# No-Interaction Theorem

1. Mar 29, 2008

### J.F.

"No-Interaction" Theorem

THEOREM. "No-Interaction" Theorem.
Suppose we seek a many-particle direct-interaction theory with the
following properties:
1. the theory is Lorentz invariant,
2. the theory is based on a Hamiltonian dynamics, and
3. the theory is based on independent (canonical) particle variables.
Then such a theory is only compatible with noninteracting particles.

"No-Interaction" Theorem in Classical Relativistic Mechanics

http://prola.aps.org/abstract/PR/v182/i5/p1397_1

Relativistic particle dynamics—Lagrangian proof of the no-interaction theorem

http://prola.aps.org/abstract/PRD/v30/i10/p2110_1

G. Marmo and N. Mukunda
Instituto di Fisica Teorica, Universita di Napoli, Napoli, Italy and Istituto Nazionale di Física Nucleare, Gruppo Teorico, Sezione di Napoli, Napoli, Italy
E. C. G. Sudarshan
Center for Particle Theory, Department of Physics, The University of Texas at Austin, Austin, Texas 78712

An economical proof is given, in the Lagrangian framework, of the no-interaction theorem of relativistic particle mechanics. It is based on the assumption that there is a Lagrangian, which if singular is allowed to lead at most to primary first-class constraints. The proof works with Lagrange rather than Poisson brackets, leading to considerable simplifications compared to other proofs.

2. Mar 29, 2008

### meopemuk

This theorem was first formulated and proven in a nicely written paper

D. G. Currie, T. F. Jordan and E. C. G. Sudarshan, "Relativistic invariance and Hamiltonian theories of interacting particles", Rev. Mod. Phys., 35 (1963), 350

Eugene.

3. Mar 6, 2010

### aspidistra

Re: "No-Interaction" Theorem

After reading few pages of this paper, I got stuck.
Before posting my question, let me summarize the common abstract mathematical structure described in section 2 of this paper for both classical and quantum mechanics.
1)
Let R be a real inear space with lie braket define on it ie. R is a lie algebra. Any $$A \in R$$ represents the quantities descriptive of some pysical system. ( I think what the authors mean is A represents measurable quantity or observable, am I right?)
2)
Let $$F \in S \subset R$$, there exist a real bilinear functional ( , ) defined for all pairs of A, F such that (A,F) is a real number. The real linear functional ( , F) on R represents a particular state of the system, it maps any $$A \in R$$ to an expectation value (measurement value) $$\langle A \rangle\equiv (A,F)$$

combine 1) and 2) The real lie algebra and A linear functional on R specified by An elements of S provied a complete description of one possible instantaneous state fo the system.

3) Transformations of reference frame is represented by linear automorphisms of R generated by elements of $$L \subset R$$. To be specific, the automorphisms are defined by $$e^{[H]t}(A):=A+[A,H]t+\frac{1}{2!}[[A,H],H]t^2+...$$
, where $$A \in R, H \in L$$ and t is real number for which the series is meaningful.

4)Two further postulates needs to be state.
i)$$e^{[H]t}(F) \in S$$ if $$F\in S$$
ii)$$(e^{[H]t}(A),e^{[H]t}(F))=(A,F)$$ for any A in R and F in S

With the above statement, the authors have the following assertion in pp.354:
"we can generate a description of the system at time zero with respect to a second reference frame translated an amount t in time by letting F represent the state of the system in the second description as well as in the first, and by letting $$e^{[H]t}(A)$$ represent, in the second description, the quantity that was represented by A in the first description."

This statement contradict with what I have learnd about the notion of time translation. See
pp.17-18 of the following lecture note:
ed.mvps.org/physics/files/Classical%20Mechanics.pdf

Take passive point of view (the reference frame itself transform), the system should not at time zero with respect to the second reference frame.

Maybe I misunderstand the notion of time translation, hope someone can explain it to me or recommend some good reference. The two undistinguishable terminology "time translation" and "time evolution" especially troubles me.

Any help will be appreciate, thanks.

4. Mar 6, 2010

### meopemuk

Re: "No-Interaction" Theorem

It seems that some words are missing in this sentence. Could you please clarify what is your question?

Eugene.

