Momentum operator in "open" space

[ShayanJ]

Consider the Hilbert space $H=L^2([0,1],dx)$. Now we define the operator $P=\frac \hbar i \frac{d}{dx}$ on this Hilbert space with the domain of definition $D(P)=\{ \psi \in H | \psi' \in H \ and \ \psi(0)=0=\psi(1) \}$.
Then it can be shown that $P^\dagger=\frac \hbar i \frac{d}{dx}$ with $D(P^\dagger)=\{ \varphi \in H | \varphi ' \in H \}$.
So the operator P is Hermitian but no self-adjoint. But its possible to make P self-adjoint by enlarging its domain to the following:
$D(P)=\{ \psi \in H | \psi ' \in H \ and \ \psi(0)=e^{i \alpha} \psi(1) \} \ \ \ \ \ (\alpha \in \mathbb R)$.
My question is, how does this procedure work if the Hilbert space is $H=L^2((-\infty,\infty),dx)?$(In "open" space, which is of course not in its mathematically rigorous meaning.)
Thanks

 

[dextercioby]

In <open space>, there's no restriction on the domain of the derivative operator other than the norm of the image be finite: $D(P)=D(P^{\dagger})=\{φ∈H|φ′∈H\}$

 

[ShayanJ]

In <open space>, there's no restriction on the domain of the derivative operator other than the norm of the image be finite: $D(P)=D(P^{\dagger})=\{φ∈H|φ′∈H\}$

I don't think that works. How should we kill the boundary terms when we do the integration by parts then?

 

[dextercioby]

That $\phi(-\infty) = 0$ and $\phi(\infty) = 0$ follows from the requirements that the functions in the maximal domain of P be absolutely continuous on any finite interval of $\mathbb{R}$ and moreover, as I said, both $\phi(x)$ and $\phi'(x)$ belong to $\mathcal{L}^2(\mathbb{R})$. This is firstly stated on page 106 of Akhiezer and Glazman's <Theory of Linear Operators in Hilbert Space> and later on page 111. Alternatively, check out page 18 of http://arxiv.org/abs/quant-ph/0103153v1.