Are clicks proof of single photons?

[harrylin]

No, QM is not that kind of theory. Even if you know your input observables as well as possible QM allows you to predict the output observables only probabilistically.

That is what I would call a yes, although you said "no". Or do you claim that probability theory makes no predictions about the results of throwing dice?

[..] QM does not explain this random behavior of nature. [..]

Exactly.

No other theory can explain this.

That claim is not a postulate of QM, or is it?

 

[meopemuk]

Or do you claim that probability theory makes no predictions about the results of throwing dice?

The probability theory predicts that in 50% of cases you'll see head up and in 50% of cases you'll see tail up. Similarly, QM predicts that a polarized photon will pass through the filter in 50% of cases and not pass in other 50% of cases. If you call this "prediction of output observables based on input observables", then yes, we agree.

The important thing is that both theories cannot tell you what will be the outcome in each individual case. So, there are some aspects of nature, which are totally unpredictable. That was my point.

Eugene.

 

[strangerep]

Please remind me what was the field theory explanation for the interference experiment
with atomic deposition? In particular, how the atomic "field" collapses to a specific position
on the substrate? Isn't it the same dreadful Copenhagen collapse?

I hope in this explanation you [Arnold] wouldn't invoke the quantum nature of the
substrate as M&W did in their Chapter 9.

I don't see anything fundamentally wrong with a similar calculation as in M&W ch9.

For the benefit of other readers, I'll quickly review sections 9.2 and 9.3 of Mandel & Wolf:

M&W work in the interaction picture, with the Schroedinger equation

$$\partial_t \; |\psi(t)\rangle ~=~ \frac{1}{i\hbar} \, \hat{H}_I(t) \, |\psi(t)\rangle ~~~~~~~~(9.2-1)$$

where $$|\psi(t)\rangle$$ is the state of an electron in the detector,
and $$\hat{H}_I(t)$$ is the interaction part of the Hamiltonian.
For such a quantized detector interacting with a classical EM field, the latter is

$$\hat{H}_I(t) ~=~ - \frac{e}{m}\, \hat{p}_i(t) \hat{A}^i(r,t) ~. ~~~~~~~~(9.2-3)$$

where $$\hat{p}_i(t)$$ is the electron momentum,
$$\hat{A}_i(r,t)$$ is the (classical, c-number) EM vector potential,
and the electron is initially in a tightly bound state at position r.

By standard techniques, equation (9.2-1) is formally integrated
iteratively (as usual for Volterra-type integral equations) to obtain:

$$|\psi(t)\rangle ~=~ |\psi(t_0)\rangle ~+~ \frac{1}{i\hbar} \int_{t_0}^t dt_1 \hat{H}_I(t_1) \, |\psi(t_0)\rangle ~+~ \mbox{(higher order terms in } \hat{H}_I) \dots ~~~~~~~~(9.2-8)$$

In the situation being considered, truncation of the higher order terms
is acceptable. The transition probability from an initial state
$$|\psi(t_0)\rangle$$ to some new state $$\Phi$$ at time t
which is orthogonal to $$|\psi(t_0)\rangle$$ is then approximately:

$$\left| \mbox{Transition probability} \right| ~=~ \left| \langle \Phi | \psi(t)\rangle \right|^2 ~=~ \frac{1}{\hbar^2} \left| \int_{t_0}^t dt_1 \langle \Phi | \hat{H}_I(t_1) |\psi(t_0)\rangle \right|^2 ~~~~~~~~(9.2-9)$$

Since we're working in the interaction picture, and the EM field is
a c-number here, the interaction part of the Hamiltonian at $$t_1$$
can be expressed in terms of an arbitrary initial time $$t_0$$ as

$$\hat{H}_I(t) ~=~ - \frac{e}{m} ~ e^{i H_0 \Delta t/\hbar} \, \hat{p}_i(t_0) \, e^{-i H_0 \Delta t/\hbar} ~ \hat{A}^i(r,t)$$

where $$\Delta t := t_1 - t_0$$.

With an extra (analytic signal representation) assumption about the
incident EM field, this is sufficient information to evaluate formula
(9.2-9) above, giving the probability that a transition occurs (i.e., a
photoelectron is produced) within a small time interval $$\Delta t$$.

In summary, one calculates the probability from an initial product state
consisting of quantized bound electron and incident EM field to a final
state in which the electron has been excited into the conduction band
under the interaction $$\hat{H}_I(t)$$.

The important point from the above is that similar computations
could be done for other kinds of interactions, involving some field
incident on a many-body plate, with other possibilities for the final
state(s). For atom deposition, we could consider an field of atom type
"I" incident upon a lattice of atoms of type "L", where the final state
consists of a new bound state between a "L" atom and an "I" atom.
All that matters for the purposes of this thread is that a different final
state be possible, resulting from an interaction. Though the numbers
may differ, we still get the result that there's a certain finite transition
probability for an atom deposition at any particular point in a finite time.

