quantum mechanics - Domain of symmetric momentum operator vs self-adjoint momentum operator

Is there an example of a function that is not in the domain of the 'naive' symmetric (but not self-adjoint) momentum operator $p:=-i\frac{d}{dx}$ but is in the 'true' self-adjoint momentum operator $p:= \left(-i\frac{d}{dx}\right)^\dagger$.

I am trying to understand the mathematical differences b/w symmetric and self-adjointness and thought that this would be an enlightening example. Could you also show why this example in the domain of the self-adjoint momentum operator using the definition of the adjoint domain:

$$D(A^\dagger) := \{ \phi \in {\cal H}\:|\: \exists \phi_1 \in {\cal H} \mbox{ with} \: \langle \phi_1 |\psi \rangle = \langle \phi | A \psi\rangle \:\: \forall \psi \in D(A)\}$$

This question was motivated by reading this fantastic answer by Valter Moretti.

Answer

Define

$$P \equiv -i\frac{d}{dx} \tag{1}$$ and $$\psi(x) = |x|\,\exp(-x^2). \tag{2}$$ Let $D(P)$ denote the domain of $P$. Clearly, $\psi$ is not in $D(P)$, because it is not differentiable at $x=0$. However, $\psi$ is in the domain of $P^\dagger$, because $$\langle \psi|P\phi\rangle \equiv -i\int^\infty_{-\infty}dx\ \psi^*(x)\frac{d}{dx}\phi(x) \tag{3}$$ is well-defined for all $\phi\in D(P)$, and $P^\dagger$ is defined by the condition $$\langle P^\dagger\psi|\phi\rangle = \langle \psi|P\phi\rangle \tag{4}$$ for all $\phi\in D(P)$. The domain $D(P)$ is dense, so $\psi$ can be arbitrarily well-approximated by a function in $D(P)$, just by smoothing out the "kink" in an arbitrarily small neighborhood of $x=0$, but $\psi$ itself is not in $D(P)$, not even after accounting for the fact that vectors in this Hilbert space are represented by functions modulo zero-norm functions. We can't smooth out the "kink" at $x=0$ in $\psi$ by adding any zero-norm function.

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Edit:

The the value of $P^\dagger$ acting on the example (2) is

$$P^\dagger\psi = -i\big(s(x)-2x|x|\big)\exp(-x^2),$$ where $s(x)=\pm 1$ is the sign of $x$. This can be checked by checking that it satisfies (4) for arbitrary differentiable functions $\phi$, which is possible because the point $x=0$ can be omitted from the integrand without changing the value of the integral.