## Monday, March 26, 2012 ... //

### Mass spectrum of all objects in the Universe

Whenever it makes sense to define the Hamiltonian, i.e. the operator of total energy, this Hamiltonian knows everything about the rules according to which all physical systems and any physical systems evolve in time. That's because the evolution of observables (quantities that may be measured and that are always associated with linear operators) is given by the Heisenberg equations of motion$\ddfrac{\hat L(t)}{t} = i\hbar [\hat L(t), \hat H]$ Equivalently, one may keep all the operators constant and evolve the state vector $\ket\psi$ according to Schrödinger's equation$i\hbar \frac{\dd}{\dd t} \ket \psi = \hat H \ket \psi$ This fact makes the Hamiltonian $H$ very important. Because we may always redefine the basis of the Hilbert space by a unitary transformation i.e. diagonalize the Hamiltonian, the spectrum of the Hamiltonian knows "everything" about the laws of physics.

You may be surprised by this assertion. Don't we also need to know the "shape" of the objects or the "shape" of their wave functions? It depends what you want to know. Everything that is physical about them is actually encoded in the Hamiltonian. Of course, to prove that the Hamiltonian is this powerful, we also have to study the action of the Hamiltonian on multiparticle states.

Locality of the Hamiltonian is what makes positions useful

But a point of mine is that e.g. the usefulness of the notion of "positions of objects" isn't an additional structure that has to be added to physics. Instead, "positions of objects" are useful and (partly) well-defined quantities because of a property of the Hamiltonian, namely its locality. In quantum field theory, this statement is exactly true; in non-relativistic quantum theories, this statement is either true for different but related reasons, or it's true because these theories are limits of a quantum field theory. The Hamiltonian may be written as an integral of some energy density,$\hat H = \int\dd^3 x~\hat T_{00}(x,y,z).$ From this moment, I will assume that the reader has reconciled himself or herself with the fact that ours is a quantum world so all observables – when analyzed accurately enough – are operators and I don't have to add the redundant hats above them. Because the Hamiltonian $H$ above is the integral over space and the energy density $T_{00}(x,y,z)$ commutes with all operators associated with totally different points or regions $(x',y',z)$, it follows that it is often possible to associate positions to objects, either exactly or approximately, and the evolution at distant enough points will be independent.

I didn't have to add the notion of a "position" independently; it's included in the Hamiltonian. On the other hand, if the Hamiltonian were not local, not even approximately, it would probably be useless to talk about positions of objects even if the parameterization of operators according to the positions were possible.

Energy and momentum: relativity has its say

Another important pillar of modern physics is the special theory of relativity. (The general theory of relativity is also important but it makes any description in terms of a "Hamiltonian" more subtle and I will avoid these topics in this particular blog entry.) Independently of special relativity, Emmy Noether figured out that the energy conservation is associated with the translational symmetry of the laws of physics in the temporal dimension. After all, this fact is related to the role that the Hamiltonian plays in the dynamical equations from the beginning of this text.

However, Einstein realized in his 1905 special relativity that the properties of time and space have to be interlinked; they're parts of the spacetime. Because the energy is linked to translations in time, this energy has to be clumped with the operator linked to translations in space. And that's, of course, the momentum. Relativity therefore implies that the energy itself is the temporal component $P^0$ of the energy-momentum vector $P^\mu$. Whenever the energy is conserved and the Lorentz invariance underlying special relativity is unbroken, the momentum $P^i$ has to be conserved as well.

Moreover, special relativity says that for any transformation $U\in SO(3,1)$ and for every state $\ket\psi$, there has to exist the state $U \ket\psi$ whose momentum and energy is transformed correspondingly. So if you start with the state whose$P^\mu = (E,0,0,0)$ i.e. whose momentum vanishes, it is always possible to "boost" this state so that the momentum will be arbitrary and$P^\mu P_\mu = E^2 - \abs{\vec p}^2$ will be equal to the original value, $E^2$. It's therefore a characteristic value associated with the object, one that is independent of boosts, and it's known as $m_0^2$, the squared rest mass. I will also assume that the reader is a mature physicist who can work in units with $c=1$.

