1
$\begingroup$

It is my understanding that in quantum mechanics we use self-adjoint operators (that is an axiom of the theory). This operators can be either bounded or unbounded, being the latter the more general case.

Has the spectral theorem had been proven for the unbounded case? And if so, what references discuss this?

$\endgroup$

Before answering, please see our policy on resource recommendation questions. Please write substantial answers that detail the style, content, and prerequisites of the book, paper or other resource. Explain the nature of the resource so that readers can decide which one is best suited for them rather than relying on the opinions of others. Answers containing only a reference to a book or paper will be removed!

  • $\begingroup$ Yes it has. The wikipedia entry on the Spectral Theorem (General Self-adjoint Operators) references Section 10.1 of Quantum Theory for Mathematicians by Brian Hall. $\endgroup$ – J. Murray Mar 23 '18 at 18:33
  • $\begingroup$ If you'd prefer a free video reference, this lecture (and the ones preceding it) by Frederic Schuller does an excellent job. $\endgroup$ – J. Murray Mar 23 '18 at 18:36
1
$\begingroup$

The original source of the spectral theorem for unbounded self-adjoint operators was John von Neumann, who was a student of Hilbert. His formulation and proof can be found in the English translation Mathematical Foundations of Quantum Mechanics.

von Neumann uses the suggestive notation$$ H=\int \lambda \, \mathrm{d}{E \left( \lambda \right)}$$ for his spectral integral. von Neumann discusses the difference between symmetric operators and self-adjoint operators; he formulates abstract boundary conditions and discusses how to obtain a self-adjoint operator from the adjoint of a symmetric operator by imposing such conditions.

von Neumann's treatment is the original source and readable.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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