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klw289
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How does the collapse of a wave function influence the results of an energy measurement or a position measurement taken immediately after another energy or position measurement?
vanhees71 said:If you measure a photon's energy with a usual photodetector, it is absorbed. So there is not a single photon left with a state that's collapsed into an energy eigenstate (or a true state with a narrow energy spread). This is only one very simple example which shows that this collapse hypothesis is very misleading.
Further, it is inconsistent with quantum theory to assume that there is a classical dynamics on top of quantum dynamics, leading to the collapse. So far nobody could find a clear physical distinction when classical dynamics should be applicable and not quantum dynamics. Nowadays, one can demonstrate the superposition principle, particularly also entanglement, for macroscopic objects, and there is no natural scale (in whatever kind of system size) where you can make this "quantum-classical cut".
Last but most importantly an instantaneous nonlocal collapse of the state is incompatible with the very foundations of relativistic space time and its causality structure. If at all, you can take the collapse as a short-hand expression for "adapting ones knowledge because of new information on the system", i.e., in an epistemic instead of an ontological way of interpreting the state, but then you can as well simply adapt the minimal interpretation, taking Born's rule as another postulate rather than something derivable from the other postulates, particularly quantum dynamics. As nicely demonstrated by Weinberg in his "Lectures on Quantum Mechanics" such a derivation is anyway still not known.
bhobba said:Vanhees is of course correct.
Collapse is not a part of QM - its only a part of some interpretations:
https://en.wikipedia.org/wiki/Interpretations_of_quantum_mechanics
In some interpretations the wave-function is simply a state of knowledge. That it collapses is of zero concern.
Thanks
Bill
Nugatory said:Uh, guys... Anyone want to take a try at the original poster's question? Something about how two successive position or energy measurements behave?
klw289 said:How does the collapse of a wave function influence the results of an energy measurement or a position measurement taken immediately after another energy or position measurement?
atyy said:Contrary to what vanhees71 says, the collapse is a central part of the orthodox interpretation of quantum mechanics. It is used in the Schroedinger picture to calculate joint probabilities.
stevendaryl said:Well, you could split it into two different assertions:
Only the second assertion is a "collapse". Of course, you might argue that you need collapse to give predictions for a sequence of measurements, but you don't, really; you can just view the sequence of measurements as a single, compound measurement.
- When you measure an observable, you get an eigenvalue, with probabilities given by the Born rule.
- After a measurement, the wave function is in the eigenstate corresponding to the eigenvalue measured.
martinbn said:Isn't a collapse needed for state preparatio?.
atyy said:But would you call the compound measurement a "sequence" of measurements? The Born rule applies to measurements made at one slice of simultaneity. In the compound measurement, one does not measure A followed by B, rather one measures AA and B simultaneously, and p(AA,B) has the same form as p(A,B) so that measuring AA is taken to be the "same" as measuring A. However, the difference is that AA and B are real in the compound measurement, whereas A then B is real in the sequence of measurements.
bhobba said:In modern times states are the equivalence class of preparation procedures. The concept of collapse need not even be introduced.
klw289 said:How does the collapse of a wave function influence the results of an energy measurement or a position measurement taken immediately after another energy or position measurement?
stevendaryl said:In principle, you can get away with a single measurement (or a single application of the Born rule) in the following way:
You have some system, [itex]S_1[/itex], that you'd like to make repeated measurements on. So you create a second system, [itex]S_2[/itex], which consists of a device that makes repeated measurements on [itex]S_1[/itex] and records the results (say, on a DVD). Then the creation of the DVD does not, in principle, need to involve the Born rule at all. The various possible sequences of measurement results correspond to orthogonal states of the resulting DVD, and the Schrodinger equation for [itex]S_1 \otimes S_2 \otimes DVD[/itex] would give you the probabilities for each sequence. So it's only necessary to invoke the Born rule once, and it's not necessary to consider what state the DVD is in after the measurement.
stevendaryl said:But it seems to me that the idea that a certain "preparation procedure" results in a specific state (mixed or pure) of the system being prepared is equivalent to the collapse hypothesis.
klw289 said:How does the collapse of a wave function influence the results of an energy measurement or a position measurement taken immediately after another energy or position measurement?
