Eric Forgy Ian Durham

category: people


Scratch pad.


Geometrodynamics is the antecedent to loop quantum gravity (didn’t recent experimental evidence suggest that LQG is probably not correct?).

Rainich conditions for the energy-momentum tensor to be sourced by the electromagnetic field:

A1. T i i=0T^i{}_i=0

A2. T i kT k jδ i jT^i{}_k T^k{}_j \cong \delta^i{}_j

A3. T ijv ivj0T_{ij}v^iv^j\geq 0

for any timelike vector v iv^i

Rainich’s differential condition:

D0. iS j= jS i,\nabla_i S_j=\nabla_j S_i,


S j:=ϵ ij klT mk lT imT rsT rs S_j:=\frac {\epsilon_{ij}{}^{kl}T_{mk}\nabla_l T^{im}} {T_{rs}T^{rs}}

Relational quantum mechanics

Relational quantum mechanics is a view of quantum mechanics first developed by Carlo Rovelli in 1997 (pdf). It is closely related to quantum Bayesianism in that it rejects the absolute similitude, or observer-independent, state of a system as well as rejecting all observer-dependent measured values. Rather, relational quantum mechanics abandons these notions in favor of relative quantum states and measured values, i.e. states and values that are relative to something. In short, it replaces the notion of absolute states with one that refers to the relation between physical systems. In many ways Rovelli’s argument bears resemblance to Einstein’s discussion of special relativity and, indeed, Rovelli mentions this analogy though warns readers not to take it too far.

Since relations between systems are at the core of this POV, categories would seem to be ideally suited as a formalism for describing relational quantum mechanics. On a rather simple level, for instance, the relations would simply become morphisms. In Mac Lane’s parlance, some of the relations would be arrows within a catgory or meta-category while others would be morphisms between categories, i.e. functors. It is largely context-dependent.

Ian: This seems awfully similar to Coecke’s categorical quantum mechanics in some sense.

Relational empiricism

Scratch pad for extension of above + QBism

What I am calling relational empiricism is inspired by stripping away much of the philosophical “baggage” of physics and proceeding from the simple facts that a) physics is inherently empirical since it is supposed to describe “reality” (or at least what we perceive as reality) and b) information only makes sense in context. In addition, it seeks to find a “common language” of sorts in which to discuss unification and emergence since many of the problematic aspects of the quantum/classical divide could be traced to the differing formalistic ways in which we implement each (e.g. time is generally considered to be absolute in quantum mechanics, and though this is largely out of convenience, it has created a formalism in which this idea is firmly embedded).

Guiding principles

  1. Empiricism: Science and particularly physics is inherently empirical in nature in that it is concerned with our perception of how the world works. As such we are concerned with measurements and observations.

  2. Relationalism: Measurable and/or observable quantities only possess a useful meaning when taken in some context. In other words, numerical results of measurements and observations are meaningless unless the are in relation to something. This is, of course, very close to relativism and, in fact, we apply the same basic principles except we go beyond mere reference frames. Note that any quantity that has units is automatically relational in nature. Frequently, however, quantities that are unitless are also relational, particularly if they describe a ratio. Another way to look at this is that every measurement and/or observation is ultimately an act of transforming or mapping information or data.


In seeking a common formalism from which both relativity and quantum mechanics can be either derived in their standard forms or in which their results can be found independently, the embodiment of the guiding principles is that such a formalism should somehow allow for numerical calculations while also embodying the relational nature of the quantities represented by the formalism. This is fairly simple in that it merely needs to include a set or sets of transformation rules. It seems as if category theory would be a naturally good fit for the relational portion, but it is unclear at this point how this might be adapted to include the calculational/numerical aspect.

One of the tricks would seem to be finding a formalism that embodies the deterministic aspect of classical physics while also capturing the probabilistic aspect of quantum physics. It appears that a portion of some common structure is embodied in the so-called quantal algebras (see Moldoveanu for an overview).


Both empiricism and relationalism can be related to the notion that relations often involve interactions and these interactions transfer some kind of information (momentum, energy, etc.).

Add critique of GR, etc…

Bell’s Theorem

Original derivation

Let us denote the result A of a measurement that is determined by a unit vector, a\vec{a}, and some parameter λ\lambda as A(a,λ)=±1A(\vec{a},\lambda)=\pm 1 where we further suppose that the outcome of the measurement is either +1 or -1. Likewise, we may do the same for the result B of a second measurement, i.e. B(b,λ)B(\vec{b},\lambda). We further make the vital assumption that the result B does not depend on a\vec{a} and likewise A does not depend on b\vec{b}.

