Graduate Realism from Locality? Bell's Theorem & Nonlocality in QM

[martinbn]

Well, it depends on what exactly one means by "relativity". See e.g. Sec. 2 of http://de.arxiv.org/abs/1002.3226 to see in what sense nonlocality can be compatible with relativity.

I'll take a look, but I don't think that there should be any freedom of choice on what is meant by relativity. And if it is compatible with relativity, then why call it non-locality! Why not globality, or global correlationality or something less confusing.

 

[Auto-Didact]

Well, that's of course not what we'd consider science. Then I can as well just stick to the observations themselves. A black box providing answers is not what we aim at in science. We want to reduce it to some (as few as possible) fundamental laws, which then however are indeed a "black box".

I agree with most of what you say, but disagree with this. Determining numerical parameters from a deeper mathematical theory is certainly a form of science, even if in the past a lot of time was wasted on doing this by physicists. Usually once such an explanation is found it tends to constitute a complete revolution in mathematics - e.g. an extended theory of analysis - which is capable of turning existing physical theory upside down if the physicists realize that this is possible.

The problem is that the realization that some new idea constitutes fundamental progress - e.g. the mathematical necessity of non-locality for physics, which will probably require the invention of a new form of mathematics in tandem in order to describe it - usually through some novel form of unification of concepts, occurs at a vastly different rate in physics compared to in mathematics.

Another problem is that many different factions of physicists in different subfields spontaneously arise which each take the same validated mathematical methods (e.g. perturbative methods) as well as their own personally preferred underlying model far too seriously, such that the field of physics as a whole has become terribly fractured and horrible at effectively communicating about most partial actual new results.

 

[Demystifier]

If their statement of locality is wrong, I don't understand how this is consistent with the microcausality constraint of relativistic QFT.

But I explained it to you like 1000 times. The nonlocality is the property of the (hypothetic) more fundamental variables, called also "hidden variables". Those more fundamental variables are not described by relativistic QFT, so they do not need to obey the microcausality constraint of relativistic QFT.

A possibly useful analogy is thermodynamics vs atomic physics. Thermodynamics obeys the second law (the entropy increase law), but the more fundamental theory of atoms does not need to obey the second law. Just as there is no contradiction in the fact that the second law can emerge from a more fundamental theory which does not obey the second law, likewise there is no contradiction in the possibility that microcausality constraint can emerge from a more fundamental theory which does not obey the microcausality constraint.

 

[Demystifier]

There's indeed no need for HV's as long as none are observed

For me, that's like saying that Boltzmann had no need to introduce his theory of atoms as a statistical explanation of thermodynamics.

 

[Demystifier]

Of course, I'm not converted to Bohmianism as long as there is no convincing relativistic version of it.

Does it mean that you would be converted to Bohmianism if a relativistic version existed? If so then why? Why would Bohmian version of relatvistic QFT, if existed, would be better then ordinary relativistic QFT?

 

[Demystifier]

I think what's also often missing in all these discussions (including the ones running in the parallel thread on entanglement swaping a la Zeilinger et al) is that never the complete state of the entangled photons is considered. It's just written down the polarization part but not the momentum part, which is however important in the considerations of locality arguments. I think this deserves another FAQ/Insight article. Maybe then my point of view becomes clearer too.

It's in fact straightforward to add the momentum part, but it doesn't help to understand the issue of (non)locality. Usually only the polarization part is analyzed explicitly because only the polarization is measured in experiments that violate Bell-like inequalities.

 

[A. Neumaier]

I think what's also often missing in all these discussions (including the ones running in the parallel thread on entanglement swaping a la Zeilinger et al) is that never the complete state of the entangled photons is considered. It's just written down the polarization part but not the momentum part, which is however important in the considerations of locality arguments. I think this deserves another FAQ/Insight article. Maybe then my point of view becomes clearer too.

The momentum part is trivial; just take for each particle's part the tensor product with a smeared momentum state. Up to a joint phase, these remain constant throughout, hence can be factored out.

