We know that quantum mechanics requires self-adjoint operators, not only symmetric. Can we say that this follows ONLY from the two following axioms of quantum mechanics, namely that

  1. each observable $a$ corresponds to a linear operator $A$


  1. an expectation value of a measurement of $a$ must be real


I thought these two imply only the need for a Hermitian (i.e. symmetric) operators (because a linear Hermitian operator has real eigenvalues) and that the need for self-adjointness was somehow connected to an additional requirement such as unitarity of the time-evolution operator. What is the missing piece?

(I know how the two terms are defined, e.g. here.)


In QM, a real valued observable $A$ is mathematically represented by a projector valued measure over $\mathbb{R}$, $P^{(A)}$, i.e., if $E$ is a Borel subset of the real line, then $P^{(A)}(E)$ is a projector representing the proposition "the outcome of measuring $A$ falls in $E$". In principle, that's all you need for representing, mathematically, observables in QM (I'm assuming the fundamental formulation of the theory via the non-distributive lattice of propositions).

But, by the Spectral Theorem for unbounded self-adjoint operators (proved by von Neumann), we know that given an observable $A$ represented by the projector valued measure $P^{(A)}$, there's a self-adjoint operator, also called $A$, such that the following decomposition is unique


The spectrum $\sigma(A)\subset\mathbb{R}$ coincides with the support of $P^{(A)}$.

This is how and why we get the usual one to one correspondence between observables and self-adjoint operators in QM.

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  • $\begingroup$ So, to digest the first part of your answer I will have to check some notions in it (Borel element, ...). $\endgroup$ – wondering May 3 '15 at 11:00
  • $\begingroup$ Those are terms from measure theory. In the world of the mathematical foundations of QM, measure theory is ubiquitous. I should have said "Borel subset of the real line", "element" may be misleading (I was thinking in "element of the Borel sigma-algebra of subsets" but wrote "element" instead!). I'll edit the answer. $\endgroup$ – QuantumLattice May 3 '15 at 11:46

Unitarity of the time-evolution operator is exactly the point:
Stone's theorem (see e.g. Reed, Simon: Theorems VIII.7, VIII.8) tells us

  • If $A$ is self-adjoint, the spectral theorem holds. This gives us a functional calculus which makes it possible to define $U(t) = e^{itA}$ in the first place.
  • A such defined $U(t)$ is a strongly continuous unitary group.
  • If $U(t)$ is a strongly continuous unitary group, then there exists a self-adjoint $A$ such that $U(t) = e^{itA}$.

Edit: This only tells us why the Hamiltonian should be self-adjoint. QuantumLattice's answer is better.

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  • $\begingroup$ But I like your answer too. +1 $\endgroup$ – wondering May 3 '15 at 10:54

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