5. Mar 7, 2010

### aspidistra

Re: "No-Interaction" Theorem

Un...I am sorry Eugene, my english is not good. I try my best to clarify my question.
The authors said:
"we can generate a description of the system at time zero with respect to a second reference frame translated an amount t in time by letting F represent the state of the system in the second description as well as in the first, and by letting e^{[H]t}(A) represent, in the second description, the quantity that was represented by A in the first description."

My understanding of this paragraph paragraph is that we transform our reference frame which means we reset the origin of our time coordinate. If I understand it correctlly, the situation of our physical system and the transformed reference frame should be like fig.5(a) in pp.18 of this link:
http://ed.mvps.org/physics/files/Classical Mechanics.pdf

Since the authors said (in pp.354 of the paper) : "we may think of A and F as being part of the description of the system at time zero with respect to a given reference frame", I think the transformed description of the system should not still at time zero with respect to the transformed reference. Instead of labeling the system at t'=0, the transformed reference frame should label the system at t'=b.

6. Mar 7, 2010

### Fredrik

Staff Emeritus
Re: "No-Interaction" Theorem

I don't understand the theorem and I don't understand its proof, but I noticed that one of the authors of the second article referenced in the OP is Sudarshan, one of the people who authored the original article, and that the new article claims to have a much simpler proof. So maybe it's better to try to read that one instead of the original.

I think I'm going to have to read one of these articles one of these days, but it won't be today.

7. Mar 7, 2010

### meopemuk

Re: "No-Interaction" Theorem

aspidistra,

I think I understand the origin of your confusion. In order to clear things up I would suggest you to consider "instantaneous" observers, which exist and make their measurements only during very short time intervals.

To make it more clear, let me consider two instantaneous observers. One of them is "John now". Another one is "John 10 minutes later". These two instantaneous observers are related to each other by a time translation transformation (t=10). Suppose that these observers look at the same physical system - for example a kettle on the stove. From the point of view of the observer "John now" the water in the kettle is cold. From the point of view of "John 10 minutes later" the water is boiling. To describe this situation mathematically we can consider water temperature, which is an observable pertinent to the observed system - the kettle. I will denote this observable T. The temperature measured by "John now" will be denoted T(0). The temperature measured by "John 10 minutes later" is T(10). The time evolution of this observable (i.e., the transition from T(0) to T(10)) can be described in the Hamiltonian formalism. The physical system (the kettle on the stove) is described by the Hamiltonian H, which is a generator of time translations. So, in order to connect observations of the two instantaneous observers we need to apply an exponent of the Hamiltonian to the observable. Then in your notation we can write (t=10)

T(10) = e^{[H]t}T(0) = e^{[H] 10}T(0).....................(1)

The chosen physical system (stove + kettle + water) is very complex. So, its Hamiltonian H cannot be written in a simple form, and it is very difficult to perform calculation (1) in any reasonable approximation. However, the same Hamiltonian rules of time evolution apply to all isolated physical systems both simple and complex.

The good thing about using "instantaneous" observers is that the same approach that we've used above for time translations works for all other 9 types of transformations of the Poincare group. For example, consider a third observer "Bob now in another room". So, we can ask what are the results of measurements performed by this observer? This observer is displaced in space with respect to "John now". So, in order to answer the question we need to apply a space translation transformation to our observable - temperature. In order to do that we need to know the space translation operator (total momentum) P that is pertinent to our physical system. Then if the distance between John and Bob is 5 meters (x=5). Then the temperature measured by Bob can be found from formula

T(x=5) = e^{[P]x}T(0) = e^{[P] 5}T(0).....................(5)

If operator P is chosen correctly, then we should obtain the obvious result T(x=5) = T(0).

Similarly, we can introduce the generator of rotations J and the generator of boosts K pertinent to the observed system (the kettle on the stove). The ten generators (H, P, J, K) must form a representation of the Lie algebra of the Poincare group. Their exponents e^{[H]t}, e^{[P]x}, e^{[J]a}, e^{[K]v} must form a representation of the Poincare group itself.

Eugene.

8. Oct 20, 2010

### Federation 2005

Re: "No-Interaction" Theorem

The simplest statement of the theorem is that in any system whose components are "particles" (i.e. irreducible representations of the Poincare group), if the angular momentum and linear momentum are additive, then so must be the energy (and mass moment). Ergo, the potential energy must be 0 and there is no interaction.