The only difference from the photodetection case is that we must
think of the incident atom beam as a field, not a collection of
particles.

In any case, more accurate results are given by also quantizing the incident
field (c.f. M&W ch14) -- we still get atoms being deposited at random
places according to a transition probability per unit time.

Minimal QM (without the extra baggage of a "collapse" interpretation)
thus seems quite adequate.

 

[meopemuk]

Minimal QM (without the extra baggage of a "collapse" interpretation)
thus seems quite adequate.

strangerep,

what do you mean by "Minimal QM (without the extra baggage of a "collapse" interpretation)"? QM cannot be formulated without a collapse in some form. Once you started talking about Hilbert spaces and superpositions of states you have tacitly assumed the presence of a collapse.

The weirdness of the M&W model is that it places the collapse in an unusual place. Namely, in their model the collapse occurs when the photoelectron jumps to a continuous spectrum state. The probability of this jump is described by a square of a certain time-dependent wave function. This is exactly what is called "collapse" in quantum mechanics.

In the usual treatment of the double-slit experiment (see Feynman) the projectile is treated with a QM wave function. This treatment gives a full explanation of the shape of the interference picture and of the collapse effect. There is no need to build a quantum-mechanical model of the screen.

Eugene.

 

[strangerep]

what do you mean by "Minimal QM (without the extra baggage of a "collapse" interpretation)"?

I mean QM with an absolutely minimal amount of interpretation layered on top.
In particular, I have in mind the statistical interpretation as described by
Ballentine in his textbook and papers.

QM cannot be formulated without a collapse in some form. Once you started talking
about Hilbert spaces and superpositions of states you have tacitly assumed the
presence of a collapse.

Ballentine's statistical interpretation shows that this is not correct.

The weirdness of the M&W model is that it places the collapse in an unusual place. Namely, in their model the collapse occurs when the photoelectron jumps to a continuous spectrum state. The probability of this jump is described by a square of a certain time-dependent wave function. This is exactly what is called "collapse" in quantum mechanics.

I no longer see a need to talk about collapse at all, but only probability distributions.

Although the notion of collapse makes some semi-heuristic treatments of
certain situations easier to "explain", it comes at the cost of undesirable
features/paradoxes such as Schroedinger's cat.

In the usual treatment of the double-slit experiment (see Feynman) the projectile is
treated with a QM wave function. This treatment gives a full explanation of
the shape of the interference picture and of the collapse effect. There is no
need to build a quantum-mechanical model of the screen.

If one does utilize a simple quantum-mechanical model of the screen,
one no longer needs to postulate a collapse effect in order to account
for the observed phenomena.

 

[meopemuk]

I no longer see a need to talk about collapse at all, but only probability distributions.

Perhaps I am using a non-standard terminology, but for me "probability distribution" and "collapse" is the same thing. When you are talking about an ensemble you can say "probability distribution". When you are talking about individual event there is a "collapse". I didn't read Ballentine lately, but, as far as I remember, he didn't contradict this view.

Although the notion of collapse makes some semi-heuristic treatments of
certain situations easier to "explain", it comes at the cost of undesirable
features/paradoxes such as Schroedinger's cat.

If you follow my advice not to make statements about non-observable quantum things, then Schroedinger's cat will not pose any paradox.

If one does utilize a simple quantum-mechanical model of the screen,
one no longer needs to postulate a collapse effect in order to account
for the observed phenomena.

When M&W considered the quantum mechanical model of the screen they simply shifted the boundary between the quantum system and the measuring apparatus. Previously (e.g., in Feynman's approach), the screen played the role of the measuring apparatus. Now this role is played by the detector, which counts emitted photo-electrons. So, the collapse is still there, though not mentioned by name. The collapse occurs when the electron detector clicks.

Somewhere along the way M&W forgot that the incident photon should also have a quantum description. They've substituted photon's wave function with a classical field, which mimics quantum interference properties. This is OK as a rough approximation for the full quantum mechanical description of this experiment.

Eugene.

 

[A. Neumaier]

The weirdness of the M&W model is that it places the collapse in an unusual place. Namely, in their model the collapse occurs when the photoelectron jumps to a continuous spectrum state.

There is nothing weird about the M&W setting. It is the _standard_ view in quantum optics. They place the collapse precisely where it belongs: Since it only happens in a measurement, it must be due to the interaction with the measurement device.

The probability of this jump is described by a square of a certain time-dependent wave function. This is exactly what is called "collapse" in quantum mechanics.

M&W assume that an electron is actually ejected according to the probabilities computed by the QM formulas. This is assumed in any interpretation of quantum mechanics, including the most weird ones. In particular also by no-collapse interpretations such as MWI or consistent histories.