It's important to notice that if you can have a state with the energy $m_0$, you may also easily get states with an arbitrary higher energy, just by boosting the original state. The energy will increase to$E = m_0\gamma = \frac{m_0}{\sqrt{1-v^2/c^2}}$ i.e. it will be enhanced by the relativistic Lorentz factor. So even if the list of elementary particles is "discrete" if not finite (at least at low energies), the spectrum of the Hamiltonian is inevitably continuous. However, you may focus on states whose total spatial momentum vanishes and those can have a discrete spectrum. Alternatively, you may try to look for possible eigenvalues of the operator $P^\mu P_\mu$ in our Universe. This could be discrete.

Well, even this is not discrete because you may have systems composed of decoupled particles or objects and by adjusting their relative speed, you may continuously change the "invariant mass" as well. But note that this continuous spectrum may only be obtained for states that are not bound. For bound states, the spectrum of $M^2 = P^\mu P_\mu$ may be discrete and it actually is discrete in realistic situations, i.e. in the real world of particle physics. Let's look at it.

Elementary particles

The most natural oversimplified way to describe the world is to say that it's composed out of elementary particles. To discuss the spectrum of $M^2$, and I will mostly talk about the spectrum of $M \equiv \sqrt{M^2},$ it seems as though we should only list the masses of elementary particles. Let's begin with the lightest types of elementary particles we know.

Tachyons and massless particles

Well, the lightest particles are the massless ones. If $M^2$ were negative, we would have tachyons and they would make the vacuum unstable. It seems that tachyons have to be scalars in a consistent theory of quantum gravity but any hypothetical Universe with a tachyon in the spectrum (i.e. our Universe "before" the electroweak symmetry breaking) will roll to a nearby minimum around which tachyons don't exist. Note that the masses are related to the second derivative of the potential as a function of the fields expanded around the vacuum expectation values in the relevant vacuum.

So the lowest possible eigenvalue of $M$ is $M=0$, the massless particles. In our world, only some gauge bosons are massless. Photons are massless because the electromagnetic $U(1)$ gauge symmetry is unbroken; gravitons are massless because the diffeomorphism symmetry underlying the general theory of relativity (which we treat as a theory of spin-2 fields on a Minkowski background in this text) is unbroken as well. The masslessness of these particles is related not only to the unbroken symmetries; it also has physical consequences.

The main physical consequence is the infinite range of the electromagnetic and gravitational forces. The forces really drop as $F\sim 1/r^2$ in both cases if we talk about forces in the Newtonian sense. This decreases rather slowly and the total cross sections may easily be infinite. On the other hand, short-range forces have to be mediated by massive particles. They decrease faster at long distances, with an exponential factor, and the total cross section is always finite.

There's one more particle that is "exactly massless", the gluon, the particle that mediates the force between colorful quarks. However, the gluon is charged itself, so it interacts with other gluons just like quarks do. The force is confining and it confines not only quarks but also gluons because they carry some (bi)color. For this reason, gluons can't be isolated and if we only talk about the mass spectrum of objects that can exist in isolation, the proper Hilbert space, there won't be any gluons in it, massless or otherwise.

Leptons

We have discussed the $M=0$ particles. Let's begin with positive masses. The lightest positive masses are carried by neutrinos. A few decades ago, people were considering the possibility that the neutrinos were exactly massless. However, for more than a decade, we've known from "neutrino oscillations" that the masses can't be zero. They're small but positive. Because the neutrino oscillations – operating at very long distances – are the only method we have at this moment to measure the small masses, we actually only know the mass differences.

More precisely, we only know the differences between the different eigenvalues of $M^2$. The differences are$\eq{ M_2^2 - M_1^2 &= \Delta M_{21}^2 =\\ &= \Delta M_\text{solar}^2 \approx (7.6\pm 0.2) \times 10^{-5}\eV^2\\ M_3^2 - M_1^2 &= \Delta M_{32}^2\approx \Delta M_{31}^2 \approx\\ &\approx \Delta M_\text{atmos}^2 \approx (2.43\pm 0.13)\times 10^{-3}\eV^2 }$ The neutrino mass matrix hides three eigenvalues of $M^2$ because there are three generations. However, we only know two differences between them but not the overall additive shift for all of them. Moreover, the heaviest "third" eigenvalue is much larger than the other two, so the difference between the "third" and either of the lower two eigenvalues is almost the same. The differences between the squared masses are the squares of a "few millielectronvolts".

Of course, you might imagine that all the neutrino masses are actually much heavier than that, e.g. close to an electronvolt; in that case, these eigenvalues would be nearly identical. That's unlikely. More likely, the lightest eigenvalue $M_1^2$ is very close to zero, either comparable to or perhaps much smaller than the square root of $\Delta M_{12}^2$.

