Why is Quantum Field Theory Local?

[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?

 

[A. Neumaier]

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

The classic is by Currie, Jordan and Sudarshan.

 

[A. Neumaier]

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.

? Existence or nonexistence does not figure in my proof.

Like Bell I assume some natural properties and prove that they imply things contradicting quantum mechanics.

Just as Bell in his work; his local hidden variable objects cannot interfere either. Hence in your opinion, his work should be equally uninteresting.

Interference is the reason why a Bell inequality can be violated.

Bohmian theory is not classical in that sense, so your analysis does not exclude the Bohmian interpretation.

Why do you mention this triviality?

Bohmian theory is also not excluded by Bell's work.

 

[vanhees71]

The classic is by Currie, Jordan and Sudarshan.

Yes, and the no-go theorem holds for any finite number of degrees of freedom:

https://link.springer.com/article/10.1007/BF02749856

So the "natural" relativistic dynamics is field-theoretical rather than point-particle like.

 

[A. Neumaier]

Actually, this is more or less done in the book by Mandel and Wolf cited in post #154.

However, their exposition is not complete enough to enable me to gain full understanding of what is happening, and why. The reason is that there is no connection between the QFT discussion of coincidence counts using 2-point correlations in Chapter 14 and the nonrelativistic QM discussion of Bell's inequality in Chapter 12.14. A good exposition should connect the two. In particular, what is missing is a discussion of how the two dichotomic observables $A(a)$ and $B(b)$ introduced in Section 12.4.2 are realized in QFT. They are informally postulated but nowhere shown to exist in terms of the QFT machinery introduced.

I am convinced that any construction of these would reveal that they are horribly nonlocal expressions in the quantum fields. This would constitute the natural explanation why Bell nonlocality is experimentally seen.

 

[A. Neumaier]

Yes, and the no-go theorem holds for any finite number of degrees of freedom:

https://link.springer.com/article/10.1007/BF02749856

So the "natural" relativistic dynamics is field-theoretical rather than point-particle like.

Yes, the only well-defined one.

 

[Demystifier]

Just as Bell in his work; his local hidden variable objects cannot interfere either.

I don't think it's true. As far as I can see, his proof does not have an analog of the first line in your 3-line equation.

Let me also add that the experiment you describe can be explained by a (Bell) local hidden-variable theory. Namely, 1-particle Bohmian mechanics is a local theory.

 

[vanhees71]

However, their exposition is not complete enough to enable me to gain full understanding of what is happening, and why. The reason is that there is no connection between the QFT discussion of coincidence counts using 2-point correlations in Chapter 14 and the nonrelativistic QM discussion of Bell's inequality in Chapter 12.14. A good exposition should connect the two. In particular, what is missing is a discussion of how the two dichotomic observables $A(a)$ and $B(b)$ introduced in Section 12.4.2 are realized in QFT. They are informally postulated but nowhere shown to exist in terms of the QFT machinery introduced.

I am convinced that any construction of these would reveal that they are horribly nonlocal expressions in the quantum fields. This would constitute the natural explanation why Bell nonlocality is experimentally seen.

But isn't it enough that the correlators describe the experiments correctly, which violate Bell's inequality? Also, if the however constructed observables are nonlocal, then they are not in accordance with Bell's class of "local realistic" HV theories.

Unfortunately I don't have Mandel and Wolf's book at hand. I've to check the details tomorrow.

 

[Demystifier]

Yes, the only well-defined one.

String theory can also describe interactions in a Lorentz covariant manner.

 

[A. Neumaier]

Bell's argument is applicable to any local beables, ...

In Bell's words (Speakable and unspeakable..., p.8/9): 'That so much follows from such apparently innocent assumptions leads us to question their innocence. Are the requirements imposed, which are satisfied by quantum mechanical states, reasonable requirements on the dispersion free states? Indeed they are not. [...] The danger in fact was not in the explicit but in the implicit assumptions. It was tacitly assumed that [...]''

Indeed, Bell tacitly assumes a lot - whenever he talks about beable he assumes that all beables are local!

Since any local hidden variable theory contains lots of nonlocal beables, the experimental violation of Bell's inequalities does not rule out local hidden variable theories. It only proves that the beables measured are not local.

Indeed, my paper exhibited an explicit case of a local hidden variable field theory that violates Bell inequalities - by the same arguments used in the paper to show the violation of (2).

... 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.)

Bell's argument proves that bounded local beables in local hidden variable theories satisfy a Bell inequality. This includes both pointlike particles and fields. (But it excludes multi-local beables that appear in my thermal interpretation.)

However, local beables are not characterized by being variables defined on spacetime positions. There are many possible theories with nonlocal fields defined on spacetime positions, the prime example being Newtonian gravity.

Instead, local beables are characterized by Bell's locality definition - that they are influenced only by their past light cone. This is a much more restrictive condition!

 

[A. Neumaier]

I don't think it's true. As far as I can see, his proof does not have an analog of the first line in your 3-line equation.

He adds up probabilites in (6) p.37 of 'Speakable...' This is not permitted if there is interference.

Let me also add that the experiment you describe can be explained by a (Bell) local hidden-variable theory. Namely, 1-particle Bohmian mechanics is a local theory.

The free Maxwell field discussed in my paper predated quantum mechanics, is also a (Bell) local hidden-variable theory, and also explains the experiment.