# Resolvent of momentum operator

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.

• 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) – Cppg Mar 5 at 19:51