Before proceeding, we should note that λ\lambda here plays the role of a “hidden” parameter or variable. We say it is “hidden” because its precise nature is not known. However, it is still a very real parameter with a probability distribution ρ(λ)\rho(\lambda). The expectation value of the product of the two measurements is

(1)P(a,b)=dλρ(λ)A(a,λ)B(b,λ). P(\vec{a},\vec{b})=\int d\lambda\rho(\lambda)A(\vec{a},\lambda)B(\vec{b},\lambda).

Because ρ\rho is a normalized probability distribution,

dλρ(λ)=1 \int d\lambda \rho(\lambda) = 1

and because A(a,λ)=±1A(\vec{a},\lambda)=\pm 1 and B(b,λ)=±1B(\vec{b},\lambda)=\pm 1, P cannot be less than -1. It can be equal to -1 at a=b\vec{a}=\vec{b} only if A(a,λ)=±1=B(a,λ)=±1A(\vec{a},\lambda)=\pm 1 = -B(\vec{a},\lambda)=\pm 1 except at a set of points λ\lambda of zero probability. Thus we can write (1) as

(2)P(a,b)=dλρ(λ)A(a,λ)A(b,λ). P(\vec{a},\vec{b})=-\int d\lambda\rho(\lambda)A(\vec{a},\lambda)A(\vec{b},\lambda).

If we introduce a third unit vector c\vec{c} we can find the difference between the correlation of a\vec{a} to the two other unit vectors,

(3)P(a,b)P(a,c)=dλρ(λ)[A(a,λ)A(b,λ)A(a,λ)A(c,λ)]. P(\vec{a},\vec{b})-P(\vec{a},\vec{c})=-\int d\lambda\rho(\lambda)[A(\vec{a},\lambda)A(\vec{b},\lambda)-A(\vec{a},\lambda)A(\vec{c},\lambda)].

Rearranging this we may write (3) as

(4)P(a,b)P(a,c)=dλρ(λ)A(a,λ)A(b,λ)[A(b,λ)A(c,λ)1]. P(\vec{a},\vec{b})-P(\vec{a},\vec{c})=-\int d\lambda\rho(\lambda)A(\vec{a},\lambda)A(\vec{b},\lambda)[A(\vec{b},\lambda)A(\vec{c},\lambda)-1].

Given the limitations we have placed on the value of A, we may write

(5)|P(a,b)P(a,c)|dλρ(λ)[1A(b,λ)A(c,λ)]. |P(\vec{a},\vec{b})-P(\vec{a},\vec{c})| \le \int d\lambda\rho(\lambda)[1-A(\vec{b},\lambda)A(\vec{c},\lambda)].

But the second term on the right is simply P(b,c)P(\vec{b},\vec{c}) and thus

(6)1+P(b,c)|P(a,b)P(a,c)| 1 + P(\vec{b},\vec{c}) \ge |P(\vec{a},\vec{b})-P(\vec{a},\vec{c})|

which is the original form of Bell’s inequality. Note that this may be written in terms of correlation coefficients,

(7)1+C(b,c)|C(a,b)C(a,c)| 1 + C(b,c) \ge |C(a,b)-C(a,c)|

where a, b, and c are now settings on the measurement apparatus.

Quantum mechanical violations

The original derivation of Bell’s inequalities involved the use of a Stern-Gerlach device that measures spin along an axis. Suppose σ 1\sigma_{1} and σ 2\sigma_{2} are spins. The result, A, of measuring σ 1a\sigma_{1}\cdot\vec{a} is then interpreted as being entirely determined by a\vec{a} and λ\lambda. Likewise for B and σ 2b\sigma_{2}\cdot\vec{b}. It is also important to remember that the result B does not depend on a\vec{a} and likewise A does not depend on b\vec{b}.

For a singlet state (that is a state with total spin of zero), the quantum mechanical expectation value of measurements along two different axes (see the Wigner derivation below for a more intuitive explanation of the physical nature of this) is

(8)σ 1a,σ 2b=ab. \langle\sigma_{1}\cdot\vec{a},\sigma_{2}\cdot\vec{b}\rangle = - \vec{a}\cdot\vec{b}.