I think the additivity requirements correctly state the technical assumptions.

The reason for the problem rests with Newton's Third Law. The explicit statement of the law makes it immediately clear what the problem is: "to each action, there exists an equal and opposite reaction AT THE SAME TIME." If the two interactions are separated in space, this is no problem in non-relativistic theory, since "at the same time" is absolute. But in Relativity -- huge problem. There is no longer an absolute "at the same time". The rug is pulled out from under the Third Law.

That law is the underpinning to all interactions in non-relativistic theory. Is there a way out, you ask? The folklore in the Physics community goes on to assert that this "proves fields are necessary".

It proves no such thing, however! What it does, instead, is prove that the very concept of interaction for many-body systems is impossible -- and this includes what we normally refer to as quantum field theory!

Indeed, this problem is not isolated to classical theory. It ramifies to quantum theory, and the Leutwyler theorem (which is the name of the "no-interaction" theorem) morphs into its quantum version -- the Haag Theorem. The Haag Theorem states that in quantum field theory there is no non-trivial interacting many body system (i.e. no Fock space with a non-trivial interaction); the same as the Leutwyler Theorem states. For quantum fields, this specifically excludes the very setup that perturbation theory assumes!

The reason fields do not solve the problem should be clear. Think of a system composed of two isolated regions surrounded by space. We know from experience (e.g. a star with a large body orbiting it) that there is, indeed, a third-law action-reaction going on. The action of the "planet" is offset simultaneously by the recoil of the "star".

So, how would this be communicated by a field, you ask? The action-reaction pair happen at events that are at a spacelike separation. Any communication of one to the other would be going faster than light speed.

So, think of the situation where you transform to a frame of reference where have the action and reaction not being simultaneous. When one body acts, before the other reacts, what is the condition of the system and where did the extra impulse go? It's as if the impulse shot across space faster than light, leaving one body and coming into the other.

If you want to employ a field to get the impulse to communicate, it's not enough to have wave propagation. Waves do not communicate this aspect of the interaction. What they DO communicate is the kind of shock-wave or wiggling that would occur if you abruptly alter the position of one body. They communicate the 1/r "radiative" part of the field. However, they do NOT communicate the 1/r^2 "Coulomb" part of the field.

If you go into quantum field theory, and carefully analyze who's contributing to what, what you'll end up finding is that for a field such as electromagnetism, there are 4 modes. Two modes are "transverse" and are associated with the two helicities of a light-speed carrier (the photon). One mode is "longitudinal", while the last is "scalar". In effect, the longitudinal photon is a tachyon, while the scalar photon is a bradyon (slower than light). Their observable effects cancel one another out (when you take expectation values), so you don't see anything actually being communicated by them. More precisely, they contribute nothing to the expectation value of the energy, so correspond to no propagation of energy at all. Instead, what they conspire to give you the Coulomb part of the field. For a source-free region of space, the Coulomb part of the field is 0 and they two extra modes contribute nothing at all to the force. (That is: the free field has 0 Coulomb part and is purely radiative).

In particular, when we say the longitudinal mode is "off shell" and "virtual", the upshot of what this is really saying is, is that it is not a luxon at all. The sense in which this mode is off shell is that the total value of P^2 - 2MT + (1/c)^2 T^2 is non-zero, where T = E is its kinetic energy, M = E/c^2 its "relativistic mass" and P their momentum.

For ordinary slower-than-light systems, one has M = m + (1/c)^2 T and E = M c^2. So the quadratic invariant reduces to P^2 - (E/c)^2 + (mc)^2 = 0. For luxons, like the photon, E = T = M c^2, M = (1/c)^2 T, and (in effect) m = 0. So, once again, the invariant reduces to 0. For the longitudinal modes, the invariant is NOT zero. Hence, it is termed "off shell".

What this means becomes perfectly clear once you fall back to non-relativistic theory.

In non-relativistic theory, (1/c)^2 is replaced by 0. The analogue to "off-shell longitudinal mode" then corresponds to a mass M = 0 representation, with non-zero momentum P (i.e. an infinite speed "action at a distance" mode). This type of representation has never been given a name cut from the same cloth as "tachyon", "tardion" (or "bradyon") and "luxon". So, you can coin the term "synchron" for it.