 

[vanhees71]

Sure, writing it in the form of $|\text{momentum part} \rangle \otimes |\text{polarization part} \rangle$ it's (here for the singlet Bell state in the polarization part)
$$(|\vec{p}_1\rangle \otimes |\vec{p}_2 \rangle - |\vec{p}_2 \rangle \otimes |\vec{p}_1 \rangle) \otimes (|H \rangle \otimes |V \rangle -|V \rangle \otimes H \rangle).$$
For wave packets just integrate over the momenta with some $L^2$-function weight (e.g., a Gaussian).

 

[zonde]

Sure, writing it in the form of $|\text{momentum part} \rangle \otimes |\text{polarization part} \rangle$ it's (here for the singlet Bell state in the polarization part)
$$(|\vec{p}_1\rangle \otimes |\vec{p}_2 \rangle - |\vec{p}_2 \rangle \otimes |\vec{p}_1 \rangle) \otimes (|H \rangle \otimes |V \rangle -|V \rangle \otimes H \rangle).$$
For wave packets just integrate over the momenta with some $L^2$-function weight (e.g., a Gaussian).

Entanglement is between particles not observables. The state you wrote is just meaningless as polarization part already includes momentum:
$|H\rangle_{\vec{p}_1}\otimes|V\rangle_{\vec{p}_2} -|V\rangle_{\vec{p}_1}\otimes|H\rangle_{\vec{p}_2}$

 

[vanhees71]

Entanglement is a property of states.

What I wrote down is
$$|\vec{p}_1,H \rangle \otimes |\vec{p}_2,V \rangle - |\vec{p_1},V \rangle \otimes |\vec{p}_2,H \rangle - |\vec{p}_2,H \rangle \otimes |\vec{p}_1,V \rangle +|\vec{p}_2,V \rangle \otimes |\vec{p}_1,H \rangle
= [\hat{a}^{\dagger}(\vec{p}_1,H) \hat{a}^{\dagger}(\vec{p}_2,V) - \hat{a}^{\dagger}(\vec{p}_1,V) \hat{a}^{\dagger}(\vec{p}_2 H)] |\Omega \rangle.$$
The corresponding normalized state ket is given by this vector $\times 1/2$.

It takes into account the Bose nature of photons. Your state is not symmetrized as a two-particle (two-photon) bosonic state should be.

 

[PeterDonis]

polarization part already includes momentum

I have no idea what you mean by this, since the polarization degrees of freedom are different from the momentum degrees of freedom; neither one can "include" the other.

It might be the case that the momentum and polarization degrees of freedom are entangled, but that is not the same as polarization "including" momentum.

 

[zonde]

I have no idea what you mean by this, since the polarization degrees of freedom are different from the momentum degrees of freedom; neither one can "include" the other.

It might be the case that the momentum and polarization degrees of freedom are entangled, but that is not the same as polarization "including" momentum.

Well, maybe "included" is not the right word. Polarization modes are indexed by spatial directions. In entaglement measurements interference happens between polarization modes in the same spatial direction. No interference is created between different spatial directions. That relationship between different spatial directions is revealed only indirectly by comparing outcomes of polarization mode interference measurements.
So it's all tied together. Two polarization modes and two spatial directions makes four combined modes that are measured pairwise and then compared.

 

[vanhees71]

I've no clue what you are talking about. Let's talk about two photons with momenta $\vec{p}_1$ and $\vec{p}_2$. The natural complete one-photon basis is the momentum-helicity-eigenbasis $|\vec{p},\lambda$ where $\vec{p} \in \mathbb{R}^3$ and $\lambda \in \{-1,1\}$. The helicity is the projection of the total angular momentum (sic!) to the direction of momentum.

