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In the book Quantum Mechanics: A Modern Development by Leslie E. Ballentine the Spectral Theorem is defined as follows (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 $\lambda \rightarrow +\infty$

$\displaystyle \int_{-\infty}^{\infty}\lambda \, \mathrm dE(\lambda)=A$

However, if we define $A$ to be a symmetric matrix (So $A=A^T$) then we can decompose $A$ as $A=T\Lambda T^{-1}$ with $T$ containing the eigenvectors of $A$ and $\Lambda$ being a diagonalized matrix containing the eigenvalues of $A$. How does this description of the spectral theorem for symmetric matrices apply to the definition given by Ballentine? I mean how do each step for defining the spectral theorem given by Ballentine support (or rather expand) the definition of the spectral theorem for symmetric matrices?

Tobias Fünke
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Rasmus Andersen
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1 Answers1

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Consider a finite-dimensional, complex Hilbert space $H$ and let $A$ denote a hermitian (self-adjoint) operator. For each eigenvalue $a\in\sigma(A)\subset \mathbb R$, let $P_a$ denote the (orthogonal) projection on the corresponding eigenspace, such that $$A=\sum\limits_{a\in \sigma(a)} a\, P_a \tag 1 \quad $$

and $$\sum\limits_{a\in \sigma (A)} P_a = \mathbb I \tag 2 \quad ,$$ with $P_a P_{a^\prime}=P_a\delta_{aa^\prime}$, which summarizes one version of the spectral theorem in finite dimensions. Now define $$E(\lambda):=\sum\limits_{\sigma(A)\ni a\leq \lambda} P_a = \sum\limits_{a \in \sigma(A) } \theta(\lambda-a)\, P_a \quad .\tag 3 $$

This gives the desired connection: By making use of the properties of the projection operators $P_a$, one can prove that the $E$ in $(3)$ is indeed a spectral family.

Conversely, starting from the more general version of the spectral theorem sketched in Ballentine's book, the "usual" spectral theorem in finite-dimensions can be recovered and it also follows that the (unique) spectral family is indeed given by $(3)$.

Tobias Fünke
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