Understanding spectral measures Let $e_1=(1,0,0,....) \in \ell ^2 (\mathbb{N})$, and $A$ be the infinite matrix 
\begin{pmatrix}0 & 1 & 0\\1 & 0 & 1 & 0 & ...\\0 & 1 & 0 & 1 & 0\\0 & 0 & 1 & 0 & ...\\.. & ... &  & ... & ...\end{pmatrix}
I would like to calculate explicitly the functional calculus $f(A)$ for functions that are not polynomials, such as continuous functions and indicator functions., and/or to understand the spectral measure $\mu_{e_1}$, i.e. being able to calculate the integral $\int f d\mu_{e_1}$ for the functions mentioned above.
Is such a thing possible? 
 A: This defines a right shift operator $S$:
$$
            Se_n = e_{n+1},\;\; n \ge 1.
$$
This operator has a standard representation on $H^2(\mathbb{T})$, consisting of holomorphic functions in the unit disk with square integrable boundary functions. On this space, $S$ is represented in function form by
$$
                      (Sf)(z) = zf(z).
$$
Here $1$ is identified with $e_1$, and $z^{n}$ is identified with $e_n$. The adjoint $S^*$ of $S$ is given by
$$
       (S^*f)(z)=\frac{f(z)-f(0)}{z}.
$$
Your operator is $S+S^*$, which is given by
$$
           (S+S^*)f = zf(z)+\frac{f(z)-f(0)}{z}.
$$
The resolvent of this operator $R(\lambda)=(S+S^*-\lambda I)^{-1}$ is defined at least for $\lambda\notin\mathbb{R}$, and $g=R(\lambda)f$ is determined by the equation:
$$
             (S+S^*-\lambda I)g=f.
$$
So the resolvent equation becomes
$$
         \left(z+\frac{1}{z}-\lambda\right)g(z)-\frac{g(0)}{z}=f(z).
$$
The singular terms on the left cancel, leaving the limit as $z\rightarrow 0$ of both sides to be
$$
             -\lambda g(0) = f(0), \\
                 g(0) = -\frac{1}{\lambda}f(0).
$$
Then you can obtain an expression for the resolvent $g=R(\lambda)f$.
A: You must excuse my astonishment, when reading pages $2$ and $3$ of Paul-Andre Meyer's "Quantum Probability for Probabilists" I encountered just the same question that you posed. I post this, because the above answer does not completely answer your question, and because the answer is truly surprising.
Let me spell out the (simple) techniques involved in finding the measure $\mu_{e_1.}$
Consider the operator $B = (S + S^*)/2 = A/2,$ where $S$ is the right shift operator and $S^*$ is it's adjoint. Then consider the "probability law" $\rho$ on the algebra of bounded linear operators defined by $$\rho(C) = \langle e_1 | C|e_1 \rangle.$$ For the definition of a probability law on a $^*$-algebra consult the reference I gave. What is important is that for selfadjoint operators this is really gives rise to a probability measure $\pi_C$ which is characterised by the moments:
$$
\int x^k\pi_C(dx) = \rho(C^k).
$$
All this to say that the measure $\pi_{B}$ Meyer is considering is really the spectral measure $\mu_{e_1}^B$ associated to $B$. Now one can simply compute these moments: a combinatorial argument which I will not explain since it's contained in the reference shows that:
$$
\mu_{e_1}^B = \text{ Wiegner semicircle law with radius } r = 1.
$$
Now to find the measure $\mu_{e_1}^A$ we simply rescale the above measure, so
$$
\mu_{e_1}^A = \text{ Wiegner semicircle law with radius } r = 2.
$$