A basis for the two-photon states are
$$|1_{\vec{p}_1,\lambda_1},1_{\vec{p}_2,\lambda_2} \rangle = \frac{1}{\sqrt{2}} \hat{a}^{\dagger}(\vec{p}_1,\lambda_1) \hat{a}^{\dagger}(\vec{p}_2,\lambda_2) |\Omega \rangle.$$
By definition these states describe non-entangled photons. Note however that these are not simply product states of single photons, but the creation operators automatically imply the symmetrization necessary to describe the photons as indistinguishable bosons.

A basis of maximally entangled two-photon states is given by
$$|\phi_{12}^{\pm} \rangle = \frac{1}{2} \left [\hat{a}^{\dagger}(\vec{p}_1,1) \hat{a}^{\dagger}(\vec{p}_2,1) \pm \hat{a}^{\dagger}(\vec{p}_1,-1) \hat{a}^{\dagger}(\vec{p}_2,-1) \right]$$
and
$$|\psi_{12}^{\pm} \rangle = \frac{1}{2} \left [\hat{a}^{\dagger}(\vec{p}_1,1) \hat{a}^{\dagger}(\vec{p}_2,-1) \pm \hat{a}^{\dagger}(\vec{p}_1,-1) \hat{a}^{\dagger}(\vec{p}_2,1) \right].$$
For details about entanglement of indistinguishable particles, see e.g.

J. C. Garrison, R. Y. Chiao, Quantum Optics, Oxford University Press (2008)

 

[PeterDonis]

Polarization modes are indexed by spatial directions.

No, they aren't. I think you are confusing photon polarization with spin measurements on spin-1/2 particles like electrons. Both lead to the Hilbert space of a qubit for the relevant degrees of freedom, but the physical measurements are not the same.

 

[zonde]

No, they aren't. I think you are confusing photon polarization with spin measurements on spin-1/2 particles like electrons. Both lead to the Hilbert space of a qubit for the relevant degrees of freedom, but the physical measurements are not the same.

Of course they are. Look here: Entangled photons, nonlocality and Bell inequalities in the undergraduate laboratory, page 2
Here spatial directions are called "idler" and "signal". If you have any doubts that "idler" and "signal" are spatial directions in page 3 you can read sentence:
"By placing polarizers rotated to angles $\alpha$ and $\beta$ in the signal and idler paths, respectively, we measure the polarization of the downconverted photons."

 

[PeterDonis]

Here spatial directions are called "idler" and "signal".

Yes, those are spatial directions, which are different degrees of freedom from the polarization degrees of freedom. Different polarization measurements are performed on the photons going in those two spatial directions, but that does not mean the polarization degrees of freedom are the same as the spatial direction degrees of freedom.

 

[zonde]

Yes, those are spatial directions, which are different degrees of freedom from the polarization degrees of freedom. Different polarization measurements are performed on the photons going in those two spatial directions, but that does not mean the polarization degrees of freedom are the same as the spatial direction degrees of freedom.

Of course polarization degrees of freedom and spatial direction degrees of freedom are different. There are both orthogonal polarization modes in each spatial direction for entangled state. And that's what goes into description of entangled state. There is $|H_s\rangle$ and there is $|H_i\rangle$. So when someone writes entangled state omitting these subscripts, like: $|\psi^+\rangle=\frac{1}{\sqrt{2}}(|H\rangle\otimes|V\rangle\ + |V\rangle\otimes|H\rangle)$ different spatial directions are implied for the same polarization modes.

 

[zonde]

I've no clue what you are talking about.

You can take a look at this paper: Entangled photons, nonlocality and Bell inequalities in the undergraduate laboratory

 

[vanhees71]

Well, this paper uses the usual simplification, treating the photons as if they were distinguishable. The correct polarization-singlet state, idealized as momentum eigenstates is
$$|\Psi^{-} \rangle = \frac{1}{2} [\hat{a}^{\dagger}(\vec{p}_1,H) \hat{a}^{\dagger}(\vec{p}_2,V) -\hat{a}^{\dagger}(\vec{p}_1,H) \hat{a}^{\dagger}(\vec{p}_2,V) ] |\Omega \rangle.$$
The location of the registration events of these photons comes in by placing the detectors at their locations and then what's well defined are the two-photon-detection and/or single-photon probabilities given by the expectation values of appropriate electric-field operators.

