit plate of the double slit experiment, you may say you have the measurement given by this projector: P_slit= |slit1><slit1|+|slit2><slit2| and you still have the interference pattern that is given by the projector P_int=|interference><interference|.
These 2 projectors do not require the introduction of additional potentials to the ones already present (the interaction of the plate and of the screen). However, I may say: I have the results P_slit and P_int for each electron or P_int or P_slit.

With QM formalism, you are always able to do that: a clear separation between the [quantum] interactions of the system (the unitary evolution) and the measurement results (i.e. the projectors). To break the unitary evolution [of a closed system], you need to define a new theory that is no more QM theory: that's what Patrick tries to tell you.

Seratend.

 

[ZapperZ]

My model is this:-a measurement process is like the introduction of a potential.In absence of the potential(i.e. when we do not make a measurement) of course we see an interference pattern and the wavefunction of the electron(or whatever) spans both the slits.This broad wavefunction gets affected when we introduce the potential--what I am proposing here is that the potential is such that the wavefunction shrinks to span only one slit(of course the interference pattern is destroyed).The question being discussed is if this model is right or not.

That question in your last statement cannot be answered until you explicitly describe the nature of your potential. Just throwing out the word "potential" is meaningless. Does this potential have a profile? Or is it a delta function? Until you describe such things, this "model" remains vague.

Zz.

 

[ZapperZ]

It is the essence of the Copenhagen view, to which I do not subscribe AS SUCH. Of course, in practice, everybody who uses QM works with it this way. But I find it fundamentally disturbing that quantum mechanics somehow stops to be valid, and classical theory takes over, and that's it. I'd rather see then both theories as limiting cases of a more fundamental theory.

I should have just let this slip by, but I can't help myself. :)

Why do you find this "disturbing"? Discontinuity in understanding and discontinuity in description happens all the time. At a first order phase transition, practically all bets are off. What you used to describe in one phase simply cannot be extrapolated into the other phase. There's an abrupt discontinuity in a number of state variables. So you switch gears and adopt a different set of description and properties.

My point here is that if you live by a principle where you are disturbed by this abrupt change, then you should also be disturbed by many more phenomena than just QM-classical boundary. So are you? :)

Zz.

 

[gptejms]

I'm sorry but it just doesn't make sense.

ok,my formula applies to the transformation of a hamiltonian under a time dependent unitary transformation.

 

[gptejms]

QFT and the measurement problem

You can do this, but only if you allow for a stochastic potential. Some people work on this.

Why stochastic?Can you give the references?

I am now throwing in another idea.In terms of QFT:-the principle of microscopic causality tells you that the measurement of the field at place 1 and subsequently at place 2(time-like separated) doesen't give you the same result as measurement at place 2 first and then at place 1(non-commutation).What this implies is that when I make a measurement at place 1,I create a disturbance which travels at the speed of light and the whole field readjusts to some new values.This is precisely what is happening in the 2-slit experiment(or for that matter any other experiment)--you measure the field or find an electron at slit 1,the field readjusts at slit 2 to no electron(if one electron comes at a time) and the field at the screen also readjusts to 'no interference pattern'.

 

[vanesch]

My point here is that if you live by a principle where you are disturbed by this abrupt change, then you should also be disturbed by many more phenomena than just QM-classical boundary. So are you? :)

What is disturbing in the QM-classical boundary is that it is "slippery". It is not defined, and you cannot deduce its location (at least, according to standard QM - I'm not talking about speculative extensions). If QM, in its unitary version, applies "all the way up", it should be *in principle* possible to UNDO about just any measurement result, and have "notebooks, people and all that" interfere quantummechanically. For the moment, we are technologically only able to do that for systems of 2 or 3 photons or so.
So wherever you put physically the QM - classical boundary, it is *in principle* possible, if QM applies all the way up, to transgress it, which would then show that the QM-classical boundary is not yet reached.

