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I was reading the book by Roman where he discusses the linear momentum operator on the half line: $pf= -if’$, this operator is densely defined and in the book he finds the adjoint by using the inner product on $L^2(0,\infty)$, one wishes to find the class of $h$ such that: $$ \int _0 ^\infty h^*(-ig’) dx = \int _0 ^\infty \psi^* g \: dx $$ for some $\psi$ and all $g$ in the domain of $p$. So clearly this is easy to do if $h$ is differentiable and with square integrable derivative by integration by parts. The question is: could there be non differentiable $h$ that satisfy this? I have been trying to show that there aren’t any such functions but can’t get anywhere and the book assumes this.

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  • $\begingroup$ In the context of quantum mechanics, don't functions have to be continuous? And usually they're differentiable as well, right? Certainly wave functions are, or they couldn't be solutions of the Schrodinger equation. $\endgroup$ – Adrian Keister Mar 28 at 15:50
  • $\begingroup$ no strictly that’s not true, in that what is of interest here is the momentum operator on a certain space and finding its adjoint, so basically my question is what the domain of this adjoint is, momentum is not an observable on this space because it’s not self adjoint $\endgroup$ – Michele Galli Mar 28 at 15:57
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Suppose your equation holds for all $g\in\mathcal{D}(p)$. Let $g_y$ be the function that is $1$ on $[0,y]$, is linear on $[y,y+\epsilon]$ and is $0$ at $y+\epsilon$, as well as for all $x > y+\epsilon$. Then your adjoint equation would require $$ \int_0^{\infty}h^* (-ig_y')dx=\int_0^{\infty}\psi^* g_ydx \\ -i\frac{1}{\epsilon}\int_{y}^{y+\epsilon}h^*dx=\int_0^{\infty}\psi^*g_ydx $$ Now, letting $\epsilon\downarrow 0$ gives $$ -ih^*(y)=\int_0^y\psi^*dx $$ Hence, $h$ must be differentiable a.e., $h(0)=0$, and $$ h'(y)=-i\psi(y). $$

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