Canonical Transformations in Quantum Field Theory

In his lecture notes on canonical transformations in quantum field theory, Massimo Blasone defines the boson translation transformation to be

begin{equation}
a_k rightarrow a_k(theta) = a_k + theta_k
end{equation}
for $theta_kinmathbb C$, and notes that the transformation can be thought of as the action of some operator $U(theta)$:
$$a_k(theta) = U(theta) a_k U^dagger(theta) = a_k + theta_k$$
$$U(theta) = exp[iG(theta)]$$
$$G(theta) = -iint d^3k left(theta^*_k a_k – theta_k a^dagger_kright)$$

He goes on to say that because the $a_k(theta)$‘s do not annihilate the vacuum $|0rangle$, we must define a new vacuum state $|0(theta)rangle$ such that $a_k(theta)|0(theta)rangle = 0$. In terms of the old vacuum, we have

$$|0(theta)rangle = U(theta)|0rangle = expleft[-int d^3k |theta_k|^2right] expleft[-int d^3k theta_k a^dagger_kright] |0rangle tag{*}label{*}$$

and so

$$langle 0 | 0(theta)rangle = expleft[-int d^3k |theta_k|^2right] $$

which means that if $int d^3k |theta_k|^2 = infty$ then $langle 0 | 0(theta)rangle = 0$. This then means that this transformation leads out of the original Hilbert space and that the new representation of the CCRs is unitarily inequivalent to the original one.

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My question is regarding eqref{*}. It seems strange to me that the two exponentials can be split in that way – if the integral in the first exponential evaluates to $infty$, would not that just kill the entire state? Should it not instead be written

$$ |0(theta)rangle = expleft[-int d^3k left( |theta_k|^2 + theta_k a^dagger_kright)right]$$

instead?

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Related:
Unitarily Inequivalent Representations