In theory this ought to equal P(a,b)P(\vec{a},\vec{b}) but in practice it does not. It is important to remember that we are using classical reasoning throughout our derivations of the various forms of Bell’s inequalities.

The setup envisioned here consists of pairs of spin-1/2 particles produced in singlet states that then each pass through separate Stern-Gerlach (SG) devices. Since they are in singlet states, if we measured the first particle of a pair to be aligned with a given axis, say a\vec{a}, then the second should be measured to be anti-aligned with that same axis, giving a total spin of zero.

In practice we are dealing with beams of particles and thus we can never be absolutely certain that correlated pairs are measured simultaneously and so we ultimately are making statistical predictions. Nevertheless, in a given sample consisting of a large-enough number of randomly distributed spin-1/2 particles, we can be certain that, for example, a definite number are aligned with an axis a\vec{a} while a definite number are aligned with an axis b\vec{b}.

Now take an individual particle and suppose that, for this particle, if we measured σa\sigma\cdot\vec{a} we would obtain a +1 with certainty (meaning it is aligned with a\vec{a}) but if we instead chose to measure σb\sigma\cdot\vec{b} we would obtain a -1 with certainty (meaning it is anti-aligned with b\vec{b}). Notationally we refer to such a particle as belonging to type (a+,b)(\vec{a}+,\vec{b}-). Clearly for a given pair of particles in a singlet state, if particle 1 is of type (a+,b)(\vec{a}+,\vec{b}-), then particle 2 must be of type (a,b+)(\vec{a}-,\vec{b}+).


For beams of correlated particles measuring along only two axes, we should expect to get a roughly evenly balanced distribution of types as follows:

Particle 1 Particle 2 (a+,b) (a,b+) (a+,b+) (a,b) (a,b) (a+,b+) (a,b+) (a+,b) \array{ \text{ Particle 1 } & & \text{ Particle 2 } \\ (\vec{a}+,\vec{b}-) & \leftrightarrow & (\vec{a}-,\vec{b}+) \\ (\vec{a}+,\vec{b}+) & \leftrightarrow & (\vec{a}-,\vec{b}-) \\ (\vec{a}-,\vec{b}-) & \leftrightarrow & (\vec{a}+,\vec{b}+) \\ (\vec{a}-,\vec{b}+) & \leftrightarrow & (\vec{a}+,\vec{b}-) }

There is a very important assumption implied here. Suppose a particular pair belongs to the first grouping, that is if an observer A decides to measure the spin along a\vec{a} for particle 1, he or she necessarily obtains a plus sign (corresponding to it being aligned with a\vec{a}) regardless of any measurement observer B may make on particle 2. This is the principle of locality: A’s result is predetermined independently of B’s choice of what to measure.

Wigner’s derivation

Now suppose we introduce a third axis, c\vec{c}, so that we can have, for example, particles of type (a+,b+,c)(\vec{a}+,\vec{b}+,\vec{c}-) corresponding to being aligned if measured on a\vec{a} and b\vec{b} and anti-aligned on c\vec{c}. Further let us “count” the pairs that fall into the various groupings and label the populations as follows:

Population Particle 1 Particle 2 N 1 (a+,b+,c+) (a,b,c) N 2 (a+,b+,c) (a,b,c+) N 3 (a+,b,c+) (a,b+,c) N 4 (a+,b,c) (a,b+,c+) N 5 (a,b+,c+) (a+,b,c) N 6 (a,b+,c) (a+,b,c+) N 7 (a,b,c+) (a+,b+,c) N 8 (a,b,c) (a+,b+,c+) \array{ \text{ Population } & \text{ Particle 1 } & \text{ Particle 2 } \\ N_{1} & (\vec{a}+,\vec{b}+, \vec{c}+) & (\vec{a}-,\vec{b}-,\vec{c}-) \\ N_{2} & (\vec{a}+,\vec{b}+, \vec{c}-) & (\vec{a}-,\vec{b}-,\vec{c}+) \\ N_{3} & (\vec{a}+,\vec{b}-, \vec{c}+) & (\vec{a}-,\vec{b}+,\vec{c}-) \\ N_{4} & (\vec{a}+,\vec{b}-, \vec{c}-) & (\vec{a}-,\vec{b}+,\vec{c}+) \\ N_{5} & (\vec{a}-,\vec{b}+, \vec{c}+) & (\vec{a}+,\vec{b}-,\vec{c}-) \\ N_{6} & (\vec{a}-,\vec{b}+, \vec{c}-) & (\vec{a}+,\vec{b}-,\vec{c}+) \\ N_{7} & (\vec{a}-,\vec{b}-, \vec{c}+) & (\vec{a}+,\vec{b}+,\vec{c}-) \\ N_{8} & (\vec{a}-,\vec{b}-, \vec{c}-) & (\vec{a}+,\vec{b}+,\vec{c}+) }

