# "Wave function Collapse" 'proof' articles

1. Dec 21, 2015

### _PJ_

I have seen a number of references to apparent experimental "proof" of wavefunction collapse
www.nature.com/articles/ncomms7665

However, I am still seeing propagation of the "Many Worlds" theory, which, and I admit that my understanding is limited, but the MW hass at its very core, a necessary foundational requirement that the wavefunction DOES NOT collapse.

Is this just an issue with terminology whereby the words "wavefunction collapse" are being used on an assumption that this is the mechanism which belies the measurement problem, thereby the misleasding titles should perhaps claim "quantum state superposition resolved to definite" rather than any actual reference to actual collapse of a wavefunction?

2. Dec 21, 2015

### Strilanc

I assume they're comparing Copenhagen-with-collapse against Copenhagen-without-collapse, experimentally rejecting Copenhagen-without-collapse, and concluding "therefore collapse".

Other interpretations explain the results with different mechanisms, so it's certainly misleading to call it a "proof of collapse" without tacking on "within the framework of the Copenhagen interpretation".

3. Dec 21, 2015

### zonde

I would say that emphasis is on "nonlocal" i.e. this experiment falsifies "local collapse" as opposed to "nonlocal collapse".

4. Dec 21, 2015

### DrChinese

5. Dec 21, 2015

### DrChinese

I would say that if you are looking to exclude interpretations of QM, this experiment will not do it for you any more than other tests of non-locality (using entangled pairs).

On the other hand, this is a great experiment showing one of the important aspects of individual particles (i.e. not members of an entangled pair). Just as a double slit experiment shows a particle does not take a single path (when there is interference), this shows a particle's position is not limited to a single place (prior to detection).

I don't think there is anything misleading about the title. It is just a title, and makes perfect sense given the excellent article itself.

6. Dec 23, 2015

### vanhees71

I'm already stuck at Eq. (1). In the text they say, it's a single-photon state, but the equation looks like an entangled two-photon state, how should the tensor products like $|1 \rangle_A |0 \rangle_B$ be understood?

Usually the resolution of the paradox is in noting that the observed correlations described by quantum entanglement used to violate Bell-type inequalities is due to the preparation of the state in the entangled state before any measurement on the system has been performed but not due to these measurements. Here, I'm already unable to understand the claim due to unexplained sloppy notation! It's really surprising what goes through the strict review process at Nature!

7. Dec 24, 2015

### zonde

This is false and by repeating it you won't make it true. It's fine to have a doubt but try to reformulate it as a question instead of statement.
Btw have you noticed that in recent loophole free Bell inequality tests they don't use maximally entangled state?

Yes, "experimental proof" is a very strange phrase to see in the title of scientific paper.

8. Dec 24, 2015

### vanhees71

Well, I'd use the usual formalism as in quantum-optics textbooks (I'm not an expert in quantum optics though). You start with the mode decomposition of the electric field operator,
$$\vec{E}^{(+)}(\vec{r},t) = \sum_{\lambda=\pm 1} \int \mathrm{d}^3 \vec{k} \vec{\epsilon}(\vec{k},\lambda) A(\vec{k},\lambda) \hat{a}(\vec{k},\lambda) \exp(-\mathrm{i} \omega_k t + \mathrm{i} \vec{k} \cdot \vec{r}).$$
Then if yo have a symmetric beam splitter where you put a single photon (for simplicity I write down only the annihilation operators of a single mode) with the annihilation operator $\hat{a}$. Then at the detector in the one direction you have $\hat{d}_1=\sqrt{T} a$ (transmitted part) and at the other $\hat{d}_2=\mathrm{i} \sqrt{1-T}$ (reflected part), taking into account the phase shift between transmitted and reflected waves, and $T$ is the transmittivity of the beam splitter.

It would be great, if you could answer my question about the meaning of the tensor products in Eq. (1) of the above paper, because there it reads as if there is a two-photon state considered rather than a one-photon state, which I thought they describe there (according to the text). Also later on they write down single-photon states again. Perhaps it's just my ignorance, but this seems to be somehow inconsistent. So far, I had no big trouble to understand experiments of this kind (at least in principle) as described in physics journals (including Nature of course).