 

[zonde]

Well, this paper uses the usual simplification, treating the photons as if they were distinguishable. The correct polarization-singlet state, idealized as momentum eigenstates is
$$|\Psi^{-} \rangle = \frac{1}{2} [\hat{a}^{\dagger}(\vec{p}_1,H) \hat{a}^{\dagger}(\vec{p}_2,V) -\hat{a}^{\dagger}(\vec{p}_1,H) \hat{a}^{\dagger}(\vec{p}_2,V) ] |\Omega \rangle.$$

Does this "correct" state description produce the same predictions for entanglement measurements as the "incorrect" one?

 

[zonde]

There is an experiment demonstrating polarization entangled photons with different wavelength:
Three-color Sagnac source of polarization-entangled photon pairs
It seems that polarization entangled photons can be quite distinguishable.

 

[PeterDonis]

when someone writes entangled state omitting these subscripts, like: $|\psi^+\rangle=\frac{1}{\sqrt{2}}(|H\rangle\otimes|V\rangle\ + |V\rangle\otimes|H\rangle)$ different spatial directions are implied for the same polarization modes.

Different spatial directions are implied for the first ket vs. the second ket in each term. That's because each ket (first and second in each term) refers to a distinct photon, which is distinguishable by the direction in which it is moving. The more complete expressions that @vanhees71 is writing down make all this explicit.

I don't know if this is what you mean by "the same polarization modes" or not, but if it is, your choice of words is very poor, and your posts are only going to cause confusion.

 

[vanhees71]

Does this "correct" state description produce the same predictions for entanglement measurements as the "incorrect" one?

Fortunately yes.

 

[vanhees71]

There is an experiment demonstrating polarization entangled photons with different wavelength:
Three-color Sagnac source of polarization-entangled photon pairs
It seems that polarization entangled photons can be quite distinguishable.

Well, what I wrote is a general two-photon state with different momenta. Of course you can also have $E_j=\hbar \omega_j=c |\vec{p}_j|$ different.

Nevertheless the photons are indistinguishable in the sense that it's a bosonic and not a product state. Already
$$
|1_{\vec{p}_1,\lambda_1},1_{\vec{p}_2,\lambda_2} \rangle
=\frac{1}{\sqrt{2}} \hat{a}^{\dagger}(\vec{p}_1,\lambda_1)
\hat{a}^{\dagger}(\vec{p}_2,\lambda_2) |\Omega \rangle =
\frac{1}{\sqrt{2}} (|\vec{p}_1,\lambda_1 \rangle \otimes
|\vec{p}_2,\vec{\lambda}_2 \rangle + |\vec{p}_2,\lambda_2 \rangle
\otimes |\vec{p}_1,\lambda_1 \rangle)$$
is not a product state.

Of course the photons are distinguishable in some sense by their momenta, and what's indeed entangled for this state is that in this state the photon with momentum $\vec{p}_1$ necessarily carries helicity $\lambda_1$ and that with $\vec{p}_2$ carries $\lambda_2$.

The "different spatial directions" discussed in other postings are of course the different directions of the photons' momenta.

 

[zonde]

Nevertheless the photons are indistinguishable in the sense that it's a bosonic and not a product state. Already
$$
|1_{\vec{p}_1,\lambda_1},1_{\vec{p}_2,\lambda_2} \rangle
=\frac{1}{\sqrt{2}} \hat{a}^{\dagger}(\vec{p}_1,\lambda_1)
\hat{a}^{\dagger}(\vec{p}_2,\lambda_2) |\Omega \rangle =
\frac{1}{\sqrt{2}} (|\vec{p}_1,\lambda_1 \rangle \otimes
|\vec{p}_2,\vec{\lambda}_2 \rangle + |\vec{p}_2,\lambda_2 \rangle
\otimes |\vec{p}_1,\lambda_1 \rangle)$$
is not a product state.