In the usual treatment of the double-slit experiment (see Feynman) the projectile is treated with a QM wave function. This treatment gives a full explanation of the shape of the interference picture and of the collapse effect. There is no need to build a quantum-mechanical model of the screen.

The quantum mechanical model is needed to explain the photoelectric effect. It explains why a photographic plate acts as a measurement device while a candle light doesn't.

While the usual treatment explains nothing but simply _postulates_ that hitting the screen produces an event.

 

[A. Neumaier]

$$\hat{p}_i(t)$$ is the electron momentum,
$$\hat{A}_i(r,t)$$ is the (classical, c-number) EM vector potential,

and the case of an incident quantum field would be very similar, having here the vector potential operator. To a first approximation, this can be replaced by its expectation value, which is why the treatment as a classical field is often adequate. For a nonclassical incident field, one gets minor corrections; qualitatively nothing changes but quantitatively one has the usual O(hbar) differences.

In summary, one calculates the probability from an initial product state
consisting of quantized bound electron and incident EM field to a final
state in which the electron has been excited into the conduction band
under the interaction $$\hat{H}_I(t)$$.

The important point from the above is that similar computations
could be done for other kinds of interactions, involving some field
incident on a many-body plate, with other possibilities for the final
state(s). For atom deposition, we could consider an field of atom type
"I" incident upon a lattice of atoms of type "L", where the final state
consists of a new bound state between a "L" atom and an "I" atom.
All that matters for the purposes of this thread is that a different final
state be possible, resulting from an interaction. Though the numbers
may differ, we still get the result that there's a certain finite transition
probability for an atom deposition at any particular point in a finite time.

The only difference from the photodetection case is that we must
think of the incident atom beam as a field, not a collection of
particles.

In any case, more accurate results are given by also quantizing the incident
field (c.f. M&W ch14) -- we still get atoms being deposited at random
places according to a transition probability per unit time.

Minimal QM (without the extra baggage of a "collapse" interpretation)
thus seems quite adequate.

Yes. Thanks for the labor of writing out things explicitly!

 

[A. Neumaier]

The probability theory predicts that in 50% of cases you'll see head up and in 50% of cases you'll see tail up. Similarly, QM predicts that a polarized photon will pass through the filter in 50% of cases and not pass in other 50% of cases. If you call this "prediction of output observables based on input observables", then yes, we agree.

The important thing is that both theories cannot tell you what will be the outcome in each individual case. So, there are some aspects of nature, which are totally unpredictable. That was my point.

And you call both collapse. You are the first person I heard about who postulates the collapse of coins upon throwing them.

 

[A. Neumaier]

Yes, the transition between the random quantum regime and the predictable classical regime occurs gradually as N grows.

How should this come about in your interpretation? For N=1 you can say nothing without measuring it. But N times nothing is still nothing. How do you explain that you can gradually say more and more, until when N reached the size of the pointer of a measuring apparatus you can say definitely which position it has?

There is no unique threshold value. Some systems show quantum behavior even for very large N. The usual examples are superconductors and superfluid helium.

Ah. So you want to say we can't predict anything about superconductors and superfluid helium unless we have measured it? But these are used in industrial practice, where people have to rely on things working very predictably even without constantly measuring them!

Considered as a microscopic system, the detector is a highly chaotic dynamical system. Therefore, the individual binding sites for the atomic field behave like random qubits responding to the intensity of the incident atom field, just as the outer electrons in a photodetector behave like random qubits responding to the intensity of the incident electromagnetic field. The response rate is proportional to the incident energy. Once one of the qubits fires, it uses up the whole energy and matter of the atom field, and the choice has been made. No other qubit can fire, by a most likely existing analogue of M&W's formula (14.8-16).

In quantum field theory, causality is therefore as well-behaved as in any classical stochastic process.

I see. So, you are a follower of Einstein's "God does not play dice" camp. So, you are saying that quantum uncertainties result from classical "hidden variables"?

If you read again what you quoted, you'll see no reference to hidden variables.
The quantum dynamics of the detector becomes, in the approximation of a system of qubits coupled to a semiclassical background formed by the remainder of the photographic plate, a quantum-classical system where the classical part is described by a chaotic
dynamics or - with less resolution - by a stochastic process. (Already few-particle quantum systems become chaotic in such a semiclassical description.)

So, if we could entangle all these stochastic chaotic classical processes, we would be able to predict the exact location of the atom deposition or pixel firing in each case. Is it right?

No, since already the classical model involves approximations, and even tiny errors in the model preclude deterministic predictions for a chaotic process.

On the other hand, your position is not pure as you mix classical fields and quantum mechanics in your version of quantum field theory. I find it inconsistent.

Classical fields are only for convenience, since they let one see better what is going on.
As mentioned in a previous post, quantum corrections are small and do not change the classical picture drastically. As long as the field expectation dominates the field fluctuations, one can safely ignore the quantum aspects.

