# As written, how is the spectral theorem useful?

I am self-teaching myself QM using Brian Hall's Quantum Theory for Mathematicians. He devotes a good chunk of the book to motivating, explaining, and proving the spectral theorem in multiple ways. The two main motivating reasons he gives for the spectral theorem is to provide us with

1. Allow us to perform functional calculus
2. Provide a probability distribution for the result of measuring a self-adjoint operator $$A$$ by way of projection operators.

For simplicity, I will only be considering bounded operators. Let $$H$$ be an infinite dimensional separable Hilbert space, and let $$B(H)$$ denote the set of all bounded linear operators from $$H$$ to $$H$$. Using this notation, Hall provides two versions of the spectral theorem:

1. If $$A \in B(H)$$ is self-adjoint, then there exists a unique projection-valued measure $$\mu^A$$ on the Borel $$\sigma$$-algebra in $$\sigma(A)$$, with values in projections on $$H$$, such that $$\int_{\sigma(A)} \lambda\,d\mu^A(\lambda) = A.$$
2. If $$A \in B(H)$$ is self-adjoint, then there exists a $$\sigma$$-finite measure $$\mu$$ on $$\sigma(A)$$, a direct integral $$\int_{\sigma(A)}^\oplus H_\lambda \,d\mu(\lambda),$$ and a unitary map $$U$$ between $$H$$ and the direct integral such that $$[UAU^{-1}(s)](\lambda) = \lambda s(\lambda)$$ for all sections $$s$$ in the direct integral.

The first approach uses projection-valued measures, the second uses direct integrals. Hall mentions how the second version is a generalization of the multiplication operator, and that this somehow is similar to viewing each vector in $$H_\lambda$$ as being a generalized eigenvector for the self-adjoint operator in question.

Unfortunately none of this is quite clicking. I understand what the theorems are saying but I cannot see why they are of particular use or interest, nor how any useful conclusions follow. I have taken a look at Applications of the Spectral Theorem to Quantum Mechanics, but I am still left feeling unsatisfied. Would anyone be able to explain this in further detail?

• It says that self-adjoint operators (and, most importantly, the Hamiltonian) can be diagonalized. That is immense theoretical importance; whether it is of practical importance is more debateable.
– Buzz
Commented Jul 6, 2022 at 2:45
• @Buzz I suppose what is bothering me is I cannot see how that follows from the theorem as Hall has written it. Commented Jul 6, 2022 at 3:10

It is a good idea to first consider bounded operators, or even operators on finite-dimensional Hilbert spaces. In this case, the spectrum is discrete (there are only eigenvalues, no continuous spectrum), let us denote the eigenvalues by $$\lambda_n$$. In Physicists' terms, the projection-valued measure is then simply $$\mathrm d\mu^A(\lambda) = \sum_n \delta(\lambda - \lambda_n)\, |\lambda_n\rangle\langle\lambda_n|\, \mathrm d\lambda$$ and the expression you have given reduces down to the familiar $$A = \sum_n \lambda_n\, |\lambda_n\rangle\langle\lambda_n| .$$ We see that $$A$$ is diagonal in the basis of eigenvectors.

The integral $$\int_{\sigma(A)} \lambda\,d\mu^A(\lambda)$$ generalizes this to a scenario with operators with continuous spectrum. In the case of continuous spectrum, $$\int_a^{a+\epsilon} \mathrm d\mu^A(\lambda)$$ goes to zero (i.e., the zero-operator) for $$\epsilon \to 0$$, just like with any "normal" integral with non-atomic measures (no delta functions).

I am not so familiar with direct integrals, but it is probably helpful to also think about this formulation in the case of a finite-dimensional space.

Edit in response to the comment

Suppose we can rewrite some bounded self-adjoint operator in terms of an integral taken with respect to a projective measure, so what? How does that really help us or show that our operator is "diagonalizable" in some sense?

In a finite-dimensional system, expanding an operator $$A$$ in the form $$A = \sum_n \lambda_n\, |\lambda_n\rangle\langle\lambda_n|$$ is the diagonalization, expressed in a basis-independent way. You can immediately see that the matrix representation of $$A$$ in the $$|\lambda_n\rangle$$-basis is diagonal, and what the eigenvalues and eigenvectors are. What I was trying to explain above is that the expansion $$A = \int_{\sigma(A)} \lambda\,d\mu^A(\lambda)$$ generalizes that, so $$A$$ is diagonalizable in this sense.

It is useful mostly in the same way that the expansion $$A = \sum_n \lambda_n\, |\lambda_n\rangle\langle\lambda_n|$$ is useful in finite dimensions. For example, we can easily calculate functions $$f(A)$$: $$f(A) = \int_{\sigma(A)} f(\lambda)\,d\mu^A(\lambda).$$

• Thank you. Perhaps I have not phrased my question properly, which is probably better worded as suppose we can rewrite some bounded self-adjoint operator in terms of an integral taken with respect to a projective measure, so what? How does that really help us or show that our operator is "diagonalizable" in some sense? Commented Jul 6, 2022 at 4:33
• @CBBAM I have edited my answer Commented Jul 6, 2022 at 5:46
• Thank you, that helps clear things up! Commented Jul 6, 2022 at 6:34