What I mean is the following:

If you have a system S, in the famous state a|1> + b|2>, and you have a measurement apparatus M and we do the "pre-measurement" interaction thing a la von Neumann, we end up with
|psi> = a |1> |M1> + b |2> |M2>

Now, as long as M1 and M2 are very complicated states, which are going to stay essentially orthogonal (so that the entanglement is "for ever"), if you take a "relative observer position" to be the measurement apparatus, then you can say that the measurement apparatus has "irreversibly" recorded the result.
But if we are technologically advanced enough to subject this complicated apparatus M to such an evolution that at a certain time M1 and M2 evolve towards the same state MX (this is unitarily possible, because we can "compensate" on the system: it is in essence the INVERSE measurement interaction), then we arrive back at |psi> = (a |1'> + b|2'>) |MX>. We could now perform an experiment on the system S IN ANOTHER BASIS, and observe quantum interference.
This is the essence of a quantum erasure experiment (although all papers about it formulate it in a much more spooky way about "forgetting" and "not violating the Heisenberg uncertainty relations" etc...).
We can do it with photons and even with more sophisiticated stuff.
If we would have thought that the interaction of M with the system was a "measurement" (was the "phase transition" between quantum and classical), we would have been proven wrong, because this quantum erasure could not happen anymore. In density matrix language: once the non-diagonal elements have been set to 0 (the quantum-classical transition: the Born rule), there's no way to put them back.
There is *in principle* no indication in current quantum theory that M cannot include the laboratory, the experimenter, the solar system, the galaxy etc... (except, except... for gravity - but I stick to non-speculative, current, quantum theory). So you can push back this "quantum-classical" transition in principle to beyond "M" systems the size of our galaxy.
To do that experimentally of course would require a mind-bogglingly sophisticated technology, but there is nothing in current quantum theory (except for our ignorance of how to treat gravity) that indicates that it cannot be done in principle.
Once you've pushed back the QM-classical transition far beyond the scale of humans and laboratories, it doesn't matter anymore: we clearly have then bodystates in a superposition, of which we consciously only observe ONE term.
This infinite possible "pushing back" is not present in the thermodynamical examples of phase transitions.

cheers,
Patrick.

 

[vanesch]

ok,my formula applies to the transformation of a hamiltonian under a time dependent unitary transformation.

Can't be !
Let us have for simplicity a time-independent Hamiltonian (it should be valid for that case too, right ?)

So you claim that:

H_H = U_dagger H_S U + U_dagger H_S U

(because we still have: hbar U-dagger dU/dt = U-dagger (i hbar dU/dt) = U-dagger H U)

So H_H = 2 U_dagger H_S U

However, in the case of a time-independent hamiltonian, you can easily find out that U = exp(- i hbar H_S), which commutes of course with H_S

So we find that H_H = 2 H_S ?

Expectation values of energy in the Heisenberg picture are twice as big as those in the Schroedinger picture ??

cheers,
Patrick.

 

[vanesch]

Why stochastic?Can you give the references?

Have a look at "decoherence and the appearance of a classical world in quantum theory" (by Joos, Zeh et al). The last 3 chapters (which I haven't studied in detail myself) are: 7. Open quantum systems (by Kupsch), 8. Stochastic collapse models (by Stamatescu) and 9. Related concepts and methods (by Zeh).
Look for the model of Ghirardi, Rimini and Weber (non-hamiltonian evolution) or Barcheilli, Lanz and Prosperi who build a model with "continuous approximate position measurement".

But I'm not very knowledgeable of all this stuff - I don't like the approach: it is introducing small "fudge terms" in the Schroedinger equation which aren't there in standard QM, so that the evolution is non-unitary, and then trying to find values for the free parameters so that at small scales, the effect of the terms is neglegible, and at large scales, it introduces a projection. It is not based upon some grand a-priori principle, and that's what I don't like about it.

you measure the field or find an electron at slit 1,the field readjusts at slit 2 to no electron(if one electron comes at a time) and the field at the screen also readjusts to 'no interference pattern'.