Well, you might say that quantum field theory is mathematically not strictly well defined in the sense of an exactly solvable theory, but perturbative relativistic local QFT is a very successful description of nature. It describes all Bell experiments with very high accuracy. Since by construction this class of theories does not contain any nonlocal interactions, it is compatible with the relativistic space-time structure (by construction!). It's also not what Bell's theorem says, because Bell's theorem refers to local classically realistic theories, and quantum theory is not such a theory. That's the whole point of Bell's work: You have an objective way to decide between classical realistic local theories and quantum theory, with the result of ##N## experiments to 0, where ##N## is a very large number ;-)).atyy said:This is wrong, because quantum mechanics is incompatible with the very foundations of relativistic spacetime and its causality structure. That is what Bell's theorem says.
What is a "collapsed" or "non-collapsed" wave function? There's no such thing in standard quantum mechanics. The wave function describes a state in a certain basis, namely the (generalized) eigenbasis of the position operator (and perhaps other observables compatible with position like spin to have a complete basis). That's it. There's no way to say a wave-function is "collapsed" or "non-collapsed". It simply doesn't make any sense!DuckAmuck said:Imagine you measure position, and collapse the position wave function. Now imagine you keep measuring it over and over every second. You'll just keep getting the same result, since the wave function will remain collapsed. But if you stop and give it some time, the collapsed wave function will evolve back into a non-collapsed wavefunction. So the next time you measure it after waiting, you will get a different result, in general.
vanhees71 said:Well, you might say that quantum field theory is mathematically not strictly well defined in the sense of an exactly solvable theory, but perturbative relativistic local QFT is a very successful description of nature. It describes all Bell experiments with very high accuracy. Since by construction this class of theories does not contain any nonlocal interactions, it is compatible with the relativistic space-time structure (by construction!). It's also not what Bell's theorem says, because Bell's theorem refers to local classically realistic theories, and quantum theory is not such a theory. That's the whole point of Bell's work: You have an objective way to decide between classical realistic local theories and quantum theory, with the result of ##N## experiments to 0, where ##N## is a very large number ;-)).
It is very important to distinguish between nonlocal interactions (which are by construction not present in standard relativistic QFT) and longranged correlations, which are a specific consequence of any quantum theory, including relativistic QFT. It is not that measurement of one photon's polarization in an entangled two-photon state that causes the correlation but the preparation of the entangled two-photon state in the very beginning of the experiment. Although the single-photon polarizations are maximally indetermined, the 100% correlation between them is there all the time due to the preparation procedure in an entangled state. There is no non-local interaction between the single photons and the polarizers and photo detectors at A's and B's far-distant locations whatsoever! Consequently, according to local relativistic QFT it cannot be nonlocal interactions caused by the local measurements at the local detectors. This shows that these experiments do not contradict standard QED and thus there's no need for nonlocal interactions!
I'd use a magnetic field like in the Stern-Gerlach experiment. There's nothing collapsing there ;-).martinbn said:Isn't a collapse needed for state preparatio?. If you need a system of half spin in the state |spin up>, how would one prepare it without collapse?
atyy said:Let's take QED with just photons. It's a mathematically well-defined theory, and it has collapse.
It is wrong to say that collapse has to be rejected because it is not consistent with "relativistic causality". There are two possible meanings of "relativistic causality".
1. Signal causality - collapse does not violate signal causality, and it does not need to be rejected for being inconsistent with this type of relativistic causality, so this cannot be what you mean if your statement is to be correct.
