## Norm and correlations

Continuing the discussion from last time, today I want to talk about the norm of a linear operator and it's implication for the maximum correlations which can be achieved in nature: Tsirelson's bound. The very same norm definition would later on play a key role in what would unexpectedly become in the end a "geometrization" of the Standard Model coupled with (unquantized) gravity.

In a Hilbert space, the definition of the norm of a bounded linear operator is:

$$||A|| = sup (||Au||/||u||)$$ for $$u\ne 0$$

The most important properties of the norm for bounded operators are the triangle inequality:

$$||A+B|| \leq ||A|| + ||B||$$

and a multiplication identity which guarantees the continuity of multiplication:

$$||AB|| \leq ||A|| ||B||$$

(can we call this a triangle inequality for multiplication?)

On the basis of the triangle inequality, one may be tempted to explore the association of the notion of physical distance with the notion of the norm of an operator in a Hilbert space, but this is a dead end. The triangle inequality for operators is essential for quantum mechanics because it ensures the usual notions of convergence in functional analysis (most of functional analysis follows from it). The name of this blog is elliptic composability, and the "elliptic" part follows from the triangle inequality above. If one imagines a quantum mechanics where the triangle inequality is reversed, then one arrives at the unphysical hyperbolic quantum mechanics based on split-complex numbers which violates positivity which in turns prevents the usual definition of probability as a positive quantity.

There turns out however to be a deep and completely counter-intuitive relationship between the "sup" in the norm definition and the notion of physical distance, but more on this in subsequent posts-don't want to spoil the surprise, I only want to whet the (mathematical) appetite a bit.

Now back to correlations. Suppose we have four operators $$\sigma_\alpha, \sigma_\beta, \sigma_\gamma, \sigma_\delta$$ such that:

$${\sigma_\alpha}^2 = {\sigma_\beta}^2 = {\sigma_\gamma}^2 = {\sigma_\delta}^2 = 1$$
and
$$[\sigma_\alpha, \sigma_\beta] = [\sigma_\beta, \sigma_\gamma] = [\sigma_\gamma, \sigma_\delta] = [\sigma_\delta, \sigma_\alpha] = 0$$

If we define an operator $$C$$ as follows:

$$C= \sigma_\alpha \sigma_\beta + \sigma_\beta \sigma_\gamma + \sigma_\gamma \sigma_\delta - \sigma_\delta \sigma_\alpha$$

Then it is not hard to show that:

$$C^2 = 4 + [\sigma_\alpha, \sigma_\gamma][\sigma_\beta, \sigma_\delta]$$

From the triangle inequalities we have in general that:

$$||[A, B]|| = ||AB - BA|| = ||AB + (-B)A|| \leq ||AB|| + ||-BA||$$
$$= ||AB|| + ||BA|| \leq ||A||||B|| + ||A||||B|| = 2||A||||B||$$

and so

$$|| [\sigma_\alpha, \sigma_\gamma]|| \leq 2 ||\sigma_\alpha|| ||\sigma_\gamma|| = 2$$
$$|| [\sigma_\beta, \sigma_\delta]|| \leq 2 ||\sigma_\beta|| ||\sigma_\delta|| = 2$$

Therefore:

$$||C^2|| \leq 4+4$$
$$||C|| \leq 2 \sqrt{2}$$

And this is Tsirelson's bound because $$C$$ appears in the left hand side of the CHSH inequality.

Now one can read about textbook derivation of Tsirelson bound in many places, but the key point is that quantum correlations have their origin in the notion of operator norms in a Hilbert space. Nowhere in the derivation we have used the notion of distance or causality. Quantum correlations are a mathematical consequence of the quantum formalism which are in turn is a consequence of considerations of composition and information. Quantum mechanics is nothing but composition and information, and correlations (both quantum and classical) are nothing but considerations of composition and information as well.

The usual way we understand classical correlations as generated by a common cause is a parochial view due to our classical intuition. Yes, a common cause can generate correlations, but correlation does not imply causation

So what does all of this have to do with the notion of  distance and that of space-time? To uncover the link we would need first to generalize the notion of a space in geometry. Crazy talk? Not when it is based on von Neumann algebras research which led to a Fields medal. Please stay tuned...

1. Florin,

Thanks for this blog entry. It reduces the matter of the Tsirelson bound to simple terms. In fact your definition of the operator C is done in a way that is similar to correlation coefficients in Bell's theorem.

LC

1. You are welcome, but it is not my derivation: I got this from Asher Peres: Quantum Theory: Concepts and Methods.

2. Hi Florin, This is impressive.But I do not have Peres's book. What are these 4 sigma matrices (or operators) which commute with each other and their square is 1?

1. The 4 operators correspond to the observables in Aspect's experiment:

a common source of photons emits pairs of photons to the left and right measuring stations. At the measurement station a rapidly moving mirror pushes the photons to be measured for polarization either on direction alpha or gamma for the left detectors, or beta or delta for the right detector. The outcomes are +1 or -1.

Sigmas square are one because the outcome for the polarization analyzers are either +1 or -1.

Alpha and gamma commutes with beta and delta because they are spatially separated.

2. OK. Thanks. I am wondering how QM would violate this inequality.Is it by non commutativity of the corresponding Sigmas or something else?

3. Kashyap,

QM does not violate THIS inequality. This is the Tsirelson bound of maximum QM correlations. QM violates Bell's inequalities which put lower bounds on maximum correlations.