Some weeks ago, I discussed the PMNS mixing matrix.

Neutrinos are not too important for our current lives because their interactions with the normal matter are almost negligible. The charged leptons are much more important. The masses in multiples of an electronvolt are$\eq{ m_e &= 511~\keV\\ m_\mu &= 105.7~\MeV\\ m_\tau &= 1.777~\GeV }$ The electron $e$ is the lightest charged particle in the Universe which is why it's the most important charged particle for life, electrical engineering, and for all materials we know in general. Its heavier cousins, the muon and the tau, are less important because they're unstable. But even if they were not decaying, their high mass would make them less important.

Note that we have gone from millielectronvolts to gigaelectronvolts, around twelve orders of magnitude. There's quite some hierarchy.

Quarks

I've already discussed the confinement in the context of massless gluons. Quarks are the most famous colorful particles in particle physics and they're confined as well. You can't isolate them; it's like trying to isolate just the North pole of a bar magnet or just one endpoint of a piece of rope.

Still, at short enough distances, the glue that confines them is weak enough and quarks look like free particles with a mass. We are discussing the mass spectrum. The quarks have the following masses (in their case, it becomes really important to discuss the renormalization scales and schemes but I won't go into that):$\eq{ m_u &= 2\sim 3~\MeV\\ m_c &= 1.3~\GeV\\ m_t &= 173~\GeV\\ m_d &= 4\sim 6~\MeV\\ m_s &= 80\sim 130~\MeV\\ m_b &= 4\sim 5~\GeV }$ You see that the accuracy is poorer here. It's partly because the strongly interacting objects are messier; some of the uncertainty above may be fixed and is related to the need to define the renormalization scales and schemes. One could make the figures more accurate.

If you try to "construct" a proton from two up-quarks and one down-quark, you will only get something like $10~\MeV$. So why is the proton mass equal to $938~\MeV$? Where are the remaining 99 percent? The answer is that 99 percent of the mass of the proton (and more than 98 percent of the mass of the neutron) is actually carried by something else than the rest mass of the three "valence" quarks. A part of it comes from the relativistic increase of the mass from the motion – the quarks are moving inside the protons and neutrons by speeds that are comparable to the speed of light so the relativistic increase matters. However, most of the mass comes from the "glue", the "potential energy" between the valence quarks, and masses of gluons and additional quark-gluon pairs that are really included inside the protons and neutrons as well. These particles have a complicated internal structure.

In this case, the mass of the bound state is dramatically different from the sum of the pieces, the three quarks' masses. Because the naive "the whole is a sum of pieces" totally fails here, we say that the system is "strongly coupled". The strong nuclear force carries the name "strong" as well and it is called in this way because it is an example of a strongly coupled interaction. Note that the adjective "strong" is used both in a general way (a force so strong that the whole isn't the sum of pieces at all) and a particular way (the force that attracts quarks and gluons i.e. colorful elementary particles).

Nuclei

Protons and neutrons are therefore complicated animals if you want to "model them" out of the most elementary building blocks we know in Nature. The nuclei are a bit easier because in some sense, they may be thought of as composites of protons and neutrons (although the really accurate description has to involve quarks and gluons as well). The "residual" interactions between protons and neutrons may still be "strong" but they're much weaker than the forces between quarks. More importantly, and it is related, they're not confining; the residual force between protons and neutrons is a short-range force and certainly doesn't prevent protons and neutrons from existence in isolation.

One may create nice bound states of protons and neutrons, the nuclei. Some of their properties may be pretty nicely understood by approximate models. For example, large enough nuclei behave as droplets of a "liquid" – a nuclear liquid much much denser than water. This liquid has a "volume density", some "surface tension", punishment for the asymmetry between neutrons and protons, and the electrostatic repulsion energy for the protons (which is what favors many neutrons in the nuclei over protons).

Of course, such a "liquid" model can't be and isn't perfect. Some nuclei are much more stable than others because the neutrons and protons fully occupy "shells" that are analogous to the atoms – but with energies that are a million times higher, at order many $\MeV$'s, a fraction of a percent of the rest mass of the proton or the neutron.

