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  Are there any non-physical eigenstates of interacting QFT?

+ 2 like - 0 dislike

While reading about the Källén-Lehmann representation I came across the definition of eigenstates in general QFT. As $\vec{p}$ (total momentum) and $H$ commute they can be simultaneously diagoalized, thus one obtains:
$$ H | {\lambda_{\vec{p}}} \rangle = E_{\vec{p}} |{\lambda_{\vec{p}}} \rangle$$ 
$$ \vec{p} |{\lambda_{\vec{p}}} \rangle = \vec{p} |{\lambda_{\vec{p}}} \rangle$$
Given a ket $|{\lambda_{\vec{0}}} \rangle$ one can go to all the kets $|{\lambda_{\vec{p}}} \rangle$ by a Loretz transformation. We can then partion the set of all possible eignestates grouping those which are related by a Loretz transformation and giving them the symbol $\lambda$ (as I have actually already done). We would expect physical states to have an invariant mass, so there should exist $m$ such that $m^2 =E_{\vec{p}} - \vec{p}^2$, but it doesn't seem obvious to me that all the eigenstates admit such a relation between eigenvalues. Do I have to sum up only on the physical states (if there is any that is non-physical) in the completeness relation? 
$$\mathbb{1} = | \Omega \rangle \langle \Omega | + \sum_{\lambda}\frac{d^3 p}{(2 \pi)^3} \frac{1}{2 E_{\vec{p}}} |\lambda_{\vec{p}} \rangle \langle \lambda_{\vec{p}} | $$ 
Thus effectively defining the Hilbert space as the one generated by physical states?

asked Aug 31, 2017 in Theoretical Physics by MrRobot (20 points) [ revision history ]

1 Answer

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The relevant Hilbert space in which to diagonalize the quantum field Hamiltonian and get a partially discrete spectrum is the space in which the system under consideration is in its rest frame; hence the spatial momentum vanishes. Then the discrete eigenvalues of H are the bound state masses $m$ (times $c^2$).

Extending the Hilbert space to arbitrary Lorentz frames then amounts to allowing eigenstates with arbitrary spatial momentum, obtained by an appropriate Lorentz transform, and this changes the energy $E$ to satisfy the relation $E^2=m^2+p^2$ (if $c=1$ or $E^2=(mc^2)^2+(cp)^2$ in arbitrary units. Thus each mass becomes a mass shell. Arbitrary states are then given by a momentum integral over spatial momentum of a sum over bound state masses, including an integral over the mass for the continuum part of the mass spectrum.

answered Sep 12, 2017 by Arnold Neumaier (15,757 points) [ no revision ]

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