I also don't understand the other point. Take a very simple Bell test with A and B measuring a polarization entangled pair of photons in the same polarization direction, say $H$. Let the initial state of the photon pair be $|HV-VH \rangle$, then both A's and B's single photons are totally unpolarized. Still, when A' photon goes through her polarizer (i.e., it's measured to have polarization in $H$ direction), then B won't see a photon, i.e., it's polarized in $V$ direction and vice versa. Although the single-photon polarizations at A and B were totally unpolarized the preparation in the entangled state however already contains this 100% correlation. It's not A's measurement that determines B's photon's polarization but the correlation exists because of the preparation of the photon in that entangled state. I don't see, what's wrong with this minimal interpretation, just using Born's rule. There's no need for an instantaneous non-local collapse, which contradicts the very foundations of QED, which is a local relativistic QFT, and A's and B's measurements base on local interactions of single photons with their measurement devices. As long as there is not a clear contradiction of such experiments to QED, I tend to believe that QED is the correct description of the electromagnetic interaction between electrically charged particles and the electromagnetic field.

9. Dec 24, 2015

### atyy

http://arxiv.org/abs/quant-ph/0507189 (published in PRA, so they are just as bad as Nature Communications)[/PLAIN] [Broken]

Last edited by a moderator: May 7, 2017
10. Dec 24, 2015

### Staff: Mentor

Since collapse is not part of the QM formalism I can say without even reading it a claim like the above isn't correct.

The only way it could be is using a different definition of collapse than usually used.

Thanks
Bill

Last edited: Dec 24, 2015
11. Dec 24, 2015

### zonde

One clear problem with this minimal interpretation is that it contradicts Bell type inequalities (I would refer to Eberhard inequality).
But of course then there should be something specific in QED that is non-local and the place of that non-locality should be clearly identifiable.
And yes, I tend to agree that non-locality is not due to the measurement (if the first measurement is performed in the same basis as preparation basis).
But it can't be preparation either, as per Bell type inequalities. And if you look at Bell inequality it requires at least two measurement angles (with non 90deg difference) at each end to derive inequality. So it means you would have to perform unitary transformation at least once to keep first measurement in preparation base.
So my take on it is that unitary transformation of photon polarization (say by rotating half-wave plate in the path of the beam) should be responsible for non-locality.

12. Dec 24, 2015

### vanhees71

[/PLAIN] [Broken]

Indeed, (1) is an abuse of notation, and it's not the same (at least not in the usual notation) as (2). How can this go through in highly respected journals? As I referee, I never let through something like this. What is reviewing good for if not to make the content of the paper clear (provided it's not flawed; then you have to reject it anyway)?

Obviously (1) doesn't make sense, because the vacuum has no position information. If I understand the author right, what he labels with subscripts $A$ and $B$ are locations of observers Alice and Bob, but how can that make sense for the vacuum state, which is Poincare (or Galilei for non-relativistic QT) invariant? So again, what's the meaning of such notation?

A single particle state is always created from the vacuum with a creation operator acting on the vacuum. If you want to describe a single-particle state which is peaked at different locations, e.g., prepared by sending a particle through a potential well, where the "matter wave" gets partial reflected and transmitted, the correspoding annihilation operator reads (in the Heisenberg picture)
$$\hat{\psi}^{(+)}(t,x)=\int_{\mathbb{R}} \mathrm{d}^3 k \hat{a}(k) [A(k) \exp(-\mathrm{i} E_k +\mathrm{i} k x + B(k) \exp(-\mathrm{i} E_k-\mathrm{i} k x)],$$
and the corresponding state (time-independent) state is given by
$$|\Psi \rangle=\hat{\psi}^{(+) \dagger}(0,x)|\Omega \rangle,$$
where $|\Omega \rangle$ is the vacuum. If you choose $A(k)$ and $B(k)$ both to peak around $k=k_0$ at long times you have a single particle operator which is peaked at opposite positions.

If you put A somewhere far to the "right" and B somewhere far to the left, indeed only one of them will detect a particle, i.e., if A detects a particle, then necessarily B won't detect it and vice versa. I guess, that's what the author wanted to express with his Eq. (1), but to get the probabilities for finding the particles by A and B you must use simply a single-particle position eigenstate $|x,t \rangle$ (note that in the Heisenberg picture the eigenstates of operators are time dependent). Then the probability distribution to find the particle around $x$ is simply given by
$$\psi(t,x) = \langle x,t|\Psi \rangle.$$

You can of course as well use the Schrödinger picture. Then
$$|\Psi,t \rangle=\hat{\psi}^{(+)\dagger}(t,x)|\Omega \rangle$$
and
$$\psi(t,x)=\langle x|\Psi,t \rangle,$$
where now $|x \rangle$ is the time-independent Schrödinger-picture position eigenstate and $|\Psi,t \rangle$ the time-dependent Schrödinger-picture state. Now $\psi(t,x)$ for large times is peaked on opposite positions (with little overlap of the two parts, if $A(k)$ and $B(k)$ are appropriately chosen).