The rest mass of a proton or a neutron is almost $1~\GeV$. However, if you combine them into nuclei and fuse the nuclei or make them decay, the total mass/energy carried by the system is nearly additive, up to defects of order those several $\MeV$'s, as I mentioned. Those megaelectronvolts were constructively used both in Hiroshima and Fukushima. By the way, only 1 of the previous 50+ Japanese reactors is running at this moment. The psychologically induced financial effect of the Fukushima non-events are clearly dramatic.

When we talk about nuclear technologies, Iran is developing a 20-80 percent enriched uranium, building multi-billion-dollars fortified caves for this research, and risking the health of its economy in the wake of sanctions as well as hundreds of thousands of lives of its citizens who might die in a possibly looming Israeli/U.S. attack in order to bring some isotopes to five doctors in Tehran who say that a 20-80 percent enriched uranium could perhaps be useful to improve some treatment and save a few people in a hospital although they're not sure how it could be useful yet. At least, that's the explanation of their enrichment activities accepted by those observers who take the Iranian authorities seriously.

Bound states: the Hydrogen atom

While the strong force between quarks is really strong, there exists a weaker interaction, electromagnetism. The electromagnetic bound states only subtract a small interaction mass/energy from the mass/energy of the constituents.

Take the Hydrogen atom. An electron whose rest mass is $511~\keV$ when in isolation is orbiting around a proton whose mass is $938~\MeV$. And the resulting atom is only $13.6~\eV$ lighter than the sum of the masses of the two particles in the system. That's a very tiny binding energy. Why is it so tiny? Why it's not comparable to the rest masses of both particles?

It's because the electrostatic force responsible for the binding of the two particles is a rather weak interaction, one described by the dimensionless fine-structure constant$\alpha \sim \frac{1}{137.03604\dots}.$ In fact, this small number is squared once again and the small constant of order $10^{-4}$ has to multiply the rest mass of the electron, the lighter particle in the Hydrogen atom, to get a good estimate of the binding energy of the Hydrogen atom. Exactly because the fine-structure constant is so much smaller than one, it's possible to imagine that electromagnetically bound objects are "almost the sums of their parts". That's why it's always helpful, whenever you study electromagnetism, to begin with the idea of "free objects" and the interactions are added as a small perturbation that affects their motion. This strategy is much less useful for strongly coupled interactions such as the strong nuclear force.

By the way, the smallness of the fine-structure constant is also the reason why we can use non-relativistic physics to describe the Hydrogen atom and, similarly, other atoms. One may show that the velocity of the electron divided by the speed of light $v/c$ in the Hydrogen atom – and similarly in other atoms – is of order the fine-structure constant. The relativistic corrections start at $v^2/c^2$ so they are suppressed at least by four orders of magnitude relatively to the main non-relativistic result.

Again, this non-relativistic strategy can't be accurately used for the bound states of quarks because the speed of quarks in the nucleons is comparable to the speed of light because the relevant "strong fine-structure constant" is much closer to one than the electromagnetic one.

Other atoms and molecules

One may discuss the bound state of more complicated nuclei which naturally attract a larger number of electrons to get neutral. And ions. Some of the interaction energies scale with $Z$, the number of protons in the nucleus, in various approximate ways. But the calculation can't be done as exactly as it can be done for the non-relativistic Hydrogen. Already three charged objects, e.g. a nucleus and two electrons, that interact with each other leads to a mathematical problem that can't be analytically solved in terms of elementary (and even "not quite elementary") functions.

One may say that the electron-electron interactions in the atoms make the problem complicated. If there were only nucleus-electron interactions, we would be simply filling the Hydrogen states for the electron by many electrons. That's roughly how the periodic table of the elements emerges. However, the electron-electron interactions distort the spectrum, change the numbers, and push even the qualitative picture and scalings with $Z$ somewhere so that you don't want to believe every detail extracted from the model where the electron-electron interactions are neglected.

Still, the binding or ionization energies in the atoms are of order $1~\eV$.

Molecules

When you go to molecules, it's useful for a qualitative understanding of what happens to appreciate that the nuclei are much heavier than the electrons. In the so-called Born-Oppenheimer approximation, their kinetic energy is negligible and they sit at fixed points because the uncertainty principle isn't too constraining for heavy particles. The electrons are moving in the external potential of these classical nuclei. You get some spectrum for the electrons.

