Bosonic Field operator in this exponential

[QFT1995]

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]

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]

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.