For a synchron, one can always find a (non-unique) frame of reference in which its kinetic energy T = 0. Then the quadratic invariant P^2 - 2MT + (1/c)^2 T^2 reduces to P^2. This invariant is just the impulse transferred by the synchron. A synchron is an action-at-a-distance transfer of impulse across space. Denoting the impulse by Pi, the invariant says P^2 - 2MT + (1/c)^2 P^2 = Pi^2, which is non-zero. This is the non-relativistic version of being "off-shell".

So, when you pass over to relativistic theory, the analogous situation is a system whose momentum P, kinetic energy T and "relativistic mass" M give you a non-zero total for the quadratic invariant. That "off-shell" total is just Pi^2, the square of the impulse transferred.

Thus, the virtual mode is just the non-relativistic version of the synchron. Its non-zero value for the quadratic invariant (which is what makes it off-shell) is the relativistic analogue of impulse squared. It's the instantaneous transfer of impulse across space.

Thus, we find that the community folklore that "fields got rid of action at a distance" and "fields resolved the problem with the Third Law" is just a myth. Field theory didn't get rid of the Coulomb part of the field nor its "simultaneous" action, but simply called it by a different name and buried it under disguise.

To date, there is no clear cut integration of the notion of tachyon and virtual mode, and no clear cut integration of the tacyhon + luxon with the non-relativistic analogue, the synchron, in the literature. So, part of the problem with the no interaction theorem may simply be that we have not fully worked out the basic concepts in the right way and have not correctly accounted for how relativistic concepts are to be linked up with their non-relativistic analogues.

So, there MAY be a way out of the no-interaction theorem that somehow "reverse engineers" Newton's Third Law and the related concept of "action at a distance synchrons" into relativistic form. And it may give more cohesive interpretation of the Coulomb part of the field and how it's propagated by the virtual modes. But this framework would represent a slight upgrade beyond what we presently call relativistic theory.

So, my point of view is that the No Interaction Theorems (both Haag and Leutwyler) signal the presence of a gap or oversight in what we presently call Relativity and point the way toward the upgrades required to fill in this incompleteness and to salvage or recover a Relativistic form of Newton's Third Law.

9. Oct 20, 2010

### bcrowell

Staff Emeritus
Re: "No-Interaction" Theorem

Here's a recent talk on this topic that I found interesting: http://streamer.perimeterinstitute.ca/Flash/1a7787fa-5478-49ca-82c2-4b7a342117c8/index.html [Broken]

Hmmm...this was the first place where you lost me. How does it follow from this that the PE is zero?

Are you asserting that even classically, it's impossible to have an interacting many-body system, and that it doesn't matter whether the theory describes particles or fields? I don't see how that can be right. For example, take a system of point charges interacting electromagnetically in special relativity. You have conservation of energy, momentum, and angular momentum. What is wrong with this theory?

[EDIT] Maybe I'm misunderstanding your references to classical theories...? Haag's theorem http://en.wikipedia.org/wiki/Haag's_theorem is a theorem about quantum field theories.

My understanding is that results like the CJS no-interaction theorem are purely about quantum field theories, and that they basically just say that you can't use the theory to predict the time-evolution of observables -- but you can still find quantities like S-matrices and energies of bound states.

Last edited by a moderator: May 5, 2017
10. Oct 21, 2010

### Fredrik

Staff Emeritus
Re: "No-Interaction" Theorem

I understand almost nothing about that theorem, but one thing I've been told is that it applies both to classical and quantum theories. They used some kind of clever notation to make sure that a single proof would cover both cases.

I still haven't read Federation 2005's post, but I will.

11. Oct 21, 2010

### Fredrik

Staff Emeritus
Re: "No-Interaction" Theorem

@Federation 2005: That's an interesting post, but unfortunately there's a lot I don't understand in it. Can you suggest some articles or books that cover the stuff you're talking about in more detail? I'm going to have to study this stuff sooner or later, and I think I will find your summary of the key points useful when I do, but I'm still hoping there's something I can read that's easier than the original articles.

By the way, a Google search for "Leutwyler theorem" returns your post as #1, so it doesn't seem to be a well-known term. What do you mean by it? Is it what many others call the Currie-Jordan-Sudarshan theorem, or is it a different theorem?

12. Oct 21, 2010

### bcrowell

Staff Emeritus
Re: "No-Interaction" Theorem

I'm sure Fredrik is right about it applying to classical theories as well as quantum-mechanical ones.

