Physics Asked by J. Murray on November 12, 2020
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.
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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