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If the S-Matrix is the only observable, that rules out both generalized free fields and Wick-ordered polynomials of generalized free fields as interesting Physical models, because both result in a unit S-matrix. Neither possibility has been developed since the 1960s when these results were proved, and when the S-Matrix was ascendant as the only observable in Particle Physics.

If the S-matrix is not the only quantum field observable, which it certainly seems not to be in Condensed Matter Physics and in Quantum Optics, to name just two fields in which Wightman or correlation functions play a large part in modeling, does that encourage us to construct Wick-ordered polynomials of generalized free fields as algebraic, non-dynamical deformations of free fields? In such a construction, the point is to avoid deforming a Hamiltonian or Lagrangian evolution, instead working with deformations of the observables, requiring for example that all observables and all states will be constructed as Wick-ordered polynomials in a Wick-ordered operator-valued distribution such as $\hat\Phi(x)=:\!\!\hat\phi(x)+\lambda\hat\phi^\dagger\hat\phi(x)^2\!\!:$. The Hamiltonian is taken to be a derived quantity in this approach.

The S-Matrix seems remarkable, in that it requires us to set up a hyperplane at an initial time $t_I$ and another at a final time $t_F$, and another two hyperplanes between which the interaction will be nontrivial, at times $t_I<t_A<t_B<t_F$, all of which is not Lorentz-invariant, then take the limits $t_I\rightarrow-\infty,t_A\rightarrow-\infty,t_B\rightarrow+\infty,t_F\rightarrow+\infty$. It seems that in no other branch of Physics would we construct such an idealization and say it's the only possible way to report the experimental data. In other words, I wonder, going further, whether the S-matrix even is an observable? Particle Physics seems to be the only field in Physics that seems to think it is.

post-Answer addition. It's Greenberg who proves in JMathPhys 3, 31(1962) that Wick-ordered finite polynomials in the field have a trivial S-matrix, and I think, without looking it up, that it's also Greenberg who introduced and ruled out generalized free fields. I've resisted doing something else because Wick-ordered polynomials in a generalized free field seem to give such a lot of freedom to create models that surely they can be useful, which I know is not a good reason, but I had to feel how bad it is before I could move on. I suppose I was almost ready to give up this way of doing things before I asked this question. While Tim van Beeks' response is definitely of interest, and Matt's Answer is clearly the "right" answer, but doesn't go into the periphery of the question in the right way to help me, it was Marcel's response that particularly pushed me.

My comment to Marcel indicates the way I'm now going to take this, to functions of the field such as $\hat\Phi(x)=\tanh{(\hat\phi(x))}$. Insofar as we can say that the measured value of $\hat\phi(x)$ is almost always $\pm\infty$ in the vacuum state, because the expected value $\left<0\right|\hat\phi(x)\left|0\right>$ is zero and we can say, loosely, that the variance is $\infty$ in the vacuum state, presumably $\left<0\right|\delta(\hat\Phi(x)-\lambda)\left|0\right>$ is non-zero only for $\lambda=\pm 1$ (though with some worries about this construction). It's not clear that we even have to introduce normal-ordering. $\tanh{(\hat\phi(x))}$ is of course a bounded operator of the unbounded Wightman free field, whereas no nontrivial polynomial in the field $\hat\phi(x)$ can be a bounded operator. The particular choice of $\hat\Phi(x)=\tanh{(\hat\phi(x))}$ is clearly a particular coordinatization; if we change the coordinatization by taking a function of $\hat\Phi(x)$, we in general get a measurement operator that results in two discrete values of the field, mapping $\pm 1$ to $a,b$ respectively. If we take something different from the real Klein-Gordon field, my aim as I envisage it four hours after first imagining it is to map something like an $SU(3)$ invariant Wightman field, say, to a finite number of discrete values. If such an $SU(3)$ symmetry is unbroken, the relationships between each of the discrete values will all be the same, but if the $SU(3)$ symmetry is broken, there presumably has to be a coordinate-free way in which the relationships between the different discrete values are different. There will, I now suppose, have to be enough distinction between raising and lowering operators between different sectors of the theory and measurement operators to allow there to be a concept of an S-matrix.

