Spectral measure associated to eigenvector of self-adjoint operator

Question

Let $A$ be a self-adjoint operator on .the Hilbertspace $H$ and let $\lambda_0$ be an eigenvalue of $A$ with corresponding eigenvector $\psi$. The spectral theorem tells us,that there is a projection-valued measure $P_A$ such that $A = \int \lambda dP_A(\lambda)$. To each vector $x \in H$ we associate a finite Borel measure $\mu_x$ defined by $\mu_\psi(\Omega) = \langle \psi, P_A(\Omega) \psi\rangle$.

1) I would like first to understand the very basic example of $A$ an operator on finite-dimensional $\mathbb C^n$ with eigenvalues $\lambda_1,...,\lambda_m$, $P_j$ the projection corresponding to eigenspace of $\lambda_j$. Then $A = \sum_j \lambda_j P_j$. I would like to understand why then we have $\mu_\psi(\lambda) = \sum \|P_j \psi \|^2 \delta(\lambda-\lambda_j)$.

2) Secondly, for a self-adjoint operator $A$ on $H$, if $\psi$ is an eigenvector to eigenvalue $\lambda_0$, $\|\psi\| = 1$, is it possible to give $\mu_\psi$ explicitely? Maybe someone can give a short explanation what I have to understand to find the expression by myself.

Edit: I feel stupid, but one last question: why is $\mu_\psi(\{\lambda_0\}) := \langle \psi, P_A(\lambda_0\}) \psi \rangle = \|\psi\|^2$? (in the case 2 that $\psi$ is an eigenvector to the eigenvalue $\lambda_0$)

Answer 1

1) You have, when $A=\sum_j\lambda_jP_j$, that $$ P_A(\Omega)=\int1_\Omega(\lambda)\,dP_A(\lambda)=1_\Omega(A)=\sum_j1_\Omega(\lambda_j)P_j. $$ In particular $$ \mu_\psi(\{\lambda\})=\langle\psi,P_A(\{\lambda\})\psi\rangle=\sum_j1_{\{\lambda\}}(\lambda_j)\langle\psi,P_j\psi\rangle=\sum_j\delta(\lambda-\lambda_j)\langle P_j\psi,P_j\psi\rangle\\ =\sum_j\delta(\lambda-\lambda_j)\|P_j\psi\|^2 $$

2) It depends on what you mean by "explicitly". In general, writing $A=\int\lambda\,dP_A(\lambda)$ is as much as you can say about an arbitrary normal operator. If your $A$ and $\psi$ are given in some kind of explicit form, it might be possible to say more. Note also that $\mu_\psi$ is a measure on all Borel subsets of the spectrum of $A$, so you cannot have a very explicit expression unless $A$ is very particular.

Answer 2

The spectral measure is a discrete measure concentrated on $\sigma(A)=\{ \lambda_1,\cdots,\lambda_{m}\}$, and given by $$ P(S) = \sum_{\{ j : \lambda_{j} \in S\}}P_{j}. $$ The projections are a mutually orthogonal partition of unity: $$ I = P_{1}+\cdots+P_{m}, \\ P_{j}=P_{j}^{\star}=P_{j}^{2}, \\ P_{j}P_{k}=0,\;\;\; j \ne k, \\ AP_{j} = \lambda_{j}P_{j}. $$ From the above, $(P_{j}x,x)=(P_{j}^{2}x,x)=(P_{j}x,P_{j}^{\star}x)=(P_{j}x,P_{j}x)=\|P_{j}x\|^{2}$, which gives $$ (P(S)x,x) = \sum_{\{j: \lambda_{j} \in S\}} (P_{j}x,x) =\sum_{\{j : \lambda_{j}\in S\}}\|P_{j}x\|^{2}. $$ The projection $P_{j}$ is the orthogonal projection on $\mathcal{N}(A-\lambda_{j}I)$. If $\psi$ is an eigenvector with eigenvalue $\lambda_{k}$, then $P_{k}\psi = \psi$ and, if $j \ne k$, you get $P_{j}\psi=P_{j}P_{k}\psi=0$. So $P(S)\psi = \psi$ if $\lambda_{k} \in S$ and $0$ otherwise.