8
$\begingroup$

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.

$\endgroup$
2
  • 2
    $\begingroup$ Welcome to PSE. Although necessary for quantum mechanics, your question is pure Mathematics (functional analysis). May be you must post it in Mathematics. $\endgroup$
    – Frobenius
    Feb 25, 2022 at 7:07
  • 6
    $\begingroup$ 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. $\endgroup$ Feb 25, 2022 at 10:15

1 Answer 1

13
$\begingroup$

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 $$\left\{\psi \in {\cal H} \: \left| \: \sum_{n} |\lambda_n|^2 |\langle \psi_n| \psi \rangle|^2 < +\infty\right. \right\}$$

$\endgroup$
1
  • 1
    $\begingroup$ Dear Prof. Moretti, if you have time, I'd highly appreciate if you could take a look on my related question here. $\endgroup$ May 17, 2022 at 7:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.