The discrete energy levels calculated from the quantum multi-electron problems depend on the positions of the nuclei. You may optimize the positions of the nuclei so that the total electronic energy plus the electrostatic repulsion energy from the nuclei is minimized. In this way, you get a good elementary picture of the molecules, with a focus on the electronic transitions that still change the energy by an electronvolt or so. That's what chemistry is all about.

However, we have neglected the motion of the nuclei and we may restore it now. The nuclei may vibrate around the optimum positions and they behave as quantum harmonic oscillators in some approximation. These vibrations add the tiny "vibrational spectrum" to the molecules. The splitting i.e. energy differences are much smaller than one electronvolt and they effectively split the electronic spectral lines into some bands. The rotational spectrum of the molecules (those which are not spherically symmetric) add even smaller energy differences, the "rotational spectrum". There are also mixed rotationally vibrational terms to the total energy.

So the spectrum of $M$ has many possible values. You may start from the total mass of the elementary particles. The interaction energies involving the electron change the total mass by a fixed number of order an electronvolt. The relative motion between the bound nuclei splits the energy levels and the splitting is well below an electronvolt and the rotational ones are even smaller.

Solids, crystals, metals

I won't discuss gases because they're composed of largely non-interacting, free molecules. So the spectrum of $M$ becomes continuous as the relative speeds between the unbound molecules may be continuously adjusted. Liquids are similarly continuous but the molecules are very close to each other which prevents us from saying that they're independent and non-interacting. So the calculations of properties are much harder than they are for gases but it's still true that the spectrum is messy and kind of continuous.

Even materials such as glass which we would call "solid" may be viewed as some "very slow liquids". The true solids have to be crystals – think about the diamonds and/or metals; the latter tend to be conductors (both of electricity and heat). Because a crystal may be viewed as a "single very large molecule", you may also see that it has some vibrational spectrum, just like other molecules. Because a crystal is large, you may organize the spectrum in a simplified way and find out that there may be "photons" (quasiparticles of sound) propagating through the crystal. Even the electrons are shared by all the atoms of the crystal and because the crystals are uniform, the motion of the electrons through crystals resembles the motion through empty space, at least many aspects of it do, much like for the phonons and other quasiparticles.

One could discuss specific features of many types of materials etc. but this blog entry is meant to be just a review of the big picture so let me stop with the material science and squalid state physics immediately after I began. ;-)

Heavier particles

I've jumped into nuclei, atoms, molecules, and materials right after quarks. However, I haven't completed the elementary particles list yet. What I skipped so far were the W-bosons, Z-boson, and the Higgs boson. Their masses are 81, 92, and 125 GeV, respectively. Just four months ago, we couldn't have said what the mass of the God particle was although sensible people's guesses were not far from 125 GeV.

There may be many other additional particles such as superpartners whose masses should be comparable to $1~\TeV$ within an order of magnitude or so. Nature may be unified etc. so there can be heavier particles linked to grand unification or other types of physics. Their masses may be arbitrary positive numbers up to $10^{18}~\GeV$ or so, the reduced Planck scale. As you're approaching higher energies, it becomes harder to produce the particles. Probably all those heavy particles are unstable so you have to create them artificially if you need them. If some of them are stable, e.g. magnetic monopoles, the cosmological evolution guaranteed that their density around the Earth is extremely low.

I haven't said one thing. If a particle is unstable, its mass is really complex, $M \approx M_\text{real} - \frac{i\Gamma}{2}$ where $\Gamma$ is the inverse lifetime. It's extremely natural and important to complexify the energies and momenta and ask what happens for complex values. Of course, normalizable states in the proper Hilbert space can only have real values of the mass/energy.

Strings and black holes

If you want to know the part of the spectrum of localized momentum-less objects whose mass is comparable to $10^{18}~\GeV$, you enter the realm in which quantum field theory with its independent point-like particle species breaks down. You really have to switch to a more complete theory according to which even the most elementary particles refuse to be point-like. (I recommend all physicists to avoid the verb "fail" and use "refuse" instead. The verb "fail" creates the incorrect impression that there is something wrong e.g. with objects' not being point-like.)

String theory is the only known – and quite certainly, the only mathematically possible – consistent theory of quantum gravity (theory agreeing with the postulates of quantum mechanics, local Lorentz invariance, as well as the equivalence principle for the gravitational force). But we still use the terms "string theory" and "quantum gravity" as slightly inequivalent terms. "Quantum gravity" is supposed to be all about the statements whose validity doesn't "obviously" depend on the probably inevitable stringy character of quantum gravity. (As our understanding of quantum gravity in string theory and the inevitability of some stringy features of gravitational systems improves, the boundary between "quantum gravity" and "string theory" gets fuzzier and the two concepts are gradually morphing into two synonyma.)

