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Is it correct to represent Higgs VEV as the coherent state?

I mean, suppose translational invariant coherent state

$$ |\alpha\rangle = Ne^{\alpha \hat{a}^{\dagger}_{\mathbf p =0}}|\alpha =0\rangle, \quad N: \quad \langle \alpha|\alpha\rangle = 1 $$

of massless particles with zero dispersion relation. In principle, it is possible to state that the non-shifted Higgs doublet operator has VEV on $|\alpha \rangle$ state. But is this correct?

With a cutoff, something like this holds with an appropriate Bogoliubov transformation. But a coherent state cannot be translation invariant. In the renormalized version the Bogoliubov transformation makes no longer sense.

@ArnoldNeumaier : when writing about absence of translational invariance of the coherent state, did You mean nonzero mass case? If no, could You explain why the coherent state isn't translation invariant?

With a cutoff, the translational symmetry is not present. Without a cutoff, the B-transformation that causes the VEV to shift makes no longer sense as an operator (it is divergend, not unitarily implementable).

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