I don't follow you. Twin photons generated in PDC process with known output polarizations are not entangled. From what you wrote it seems you claim that such photons are entangled. So I doubt that I understand you correctly.

Of course the photons are distinguishable in some sense by their momenta, and what's indeed entangled for this state is that in this state the photon with momentum $\vec{p}_1$ necessarily carries helicity $\lambda_1$ and that with $\vec{p}_2$ carries $\lambda_2$.

I'm not sure why you call such relationship an "entanglement". Can you use such relationship to violate Bell inequality? I don't see how. To me it seems just like classical determinism.

 

[DrChinese]

I don't follow you. Twin photons generated in PDC process with known output polarizations are not entangled. From what you wrote it seems you claim that such photons are entangled. So I doubt that I understand you correctly.

Not sure if this is the sticking point between you and PeterDonis or not. Generally PDC can EITHER produce polarization entangled pairs OR non-polarization entangled pairs. It depends on the specific setup. I think this is what you are referring to.

 

[vanhees71]

What we are talking about are pdc pairs prepared in one of the four Bell states
$$|\phi_{12}^{\pm} \rangle=\frac{1}{2} \left [\hat{a}^{\dagger}(\vec{p}_1,H) \hat{a}^{\dagger}(\vec{p}_2,H) \pm \hat{a}^{\dagger}(\vec{p}_1,V) \hat{a}^{\dagger}(\vec{p}_2,V) \right], \\
|\psi_{12}^{\pm} \rangle=\frac{1}{2} \left [\hat{a}^{\dagger}(\vec{p}_1,H) \hat{a}^{\dagger}(\vec{p}_2,V) \pm \hat{a}^{\dagger}(\vec{p}_1,V) \hat{a}^{\dagger}(\vec{p}_2,H) \right].$$
Of course I've idealized this a bit. In reality there are always proper normalizzble Hilbert-space vectors with the momenta having a distribution of finite width.

 

[zonde]

What we are talking about are pdc pairs prepared in one of the four Bell states
$$|\phi_{12}^{\pm} \rangle=\frac{1}{2} \left [\hat{a}^{\dagger}(\vec{p}_1,H) \hat{a}^{\dagger}(\vec{p}_2,H) \pm \hat{a}^{\dagger}(\vec{p}_1,V) \hat{a}^{\dagger}(\vec{p}_2,V) \right], \\
|\psi_{12}^{\pm} \rangle=\frac{1}{2} \left [\hat{a}^{\dagger}(\vec{p}_1,H) \hat{a}^{\dagger}(\vec{p}_2,V) \pm \hat{a}^{\dagger}(\vec{p}_1,V) \hat{a}^{\dagger}(\vec{p}_2,H) \right].$$
Of course I've idealized this a bit. In reality there are always proper normalizzble Hilbert-space vectors with the momenta having a distribution of finite width.

You can make careful arrangements of PDC crystal(-s) and pump beam so that two different downconversion processes contribute orthogonal polarization modes (or probability amplitudes in theoreticians language) to the same spatial modes. And in addition you can make very careful arrangements so that two polarization modes in the same spatial mode can interfere with very high visibility.
If you do these things you will get polarization entangled photon pairs.

On the other hand if you do only basic arrangements of PDC process you will get photon pairs with perfectly known polarizations. They will still show a lot of correlations, but all these can be explained using correlated LHVs that are determined at the source.
In post #55 I was talking about this second ("raw") type of PDC process.

 

[vanhees71]

What does "LHVs" mean? Of course, one has to do the right manipulations with the photons to get (approximately) the said Bell states. As a theorist (and not a quantum-optics expert) I'm only interested in the fact that the experimentalists can do that somehow with great accuracy.

 

[Demystifier]

What does "LHVs" mean?

Local hidden variables.