Question on observer created reality

[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 $$|\psi^\prime(t)> =U(t)|\psi (t)\rangle$$
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 $$|\psi^\prime(t)> =U(t)|\psi (t)\rangle$$
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.

 

[gptejms]

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.

It will be nice if you can do this exercise:-show that the atom radiates in the Heisenberg picture-should be possible.
You have so far not commented on the x-part of the plane wave.

 

[gptejms]

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.

Has an experiment of above nature been done?I find this idea so appealing--do a measurement at one place and the quantum field readjusts everywhere else within some time dictated by the velocity of light(and this arises naturally out of the commutation properties of the quantum field(microscopic causality))--I don't feel like giving it up!But yeah EPR is non-local,there may be other non-local experiments too--are QM(non-rel.) and QFT in contradiction here..?Is the principle of microscopic causality being violated in such experiments..??

 

[gptejms]

But yeah EPR is non-local,there may be other non-local experiments too--are QM(non-rel.) and QFT in contradiction here..?Is the principle of microscopic causality being violated in such experiments..??

Answering my own question--in EPR situations you are just measuring the polarization--so the principle of microscopic causality doesen't come in in any way.The other experiment where you detect a photon in one of the arms and there is not enough time for the other arm to know what you did--well you have made a measurement of the field at one of the arms--now your contention is that the message can't go fast enough to the other arm to readjust to 'no photon'--but that's not required.After all you look at the screen,see that the interference pattern is destroyed and then conclude positively that the photon passed through one arm only--for the message to pass to the screen to readjust the field to 'no interference pattern',there is enough time.

 

[vanesch]

It will be nice if you can do this exercise:-show that the atom radiates in the Heisenberg picture-should be possible.

It surely is very difficult if you want the full treatment in QFT, for many reasons. The physical picture behind it is the the E-fields in the right transition mode(s) in QFT are non-zero even in vacuum (the expectation values of the squares of the amplitudes of these modes are not zero), and it is the interaction of the electon of the atom with these modes that makes that the stationary states of the atom when calculated, taking purely into account the coulomb interaction with the nucleus, are NOT stationary states anymore when taking into account the dynamical EM field (as a quantum field). As such, the atom will get entangled with the EM field, and the transition amplitude for doing so can be calculated perturbatively, but in all these calculations, approximations are made. Even this treatment is quite complicated, and I'm not aware of a "frontal attack" of the problem in full, non-perturbative QED.
Nevertheless, there ARE calculations of the rate of decay of excited atoms following the idea I outlined very roughly above, and as far as I know, they fit with observations. I don't have references handy as such, but some stuff on this must be found in Mandel and Wolf in their "bible" on quantum optics, I'm sure.

You have so far not commented on the x-part of the plane wave.

I did, in post 81.

Incoming "plane wave" for t<t0 (free hamiltonian)

At t1 < t0, the incoming wave is exp(i k x), so solving for the time evolution in the Schroedinger picture with the free hamiltonian, will give us a time dependence which is:

exp(i k x) exp(- i omega t)
up to t = t0.
Note that for the free hamiltonian, the plane wave exp(i k x) is a stationary state.

So at t = t0, the wavefunction is exp(i k x - i omega t0).
Now, the potential is switched on, but this doesn't immediately change the wavefunction which is supposed to evolve continuously.

So simply said, for t>t0, we have to solve the Schroedinger equation with the new hamiltonian, with as a starting wave function, exp(i k x - i omega t0).
But note now that this function of x is not a stationary state anymore, so this time, the time evolution will not simply add a phase factor.
A way to solve this, is to expand exp(i k x - i omega t0) in the Hermite-Gauss functions (the new stationary states), so you will find:

exp(i k x - i omega t0) = a H0(x) + b H1(x) + c H2(x) + ...
(where H0, H1, etc... are the new stationary states)

The time evolution will now do the following thing:

at t = t2 > t0, we have that the wave function equals:

psi(x) = a H0(x) exp(-i E0 t2) + b H1(x) exp(- i E1 t2) + c H2(x) exp(- i E2 t2) ...

This will of course now not take on the aspect any more of a plane wave.

cheers,
Patrick.

 

[vanesch]

are QM(non-rel.) and QFT in contradiction here..?

No, not at all, but even experimented particle physicists let themselves mislead by this apparent conflict, and that's because they are less used to make the distinction between the unitary dynamics (say, the properties of the hamiltonian), and quantum superposition.
The unitary dynamics in QFT satisfies the SR light cone conditions, in that observables (which are function of spacetime parameters) which are on spacelike intervals, commute. This prevents a DYNAMICAL interaction that is FTL.
However, nothing stops you from considering superpositions of states which ENTANGLE parts of the fields which are spacelike separated. As such, there can be no DYNAMICAL influence, but all the EPR like stuff still remains valid in QFT. But QFT is so complicated already with all Feynman integrals, renormalization schemes and so on, that these situations are rarely considered. Moreover, in typical particle physics experiments, one works from the beginning to the end in the momentum basis, so there are usually no "quantum interference" experiments (you need to change basis to observe quantum interference) done in that domain ; so QFT as usually practiced, usually "looks" very classical. But the whole machinery is there, even if not used !

I'll try to think up a simple "analogy".
Consider the free non-relativistic hamiltonian of 2 particles. We could think of an equivalent of "relativity" that forbids us to have dynamical interaction between the two particles (instead of between spacelike separated parts of quantum fields). This would be formalized by imposing commutation between all observables that relate to particle 1 and all observables that relate to particle 2 in the Heisenberg picture. So as long as we work with product states (in the Schroedinger picture) we remain in product states.
But this doesn't stop us from considering entangled states ! And then suddenly we do find EPR like correlations between both. Is this incompatible with our imposed absence of dynamical interactions between both ?
No, it is just a property of the quantum theory.
In the same way, the absence of dynamics between different, spacelike separated parts of a quantum field do not forbid you to have them entangled, and as such find EPR correlations between them.

cheers,
Patrick.