Why is Quantum Field Theory Local?

[A. Neumaier]

I disagree. The Bell's argument is applicable to any local beables, namely variables defined on spacetime positions. This includes both pointlike particles and fields. (But it excludes multi-local beables that appear in your thermal interpretation.)

My experiment linked to in post #102 predicts violation of Bell inequalities using only classical electrodynamics in vacuum, which is a local and causal relativistic theory.

 

[A. Neumaier]

For many claims, there are proofs of both the claim and the negation. You have to take such things with a grain of salt. You're claiming to have done something that others have proved can't be done. Obviously, either someone has made a mistake, or there are subtle differences in the interpretations of key concepts.

Bell didn't claim to have done anything not involving particles - you did!

The question is, which grain of salt is needed in each case, and which one is convincing! Of course there are subtle differences in the interpretations of key concepts, but these differences are the key point.

 

[stevendaryl]

Note that you still have particles traveling, not fields!

No, that's not true. The setup doesn't say anything about whether the experiments involve particles or whatever. I was using EPR as an example. The only thing that is relevant is that Alice is performing an experiment in a localized region of spacetime $A$, Bob is performing an experiment in a spacelike separated region of spacetime $B$.

 

[Demystifier]

My experiment linked to in post #148 predicts violation of Bell inequalities

No it doesn't, your paper does not deal with Bell inequalities at all.

 

[stevendaryl]

Bell didn't claim to have done anything not involving particles - you did!

He was using particles to illustrate the concept, which doesn't have anything specifically to do with particles.

 

[A. Neumaier]

No it doesn't, your paper does not deal with Bell inequalities at all.

It is not the traditional Bell inequality, but it is of exactly the same kind. namely a constraint of the values of some linear combination of probabilities, in my case the expression (1), where local hidden variable theories predict (with Bell's arguments) even a constant value rather than an inequality only.

All of Bell's arguments also apply to my classical hidden variable reasoning!

 

[Demystifier]

Then why does my single photon experiment demonstrate apparent Bell nonlocality though it is explained by Maxwell's classical local field equations?

Are you talking about your arXiv:0706.0155? This paper only shows that a classical hidden variable theory that does not involve interference is not compatible with QM. Quite trivial and uninteresting result in my view. And since it is about single photon, it has absolutely nothing to do with Bell nonlocality.

 

[stevendaryl]

Bell's inequality is not about particles.

We have a probability of the form $P(R_A = a \wedge R_B = b | O_A = \alpha \wedge O_B = \beta)$

where $R_A$ is the result of Alice's measurement, $R_B$ is the result of Bob's measurement, $O_A$ is Alice's choice of detector setting, $O_B$ is Bob's choice of detector setting.

Bell assumed that such a probability "factors" once you know the common causal influences of Alice's result and Bob's result. In terms of the spacetime regions I mentioned above, $E$ is the common backwards lightcone of Alice's and Bob's measurements. Bell assumed that, under the assumption that there is no causal influence of Alice's measurement on Bob, nor vice-versa, then there is some fact about region $E$, call it $F(E)$ such that knowing that fact would allow us to factor the probabilities:

$P(R_A = a \wedge R_B = b | O_A = \alpha \wedge O_B = \beta)$
$= \sum_\lambda P_E(F(E) = \lambda) P_A(R_A = a | O_A = \alpha \wedge F(E) = \lambda) P_B(R_B = b | O_B = \beta \wedge F(E) = \lambda)$

where $P_E$ gives the probability of region $E$ having property $\lambda$, $P_A$ is the probability of Alice's results given her setting and the hidden variable $\lambda$, and $P_B$ is the probability of Bob's results given his setting and the hidden variable.

There is nothing about particles in the mathematical derivation.

 

[A. Neumaier]

He was using particles to illustrate the concept, which doesn't have anything specifically to do with particles.

Bell's inequality is not about particles.

We have a probability of the form $P(R_A = a \wedge R_B = b | O_A = \alpha \wedge O_B = \beta)$

The mathematics is independent of any nonlocality issues. Nonlocality only enters through its interpretation, which involves particles - something that moves from one place to another.

 

[A. Neumaier]

Are you talking about your arXiv:0706.0155? This paper only shows that a classical hidden variable theory that does not involve interference is not compatible with QM. Quite trivial and uninteresting result in my view. And since it is about single photon, it has absolutely nothing to do with Bell nonlocality.

Measured are correlations that are exactly as nonlocal (or as little nonlocal) as those in the interpretation of Bell inequalities, though the classical field propagates locally.

 

[Demystifier]

Measured are correlations that are exactly as nonlocal as those in the interpretation of Bell inequalities

In your paper I don't see any nonlocal correlations resembling those of Bell.

