$\newcommand{\ket}[1]{|#1\rangle}$$\newcommand{\bra}[1]{\langle#1|}$In Principles of Quantum Mechanics (2nd edition) by Shankar, Exercise 5.1.3 asks to find the wave function of the free particle by means of applying the propagator to an wave function in $x$-space.
The propagator $U(t)$, which satisfies $\ket{\psi(t)} = U(t)\ket{\psi(0)}$ by definition, can be shown to have the form $U(t) = \exp(-iHt/\hbar)$ from Schrödinger's equation.
Now, for the free particle, the Hamiltonian is given by $H = P^2 / 2m$. However, Shankar then says the propagator for this problem is given by $$U(t) = \exp\left[\displaystyle-\frac{it}{\hbar}\left(-\frac{\hbar^2}{2m}\frac{\partial ^2}{\partial x^2}\right)\right].$$
But $H \ne -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}$ !! Sure, this is its action on $\ket{\psi}$ in $x$-space, in the sense that $$\bra x H\ket{\psi(t)} = -\frac{\hbar^2}{2m}\frac{\partial ^2}{\partial x^2}\psi(x,t)$$ but to be a pedant about the mathematics, what Shankar used for $H$ is surely not the Hamiltonian operator acting on Hilbert space.
It turns out that Shankar's "propagator" does in fact propagate the wave function in $x$-space in the sense that $\psi(x,t) = U(t)\psi(x,0)$. So it still smells like a propagator.
Now for my actual question: (that was just context)
What kind of mathematical object is Shankar's propagator (if it's not an operator on Hilbert space)? Is it an operator on a new vector space ($x$-space, perhaps)? Also, how does it relate to the actual propagator (the one that is an operator on Hilbert space)?