This may look crazy, but I'll also put on the table here why the structures of the Feynman graphs formalism encourage me. We introduce connected Feynman graphs at different orders to calculate $n$-point connected Wightman functions. Although we typically expand the series in terms of the number of loops, we can alternatively expand the series in terms of the number of points we have introduced between the $n$ points at which we measure; the extra points ensure that there are infinite numbers of different paths between the $n$ points. With something like $\hat\Phi(x)$, we introduce an infinite number of paths directly between the $n$ points, without introducing any extra points, so we need, in effect, a transformation of the superposition of an infinite number of Feynman path integrals into a superposition of an infinite number of weighted direct paths between the $n$ points. Getting the weights on the direct paths right is of course rather important, and I also imagine the analytic structure has to be rather carefully done, particularly if I don't use normal-ordering. I'll do any amount of work to avoid renormalization, even in its modern gussied up form.

It seems to me significant that the quantum field $\hat\Phi(x)=\tanh{(\hat\phi(x))}$ is not reducible. A number of proofs concerning Wightman fields rely on this property.

If anyone else understands this (or reads this far) I'll be surprised. In any case I expect it will look very different a few years down the road if I ever manage to get it into a journal. Although I've worked in and around the Wightman axioms for the last few years, I find it interesting that I can now feel some pull towards something like the Haag-Kastler axioms. Lots of work to do! Thank you all! Good luck with your own crazy schemes!

That's the question (and the to me unexpected state of play a day later). Completely separately, as an example, to show the way I'm going with this, hopefully (which, a day later, looks as if it will be only a background concern for the next little while, but I think not likely to be completely forgotten by me), the real-space commutator of the creation and annihilation operator-valued distributions of the free field of mass $m$ is, in terms of Bessel functions, $$C_m(x)=\frac{m\theta(x^2)}{8\pi\sqrt{x^2}}\left[Y_1(m\sqrt{x^2})+i\varepsilon(x_0)J_1(m\sqrt{x^2})\right]$$ $$\qquad\qquad+\frac{m\theta(-x^2)}{4\pi^2\sqrt{-x^2}}K_1(m\sqrt{-x^2})-\frac{i}{4\pi}\varepsilon(x_0)\delta(x^2).$$ If we take a weighted average of this object with the normalized weight function $w_{\alpha,R}(m)=\theta(m)\frac{R(Rm)^{\alpha-1} {\rm e}^{-Rm}}{\Gamma(\alpha)}, 0<\alpha\in \mathbb{R},\ 0<R$, $\int w_{\alpha,R}(m)C_m(x)dm$, we obtain the commutator of a particular generalized free field, which can be computed exactly in terms of Hypergeometric functions and which at space-like separation $\mathsf{r}=\sqrt{-x^\mu x_\mu}$ is asymptotically $\frac{\Gamma(\alpha+1)R^{\alpha}}{\pi\Gamma(\frac{\alpha+1}{2})^2\bigl(2\mathsf{r})^{\alpha+2}}$, and at time-like separation $\mathsf{t}=\sqrt{x^\mu x_\mu}$ is asymptotically $$\frac{\cos{(\frac{\pi\alpha}{2})}\Gamma(\frac{\alpha}{2}+1)R^\alpha} {4\sqrt{\pi^3}\Gamma(\frac{\alpha+1}{2})\mathsf{t}^{\alpha+2}} -i\frac{R^\alpha}{4\Gamma\left(-\frac{\alpha}{2}\right)\Gamma\left(\frac{\alpha+1}{2}\right)\mathsf{t}^{\alpha+2}},$$ except for $\alpha$ an even integer. At small space-like or time-like separation, the real part of this generalized free field is $-\frac{1}{4\pi^2x^\mu x_\mu}$, identical to that of the massless or massive free particle, independent of mass, but we can tune the 2-point function at large distances to be any power of the separation smaller than an inverse square law. On the light-cone itself, the delta-function component is again identical to that of the massless or massive free particle, independent of mass.