 

[stevendaryl]

The mathematics is independent of any nonlocality issues. Nonlocality only enters through its interpretation, which involves particles - something that moves from one place to another.

You're wrong about that. But I know from past discussions in PF to just drop it at this point.

 

[A. Neumaier]

In your paper I don't see any nonlocal correlations resembling those of Bell.

My paper is about nonclassical correlations of a tensor product of two-state systems, precisely the situation considered by Bell in his paper ''On the problem of hidden variables in quantum mechanics''.

If you interpret it in terms of particles (photons) with a local hidden variable description you get wrong predictions - just as in the case Bell considered. If you interpret in terms of classical local fields you get the same predictions as quantum mechanics.

The nonlocality is in the fact that the measurements can be done arbitrarily far apart, precisely as in Bell's arguments.

 

[Demystifier]

My paper is about nonclassical correlations of a tensor product of two-state systems

I don't even know what that means, given that you have only one photon.

 

[A. Neumaier]

The mathematics is independent of any nonlocality issues. Nonlocality only enters through its interpretation, which involves particles - something that moves from one place to another.

One can check this easily by looking at a proof of the inequalities, which is just an exercise in probability theory without any physical content.

You're wrong about that. But I know from past discussions in PF to just drop it at this point.

That's the way to sweep the contradiction under the carpet.

 

[A. Neumaier]

I don't even know what that means, given that you have only one photon.

You didn't read carefully. There is an ensemble of photons emerging from the source. Otherwise one cannot make statistics.

 

[Demystifier]

You didn't read carefully. There is an ensemble of photons emerging from the source. Otherwise one cannot make statistics.

Is your $\psi$ a 1-photon state, a 2-photon state, or a many-photon state? If it is a 1-photon state (which I suspect it is), then what is the many-photon state describing your experiment?

 

[A. Neumaier]

You didn't read carefully. There is an ensemble of photons emerging from the source. Otherwise one cannot make statistics.

The point is that a field theory photon (in free space essentially an arbitrary solution of the Maxwell equations) is a much more complex object than a Bell photon (a 2 state system traveling in a particular direction).

To arrange a nonlocal measurement situation in Bell's case, one needs at least 2 simultaneous Bell photons moving in different directions, modeled by the tensor product, and has to generate an entangles ensemble of such photon pairs.

To arrange a nonlocal measurement situation in the field case, one just needs to observe that it was known long before the advent of quantum mechanics that a beam splitter splits a field concentrated along a single direction into one concentrated along two different directions. This holds for any solution of the free Maxwell equations and does not change the number of photons involved in the experiment. Once one has two directions, the state of each photon entering the beam splitter is transformed in a state along these two directions described by a tensor product.

Once one has the tensor product (whether in Bell's way or by a beam splitter), the (interpretation-independent) math of the tensor product applies, and this gives rise to all entanglement arguments that have ever been considered, with all their consequences. In particular, it predicts Bell nonlocality.

That I discussed not the original Bell inequalities was mainly because predicting an always valid equality from hidden variables is experimentally much easier to refute than predicting an inequality that holds only in special situations (preparations with a striongly nonclassical Wigner distribution).

 

[A. Neumaier]

Is your $\psi$ a 1-photon state, a 2-photon state, or a many-photon state? If it is a 1-photon state (which I suspect it is), then what is the many-photon state describing your experiment?

$\psi$ is a 1-photon state in the field theoretic sense, i.e., an arbitrary solution of the free Maxwell equations.

If the field is concentrated along a single ray (as the field produced by a strong laser), it can be approximated by a 2-state vector describing the polarization, together with giving the classically described direction of the ray.

After beam splitting, the field is concentrated instead along two rays and can only be approximated by a tensor product of 2-state vector describing the polarization, together with the classically described directions of the two rays.

The many photon aspect comes from the fact that a strong laser produces the photons (described by a coherent state) at a given (high) rate - arbitrarily many if you hve it on for a long enough time. (Making individual photons is in comparison very hard, and one of the difficulties of performing standard Bell tests.)

In the Heisenberg picture in which relativistic fields are generally described, the coherent photon state does not change at all. It describes a stationary source producing a continuously streaming ensemble of photons. The detector clicks turn these into individual observed photons.

Note that QFT does not predict the clicks themselves, only the rate of producing clicks, or the parameters of the stochastic process for the temporal distribution of clicks when there are multiple detectors.

 

[Demystifier]

Once one has the tensor product(whether in bell's way or by a beam splitter), the (interpretation-independent) math of the tensor product applies

I think it's wrong. The beam splitter does not create a tensor product. It creates a superposition.

 

[Demystifier]

$\psi$ is a 1-photon state in the field theoretic sense, i.e., an arbitrary solution of the free Maxwell equations.