This won't work in all generality: you need faster-than-light propagation to do so. Ok, Bohm's theory does exactly this, but it has instantaneous action at a distance terms in it.

cheers,
Patrick.

 

[gptejms]

Can't be !
Let us have for simplicity a time-independent Hamiltonian (it should be valid for that case too, right ?)

So you claim that:

H_H = U_dagger H_S U + U_dagger H_S U

(because we still have: hbar U-dagger dU/dt = U-dagger (i hbar dU/dt) = U-dagger H U)

So H_H = 2 U_dagger H_S U

However, in the case of a time-independent hamiltonian, you can easily find out that U = exp(- i hbar H_S), which commutes of course with H_S

So we find that H_H = 2 H_S ?

Expectation values of energy in the Heisenberg picture are twice as big as those in the Schroedinger picture ??

cheers,
Patrick.

Why don't you use tex?The math notation you write is not so readable--e.g. what is U-dagger?

Since you are insisting on knowing the origin of this formula,I have worked it out.No,you are not in the Heisenberg picture(my mistake if I gave that impression) here--make a transformation |\psi^\prime(t)&gt; =U(t)|\psi (t)\rangle
Differentiating the above(plus using the Schrodinger equation) you get
H^\prime(t) = U(t)H(t)U^\dagger(t) + i \hbar U^\dagger(t)\dot U(t)

 

[gptejms]

This won't work in all generality: you need faster-than-light propagation to do so. Ok, Bohm's theory does exactly this, but it has instantaneous action at a distance terms in it.

Why do you need ftl propagation?The field readjusts after the measurement at one place at the speed of light i.e. the disturbance is communicated to other places at the speed of light--for typical experimental dimensions this happens very very fast.

 

[vanesch]

Why don't you use tex?The math notation you write is not so readable--e.g. what is U-dagger?

I used to write in pure ASCII, from the good old days when news forums were pure ascii :-)

Since you are insisting on knowing the origin of this formula,I have worked it out.No,you are not in the Heisenberg picture(my mistake if I gave that impression) here--make a transformation |\psi^\prime(t)&gt; =U(t)|\psi (t)\rangle
Differentiating the above(plus using the Schrodinger equation) you get
H^\prime(t) = U(t)H(t)U^\dagger(t) + i \hbar U^\dagger(t)\dot U(t)

This formula is right. But that's not what you were saying ! You were claiming (post 78) that in my transformation from a Schroedinger picture operator into its Heisenberg equivalent, I missed a term, when I wrote:

A_H(t) = U^\dagger(t) A_S U(t)

The above formula is part of the definition of the Heisenberg picture, so I don't see how "a term can be missing".

Now, i \hbar dU/dt = H_S(t), it is the DEFINITION of the Hamiltonian (see chapter 2 of Sakurai for instance).

From these two equations follows:
i \hbar d A_H(t) / dt = [A_H(t), H_H(t)]

with H_H the Hamiltonian in the Heisenberg picture, and given by
H_H(t) = U^\dagger(t) H_S(t) U(t)

If all H_S(t) commute for all values of t, then H_S commutes with U and hence H_H = H_S, but this is not the case in all generality.

But I didn't need this point at all. I just wanted to show that a unitary evolution operator U exists, whether you switch on a potential or not, that this allows you to go to the Heisenberg picture, and that there, by definition, the state of the system is given by a constant ket. Hence, switching on a potential doesn't affect the constant ket at all.

 

[vanesch]

Why do you need ftl propagation?

EPR situations require this, for instance. But you do not need to set up things so complicated: imagine a light beam, hitting a beam splitter, and the two resulting beams are diverging. When they are at a reasonable distance of a few meters, say, you make the thing such that you can, or cannot, "watch the photon" (say, by guiding the beam in a local detector or not) in each arm.
Next, you let the beams (with mirrors and so on) come together again, and form an interference pattern.
If you switch the arm detectors quickly enough "in and out" the beam, there's not enough time to propagate from one arm to the other to go and tell what you did. We're in the measurable 10 ns range here for a few meters distance.

cheers,
Patrick.