There is of course an infinity of possible normalized weight functions, a half-dozen of which I have worked out exactly and asymptotically, and somewhat obsessively, by use of MAPLE and Gradshteyn & Ryzhik, though I've managed to stop myself at the moment. In a subsequent edit, I can't resist adding what we obtain if we use the weight function $w_{\mathsf{sm}[R]}(m)=\frac{\theta(m)R\exp{\left(-\frac{1}{mR}\right)}}{2(mR)^4},\ 0<R$ [using 6.591.1-3 from Gradshteyn & Ryzhik]. This function is smooth at $m=0$, and results at time-like separation in $$\left\{Y_2\!\left(\sqrt{\frac{2\mathsf{t}}{R}}\right)+iJ_2\!\left(\sqrt{\frac{2\mathsf{t}}{R}}\right)\right\} \frac{K_2\!\left(\sqrt{\frac{2\mathsf{t}}{R}}\right)}{8\pi R^2} \asymp \frac{\exp{\left(-i\sqrt{\frac{2\mathsf{t}}{R}}+\frac{\pi}{4}\right)} \exp{\left(-\sqrt{\frac{2\mathsf{t}}{R}}\right)}} {16\pi\sqrt{R^3\mathsf{t}}}$$ and at space-like separation in $$\frac{K_2\!\left((1+i)\sqrt{\frac{\mathsf{r}}{R}}\right) K_2\!\left((1-i)\sqrt{\frac{\mathsf{r}}{R}}\right)}{4\pi^2R^2} \asymp \frac{\exp{\left(-2\sqrt{\frac{\mathsf{r}}{R}}\right)}} {16\pi\sqrt{R^3\mathsf{r}}}.$$ I wish I could do this integral for more general parameters, but hey! This weight function has observable effects only at space-like and time-like separation $\mathsf{r}<R$ and $\mathsf{t}<R$, so this is essentially unobservable if $R$ is small enough. The kicker is that this decreases faster than polynomially in both space-like and time-like directions.

The generalized free field construction always results in a trivial $n$-point function for $n>2$, however by introducing also Wick-ordered polynomials of these generalized free fields, we can also tune the $n$-point connected correlation functions, which are generally non-trivial and to my knowledge finite for all $n$. All this is far too constructive, of course, to prove much. I think I propose this more as a way to report the $n$-point correlation functions in a manner comparable to the Kallen-Lehmann representation of the 2-point correlation function than as something truly fundamental, because I think it does not generalize well to curved space-time.

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5 Answers 5

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Do I understand your question correct? You ask: if the S-matrix is the only quantity which really can distinguish interesting models from free models? Then I would say no. But there are no interesting models in 4D, fullfilling let's say the Wightman axioms, are there? In lower dimension for example conformal field theories are almost like free ones, but they have a very interesting representation theory, with anyionic sectors braiding etc, which make them also mathematically interesting. But there scattering is trivial.

Your example taking quasi free fields (with a certail Kallen-Lehmann density) doesn't give something new, neither does taking Wick polynomials of these fields, because it gives you essential the same model (same Borchers class).

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Yes, they're the same Borchers class. So, I think they have the same S-Matrix. But they don't give the same Wightman functions, which are measured, so I think we can distinguish them experimentally. I don't yet understand why they're uninteresting, except that there is no discrete structure additional to the Fock space structure. [Note that the example I give has no mass gap, even for $\alpha>1$, so it falls outside the scope of many proofs in AQFT, because it has long-range correlations. I can do the integration exactly for $\exp(-\frac{1}{Rm})/R^4$, which almost-not-quite has a mass gap] –  Peter Morgan Feb 1 '11 at 12:15
    
