Second quantization of partial isometry

If we have a unitary map from Hilbert space $H$ to $H$, we get a unitary map from $e^{H}$ to $e^{H}$, where $e^{H}$ is the symmetric Fock space of $H$. But if we replace the unitary with partial isometry, will we get partial isometry at the Fock space level?

My main aim is to study $E_0$ semigroup (on type I factor) coming from semigroup of partial isometries on some Hilbert space.

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If $U$ is your partial isometry on $H$, I don't see what prevents you from defining a partial isometry $S(U)$ on $e^H$ by $S(U)(\oplus_{n=0}^\infty(\xi_1\otimes...\otimes\xi_n)=\oplus_{n=0}^\infty(U\xi_‌​1\otimes...\otimes U\xi_n)$. –  Alain Valette Jan 23 '13 at 17:14

For any contraction $X$ on $H$, the operator $\Gamma(X)$ (in Alain's notation S(X)) is a contraction. It is essentially algebraic to check that a pair of contractions $X,Y$ on $H$ will satisfy $\Gamma(X)\Gamma(Y)=\Gamma(XY)$ and $\Gamma(X)^\*=\Gamma(X^\*)$. So yes, you do get a partial isometry.
(In fact, for any closed densely defined operator $X$ you can define a second quantisation $\Gamma(X)$ which is again closed and densely defined. It is bounded precisely when $X$ is a contraction.)
If you're asking whether the map $W(f)\mapsto W(U_tf)$ defines an E$_0$-semigroup then the basic condition it has to satisfy is $Im<U_tf,U_tg>=Im<f,g>$ so it must be a symplectic map. If it's complex linear then it must be isometric. If you're asking whether the map $X\mapsto \Gamma(U_t)X\Gamma(U_t)^*$, (and other variants on this theme) defines an E$_0$-semigroup then note that the semigroup property implies $U_t$ should at least be an isometry. Interestingly this leaves space for E$_0$-semigroups from semigroups of symplectic maps –  Ollie Margetts Jan 24 '13 at 10:20