Consider the Hilbert space [itex]H=L^2([0,1],dx) [/itex]. Now we define the operator [itex] P=\frac \hbar i \frac{d}{dx} [/itex] on this Hilbert space with the domain of definition [itex] D(P)=\{ \psi \in H | \psi' \in H \ and \ \psi(0)=0=\psi(1) \} [/itex].(adsbygoogle = window.adsbygoogle || []).push({});

Then it can be shown that [itex] P^\dagger=\frac \hbar i \frac{d}{dx} [/itex] with [itex] D(P^\dagger)=\{ \varphi \in H | \varphi ' \in H \}[/itex].

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:

[itex]

D(P)=\{ \psi \in H | \psi ' \in H \ and \ \psi(0)=e^{i \alpha} \psi(1) \} \ \ \ \ \ (\alpha \in \mathbb R)

[/itex].

My question is, how does this procedure work if the Hilbert space is [itex] H=L^2((-\infty,\infty),dx)?[/itex](In "open" space, which is of course not in its mathematically rigorous meaning.)

Thanks

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# Momentum operator in "open" space!

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