I'm gonna have to give the next Answer Rep to Tim van Beek, but this is the Answer that pushed me most. Wick polynomials definitely don't help, which leaves us with $\tanh{(\hat\phi(x))}$, say, or its Wick-ordered form, which are not covered by Greenberg's proof, which applies only to Wick-ordered polynomials. This gives a discrete structure of being almost everywhere +/-1, insofar as, loosely, $\hat\phi(x)$ is almost always +/-infinity (because, loosely, the variance $\left<\hat\phi(x)^2\right>$ is infinite). More comments in a post-Answer addition to the question. Many thanks. –  Peter Morgan Feb 2 '11 at 13:27
    
The S-matrix is nontrivial for nonfree models in 2d because it includes the phase shifts. –  Ron Maimon Aug 16 '11 at 14:09
    
-1 --- not only this, but it is certain that QCD is a nontrivial solution to the Whitman axioms, or any other definition of a quantum field theory, so it is not true that the Whitman axioms have no solutions in 4d, only that we haven't proven that these solutions exist in a way that would be fully satisfying to a mathemtician. –  Ron Maimon Aug 28 '11 at 2:52
    
Dear Ron as you see I put the statement about the existence of non-trivial models in 4D as a question, because I am not aware a rigorous construction or something towards it in 4D (in 2D there are certainly some examples), it is actually one of the Clay problems, so could you give any references? –  Marcel Aug 28 '11 at 17:43

Correlation functions of local operators are the other observables that field theorists talk about most of the time: things like $\left<{\cal O}_1(x_1) {\cal O}_2(x_2) \ldots\right>$. But there's no shortage of observables in quantum field theory.

The situation where you'll hear people say that the S-matrix is the only observable is quantum gravity in asymptotically flat space; this is because local operators don't give gauge-invariant physical observables in theories with gravity.

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Also, lots of perfectly good QFTs don't have S-matrices, say any conformal field theory. But, I think the question is about which observables are possible to encode in AQFT, not really a question about "our" QFT. I might be wrong though. –  user566 Feb 1 '11 at 6:47
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@Moshe I know what my question is? Helping me clarify my question is good. I don't have the tools to do it, but I'm most interested in bringing AQFT closer to "our" QFT, because I'd like to be able to say that I understand QFT. I'd like to explain QFT to my mother, at least. This Answers' perception of the received view is very helpful, but I think Marcel has gone more to the heart of what I think now my question might have been. –  Peter Morgan Feb 1 '11 at 12:54
    
Sorry Peter, I just did not know whether this addresses your question, since I did not properly understand it. Partially I think it is because my misunderstanding if by "QFT" you refer to the informal practice of physicists, or to the particular way to attempt formalizing it known as AQFT. These two options will lead to two different sets of answers. –  user566 Feb 1 '11 at 16:03

I'm not sure I understand what your question really is, so I'll try to answer

a) Is the S-matrix the only observable in QFT?

No, other observables are for example the total values of charges like the electric charge, that is all numbers that specify a superselection sector correspond to non-trivial observables. The S-matrix may be one of the most interesting observables because most experiments in high energy physics are about scattering processes, however...

b) Is the S-matrix an observable in QFT?

Well, in traditional Lagrangian QFT one simply assumes that the S-matrix exists and has all the nice properties that one wishes for, so in this framework the question is not very interesting. It is very hard and non-trivial in axiomatic QFT, on the other hand, since you'll find that

b.1) defining the concept of a "particle" is not trivial and

b.2) the S-matrix may not be defined because the theory is not asymptotically complete: In QM this question is completely answered, in relativistic axiomatic QFT it seems to be an open problem.

A recent review paper is this:

  • Detlev Buchholz, Stephen J. Summers: "Scattering in Relativistic Quantum Field Theory: Fundamental Concepts and Tools" (arxiv).

And again all physicists are free to choose if the failure of incorporating "simple and obvious concepts" is a catastrophic failure of AQFT, or if this difficulty hints at some deeper truth that we still don't understand in QFT. Your choice.