 

[meopemuk]

The quantum mechanical model is needed to explain the photoelectric effect. It explains why a photographic plate acts as a measurement device while a candle light doesn't.

While the usual treatment explains nothing but simply _postulates_ that hitting the screen produces an event.

Yes, the usual (Feynman) treatment is interested in demonstrating quantum properties of the photon. So, the boundary between the quantum system and the measuring apparatus is drawn in such a way that the photon is on the quantum side and is represented by a wave function. The photographic plate is on the "apparatus" side of the boundary. We are basically assuming that the photographic plate serves as an ideal measuring device for position. So, in our quantum 1-photon theory we represent this ideal measuring device by the position *operator*. (Let us not argue now about the absence of a rigorous photon position operator. I could use electron projectiles in my example to avoid your criticism.) Please note that I am not saying (as many other authors do) that the measuring apparatus is a classical object. In this theory the measuring apparatus has a quantum description, but not in terms of a wave function, but in terms of an Hermitian operator of observable -- in this case the observable of position.

This treatment coupled with the collapse postulate allows us to reproduce all significant features of the single-photon interference: the randomness of photon hits and the accumulation of these hits along the constructive interference spots.


Now, you can say that you are not satisfied with the above idealized treatment of the photographic plate, and you want to include the plate in the quantum-mechanical part of the description. Fine, you are definitely allowed to do that. But this doesn't save you from the necessity to use the collapse postulate. You've simply defined the quantum part of the world differently and thus moved the boundary between the quantum system and the measuring apparatus to a different place. Now, your quantum system includes both the incident photon and the photographic plate with all its atoms. This part of the world is described by a wave function. The measuring apparatus in this case could be a detector catching emitted photo-electrons or your eye's retina that registers sunlight reflected from the photographic plate. Just as before, the collapse occurs at the "boundary" between the physical system and the measuring apparatus. This new measuring apparatus is described by a Hermitian operator.

My point is that no matter how exact you want your quantum description to be, you must always have a boundary between the quantum physical system and the measuring apparatus. There is always a collapse happening at this boundary. This separation of the world in two distinct parts is reflected in the formalism of quantum mechanics quite naturally: the quantum system is described by a state vector or a wave function; the measuring apparatus is described by a Hermitian operator of some observable. You need both these mathematical components in order to have a complete quantum-mechanical theory. It it not possible to describe the whole world by one wave function.

Eugene.

 

[meopemuk]

And you call both collapse. You are the first person I heard about who postulates the collapse of coins upon throwing them.

You've probably misunderstood. harrylin said that QM predict output observables based on input observables. I disagreed by saying that QM prediction is only probabilistic. In this sense, QM is not doing a better job at predictions than classical probability theory with coins. There is one important difference, though. When we through a coin, we can in principle predict the result (head or tail) with 100% certainty. This is very difficult, but possible within classical mechanics. On the other hand, when we let a photon to pass through a polarizer filter, there is absolutely no way to predict the outcome. This is one important difference between quantum and classical probabilities.

I've never said that classical coin experiences any kind of collapse.

Eugene.

 

[meopemuk]

No, since already the classical model involves approximations, and even tiny errors in the model preclude deterministic predictions for a chaotic process.

Could you please be more clear? Are you saying that there exists a distinct classical trajectory behind each (seemingly random) quantum event? So, you are saying that this trajectory involves many degrees of freedom, is chaotic, stochastic, etc. and for this reason we cannot make deterministic predictions. However, if we (humans) had unlimited power to control the errors, to observe and calculate things, then we would be able, in principle, to disentangle this seemingly chaotic process and to predict with 100% certainty the place of photon's landing in the double-slit experiment. Is this what you're actually saying, or I am misinterpreting your words?

Eugene.

 

[meopemuk]

...no-collapse interpretations such as MWI or consistent histories.

We disagree about the meaning of "collapse" in quantum mechanics. I would say that MWI and consistent histories need the same collapse as Copenhagen. These models simply hide the fact that they use the collapse. For example, in MWI the collapse occurs when we choose the "right" universe randomly.

In general, every time we use word "probability" we mean "collapse". The collapse does not happen in classical mechanics, because everything is predictable there and probabilities reduce to either 1 (yes) or 0 (no). If sometimes we use classical probabilities, like in coin tossing, we are simply lazy and don't want to bother to specify all necessary conditions exactly.

So, there can be only two legitimate interpretations of quantum mechanics. One is the "hidden variable" interpretation, which basically says that QM is just a branch of classical mechanics, where everything is deterministic and predictable. No probabilities involved and no collapse. The other interpretation is that quantum events are truly random. Then the collapse is needed. There is no third way.

Eugene.

 

[lightarrow]

Just a question. Let's say that I radiate some form of energy into a group of many people. Sometimes, casually, one of those persons feels "excited" because of that energy and jumps for some seconds. Where is the "collapse"?