Of course, what's said in the introduction is true. You don't need two- or more-particle states to have entanglement. You can also have entanglement between different observables of a single particle. A well-known example is the preparation of position-spin entangled states by a Stern-Gerlach apparatus: After the beam rund through the magnetic field it splits in partial beams, each with a (nearly perfectly) prepared spin component in direction of the magnetic field. So you created a state with position-spin entanglement.

Recently they even created states where position and magnetic moment of a neutron seem to be transported on separate paths through a beam splitter:

T. Denkmayr et al, Observation of a quantum Cheshire Cat in a matter-wave interferometer experiment, Nature Communications 5, Article number: 4492
http://dx.doi.org/10.1038/ncomms5492

The article is open access!

So maybe the author in the above cited PRA article had in mind the following (take a spin-1/2 particle like a neutron as an example) the state defined in the above explained way by using the annihilation operator
$$\hat{\psi}(t,x) = \int_{\mathrm{R}} \mathrm{d} k [A(k) \hat{a}(k,\sigma=+1/2) \exp(-\mathrm{i} E_k t +\mathrm{i} k x) + B(k) \hat{a}(k,\sigma=-1/2) \exp(-\mathrm{i} E_k t-\mathrm{i} k x)].$$
Then spin up and down are the properties labeled with A and B in the author's notation. Suppose $A(k)$ and $B(k)$ are both peaked around $k=k_0$ and we look at times, were the partial waves are well seperated in opposite directions, then A will measure with probabilities very close to
$$P_{\sigma=1/2,A}=\int_{\mathbb{R}} \mathrm{d} k |A(k)|^2, \quad P_{\sigma=-1/2,A}=0,$$
i.e., either a particle with spin up with the given probability but with certainty never a particle with spin down, while for B it's
$$P_{\sigma=-1/2,B}=\int_{\mathbb{R}} \mathrm{d} k |B(k)|^2, \quad P_{\sigma=1/2,B}=0.$$
Of course the single-particle state is position-spin entangled, and indeed, if A detects a particle with spin 1/2, she knows that B can't find one with spin -1/2, because she knows that a single-particle state was prepared, but I still don't understand what (1) means.

Of course you could also use the occupation-number basis. Then the tensor product makes some sense again, because then you define
$$|\{N_i \}_{i} \rangle = \prod_i \frac{1}{\sqrt{N_i!}} (\hat{a}^{\dagger}_i)^{N_i} |\Omega \rangle.$$
This is indeed a (anti-)symmetrized product state for (fermions) bosons. Then with (1) the authors means a state
$$|\psi \rangle_{A,B} = \frac{1}{\sqrt{2}} (|\{N_B=1,N_{i \neq B}=0 \} \rangle + |\{N_A=1,N_{i \neq A}=0 \} \rangle,$$
which indeed is precisely what he wrote in Eq. (2). If you interpret (1) in this way, it makes sense and is consistent, while in the Nature article it's not even clear what's meant with their Eq. (1) (at least not for me).

Last edited by a moderator: May 7, 2017
13. Dec 24, 2015

### vanhees71

Of course, it violates the Bell inequalities as quantum theory does. This is independent of the metaphysical interpretation, because it follows simply from the mathematical formalism, and sure, to find such a violation you have to use another setup than the simple one I referred to in my previous posting. That doesn't invalidate the minimal interpretation, because to claim that the minimal interpretation is wrong is the same as to claim that quantum theory is wrong, and so far all tests, also the recent ones, claimed to be loophole-free, agree with standard quantum theory (including local relativistic QFT, which excludes at least a naive collapse interpretation). Again, you just have to distinguish between correlations and causally connected influences of the measurement on the system.

I also do not see what the choice of A's and B's polarizers should have to do with the prepared state. Of course they can choose the direction of their polarizers independent from which photon state was prepared, and you can calculate all probabilities for any setup, using the prepared state.

Also rotating one photon's polarization by an ideal half-wave plate (which is a unitary operation on the photon state) is according to local interactions of the corresponding photon (localized at the position of this half-wave plate). I don't see, how the claim it should instantaneously have an influence on another far-distant photon, is compatible with the very construction of QED.

14. Dec 25, 2015

### zonde

The experiment you describe can be realized with product state ($|HV\rangle \otimes |VH \rangle$) as well. And product state does not violate Bell inequalities.
Just as well we can analyze more general state $|HV\rangle - e^{i\phi}|VH \rangle$. With $\phi$ being close to but not equal with zero you can't pretend it's the same base as you rotate both polarizers (while it still can violate Bell inequality). So you would have to add something to your minimal interpretation in order to cover Bell inequality test.