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It seems to me that the total value of charges will appear in an S-matrix because the S-matrix specifies the charges of all the incoming and outgoing particles individually. It's a matter of taking a sum. –  Carl Brannen Feb 1 '11 at 0:19
    
Thanks for being generous, Tim. I'll try to clarify. The S-Matrix is the unit matrix for all free fields, of whatever mass, and for all generalized free fields. Borchers equivalence says, I think, the same for Wick ordered multinomials of quantum fields. That is, the S-Matrix conflates a lot of things that are different if we can measure Wightman functions. Everybody except Particle Physicists happily measure Wightman/correlation functions, and refer almost all other observables back to their properties, not back to the S-matrix. Is that no more clearly restating what I already said? –  Peter Morgan Feb 1 '11 at 1:31
    
@Carl Brannan: Yes, but one can also define a charge operator in cases where there is no S-matrix –  Tim van Beek Feb 1 '11 at 8:56
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@Peter: It's not a matter of being generous :-) Sorry, I don't understand "Everybody except Particle Physicists happily measure Wightman/correlation functions". Particle physicists use detectors, which are modeled in AQFT by almost localized observables, which is then an observable that is of interest and is not associated to the S-matrix (see the paper I mentioned). –  Tim van Beek Feb 1 '11 at 8:57
    
Right, but measurements by almost localized observables have to determine not only the prepared state, but, in the long run, also the structure of the algebra of observables (noting, too, that before we know the structure of the algebra of observables, we can represent both measurements and states only hesitantly). The most important structure in question is perhaps whatever discrete structure there may be, which I think for gauge theories is expressed as the category of representations of a gauge group? –  Peter Morgan Feb 1 '11 at 12:42

The exact S-matrix elements with finite number of the initial and final photons in QED are equal to zero so such processes are unobservable (never happen). For observation you need an infinite number of photons (=inclusive picture). This makes me think that the theory should be reformulated so that this fact (dominating inelastic processes) would be taken into account automatically, like in atom-atomic potential scattering (consider the case $n >> 1$, when the threshold of excitations is negligible).

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If there are infinite numbers of photons, then I think that makes it a field, though following that thought might depend on whether you mean countable or uncountable numbers of photons. You're talking about states, however, which we then measure, which to some extent puts you in a rather classical place, whereas I try to think more in terms of measurements, which we then put into the world to see what measurement results we get. The relationships between the measurements we have then have to be sorted out. It's chicken and egg, but it changes the discourse. I Make Any Sense? –  Peter Morgan Feb 1 '11 at 12:53
    
Frankly, I do not really understand what you want, Peter, sorry. Apart from matrix elements of reactions, there are also bound states with their energies, angular momenta, parity, etc., as observables. I would say that the final results of calculations are observable. –  Vladimir Kalitvianski Feb 1 '11 at 14:59

By definition (and Wigner's Theorem) the S matrix is a unitary operator such that $|0>_{out}=S|0>_{in}$. In quantum mechanics an Observable is a Hermitian operator (Penrose has suggested changing this to a Normal Operator with complex eigenvalues). Either way an S Matrix will not be an observable (at least not in a QFT that obeys the rules of QM anyway).

This was the question of the title.

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What is your defintion of observable in a relativistic quantum theory? I would not care if something is unitary, than I take just the "real"-part of it? The reason that something is unobservable is for me another (eg. charged, non-local quantities). –  Marcel Feb 3 '11 at 21:02
    
-1 The S-matrix is not an "observable" in a quantum mechanical sense, it is an "observable" in the sense that you can physically observe it in experiments. There are ordinary QM "observables" that correspond to approximate S-matrix elements in a complicated QM system including the collision and early measuring devices. The S-matrix doesn't do eigenvectors. Infinite plane wave states of any number of particles are non-interacting in any 3+1 dimensional quantum field theory and are automatically eigenvectors, with a trivial evolution. The S-matrix has normal eigenvectors in 1d only. –  Ron Maimon Aug 28 '11 at 2:56

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