This question is a continuation of previous questions about the spectral theorem. We consider a finite-dimensional space, say $V \in \mathbb{C}^{n}$, and let $A \in V$ denote a self-adjoint (hermitian) operator in this finite dimensional space. We can then state the spectral theorem as $A=\sum_ia_iP(a_i)$, i.e. the operator $A$ can be decomposed as the summation of the product of each eigenvalue multiplied by each projection operator. I understand this representation and see how it can be coupled to the spectral theorem for matrices given by $A=U \Lambda U^{\dagger}$ where $U$ contains the eigenvectors along the columns of the matrix and $\Lambda$ contains the eigenvalues along the diagonal. I think it is a beautyful relation and imagine many advantages of decomposing the operator using this structure.
Now; For the infinite-dimensional case we consider the Hilbert space $\mathbb{H}$, and we want to express the spectral theorem within this space. Again, we let $A$ denote a self-adjoint (hermitian) operator within this Hilbert space so that $A \in \mathbb{H}$. To this operator there exists a unique family of projection operators, $E(\lambda)$, for real $\lambda$, with specific properties. To be explicit, these properties are given by Leslie E. Ballentine in the book Quantum Mechanics: A Modern Development (and I cite):
To each self-adjoint operator $A$ there corresponds a unique family of projection operators, $E(\lambda)$, for real $\lambda$, with the properties:
If $\lambda_1 < \lambda_2$ then $E(\lambda_1)E(\lambda_2)=E(\lambda_2)E(\lambda_1)=E(\lambda_1)$
If $\epsilon>0$, then $E(\lambda+\epsilon)|\psi\rangle \rightarrow E(\lambda)|\psi\rangle$ as $\epsilon\rightarrow 0$
$E(\lambda)|\psi\rangle \rightarrow 0$ as $\lambda\rightarrow-\infty$
$E(\lambda)|\psi\rangle \rightarrow |\psi\rangle$ as $\lambda \rightarrow +\infty$
$\displaystyle \int_{-\infty}^{\infty}\lambda \, \mathrm dE(\lambda)=A$
In my previous question, I was told that if we define the family of projection operators $E(\lambda)$ as follows;
$$E(\lambda):=\sum\limits_{\sigma(A)\ni a\leq \lambda} P_a = \sum\limits_{a \in \sigma(A) } \theta(\lambda-a)\ P_a \quad $$
Then we can derive a connection between the spectral theorem defined in finite- and infinite dimensions by using the properties of the projection operators $P(a_i)$ (I interpret this as the properties stated by Ballentine, given above, although these are given for $E(\lambda)$, not $P(a_i)$, -perhaps someone can clarify this as well?). I was told that the spectral theorem in finite dimensions basically states that the hermitian operator $A$ admits a complete orthonormal eigenbasis (and thus can be diagonalized), and I think this is an important aspect because, as Wikipedia states, the spectral theorem is the result when a linear operator or matrix can be diagonalized. I understand this, but yet I struggle to connect the definition of the spectral theorem in finite dimensions to its definition in infinite dimensions. Can someone explain this to me (and perhaps be as explicit as possible because I am still very new to this field)?
Related posts: What is the connection between the spectral theorem defined for Hilbert space and matrices respectively? & How to show the equivalence between different versions of the spectral theorem?