2. Classical relativistic causality - collapse does violate classical relativistic causality, but if your statement is applied here, then your statement is misleading to wrong because there is no quantum mechanics that preserves classical relativistic causality.
vanhees71 said:It is very important to distinguish between nonlocal interactions (which are by construction not present in standard relativistic QFT) and longranged correlations, which are a specific consequence of any quantum theory, including relativistic QFT. It is not that measurement of one photon's polarization in an entangled two-photon state that causes the correlation but the preparation of the entangled two-photon state in the very beginning of the experiment. Although the single-photon polarizations are maximally indetermined, the 100% correlation between them is there all the time due to the preparation procedure in an entangled state. There is no non-local interaction between the single photons and the polarizers and photo detectors at A's and B's far-distant locations whatsoever! Consequently, according to local relativistic QFT it cannot be nonlocal interactions caused by the local measurements at the local detectors. This shows that these experiments do not contradict standard QED and thus there's no need for nonlocal interactions!
stevendaryl said:In QM or QFT, Alice describes the universe (or the part of the universe under consideration) at a given moment (relativistically, I guess a "moment" means a spacelike hypersurface of spacetime) by means of a pure or mixed state [itex]\rho[/itex]. This state describes both what's happening locally to Alice, but also what's happening far away, near Bob.
vanhees71 said:IThe state describes the preparation of the two photons and probabilities for measurements of observables concerning these photons, and nothing else. You can describe the probability that A measures the polarization (say H or V) of her photon that B measures the polarization (say also H or V) of his photon. Because of the preparation of the two photons in an polarization-entangled state, it is determined (!) that if A measures H then B must measure V and vice versa although the single-photon polarizations are described (according to the usual reduced statistical operators of subsystems) by the maximally indetermined state ##\hat{\rho}_A = \hat{\rho}_B=\mathbb{1}/2##. There's also a certain probability of the registration of the photons at A's and B's place.
Nothing happens to B's photon due to A's measurement (which in fact could have been measured and thus absorbed even before A made her measurement, where I assume for simplicity that A and B are in a common inertial lab frame, i.e., at rest relative to each other, and we use the time of this common frame to make statements about temporal order of events). When A makes her measurement and if she knows that B could measure the second photon in the entangled two-photon state, of which she has measured one of the photons, she can adapt her description of the two-photon state or of the reduced state of B's photon, but that "collapse" is nothing what's happening to the photons.
I think that's more or less the same as you stated as "A updates only her knowledge about B's local situation due to her measurement (and the knowledge about the two-photon state to begin with!)", and this is the minimal interpretation. Nowhere do you need a collapse as a physical process to describe what A and B measure, finding the 100% correlation of completely indetermined single-photon polarizations due to the entanglement of the prepared 2-photon state!
vanhees71 said:I'd use a magnetic field like in the Stern-Gerlach experiment. There's nothing collapsing there ;-).
atyy said:Classical relativistic causality.
vanhees71 said:The state describes the preparation of the two photons and probabilities for measurements of observables concerning these photons, and nothing else.
Nothing happens to B's photon due to A's measurement
I think that's more or less the same as you stated as "A updates only her knowledge about B's local situation due to her measurement (and the knowledge about the two-photon state to begin with!)", and this is the minimal interpretation. Nowhere do you need a collapse as a physical process to describe what A and B measure, finding the 100% correlation of completely indetermined single-photon polarizations due to the entanglement of the prepared 2-photon state!
vanhees71 said:How can QED with only photons, which is a free-field theory have collapse? Since the photons don't interact with anything within this theory, how can their states propagate other than through the unitary time evolution according to the full Hamiltonian (take the Schrödinger picture here for ease of discussion)?
vanhees71 said:Concerning the causality debate. Ad 1) Collapse, taken as a physical process, violates "signal causality", because it claims the meausurement at A's photon instantaneously determines also the polarization of B's photon measured at a far distant place. Of course, that's a wrong interpretation, because the entanglement of A's and B's photon guarantees already the 100% correlation between their measurements of the single-photon polarizations without any necessity for this "action at a distance collapse".
vanhees71 said:I don't know, what (2) means. How do you define "classical relativistic causality"? In my opinion causality is the property of a physical theory or model describing dynamics, and so far all of physics (and all of natural sciences anyway) is based on the hypothesis that nature is describable by causal theories. Of course both classical and quantum theory are causal, i.e., if you know the state of the system at a time ##t=t_0## and you know the Hamiltonian of the system exactly, then you know its state at any time ##t>t_0## exactly (modulo the impossibility to integrate the quantum equations of motion, but that's a discussion about principles not practicalities).
martinbn said:What is classical relativistic causality?
