I have to explicitly determine the resolvent of the momentum operator, defined as follows: $$P:H^1(\mathbb{R}) \subset L^2( \mathbb{R} )\rightarrow L^2 (\mathbb{R})$$

$$\phi \rightarrow -i \frac{d}{dx} \phi$$

$H^1(\mathbb{R})$ is here the subspace of $L^2 (\mathbb{R})$ functions whose derivative (in distributional sense) is still a $L^2 (\mathbb{R})$ function.

To do this I have to determine the resolvent set of the operator (the set of $z \in C$ such that $P-z$ is a bijection, following the subsequent steps:

  1. I solved the eigenvalue equation for the operator $P$. Namely I solved the differential equation $-iT'-zT=0$ , where T is a distribution, looking for non-zero solutions. I got that the only solutions are regular distributions of the form $T=c\cdot e^{izx}$ ($c$ is a complex constant). These are not in the domain of the operator $P$. Therefore the operator has no eigenvalues and the operator $P-z$ is injective $\forall z \in \mathbb{C}$.
  2. I still have to understand for which $z \in \mathbb{C}$ the operator $P-z$ is surjective. So, given an arbitrary function $\phi \in L^2 (\mathbb{R})$, I solved for $T$ the differential equation: $-iT'-zT=\phi$. I found: $$T=c\cdot e^{izx}+i\int_{0}^{x}e^{iz(x-y)} \phi(y)dy$$

If the distribution $T$, solution of the previous equation, is an element of the domain of the operator $P$, I can say that $P-z$ is also surjective and that the inverse operator $(P-z)^{ -1}$ (which is just the resolvent of $P$ at point $z$) acts in the following way: $$\phi \in L^2 (\mathbb{R}) \longmapsto c\cdot e^{izx}+i\int_{0}^{x}e^{iz(x-y)} \phi(y)dy$$ My question is: given $z \in C$ how to prove if $T=c\cdot e^{izx}+i\int_{0}^{x}e^{iz(x-y)} \phi(y)dy$ is an element of $H^1(\mathbb{R})$ ?

Since $P$ is a self-adjoint operator I should find that if $z$ is a non real number, $P-z$ has to be a bijection. Moreover, from phyisics, I know that the spectrum of the momentum operator has to be the entire real line. So if $z$ is real, $T=c\cdot e^{izx}+i\int_{0}^{x}e^{iz(x-y)} \phi(y)dy$ will not be an element of the domain of $P$. I just can't prove this.

Of course any observation or correction to my procedure wuold be appreciated.

  • $\begingroup$ The greens functions of P - z is a compact topograph of the action around the delta function of z. This complex number is the ungraded resolvant of the operator. Use the complex fourier tranform on the nullpotent z values that make P - z = 1 and you will have the resolvant in the form of a polynomial. Classical examples are legendre, rodriguez and laguerre. (Depending on z separation variables) $\endgroup$ – Cppg Mar 5 at 19:51

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