Let’s suppose that observer A finds particle 1 is aligned with a\vec{a}, i.e. a+\vec{a}+, and that observer B finds particle 2 is aligned with b\vec{b}, i.e. b+\vec{b}+. From the above table it is clear that the pair belong to either population 3 or 4. Note that because N iN_{i} is positive semi-definite we must be able to construct relations like, for instance,

(9)N 3+N 4(N 3+N 7)+(N 4+N 2). N_{3} + N_{4} \le (N_{3} + N_{7}) + (N_{4} + N_{2}).

Now let P(a+;b+)P(\vec{a}+;\vec{b}+) be the probability that, in a random selection, A finds particle 1 to be a+\vec{a}+ and B finds particle 2 to be b+\vec{b}+. In terms of populations, we have

(10)P(a+;b+)=(N 3+N 4) i 8N i. P(\vec{a}+;\vec{b}+) = \frac{(N_{3} + N_{4})}{\sum_{i}^{8}N_{i}}.

Similarly we have

(11)P(a+;c+)=(N 2+N 4) i 8N i P(\vec{a}+;\vec{c}+) = \frac{(N_{2} + N_{4})}{\sum_{i}^{8}N_{i}}


(12)P(c+;b+)=(N 3+N 7) i 8N i. P(\vec{c}+;\vec{b}+) = \frac{(N_{3} + N_{7})}{\sum_{i}^{8}N_{i}}.

The positivity condition (9) then becomes

(13)P(a+;b+)P(a+;c+)+P(c+;b+). P(\vec{a}+;\vec{b}+) \le P(\vec{a}+;\vec{c}+) + P(\vec{c}+;\vec{b}+).

This is Wigner’s form of Bell’s inequality.

Violations and geometry

As we mentioned before, we have used purely classical reasoning to derive the two forms of Bell’s inequality that we have thusfar encountered. Recall that the context within which the above were derived was the Stern-Gerlach experiment are we are measuring along axes of the magnetic field. As such, there are angles between these various axes. Thus the quantum mechanically-derived probabilities corresponding to (10), (11), and (12) are

P(a+;b+)=12sin 2(θ ab2), P(\vec{a}+;\vec{b}+) = \frac{1}{2}sin^{2}\left(\frac{\theta_{ab}}{2}\right),
P(a+;c+)=12sin 2(θ ac2), P(\vec{a}+;\vec{c}+) = \frac{1}{2}sin^{2}\left(\frac{\theta_{ac}}{2}\right),


P(c+;b+)=12sin 2(θ cb2), P(\vec{c}+;\vec{b}+) = \frac{1}{2}sin^{2}\left(\frac{\theta_{cb}}{2}\right),

respectively. Bell’s inequality, (13), then becomes

(14)12sin 2(θ ab2)12sin 2(θ ac2)+12sin 2(θ cb2). \frac{1}{2}sin^{2}\left(\frac{\theta_{ab}}{2}\right) \le \frac{1}{2}sin^{2}\left(\frac{\theta_{ac}}{2}\right) + \frac{1}{2}sin^{2}\left(\frac{\theta_{cb}}{2}\right).

From a geometric point of view, this inequality is not always possible. For example, suppose, for simplicity that a\vec{a}, b\vec{b}, and c\vec{c} lie in a plane and suppose that c\vec{c} bisects a\vec{a} and b\vec{b}, i.e.

θ ab=2θ and θ ac=θ cb=θ. \array{ \theta_{ab} = 2\theta & \text{ and } & \theta_{ac}=\theta_{cb}=\theta. }

Then (14) is violated for 0<θ<π20 \lt \theta \lt \frac{\pi}{2}. For example, if θ=π4\theta = \frac{\pi}{4}, (14) would become 0.5000.2920.500 \le 0.292 which is absurd!

Revised on June 2, 2010 at 22:26:30 by Ian Durham