On rereading Federation 2005's #8, what strikes me is that the word "Hamiltonian" never appears.

If the CJS no-interaction theorem tells us that a certain classical theory can't be expressed in Hamiltonian form, then our reaction should probably be "so what?" The theory still works if we have to cast it in some other form, like a Lagrangian theory.

It's only a big deal if the theorem tells us that a certain quantum theory can't be expressed in Hamiltonian form, because quantum mechanics depends on Hamiltonians.

13. May 7, 2013

### Federation 2005

No Interaction Theorem: Root Cause and Resolution

Actually, to follow up on my earlier reply (a few years back), a closer examination of the theorem (both the 1963 Currie et al. version and the later Leutwyler generalization of it) do not assume additivity for angular momentum and momentum, they infer it from the transformation propeties posed for the worldline coordinates. Then, follows the slow descent into triviality.

The following also applies to a later formulation of the 2-body no interaction proof published by Jordan (1968, if I recall).

To be more precise, additivity for momentum and angular momentum only up to an adjustment of the momentum (i.e. a canonical transformation). The adjustment made by this transformation also yields certain transformation properties, as a result, for the momentum which, if assumed at the outset, would equivalently characterize the transformation made by the adjustment.

Denoting the infinitesimal forms of rotation, boost and translation respectively by ω, $\upsilon$ and ε and infinitesimal time translation by $\tau$, the transformation assumed for a worldline coordinate q is
$\delta$q = ω$\times$q + (α$\upsilon$ - $\tau$)v + ε
where v = {q, E} is the velocity, with E being the time translation generator.

If you apply this to the bracket relation {q, q'} for two different particle coordinates (noting that the bracket is the 0 dyad), then the transform should be 0. Plugging in the time translation and boost you immediately get that the inverse of the "coefficients of inertia" matrix (i.e. the matrix of all the {q, v'} brackets has zero cross terms -- or more precisely, cross terms that admit only contact terms (ones with δ(q - q') factors in it). The contact terms are not included in the analyses posed by the no interaction theorems and probably won't change the situation much at all.

This entails that E separates into a sum of kinetic terms, i.e. terms dependent on only one of the particle momenta; plus a potential term (i.e. a term independent of all the momenta).

None of this holds in the non-relativistic case. That extra factor α is 1/c2 for relativity and 0 for non-relativistic theory. That factor controls the separability.

The root of the problem is clearly seen here: the correspondence limit employed by this theory has a discontinuity at α = 0. Empirically, this is a big no no. Qualitative differences should never result from a continuous parameter adopting a specific value, because that would in principle provide a way of measuring a value with infinite precision.

In other words, we're using the wrong correspondence limit.

If you go back and take a closer -- more modern -- view of non-relativistic theory, you will see that its symmetry group is now understood to not be the 10-dimensional Galilei group, but the 11-dimensional Bargmann group. In order to have a more cohesive correspondence limit, this then requires that whatever symmetry group we adopt for relativity should have the Bargmann group as its limit. This leave no alternative but for this group to have 11 dimensions, not 10.

The modification takes place in the brackets between the boost and trnslation generators and entails a slightly different notion of time translation as a result. Let H be the new time translation generator. It will be defined so that it has the kinetic energy as its non-relativistic limit. The corresponding to the central charge m in the Bargmann group will be a trivial central charge, which we'll call $\mu$. It is understood that it will have m as its non-relativistic limit, and that the original time translation generator E will continue to be regarded as the "total" energy, with the decomposition E = H + $\mu$/$\alpha$. But to hold true to the requirement for a consistent correspondence limit, it will be necessary to replace E by the "relativistic mass" M = $\alpha$E. Then the decomposition relation will take the form M = $\mu$ + $\alpha$H. In the non-relativistic limit M also goes to m. On the other hand, E has no limit. It's divergent (as seen by the fact that the parameter $\alpha$ was sitting in the denominator). So, it drops out.

Then the Lie brackets continue to have the same form as before, but with E rewritten as M. In particular [K, P] = M I, where I denotes the identity dyad. For $\mu$, the brackets are all 0.