You can never be sure of anything. What I meant is that all Bell experiments so far can be well understood by standard QED, and as a local microcausal QFT it is a model where by construction two space-like separated events cannot be causally connected. So by construction there's no superluminal signal propagation within QED. What do you mean by "Bayesian updating"? If A measures the polarization state of her single photon AND knowing that B measures a photon that was polarization entangled with hers (in the same polarization direction), she instantly knows what B must (have) measure(d) about his photon. Then she can update her knowledge about Bob's measurement outcome to certainty for the corresponding polarization. Whether this is Bayesian or not, I don't care, but it follows standard rules of probability theory, right?atyy said:That is not the minimal interpretation. The minimal interpretation is agnostic about whether something does or does not happen. How can you be sure that nothing happens faster than light? Also, the updating of the knowledge is alone without anything happening is problematic, because the updating does not follow Bayesian updating.
Before you usually have a thermal state of a beam of silver atoms coming from an oven with a little opening, which is well approximated by a Gaussian wave packet with a momentum spread given by the thermal width. After the Stern-Gerlach apparatus you have a superposition of two wave packets (for spin 1/2), where the spin component in direction of the magnetic field is entangled with position, i.e., the spin-up and spin-down are spatially separated. This can be treated (semi-)analytically, see e.g.,martinbn said:What are the states before and after passing through the Stern-Gerlach aparatus?
What is classical relativistic causality?
According to QED, if the registration of A's and B's photons mark space-like separated events, then the one measurement cannot have affected the outcome of the other, and that's why A's update about B's photon doesn't do anything to B's photon. Since there's no known flaw of QED, I think it's a good hypothesis to think that it is the correct description. Of course, it's not a model-independent proof. One day one may find a non-local classically realistic theory which is as good as QED, and then you may interpret the experiment differently. The question, however, is if you can somehow create such a model. I'm pretty sceptical.stevendaryl said:I didn't say it described anything else. But Alice, when she makes her measurement, updates her state from [itex]\rho[/itex] (in which Bob has a 50/50 chance of measuring spin-up or spin-down along axis [itex]\vec{\alpha}[/itex]) to [itex]\rho'[/itex] (in which Bob is certain to measure spin-down along axis [itex]\vec{\alpha}[/itex]). The question is: Is this change in probabilities a physical change of Bob's particle, or is it merely a change in Alice's knowledge? Either way runs into problems. If it's merely a change in Alice's knowledge, then Bob's particle isn't changed by Alice's measurement. So if that particle is definitely spin-down along [itex]\vec{\alpha}[/itex] after Alice's measurement, then it must have been spin-down beforehand.
Why? According to the preparation in a polarization-entangled pair to the contrary the polarization of the single photons is maximally indetermined, i.e., they are perfectly unpolarized photons.That leads to the conclusion that B's particle (I was thinking spin-1/2 particles, but it doesn't matter) was a definite state of spin before Alice's measurement.
If the updating is the collapse, then I don't understand why it is still so fervently discussed, because then it's nothing what necessarily happens to the object under consideration, as in the here discussed example (at least if you accept standard QED as the valid description).The updating IS the collapse! The only issue is: what is the nature of that collapse? Is it purely a change in Alice's knowledge, or is it a change in Bob's particle's state? The first possibility contradicts Bell's theorem, and the second contradicts locality.
Let's forget about particles, and just speak in terms of measurements and probabilities of future measurement results. That's the "minimal interpretation", I think.
- Before Alice performs her measurement, she would say that Bob has a 50/50 chance of measuring spin-up or spin-down along axis [itex]\vec{\alpha}[/itex] (or H and V, if you're using photons).
- After Alice's measurement, she knows with 100% certainty what Bob will measure. She can now say: "Bob will definitely measure X" (whatever the prediction is).
- Is her statement a physical claim about the state of Bob and his local environment? It sure seems to me that it is.
[*]If it's a statement about Bob and surroundings, then you have the two possibilities: either it became true when Alice did her measurement, or it was true beforehand. The first violates locality, and the second violates Bell's theorem.