Physical meaning of the operator $\exp(-a {\hat{p}}^2)$

You are asking about the "physical meaning" of the celebrated Weierstrass transform, which is used routinely in physics, of course: $$ \bbox[yellow]{ e^{\partial_x^2}f(x) =\frac{1}{\sqrt{4\pi}} \int_{-\infty}^\infty f(x-y)~ e^{-y^2/4}\;dy}~. $$ In your case, $$\langle x |\exp(-a {\hat{p}}^2)| \psi \rangle = \exp(a \hbar^2 \partial_x^2)~\langle x|\psi \rangle\\ =e^{a\hbar^2 \partial_x^2} ~\psi(x) =\frac{\sqrt{a} \hbar}{\sqrt{4\pi}} \int_{-\infty}^\infty \!\!dy ~~\psi(x-y)~ e^{-a\hbar^2 y^2/4} ~, $$ a Gaussian smoothing, (low pass filtering) of the wavefunction.

• For the formal δ(x-x')-wavefunction you chose, you naturally smooth that to a Gaussian, $\sqrt{a/4\pi} \hbar \exp (-a\hbar^2 (x-x')^2/4)$, your matrix element. In a sense, you undo the limit of the Gaussian you took to get the δ.

There is a bevy of applications of this transform in phase-space quantization.

NB. In point of fact, through Fourier transformation, you may find the corresponding integral kernels for any power n of the momentum operator in the exponential. They are generalized hypergeometrics $_0 F_{n-2} (y)$, so, extending the Gaussian exemplified above.