# Momentum operator in "open" space!

1. Nov 25, 2014

### 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

2. Nov 30, 2014

### Greg Bernhardt

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

3. Dec 1, 2014

### dextercioby

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

4. Dec 1, 2014

### ShayanJ

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

5. Dec 1, 2014

### 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.