Question on observer created reality

[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.