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Context I read the note Light Cone Quantization and Perturbationwritten by Guillance Beuf. He gives a argument in section 3.3.2, p17, 2nd paragraph :

In particular, in a theory with a mass gap, excitations above the ground state obey $2k^{+}k^- \geq m_{gap}^2 + \mathbf{k}^2> 0$. Hence, there is no excitation of finite $k^-$ which has $k+ =0$, and vice-versa. Therefore, the vacuum can be defined uniquely as the state of minimal $P^+$, or equivalently of minimum $P^-$

Also, he adds a foot note below :

In the case of a theory without mass gap, like QED or perturbative QCD, the vacuum state (or states) still have minimum $P^+$ and $\mathbf{P} = 0$. However, this does not characterize the vacuum uniquely anymore.

In the statement, $P^{\pm}$ are the right/left going momenta operators.

Question

I can not understand the statement about "vacuum can defined uniquely". It seems to me that the arguments before this statement are irrelevant to the conclusion. Besides, I can't understand what "characterizing" the vacuum uniquely means.

From my understanding, only when one constructs the Hamiltonian and check the spectrum of the Hamiltonian explicitly, one knows whether the vacuum degenerates or not. Is there any other analysis method to see the if the vacuum is uniquely?

And why massless QED (without mass gap) doesn't have unique vacuum?

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  • $\begingroup$ One should be cautious and distinguish possibly degenerate states from points of the continuous spectrum. When the Hilbert space and the Hamiltonian are well defined this can be done with mathematical rigor. States are sometimes also called bound states, correspond to finite dimensional projectors. Points in the continuous spectrum don't have finite dimensional projector. Now bound states can also happen in the middle or at the edge of the continuous spectrum. I don't think in QFT one has a proper Hamiltonian on a separable Hilbert space, so this analysis should rather serve as a guidance. $\endgroup$
    – lcv
    Commented Jan 21 at 13:24

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