-Tsirelson follows from QM formalism
-Bell follows from local realism assumptions
-Tsirelson is higher than Bell and nature does violate Bell

Conclusion: since Bell is a mathematical theorem, the problem is with its assumptions: local realism. Nature is not locally realistic.

4. “Nature is not locally realistic.”

Or at least it doesn't admit a locally realistic description?

5. "Or at least it doesn't admit a locally realistic description?"

Is there a difference? How can nature admit a local realistic description if it is not locally realistic?

6. Florin,

"since Bell is a mathematical theorem, the problem is with its assumptions: local realism"

local realism is not the only assumption. The statistical independence or "free-will" is another assumption. This assumption is false in local-realistic theories (classical field theories) but for some reason you keep forgetting about it.

What you are in fact doing is rejecting local realism from the start (by assuming that free-will assumption is true) and then claim that it is the locality and/or realism that must be false because of Bell. This is fallacious.

The only reasonable understanding of Bell is that the free-will assumption must be false (it is also the only assumption that has nothing to do with physics, but it is based on a mixture of primitive intuition and some religious ideas).

Andrei

7. Yes, I think there is a difference and I don't understand why e.g. a Copenhagist (who I'd otherwise largely agree with) would even want to follow the ontologically conservative recognition that QM is description - “nothing more than a mathematical tool...” - with the ontologically radical assertion that “actual definite properties of a physical system "do not exist"”.

“How can nature admit a local realistic description if it is not locally realistic?”

I've no idea - probably it can't - but I don't see any reason why it couldn't be locally realistic but not admit a (useful, accurate) locally realistic description.

8. "local realism is not the only assumption. The statistical independence or "free-will" is another assumption."

Yes in theory, no in practice. I am yet to see a fully workable model of QM violating free will.

9. "I've no idea - probably it can't"

If no, then you have no foundation for your argument. Consider this:

- nature is local realistic implies a description that is locally realistic (I think there is no contention on this implication)
- locally realistic description implies locally realistic nature if one cannot construct a locally realistic description of a non-local realistic reality

So then they imply each other and they are equivalent.

The ontic vs epistemic status is an entirely distinct dichotomy without relevance here.

10. Yes - if it's true that LR(N) => ∃LRD(N) then I overreached. Either way I should've just said in the first place that the nonlocality or, preferably, nonrealism of QM doesn't imply that nature is non-local or non-real.

11. Florin,

"I am yet to see a fully workable model of QM violating free will."

I don't know what do you mean by "fully workable model" but if Bohm's theory qualifies then you have your example. It is true that the supporters of this interpretation use the non-local feature to explain Bell, but, nevertheless, free-will is also incompatible with it.

There is also a quite advanced interpretation of QM based on Couder's mechanism, where the oil bath is replaced by a classical background EM field. I am now reading the book:

"The Emerging Quantum"

that can be found here:

http://www.springer.com/us/book/9783319078922?token=prtst0416p2

The book seems serious. The quantum formalism is deduced from their local-realistic mechanism in both Schrodinger and Heisenberg version.

Andrei

3. The operator C = σ_ασ_β + σ_βσ_γ + σ_γσ_δ - σ_δσ_α, where the operators correspond to spin or polarization states measured at the positions, appears very similar to the Lorentz metric,

x*y = -x_0y_0 + x_1y_1 + x_2y_2 + x_3y_3,

which is the metric distance with Lorentz geometry or SO(3,1).. In the case of the Riemann sphere ~ CP^1 the set of conformal transformations are linear fractional transformations

z --> (az + b)/(cz + d)

where this transformation is isomorphic to PSL(2,C). The heavenly sphere is then the case of the null metric distance, or equivalently the projective light cone. The product space V of dim = n has the Jordan algebra is the v^2 = |v|^2 so that a spin factor J(V) ~ V⊕R (think of space plus time) such that

(u, α)◦(v, β) = (αv + βu, - αβ).

Then J(V) is isomorphic to Minkowski spacetime. This Clifford algebra define on the right the spacetime metric - ab.

So now suppose we let there be the spacetime V_c with the basis elements

u_1 = (σ_α, 0, 0), u_2 = (0, σ_β, 0), u_3 = (0, 0, σ_γ)

v_1 = (σ_β, 0, 0), u_2 = (0, σ_γ, 0), u_3 = (0, 0, σ_δ)

and the real line R containing the two elements (σ_α, σ_δ), it is then easy to see that the C operator can be expressed according to the Clifford algebra.

The connection between the null condition and the Tsirelson bound might be made by defining the elements of the real line R as (iσ_α + sqrt{2 sqrt{2}}, iσ_δ + sqrt{2 sqrt{2}}) so that the product is the real valued part so that this is -σ_δ σ_α + 2sqrt{2}. In that way the modified |C|^2 would be zero if it is at the Tsirelson bound, and similar to a choice of metric signature is negative if outside the Tsirelson bound.

This is a somewhat elementary or naive approach. We have though a monoid or magma connection between the Jordan algebra and the W^* algebra. This might be an interesting topic to pursue. It is my thinking that spacetime physics and quantum physics are on a deep level identical. In effect spacetime is probably built up from quantum entanglements.

1. Interesting math. However the analogy is only formal as two sigmas correspond to one measuring laboratory, and the other two to another measuring laboratory, while in Minkowski metric x,y,z,t correspond to the same point in space-time.

2. Technically the spacetime interval is evaluated between two points. The point is there is an isomorphism between these two, and that I think there is a sort of "magma" or monoid that constructs quantum mechanics and spacetime geometry.