The original time translation was {_, E}. Now it becomes {_, H}. The Leutwyler theorem still applies to the rotation generator J, boost generator K, translation generator P and E (which is now M). But H is exempt. Thus, {_, E} now represents "time translation that would occur if the bodies involve were free and non-interacting", while the difference {_, H-E} represents the interacting part of the time translation generator.

For Leutwyler, the boost transform for q is now ambiguous. There are now two forms one can state this property in:
(a) { q, $\upsilon$$\cdot$K} = $\upsilon$$\cdot$q {q, αH}​
or
(b) { q, $\upsilon$$\cdot$K} = $\upsilon$$\cdot$q {q, M}​
In the absence of the 11th parameter, both forms are equivalent with H = E = M/$\alpha$. For the extended Poincare' group, they are no longer equivalent.

Version (a) yields a Leutwyler result for the full 11-dimensional symmetry group. Version (b) breaks the no interaction theorem and allows in non-trivial interactions.

The energy difference H - E = -$\mu$/α plays the role of the potential U. The only requirement imposed on it is that all the Poisson brackets formed by the 10 other functions -- the 3 vectors J, K, P and the scalar M -- should be 0. This can be done with non-trivial results.

As a consequence, H decomposes into a sum of purely kinetic contributions from each body -- as in the non-relativistic case -- plus the potential U.

The only drawback to this revision is that this adjustment tot he correspondence limit is still not quite enough! This is best seen by considering the general solution for interacting 2-body systems. When setting α = 0, one gets for U a function of the 3 scalar 2-body invariants formed from their relative speed and relative displacement, as expected. As soon as you turn on α, allowing it to be non-zero, it drops down to only 2 invariants: namely the ones whose non-relativistic limits are relative speed and "areal speed" of the radius vector. The component of velocity collinear to the displacement is lost in the translation. So, no Kepler laws.

So, there's even more subtlety that still lies hidden with the correspondence limit than what I've brought up here. That extra term went somewhere, after turning on α to allow it to be non-zero, and I'm hunting it down to see where it got lost. But it's there somewhere.

What makes the correspondence limit so non-trivial is (at least in part) that you have a complete topological change that takes place at α = 0. You can best see this by combining all groups for all α (both positive and negative -- i.e. the 4-D Euclidean group -- as well as 0) into a single Poisson manifold. This manifold has the 11 coordinates of the duals of the Lie algebras above (which may simply be denoted J, K, P, H and $\mu$ as before), plus $\alpha$ as the 12 coordinate; and it has the Poisson bracket formed in the natural way from the Lie brackets, with it understood that {$\alpha$, _} = 0.

The Galilei group is sitting in there as the Lie group that has the submanifold for ($\alpha$, $\mu$) = (0, 0). The oridinal Poincare' group produces each of the submanifolds ($\alpha$, $\mu$) = (1/V2, 0), for all different values of V. They are immediately adjacent to one another, but the Galilei group is not simply connected, while the Poincare' group is.

This means that contraction can expose not just one order of $\alpha$, but an infinite tower of $\alpha$'s -- much like the Einstein-Infeld analysis of the gravity field. The analysis above only pushes the correction to the first order. The 3rd scalar 2-body invariant is hidden somewhere in whatever extra infrastructure is required to recover the 2nd order or higher.

A similar problem occurs when attempting to unify Einstein and Newtonian gravity as a one-parameter family of infrastructures, geometries and Lagrangian theories. Since the Bargmann and extended Poincare' groups are all little groups of the 4+1 de Sitter group, then a unified model for gravity can be formulated in 4+1 dimensions by requiring that there be an invariant covariantly constant field. The norm of that field, with respect to the 4+1 metric, is just $\alpha$. The resulting field equations include Newton's equation at $\alpha$ = 0, and Einstein's equations (for the most part) for $\alpha$ > 0. But there are important terms that get lost as soon as you get to $\alpha$ = 0 and the resulting field equations for the non-relativistic case become somewhat handicapped. The root of the problem is that the coupling coefficient in general relativity (when you do the dimensional analysis right) is actually 2nd order, not 1st -- it's proportional to 1/$\alpha2$. The conversion to a constrained 5-D geometry allows one to reduce this only by one order. But this isn't quite enough to remove the discontinuity in the correspondence limit at $\alpha$ = 0.

Although: it is enough to write non-relativistic gravity in Lagrangian form in a way that's sufficient to include Newton's equation.