 

[QuantumClue]

Just a question. Let's say that I radiate some form of energy into a group of many people. Sometimes, casually, one of those persons feels "excited" because of that energy and jumps for some seconds. Where is the "collapse"?

Lightarrow... do we meet again? Are you the man I knew from the Naked Scientists?

To answer your question, if it was possible for some quantity of energy to be tranferred, rather than simply radiated into the body of another person, then collapses may occur if there are decoherences in the stucture of the other person. These simple decoherences are collapse-like state systems.

It's a bit of an odd question, but if you are the man I remember, then it's not a great surprise :)

 

[A. Neumaier]

Now, your quantum system includes both the incident photon and the photographic plate with all its atoms. This part of the world is described by a wave function. The measuring apparatus in this case

is the photomultiplier or the chemical reaction. Both are dissipative processes that proceed locally, and essentially classically, and need because of the large number of particles involved no special treatment beyond the statistical interpretation (without the explicit collapse).

Of course, you could call any statistical element ''collapse'', but this is not the standard way of using the term. If you use your personal terminology, the only result is that nobody understand you anymore.

 

[A. Neumaier]

When we through a coin, we can in principle predict the result (head or tail) with 100% certainty.

No, we cannot, since throwing a coin is a chaotic process, especially when it first lands on the edge. You need an accuracy for the intial data that one is never able to collect.
Not even in principle can a (classical or quantum) object collect enough information about a coin to exactly determine its state. And already tiny uncertainties magnify immensely.

 

[A. Neumaier]

Could you please be more clear? Are you saying that there exists a distinct classical trajectory behind each (seemingly random) quantum event?

No, but that there is an approximate classical description behind each quantum system, and the latter is chaotic, and sufficient to describe the behavior of a photomultiplier or the approach to chemical equilibrium. Nobody models these in a full quantum mechnaical model, since it is a huge waste of effort.

However, if we (humans) had unlimited power to control the errors, to observe and calculate things

I am interested in explaining the world as it is, not a magic version of it.

 

[A. Neumaier]

We disagree about the meaning of "collapse" in quantum mechanics.

I know. I am using the standard meaning, while you mix it up with the simpler version of Born's rule, which makes no statement about what happens to the state in a measurement.

I would say that MWI and consistent histories need the same collapse as Copenhagen. These models simply hide the fact that they use the collapse.

They claim they don't. That's all I meant when citing them. I agree that they don't succeed in that, but this is a different matter.

In any case, you can't redefine the meaning of the word collapse. It means no more and no less than that one pretends to know after a measurement that the system is in an eigenstate corresponding to the measured eigenvalue.

But this is completely irrelevant for being able to read a pointer on an instrument telling that there was a nonzero current produced by the photomultiplier. This process is so macroscopic that it is universally described in terms of thermodynamics - one gets an expectation value rather than an eigenvalue!

Moreover, the collapse is provably wrong in case a single photon is measured by a photodetector. It is after the measurement not in a position eigenstate but it has no state anymore!

 

[meopemuk]

Just a question. Let's say that I radiate some form of energy into a group of many people. Sometimes, casually, one of those persons feels "excited" because of that energy and jumps for some seconds. Where is the "collapse"?

This is actually a good analogy of the photoelectron emission process as described in Chapter 9 of Mandel & Wolf. The collapse (in my non-standard terminology) happens when a person feels excited for no good reason. If there is no deterministic explanation of this excitation, then we have an unpredictable truly random effect, which I call collapse.

Before the collapse the state of this person could be described by probability, i.e., the chance to be excited is X, the chance to be not excited is 1-X. After the excitation has materialized this probability distribution has collapsed to a certain state.

Eugene.

 

[meopemuk]

Of course, you could call any statistical element ''collapse'', but this is not the standard way of using the term. If you use your personal terminology, the only result is that nobody understand you anymore.

I use the word "collapse" every time when a probability distribution (i.e., incomplete knowledge) is converted to an actual event (complete knowledge).

I wouldn't use the word collapse in the case of coin tossing, because the coin movement is described by classical mechanics, which is capable of predicting the outcome with 100% certainty if the initial state if fully specified. Yes, this is difficult to do in the case of a coin or turbulence or other seemingly untractable "chaotic" classical systems. But "difficult" or "impractical" does not mean "impossible". So, in my understanding, there is no probability associated with classical coin tossing or turbulence. So, there is no collapse.

Quantum mechanical systems are fundamentally different from the tossed coin. When electron passes through a single-slit or a double-slit there is absolutely no way to predict where it will land. This is a truly unpredictable system. Before actual landing on the screen the electron is described by a probability density (square of the wave function) which collapses after the observation is made.

Eugene.

 

[A. Neumaier]

I use the word "collapse" every time when a probability distribution (i.e., incomplete knowledge) is converted to an actual event (complete knowledge).