15. Dec 25, 2015

### vanhees71

I don't understand that. The minimal interpretation is the part of quantum theory, everybody agrees upon. It's just taking Born's rule as additional postulate and doesn't ask for any metaphysical implications as other interpretations do. Basically it's the flavor of the Copenhagen interpretation, where no collapse is assumed, and for me this is crucial. Any (naive) collapse postulate violates locality in the sense of relativistic QFT, and I'm not aware of any other class of relativistic QTs that can describe nature as successfully as local relativistic QFTs (among them the Standard Model).

Of course, the Bell inequality can be experimentally tested in many ways, and that's done. I also don't know, what you have in mind with the four-photon state, but I'm sure you could find nice Bell-violating scenarios using 4 entangled photons. Of course you can also have arbitrary relative phases as in your two-photon example, but all this does not invalidate the minimal interpretation. If you'd somehow could show with same experiment that quantum theory in the minimal interpretation is incorrect, then quantum theory (in whatever interpretation) is incorrect and must be substituted by something better. To my knowledge, such an empirical refutation of standard QT has not been seen yet.

16. Dec 25, 2015

### zonde

Yes, it does not.
In post #14 I am trying to correct my argument from post #11. Sorry for not making it clear.
I repeat what I said in post #14, your minimal interpretation alone can not violate Bell inequality. You have to add something that describes (physical) transformation of state into different basis.

Why you are asking for anything like that? (This part of) discussion is about whether QM is local or non-local.

17. Dec 25, 2015

### vanhees71

Again, the violation of the Bell inequality follows from the pure mathematics of quantum theory. You don't need any additional elements of non-minimal interpretations.

Further it is clear that contemporary standard QT is local, i.e., it is formulated as a local relativistic quantum field theory, i.e., local observables commute (by assumption!) if their arguments have space-like separated arguments, and thus no causal influence can happen with faster-than-light speed, as is demanded by the causality structure of (special) relativistic space-time. Also the Hamiltonian is local, i.e., it is the space integral of a polynomial in the field operators, their canonical field momenta, and their first (spatial) derivatives.

This, of course, does not exclude long-ranged correlations, as the many experiments with entangled photons show, but long-ranged correlations do not exclude locality in the usual sense. Again, all the violations of Bell's inequality are consistent with standard relativistic local QFT. The Bell inequalities are proven under the assumption of a local realistic (hidden-variable) theory. The key point is that QT is not such a theory, and thus one can experimentally distinguish between QT and local realistic theories. Overwhelming evidence (recently even loop-hole free experiments confirming the violation of Bell's inequality have been done) speaks against local realistic theories and for standard local QFT. So I stick with the latter!

18. Dec 25, 2015

### zonde

In post #8 you described some elements of minimal interpretation that do not require non-locality. I say that these elements are not enough to describe any experiment that violates Bell inequality. You have to include some additional element from "pure mathematics of quantum theory". Do you agree with this or you don't?

Please stop making very general statements. There is no point in that if we do not agree about them and do not look into details for the source of our disagreement.

19. Dec 25, 2015

### atyy

20. Dec 26, 2015

### vanhees71

I give up. Obviously we can't find an agreement on this very general and fundamental foundations of quantum theory :-((.

One last general statement: There are no non-local interactions in the Standard Model of particle physics by construction. There are long-ranged correlations described by entangled states. The entanglement and thus the long-ranged correlations are due to the preparation of the system under investigation in the very beginning, i.e., before any measurement of the entangled observables is done. There is no necessity to assume action at a distance ("collapse of the state") to explain the findings on these correlations, which include the violation of Bell's inequality. You can read about the standard experiments showing entanglement in Dr. Chinese's nice summary

http://www.drchinese.com/Bells_Theorem.htm

21. Dec 26, 2015

### vanhees71

For me that's indeed very strange. Usually one writes one ket of a single-particle state, not a tensor product. What they seem to mean with this notation in their Eq. (4.1) are (spoken in terms of the position-space wave function) wave packets of a single particle that have two peaks that are FAPP separated from each other by a macroscopic distance, e.g., something like
$$\psi(x)=\frac{1}{\sqrt{4\delta}}[\Theta(|x-x_1|<\delta) + \Theta(|x-x_2|<\delta)]$$
with $|x_1-x_2| \gg \delta$.

More interesting is the spin-position entangled state, that we have discussed some postings before. For the idealized "wave packets" of the kind above, it's something like
$$\psi(x)=\frac{1}{\sqrt{4 \delta}}[A \Theta(|x-x_1|<\delta) u_{\sigma=+1/2} +B \Theta(|x-x_2|<\delta) u_{\sigma=-1/2}], \quad |A|^2+|B|^2=1.$$
I'm still puzzled about this tensor-product notation, but it seems to be possible to do correct calculations with it. So why not?