How does a propagator act on a wave function in $x$-space?

Question

$\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)?

Answer 1

$\newcommand{\ket}[1]{|#1\rangle}$G. Smith's first comment got me thinking in the right direction, so I think I've figured out an answer to my question.

The propagators I'm used to are defined by $\ket{\psi(t)} = U(t)\ket{\psi(0)}$, and in order for $\ket{\psi(t)}$ to obey Schrödinger's equation, $i\hbar\frac{d}{dt}\ket{\psi(t)} = H\ket{\psi(t)}$, it follows that the propagator is given by $U(t) = \exp(-iHt/\hbar)$.

Analogously, I believe Shankar's propagator of the wave function in $x$-space must be defined by $\psi(x,t) = U(t) \psi(x,0)$. Furthermore, in order for the wave function to satisfy Schrödinger's equation in $x$-space, $i\hbar\frac{\partial}{\partial t}\psi(x,t) = -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x,t)$, it follows that this propagator must be given by $U(t) = \exp(-\frac{it}{\hbar}(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}))$, which is what Shankar used.

These propagators must be different mathematical objects (and I'm still not sure how they're directly related except by analogy), but they do the same job and now I see why they both work when applied to the relevant vector/function, so I'm content.

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Footnote:

To be clear, by "the wave function in $x$-space", I mean $\psi(x,t) = \langle x|\psi(t)\rangle$.