Quantum gravity in its general form has its consequences for the spectrum of $M$, too. If you study states with $M$ of "tightly bound and looking elementary" states so that $M$ exceeds the Planck mass, comparable to $10^{19}~\GeV$, quantum gravity tells you what these states have to be. There can of course be large objects and neutron stars but we know that they're not elementary.

On the other hand, black holes did look and still do look kind of elementary. They're not composed of anything "simpler" that would look more elementary. And even if they are, the interactions keeping these building blocks tightly bound are probably textbook examples of the "strongly coupled interactions".

Quantum gravity implies that "almost all" the mass eigenstates whose mass exceeds the Planck scale are black holes. Classically, the mass of a black hole may be any real number; it may be continuously adjusted. Quantum mechanically, black holes carry a large but finite entropy comparable to the area of the event horizons in some natural units so the number of microstates is$N \approx \exp (S_{\rm BH}) = \exp \left( \frac{A}{4 L_{\rm Planck}^2} \right)$ where "BH" stands both for "black hole" as well as "Bekenstein-Hawking" who co-discovered the formula. Because the entropy is finite, there must exist an exponentially large but finite number of microstates that are "similar" to a given classical black hole solution. It implies that even the mass spectrum has to be kind of discrete, although exponentially dense. It's so dense that the discreteness becomes unmeasurable for "really macroscopic" black holes whose mass has to be well above the Planck mass (otherwise the quantum corrections to the general relativity would still be huge, relatively speaking).

Now, "string/M-theory" in the narrow sense is everything that consistently interpolates between the spectrum of light elementary particles (much lighter than the Planck scale) and the black-hole spectrum in the opposite extreme limit. When you analyze all the consistent paths that physics may choose in between, you are mapping the "landscape of string theory".

In the landscape, the theories contain strings and branes (sometimes approximated well enough by points) that propagate in a higher-dimensional spacetime. It is only useful to say that the theory contains strings and branes if they're (relatively) weakly coupled at least in some description. Strings and branes may carry internal excitations that change the properties of the "ground state string" or "ground state brane" in a similar way as phonons change the properties of a crystal (or vibrational modes change a molecule). However, the changes to the mass arising from an added vibration to a string are comparable to the Planck mass: they are huge.

In perturbative string theory, which is a good enough description for a non-negligible part of the landscape, at least morally speaking, the spectrum of the internal excitations of the strings may be reduced to an infinite collection of harmonic oscillators and is exactly solvable. If you consider branes or strings at strong coupling, it becomes harder (or impossible) to analytically solve the problem and find the spectrum.

Many different kinds of physical phenomena contribute to the energy that influences the mass spectrum we discuss. For example, extra dimensions add reasonable multiples of $1/R$, in the units where also $\hbar=1$ used by the "really mature" physicists, where $R$ is the typical length/size of the compactified dimensions (and/or some curvature radius, but for warped geometries, the estimates are often harder, anyway). At energy scales well below the "Kaluza-Klein scale", we may neglect the extra dimensions and describe everything in terms of a four-dimensional quantum theory.

Because the string scale is typically higher (higher energies) than the Kaluza-Klein scale (the extra dimensions should be larger than the strings, otherwise they don't really behave as "ordinary geometry"), and because strings may be approximated by pointlike particles below the string scale, you see that below the Kaluza-Klein scale $1/R$, the description in terms of a four-dimensional quantum field theory (a theory based on point-like particles) will be a good approximation. Even in this regime, in which you have forgotten about the stringy character of matter as well as the extra dimensions, you find lots of new hierarchies, scales, and physical phenomena that influence the spectrum.

I have gotten to complicated waters and because I don't want too many overtaught readers to throw up, let me stop at this point. If I didn't stop, I would have to tell you that the spectrum of isolated objects is really just the 2-point-function part of the "full dynamical information" which also includes the $n$-point functions or scattering amplitudes (kind of equivalent to the detailed information in the Hamiltonian about all the multi-particle states, including the unbound ones). And that could get really messy. ;-)

This was long enough a text for me to postpone proofreading, too. Apologies for the enhanced expectation value of the number of typos above.