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One of the key elements of any quantum mechanical system is the spectrum of the Hamiltonian. But what about in quantum field theory? It seems as if nobody ever discusses the spectrum of a system at all -- or have I missed something? Just what role does the spectrum of any given potential play in field theory? To put it another way: does knowledge of a spectrum in QM help to understand the analogous problem in QFT?

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Brilliant question +1 –  user346 Jan 26 '11 at 18:42
    
I blame renormalization ;-) –  mbq Jan 26 '11 at 21:52
    
The analog of the "spectrum" solution for a field theory is the S-matrix (at least for mass-gap case--- in the non-mass-gap case you also need the asymptotic field equations that define the boundary massless fields). –  Ron Maimon Apr 15 '12 at 6:34
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Because of Lorentz Invariance, the spectrum of the Hamiltonian is always continuous. This does not mean bound state problems are not interesting, but the relevant discrete spectrum to discuss is the spectrum of (Lorentz invariant) masses. For an example think about bound state of electron and positron (positronium), in which they orbit each other. There is a discrete spectrum of masses of this composite object, to do with what is the angular momentum of their orbit around the center of mass. But of course, if you boost the system as a whole the energy of that moving object (which unlike the rest mass is not Lorentz invariance) will change and can take continuous values.

The other reason you hear less about the spectrum in QFT is because the problem of finding the spectrum in bound state problems is a lot harder (to define and to calculate). If you want an entry point to relevant discussions you can look for references for the Bethe-Salpeter equation, and to lattice QCD, two of the many approaches that discuss spectrum in QFT.

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Nice, +1. Going to look at Bethe-Salpeter equation because this term is unknown to me. –  Luboš Motl Jan 26 '11 at 18:36
    
Thanks for the edit, I should have added those links. –  user566 Jan 26 '11 at 18:50
    
OK, so the discrete mass spectrum in QFT is the relevant one. But for, say, a free scalar field, the potential is that of a harmonic oscillator -- but with a mass term put in by hand. Are you saying that were we to consider the scalar as a bound state, that the hand-inserted m would emerge from (n+½)ħω -- where ω is some energy scale of the more-fundamental theory? –  geeky Jan 26 '11 at 19:22
    
Free scalar field represent a single type of particle, and the only thing you need to specify is the number of particles (which are the levels of the harmonic oscillators you refer to) and their energy-momentum. That harmonic oscillator has nothing to do with the mass spectrum, the discreteness of the spectrum has to do with the number of particles being an integer. If you want to have the scalar as bound state, you'd have an analogous "potential" which binds your fundamental fields, but the problem is mathematically more complex that simple harmonic oscillator, or QM particle in a potential. –  user566 Jan 26 '11 at 21:17
    
OK, I think I get it: the occupation number n is the number of (in this example) identical scalar particles. Of course. But each particle brings with it an energy ħω, yes? And if so, then in the rest frame of a state with a single particle, won't 3/2 ħω=mc^2? –  geeky Jan 27 '11 at 3:00
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As the four components of the momentum vector commute, the right spectrum to consider is the joint spectrum of the momentum; the energy spectrum alone is far too degenerate. Because of Lorentz symmetry, the spectrum is a union of orbits of the Lorentz group in momentum space. Thus the spectrum is given by the set of irreducible unitary representations of the Poincare group available in the theory; each of them has a given value $m$ of the mass, and $p^2=m^2$. Thus the spectrum is given by a union of hyperboloids, the so-called mass shells. Since energy must be nonnegative in a physical theory, the only representations occuring are the vacuum ($p=0$), massless particles ($m=0$, any helicity), massive particles ($m>0$, any spin), and their scattering states (sum of single-particle momenta, and nothing else if asymptotic completeness holds); bound states count among the particle states.

If there is a mass gap (i.e., if no representation of tiny positive mass exists), the states can be restricted to their rest frame, where the spatial momentum vanishes; in this case the energy spectrum agrees with the mass spectrum, and consists of the ground state energy zero (for the vacuum), an additional discrete part for the particles (bound states), and a continuous part for the scattering states. The structure of the scattering states is further determined by the S-matrix. This tells the whole story.

Thus the spectrum tells the particle content including bound states (in a nuclear democracy, without differences between elementary and composite particles) and scattering states, though the states themselves are not visible in the spectrum but only in the associated resolvent.

To better understand what happens compare this with nonrelativistic quantum fields. There the situation is the same except that the Galilei group replaces the Poincare group, and the mass shells flatten to parallel 3-spaces parameterized by the momentum of the center of mass. In particular, the spectrum is again purely continuous. But the standard spectral analysis (for example that of the hydrogen atom) is done in the rest frame, where the center of mass motion is decoupled, resulting in a discrete + continuous spectrum of the same structure as in the relativistic case.

That this is never told in QFT textbooks (one only may find hints understandable only by those who know it already) is a terrible mistake, as it makes QFT look far more different from standard quantum mechanics than it really is.

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I just want to note that in the first chapters of your favourite texbook, whe you quantize the basic scalar field, you are actually finding a spectrum of your hamilonian, don't you?

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