From the spectral theorem to the completeness relation in quantum mechanics

I often heard that the eigenfunctions of a Hermitian operator form a completeness basis, as $$\sum_i | i \rangle \langle i | = \hat{1} \tag{1}$$

and the mathematical foundation is the spectral theorem. The spectral theorem is $$\hat{A} = \int \lambda dE_{\lambda}, \hspace{3cm} (2)$$ which in physicists' notation is (assuming discreteness) $$\hat{A} = \sum_a a | a\rangle \langle a| \hspace{3cm} (3)$$

My question is: Eq. (1) is for a unit operator, which can be inserted anywhere. Eqs. (2) and (3) are for the operator $$\hat{A}$$, which is not a unit operator. How to transform (2) and (3) into the form (1)? Does it require some kind of scaling? (is that the projection-valued measure?) Perhaps such scaling is possible for a bounded operator and requires a more dedicated proof for an unbounded operator.

• Welcome to PSE. Although necessary for quantum mechanics, your question is pure Mathematics (functional analysis). May be you must post it in Mathematics. Feb 25, 2022 at 7:07
• Yes it is pure mathematics, but usually pure mathematicians are not able to answer nor to understand well the issue (also for the esoteric jargon of physicists). A mathematical physicist instead does, so also here is a right place to post this question. Feb 25, 2022 at 10:15

1 Answer

The spectral theorem is that, if $$A: D(A) \to {\cal H}$$ is a selfadjoint operator, where $$D(A) \subset {\cal H}$$ is a dense subspace, then there exists a unique projector-valued measure $$P^{(A)}$$ on the Borel sets of $$\mathbb{R}$$ such that $$A = \int_{\mathbb R} \lambda dP^{(A)}(\lambda)\:.$$ As a consequence (this is a corollary or a definition depending on the procedure) $$f(A) = \int_{\mathbb R} f(\lambda) dP^{(A)}(\lambda) \tag{1}$$ for every $$f: {\mathbb R} \to {\mathbb C}$$ Borel measurable. Taking $$f(x) =1$$ for all $$x\in {\mathbb R}$$ we have $$I = \int_{\mathbb R} dP^{(A)}(\lambda)\:.$$ For selfadjoint operators admitting a Hilbert basis of eingenvectors $$\psi_{\lambda, d_\lambda}$$, $$\lambda \in \sigma_p(A)$$ and $$d_\lambda$$ accounting for the dimension of the eigenspace with eigenvalue $$\lambda$$, the identity above reads (referring to the strong operator-topology) $$f(A) = \sum_{\lambda, d_\lambda} f(\lambda) |\psi_{\lambda, d_\lambda}\rangle\langle \psi_{\lambda, d_\lambda} |\:, \tag{2}$$ with the special case $$I = \sum_{\lambda, d_\lambda} |\psi_{\lambda, d_\lambda}\rangle\langle \psi_{\lambda, d_\lambda} |\:. \tag{3}$$ In summary Eqs.(1) and (2) are the central identities, Eq.(3) is just a special case.

Given an orthonormal complete basis $$\{\psi_n\}_{n \in \mathbb N} \subset {\cal H}$$, one can always define ad hoc a selfadjoint operator $$A$$ (with no physical meaning in general) to implement the identities above: $$A = \sum_{n \in \mathbb{N}} \lambda_n |\psi_{n}\rangle\langle \psi_{n} |$$ for a given arbitrary choice of real numbers $$\lambda_n$$. The domain of $$A$$ is $$\left\{\psi \in {\cal H} \: \left| \: \sum_{n} |\lambda_n|^2 |\langle \psi_n| \psi \rangle|^2 < +\infty\right. \right\}$$

• Dear Prof. Moretti, if you have time, I'd highly appreciate if you could take a look on my related question here. May 17, 2022 at 7:15