I don't think that anybody else calls it a 1-photon state.

 

[A. Neumaier]

The beam splitter does not create a tensor product. It creates a superposition.

It creates a new solution of the Maxwell equations whose simplest description is that of a superposition of two 2-level states in the tensor product. Thus the tensor product machinery of entanglement applies. Whereas before passing the beam splitter, the solution is described by a single 2-level state.

I don't think that anybody else calls it a 1-photon state.

An ideal beam splitter (assumed in typical theoretical discussions of quantum optic experiments) has no dissipation and therefore does not change the total photon number, which equals the intensity of the beam. Thus if one photon goes in, one photon goes out.

This is the essential difference between particles and fields. With classical fields you can have superposition, with classical particles you cannot.

Of course, with a laser, $\psi$ is not a 1-particle state but a coherent state, but with low intensity it is very little different from a 1-particle state. The predictions with a coherent state agree with the prediction of classical electrodynamics (until the detector is hit - when one needs quantum mechanics to get the photon statistics).

 

[A. Neumaier]

Bell's inequality is not about particles.

We have a probability of the form $P(R_A = a \wedge R_B = b | O_A = \alpha \wedge O_B = \beta)$

where $R_A$ is the result of Alice's measurement, $R_B$ is the result of Bob's measurement, $O_A$ is Alice's choice of detector setting, $O_B$ is Bob's choice of detector setting.

Bell assumed that such a probability "factors" once you know the common causal influences of Alice's result and Bob's result. In terms of the spacetime regions I mentioned above, $E$ is the common backwards lightcone of Alice's and Bob's measurements. Bell assumed that, under the assumption that there is no causal influence of Alice's measurement on Bob, nor vice-versa, then there is some fact about region $E$, call it $F(E)$ such that knowing that fact would allow us to factor the probabilities:

$P(R_A = a \wedge R_B = b | O_A = \alpha \wedge O_B = \beta)$
$= \sum_\lambda P_E(F(E) = \lambda) P_A(R_A = a | O_A = \alpha \wedge F(E) = \lambda) P_B(R_B = b | O_B = \beta \wedge F(E) = \lambda)$

where $P_E$ gives the probability of region $E$ having property $\lambda$, $P_A$ is the probability of Alice's results given her setting and the hidden variable $\lambda$, and $P_B$ is the probability of Bob's results given his setting and the hidden variable.

There is nothing about particles in the mathematical derivation.

... and nothing about nonlocality. Not even anything about measurement - this only appears in the labels attached to the formulas that have nothing to do with the mathematical derivation, only with the interpretation.

Using your reasoning with respect to particles but applying it to the other concepts we see that Bell's inequality is not about nonlocality or about measurement, but just an exercise in classical probability theory.

 

[Demystifier]

It creates a new solution of the Maxwell equations whose simplest description is that of a superposition of two 2-level states in the tensor product. Thus the tensor product machinery of entanglement applies. Whereas before passing the beam splitter, the solution is described by a single 2-level state.

Fine, but my point is that your hidden variable prediction (2) rests on some additional assumptions (besides locality) that actual Bell inequality does not assume. In particular, you assume that there is no interference, which Bell inequality does not assume. So your no-go theorem is correct, but very different from Bell inequality. In fact, conceptually your no-go theorem is much more similar to the von Neumann no-go theorem, which now is generally considered to be trivial and uninteresting.

 

[A. Neumaier]

I don't think that anybody else calls it a 1-photon state.

You are misled by your habit of only working with finite-dimensional Hilbert spaces!

Mandel and Wolf introduce photons on page 479. A basis of the Hilbert space of photons is given by the special (unnormalized) 1-particle states $|1_{ks},\{0\})$ describing a state with one photon of wave vector $k$ and polarization $s$. The general 1-photon state is a superposition of these, hence an integral
$\psi:=\int \psi(k,s) |1_{ks},\{0\})\ds$; $\psi(k,s)$ equals the analytic signal (a Fourier transform) of the electric field associated with $\psi$, a solution of the free Maxwell equation. (For simplicity, I identified the two. In fact, in the Silberstein representation of the electromagnetic field, the Maxwell equation is just the zero mass, spin one analogue of the Dirac equation for a massive free spinor particle.)

An example of such a superposition is given on p. 480; they use in place of the integral the less precise but perhaps more intuitive sum notation. As stated there, ''In can be seen at once that $\psi$ is an eigenstate of the total photon operator with eigenvalue unity, so that it is also a one-photon state.'' The same holds for the general integral introduced above.

 

[A. Neumaier]

you assume that there is no interference, which Bell inequality does not assume.

?

Please justify your claim by pointing to the line in the paper where I make such an assumption!