Compare this with the conventional mainstream meaning:
''wave function collapse (also called collapse of the state vector or reduction of the wave packet) is the phenomenon in which a wave function—initially in a superposition of several different possible eigenstates—appears to reduce to a single one of those states after interaction with an observer. '' http://en.wikipedia.org/wiki/Wavefunction_collapse

I wouldn't use the word collapse in the case of coin tossing, because the coin movement is described by classical mechanics, which is capable of predicting the outcome with 100% certainty if the initial state if fully specified. But "difficult" or "impractical" does not mean "impossible". So, in my understanding, there is no probability associated with classical coin tossing or turbulence. So, there is no collapse.

But your definition that ''a probability distribution (i.e., incomplete knowledge) is converted to an actual event (complete knowledge)'' fully applies in practice. So your definition of collapse seems inconsistent.

 

[lightarrow]

Lightarrow... do we meet again? Are you the man I knew from the Naked Scientists?

Hello QuantumClue!

To answer your question, if it was possible for some quantity of energy to be tranferred, rather than simply radiated into the body of another person, then collapses may occur if there are decoherences in the stucture of the other person. These simple decoherences are collapse-like state systems.

I used that metaphor to express the idea that a quantum description is not needed for that effect.

It's a bit of an odd question, but if you are the man I remember, then it's not a great surprise :)

But quantum physics is odder, isnt'it?

 

[meopemuk]

But your definition that ''a probability distribution (i.e., incomplete knowledge) is converted to an actual event (complete knowledge)'' fully applies in practice. So your definition of collapse seems inconsistent.

I don't think our knowledge is incomplete when we are tossing a coin. Yes, it is incomplete in practice, because we are too lazy to specify all initial conditions exactly and to perform all necessary calculations.

On the other hand, when we are sending a polarized photon through a filter, the result is unpredictable. No matter how careful we are in preparing their state, the photons will behave unpredictably.

This is why coin tossing can be described (in principle, but possibly not in practice) by classical mechanics, and in order to describe photons or electrons we need quantum mechanics.

If you don't want to recognize this difference between classical and quantum mechanics, then you represent the "hidden variables" interpretation camp.

Eugene.

 

[PhilDSP]

So, there can be only two legitimate interpretations of quantum mechanics. One is the "hidden variable" interpretation, which basically says that QM is just a branch of classical mechanics, where everything is deterministic and predictable. No probabilities involved and no collapse. The other interpretation is that quantum events are truly random. Then the collapse is needed. There is no third way.

Isn't that perspective a bit too limiting? One should be allowed the option of approaching every problem with the aim of finding at least one "hidden variable". Sometimes, maybe often, those variables will be so sensitively balanced that chaotic conditions pertain to the result. But in searching and possibly finding such a variable one might ultimately find some other characteristic that isn't deterministic. In other words, we recognize the difference between deterministic-predictive variables, deterministic-chaotic variables producing effectively unpredictable results and fully non-deterministic variables where anyone situation may may be composed of variables of each type.

 

[A. Neumaier]

I don't think our knowledge is incomplete when we are tossing a coin. Yes, it is incomplete in practice, because we are too lazy to specify all initial conditions exactly and to perform all necessary calculations.

No, because it is impossible. We cannot even specify single real numbers to infinite precision (and rounding them to a rational creates already enough uncertainty to make chaos apply after a very short time.

On the other hand, when we are sending a polarized photon through a filter, the result is unpredictable. No matter how careful we are in preparing their state, the photons will behave unpredictably.

This is unpredictable only in a particle picture. In a field picture, polarization is very easy to understand. The qubit was understood classically almost 50 years before Planck discovered the first hint to quantum mechanics - see slides 6-15 of my lecture http://arnold-neumaier.at/ms/optslides.pdf

The only reason why the Schroedinger equation wasn't found by Stokes in 1852 was that there was no incentive to do so...

This is why coin tossing can be described (in principle, but possibly not in practice) by classical mechanics, and in order to describe photons or electrons we need quantum mechanics.

Of course we need quantum mechanics to describe photons and electrons and dice.
A die is a quantum mechanical object - but it behaves approximately classically to such an extent that we hardly ever regard it as a quantum object. But a correct account of its falling behavior would require that.

On the other hand, we don't need a collapse to predict the laws of elasticity and classical motion of a die from a quantum mechanical basis. We only need Ehrenfest's theorem.

 

[harrylin]

[..] If you don't want to recognize this difference between classical and quantum mechanics, then you represent the "hidden variables" interpretation camp.

Eugene.

Why do you want to push people in "camps" that they may not have joined?

