A Bosonic Field operator in this exponential

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If i had a bosonic field ##\phi(x)## and I took the exponential in the following way to get the operator $$W=e^{\imath f \phi(x)}$$ where ##f## is a parameter what effect would this have when acting on the vacuum ##|0\rangle##? Is it analogous to the space translation operator? Will it transform the vacuum into some state?
 

A. Neumaier

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Taking the exponential of a distribution is ill-defined and needs a regularization to give it a meaning. The result is a state not described by the original Fock basis.
 

Demystifier

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If i had a bosonic field ##\phi(x)## and I took the exponential in the following way to get the operator $$W=e^{\imath f \phi(x)}$$ where ##f## is a parameter what effect would this have when acting on the vacuum ##|0\rangle##? Is it analogous to the space translation operator? Will it transform the vacuum into some state?
You get a "coherent" state in which the number of particles is uncertain. To see that, expand the exponential in the Taylor series and split the field into creation and destruction operators.

As @A. Neumaier alluded, the Taylor series does not really converge, but this mathematical nuisance can be fixed by introducing a physical cutoff. See e.g. the book by Itzykson and Zuber, Sec. 4-1-2.
 
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