 

[A. Neumaier]

So your no-go theorem is correct, but very different from Bell inequality.

If you substitute for my equation (1) the Bell expression and for my (2) the Bell inequality, you can repeat Bell's argument and find that (2) should hold under his assumptions, while (3) is predicted by ordinary quantum mechanics as in Bell's setting. Thus my setup can test the standard Bell inequality as well.

 

[Demystifier]

?

Please justify your claim by pointing to the line in the paper where I make such an assumption!

I think you implicitly assume it in the first line of the 3-line equation before (2). $\psi_D$ entails interference, while the first line seems to miss it.

 

[A. Neumaier]

I think you implicitly assume it in the first line of the 3-line equation before (2). $\psi_D$ entails interference, while the first line seems to miss it.

The equality in the first line holds since a classical hidden variable particle entering beam 1 passes exactly one of F(A_1) or F(A_2). There can be no interference since the classical particle does not have wave properties. The other two equalities are simple mathematical identitites.

Thus no assumption enters beyond what was specified explicitly.

 

[A. Neumaier]

He was using particles to illustrate the concept, which doesn't have anything specifically to do with particles.

In the book of reprints, 'Speakable and unspeakable...' Bell derives on p. 18. inequality (15), now called the Bell inequality, in the setting of the Bohm-Aharonov experiment, introduced on p.14. Then he generalizes on p.20 to 'systems', the concept introduced on p.14 (measurement on one system ... operations on a distant system). These systems cannot be understood as fields since a field is everywhere and cannot be distant from itself.

 

[stevendaryl]

... and nothing about nonlocality. Not even anything about measurement - this only appears in the labels attached to the formulas that have nothing to do with the mathematical derivation, only with the interpretation.

Using your reasoning with respect to particles but applying it to the other concepts we see that Bell's inequality is not about nonlocality or about measurement, but just an exercise in classical probability theory.

Yes, once you’ve assumed that Alice’s results depend only on facts in her backwards lightcone and Bob’s results depend only on facts in his backwards lightcone, then Bell’s inequality becomes just a fact about probability distributions.

But locality is the reason for that assumption.

 

[A. Neumaier]

Yes, once you’ve assumed that Alice’s results depend only on facts in her backwards lightcone and Bob’s results depend only on facts in his backwards lightcone, then Bell’s inequality becomes just a fact about probability distributions.

But locality is the reason for that assumption.

Particle locality, not field locality! - embodied in different choices for the beables.

The difference between field theory and particle theory is in what is claimed to be measured.

In QM (and in Bell-nonlocality discussions), a photodetector is regarded as a photon detector. This means that a click is taken to be a measurement of an event, specified by a projection operator, indicating the presence of a photon that traveled from the point of creation to the point of measurement along some path in the past light cone of the point of measurement. Here Bell's probability arguments matter, since by assumption, coincidence counts have a meaning in terms of the single counts at the two counters. Thus measurement results for the presence of particles are subject to Bell inequalities if they follow a local theory; two random variables and their product are involved.

On the other hand, in QFT, a photodetector is not treated as a photon detector but as a detector of electromagnetic radiation. QFT is about predicting rates of response, not probabilities of events. (See Chapter 14 of Mandel and Wolf - there all probabilities are infinitesimal; the stuff compared to experiment are the rates.) A single count is nearly meaningless from the point of view of measuring the quantity of interest, one cannot get a rate from a single event. The observed rate of response gives a measurement of electric field intensity, and the quality of the measurement becomes better as intensity * exposure time increases. These measurements behave classically in causally separated regions, as they correspond to local, commuting observables.

The coincindence rates do not measure the product of the electric field intensities at the two detectors, hence Bell's analysis says nothing at all about them. Indeed, coincidence counting was developed not for the purpose of detecting nonlocality of photons but - several years before before Bell's work - for measuring nonclassical states of light.

 

[PeterDonis]

Moderator's note: A large number of posts that were more about infinities than locality have been moved to a separate thread:

https://www.physicsforums.com/threads/infinities-in-qft-and-physics-in-general.1002079/

 

[Demystifier]

There can be no interference since the classical particle does not have wave properties.

So by assuming non-existence of wave properties, you find that there is no interference. It's logically correct, but in my opinion too trivial to be interesting. And needless to say, Bohmian theory is not classical in that sense, so your analysis does not exclude the Bohmian interpretation.

 

[vanhees71]

Only the history of physics is the other way around. But clearly, field theory is more fundamental than particle theory (which arises in the approximation of geometric optics). Thus QFT is more fundamental than QM.

There is not even a relativistic classical theory of multiple point particles - one can even prove a corresponding no-go theorem!

There are many such theorems. Which one are you referring to?