 

[Q-reeus]

..You could as well claim that the fact that a shower emits tiny rays of water is proof that water is composed of discrete rays...The corrections obtained by using QED instead of the classical external field are very tiny and can be neglected...
Then, how on Earth can the tiny dots be interpreted as an indication that light is a flow of discrete particles rather than a continuous wave? The external field used to generate the pattern _is_ a continuous wave, as one can trivially verify by inspecting the model...This is enough to make the conclusion invalid that the tiny dots must be regarded as proof of a discrete particle structure of the incident radiation...These discussions revealed to me that real-life photons are something very different from what superficial discussions seemed to suggest.

That was from #14. Still not clear just how the QFT and classical EM pictures differ. In some places you speak of 'more or less localised' photon states, elsewhere that 'the wave spreads out beyond the slit' etc. Unless there are two or more distinctly different models of what constitutes a photon/photon state in QFT, I am assuming localization is simply the result of wave interference as for instance in an antenna array or multi-mode cavity resonator. Wrong? Consider a specific case. An excited atom in vacuo undergoes spontaneous decay into ground state, emitting a field quanta (avoiding the 'p' word). Classically, the emission might be described as a weighted ensemble of multipole fields that propagates as a spherical pulse with superposed multipole angular distributions. At large r the field gets very tenuous but never becomes 'granular'. To what extent is the QFT picture different?

 

[A. Neumaier]

That was from #14. Still not clear just how the QFT and classical EM pictures differ. In some places you speak of 'more or less localised' photon states, elsewhere that 'the wave spreads out beyond the slit' etc. Unless there are two or more distinctly different models of what constitutes a photon/photon state in QFT,

The quantum field is described by a state, defining expectation values of the field operators. By the Ehrenfest theorem http://en.wikipedia.org/wiki/Ehrenfest_theorem , these expectations correspond exactly to the classical e/m field, independent of the photon content of the state. Measurements respond to this field and/or to the associated coherence fields, given by the expectations of certain bilinear expressions in the fields - among them is the energy density of the field. To see the nonclassical character of a quantum field, one needs to make correlation experiments that exhibit the deviating statistical properties.

The notion of photon is commonly used with two different meanings:
1. as a localized wave packet of approximate frequency omega and approximate total (integrated over time) energy omega*hbar, in some cases generated by a single atomic event;
2. as synonymous to a 1-photon state. The latter are in 1-1 correspondence with classical solutions of the Maxwell equations, but they are Fock states with very nonclassical properties.
This explains why the classical and the quantum field descriptions are quite similar, even when talking about single photons.

I am assuming localization is simply the result of wave interference as for instance in an antenna array or multi-mode cavity resonator. Wrong?

Interference creates the pattern. Localization is associated with the fact that a single, localized electron responds (according to a process whose rate is proportional to the intensity pattern).

Consider a specific case. An excited atom in vacuo undergoes spontaneous decay into ground state, emitting a field quanta (avoiding the 'p' word). Classically, the emission might be described as a weighted ensemble of multipole fields that propagates as a spherical pulse with superposed multipole angular distributions. At large r the field gets very tenuous but never becomes 'granular'. To what extent is the QFT picture different?

The field expectation values are precisely what the classical picture suggests.

The quantum nature is reflected by the knowledge that in this particular situation (far from easy to produce experimentally to good accuracy and with high efficiency) the quantum field state is a 1-photon state. This is usually completely inconsequential, but can make a significant difference in special correlation experiments.

 

[Q-reeus]

..The notion of photon is commonly used with two different meanings:
1. as a localized wave packet of approximate frequency omega and approximate total (integrated over time) energy omega*hbar, in some cases generated by a single atomic event;...

And this usage is legitimately part of standard QFT? Does it imply a non-spreading entity that propagates soliton-like to any distance?

2. as synonymous to a 1-photon state. The latter are in 1-1 correspondence with classical solutions of the Maxwell equations, but they are Fock states with very nonclassical properties.
This explains why the classical and the quantum field descriptions are quite similar, even when talking about single photons.

Err... chasing around a bit like at http://en.wikipedia.org/wiki/Nonclassical_light, get the idea Fock state has this undefined phase thing, in contrast to say a coherent state. Otherwise, to say '..1-1 correspondence with classical EM...with very nonclassical properties.' leaves me scratching pate.

At any rate, taking this to mean overall that we have a physical, objectively real and continuous field whose space and time evolution is essentially classical (in most situations), this only reinforces my misgivings about detector clicks for extremely attenuated light.

