Position Operator on $l^2(\mathbb{Z})$

Question

I'm very familiar with the position operator $(Q\varphi)(x)=x\varphi(x)$ on $L^2(\mathbb{R^d})$, but I'm trying to figure out how to interpret the same operator on $l^2(\mathbb{Z})$ (the space of square summable sequences).
With a few easy calculations, it's pretty clear to see that $Q$ is self-adjoint on $D(Q)=\{\varphi\in l^2(\mathbb{Z})|Q\varphi\in l^2(\mathbb{Z})\}$.
It is well known that the spectrum of the position operator on $L^2(\mathbb{R^d})$ is purely continuous and consists of the entire real line, but I'm having trouble figuring out what the spectrum is for the operator on $l^2(\mathbb{Z})$; clearly every integer is an eigenvalue, so the discrete spectrum is $\mathbb{Z}$, but is the essential spectrum empty?
Lastly, I'm trying to figure out the spectral measure for Q on $l^2(\mathbb{Z})$. For $Q$ on $L^2(\mathbb{R^d})$, the measure is simply $\chi_{\Lambda}(x)$ where $\Lambda$ is a Borel set; what is it in this case? Wouldn't it just be the same thing? If the spectrum is just $\mathbb{Z}$, that makes sense. I've never had to deal with measures in any Hilbert space besides $L^2(\mathbb{R}^d)$.

Answer 1

Thinking of $\Bbb{Z}$ as a measure space with atomic measure for every integer $\mu(\{n\})=1$, $\forall n\in\Bbb{Z}$, then $l^2(\Bbb{Z})$ is just the space of square integrable functions $L^2_\mu(\Bbb{Z})$. The multiplication operator $M_\varphi$, where $\varphi:E\to\Bbb{R}$ is a measurable function with a Borel set $E$ as domain, is self-adjoint on the space $L^2_\mu(E)$ with domain $$\mathcal{D}(M_\varphi)=\{\psi\in L^2_\mu(E):\varphi\psi\in L^2_\mu(E)\}$$ and its spectrum is the essential image of $\varphi$. For your question, we just have to consider the identity funciton $\phi(x)=x$ and we have $Q=\mathcal{M}_x$. You already figured out the domain of self-adjointness and the eigenvalues, which are integers. To see that the spectrum is purely point, just check that the essential image (which in this case is the range of the identity function) is $\Bbb{Z}$, so $\sigma(Q)=\Bbb{Z}$. For the spectral measure, as the operator is already a multiplication one, we just have to take $\chi_\Lambda(x)$, where $\Lambda$ is a Borel set of $\Bbb{Z}$, which in this case can be any subset.