Let's consider the usual 2-slit setup, but where the detection screen is a wide and very narrow strip, total area being orders of magnitude smaller than say a hemisphere whose radius is that from twin-slit plate to detection strip. This means orders of magnitude smaller cross-section than a single field quanta (as spreading wave) presents to the screen. I share your view there is no possibility of instantaneous physical collapse of such a field quanta - what the screen 'sees' is what the screen 'gets'. OK then - let the light be so attenuated on average only one field quanta passes the slits every minute or so. Previously you have stated the detection screen electrons form a chaotic system with no memory (meaning I assume no ability to either accumulate incident energy, or retain knowledge of the intensity distribution for any reasonable length of time - ie. dissipative system). All the foregoing strongly suggests to me that by the continuous field view there will never be any clicks, or on the rare occasion a statistical fluctuation in number density allows one, there will be no final correlation with the expected interference pattern. None of this poses a problem for the corpuscular model (not necessarily 'point' photons, but at least highly localized wave packet photons). Probability of a click drops simply in direct proportion to the screen area, and the interference pattern is unaffected. And there is a ready QFT counter-argument, or have I completely misinterpreted the system?

 

[meopemuk]

This is unpredictable only in a particle picture. In a field picture, polarization is very easy to understand. The qubit was understood classically almost 50 years before Planck discovered the first hint to quantum mechanics - see slides 6-15 of my lecture http://arnold-neumaier.at/ms/optslides.pdf

The only reason why the Schroedinger equation wasn't found by Stokes in 1852 was that there was no incentive to do so...

To me this looks like a very unusual way of looking at quantum mechanics. Thanks for sharing.

Eugene.

 

[A. Neumaier]

To me this looks like a very unusual way of looking at quantum mechanics.

You can find the conventional way of looking at the same in the first Chapter of Sakurai's book.

It is classical optics made mysterious by pretending it is a particle effect...

 

[A. Neumaier]

And this usage is legitimately part of standard QFT? Does it imply a non-spreading entity that propagates soliton-like to any distance?[,QUOTE]
It is used quite a lot in practice. These photons spread, like any wave packet.

Err... chasing around a bit like at http://en.wikipedia.org/wiki/Nonclassical_light, get the idea Fock state has this undefined phase thing, in contrast to say a coherent state. Otherwise, to say '..1-1 correspondence with classical EM...with very nonclassical properties.' leaves me scratching pate.[,QUOTE]
Yes. The state vectors are in 1-1 correspondence but the normalized states are not.

At any rate, taking this to mean overall that we have a physical, objectively real and continuous field whose space and time evolution is essentially classical (in most situations), this only reinforces my misgivings about detector clicks for extremely attenuated light.

It is classical in the common situations like sunlight or laser light. It takes quantum optics ingenuity to create nonclassical states of light.

I share your view there is no possibility of instantaneous physical collapse of such a field quanta - what the screen 'sees' is what the screen 'gets'. OK then - let the light be so attenuated on average only one field quanta passes the slits every minute or so. Previously you have stated the detection screen electrons form a chaotic system with no memory (meaning I assume no ability to either accumulate incident energy, or retain knowledge of the intensity distribution for any reasonable length of time - ie. dissipative system).

The energy is absorbed collectively, of course, but the electron doesn't know that.

All the foregoing strongly suggests to me that by the continuous field view there will never be any clicks, or on the rare occasion

Rare occasion means one electron per minute, or so.

Probability of a click drops simply in direct proportion to the screen area,

No. The probability drops quadratically with the distance from the screen but grows linearly with the screen area (assuming the detector has constant thickness).

 

[Q-reeus]

It is used quite a lot in practice. These photons spread, like any wave packet.

Thanks for clearing that point up.

The energy is absorbed collectively, of course, but the electron doesn't know that...Rare occasion means one electron per minute, or so.

Not as per my scenario. Recall that the screen area was taken to be several orders of magnitude smaller than needed to fully capture an incident field quanta at some nominal distance from the slits - assuming screen effective cross-section equals it's area. For sake of argument make it a neat factor of 100 (let's not worry about interference fringes for the moment - just take an average screen intensity value). So each of the ~ 1-quanta per minute passing the slits can deposit no more than ~ 1% of their energy to the screen (I think we agree in such a setup the screen can in no way act as a resonant antenna - after all there is no monochromatic stream of radiation). Hence at best the average time between clicks will be around 100 minutes. But that's the real sticking point as I see it. This 'best case scenario' assumes the screen is not only capable of fully absorbing all incident radiation, but losslessly accumulating each hit for perhaps hours until sufficient energy is present to eject one electron. What's more in order to reproduce the interference pattern, a memory of the incident intensity is also dissipationlessly stored. This seems utterly incredible. I would expect very rapid dispersion and dissipation of incident energy to destroy any chance of even one click. This energy accumulation picture is also seemingly at odds with the known rather sharp frequency threshold for photoelectric effect. If arbitrarily small portions of a field quanta can be progressively absorbed and accumulated, shouldn't there be a very gentle dependency on frequency, with no particular cutoff frequency? But being no expert here, will defer to your much greater knowledge in this area. Is there no limit to how long this storage/accumulation/memory 'magic' can persist for?

No. The probability drops quadratically with the distance from the screen but grows linearly with the screen area (assuming the detector has constant thickness).

We actually agreed on this minor point - i was assuming fixed radius, and only screen area as variable.