4
$\begingroup$

As far as I know, two compatible observables have a complete set of common eigenvectors, and using this fact, one can prove that their corresponding operators are commutative. Well now is the converse true? Do any two commutative hermitian operators correspond to compatible observables?

Another point I have in mind is that commutativity is not transitive. For example, $[x,y]=0$, $[y,p_x]=0$, but $[x,p_x]\neq0$. Is compatibilty transitive? It seems for me that it has to be so, since a single observable can not have two different complete sets of eigenvectors. Isn't that true?

$\endgroup$
  • 2
    $\begingroup$ What is your definition of "compatible observables" if not "observables that commute"? Are you asking us to show the equivalence between "$A$ and $B$ have a common eigenbasis" and "$A$ and $B$ commute"? $\endgroup$ – ACuriousMind Jul 26 '17 at 8:26
  • $\begingroup$ For the finite-dimensional case, see Wikipedia. Related Math.SE questions: math.stackexchange.com/q/236212/11127 and links therein. $\endgroup$ – Qmechanic Jul 26 '17 at 8:51
  • 1
    $\begingroup$ @ACuriousMind Yes, that is what I want, and why compatibility seems to be transitive although it isn't. $\endgroup$ – Tofi Jul 26 '17 at 10:22
  • $\begingroup$ Please only ask one subquestion per post. $\endgroup$ – Qmechanic Jul 26 '17 at 10:29
  • $\begingroup$ The transitivity subquestion is related to physics.stackexchange.com/q/95355/2451 , physics.stackexchange.com/q/196254/2451 and links therein. $\endgroup$ – Qmechanic Jul 26 '17 at 10:30
1
$\begingroup$

This is known as the compatibility theorem. The statement as well as a proof can be found on Wikipedia:

Complete Set of Commuting Observables

However, as Griffiths says in his book about Quantum Mechanics (3rd chapter, the one about formalism; subsection Eigenfunctions of a hermitian operator), the fact that eigenfunctions of an observable operator are complete (in the QM sense, i.e. they form a basis of the Hilbert space on which this operator is defined) is just an axiom. It is provable in some cases, but not in general.

Therefore I suppose that this is a hidden assumption in the theorem quoted earlier, namely that at least one of the commuting operators has a complete set of eigenfunctions.

If you'd like to know more about when one can be sure to find a basis of an operator, there is an open question about that:

https://math.stackexchange.com/questions/1074918/why-does-the-set-of-an-hermitian-operators-eigenfunctions-spans-the-functions

Short answer is Spectral Theorem.

$\endgroup$
1
$\begingroup$

This question concerns more Mathematics than Physics, so it should be handled rigorously in order to avoid to generate even more confusion (I personally find quite confused this page Complete Set of Commuting Observables since it deals with the finite-dimensional case in the proofs and supposes that the statements are valid for the infinite-dimensional case, where instead things are much more subtle).

First of all compatibility of two observables represented by a pair of (generally unbounded) self-adjoint (not just Hermitian or symmetric) operators $A: D(A) \to H$ and $B:D(B) \to H$ in a (generally infinite-dimensional) Hilbert space $H$ means that their projection-valued measures commute.

In other words, if we focus on the spectral decompositions of the operators $$A = \int_{\sigma(A)} \lambda dP(\lambda)$$ and $$B = \int_{\sigma(B)} \lambda dQ(\lambda)$$ it must be $$P_EQ_F=Q_FP_E\quad \mbox{for all Borel measurable sets $E,F \subset \mathbb{R}$. }$$ Compatibility is the mathematical statement equivalent to the physical statement that the observables can be simultaneously measured.

If at least one, say $A$, of $A$ and $B$ is bounded (so that its domain coincide to the whole Hilbert space), compatibility is equivalent to commutativity $$AB\psi = BA\psi \quad \mbox{for every $\psi \in D(B)$.}$$ When both $A$ and $B$ are unbounded (which physically means that the outcomes of their measurements can be arbitrarily large) commutativity on every common invariant domain is not equivalent to compatibility. There are famous counterexamples due to Nelson.

Finally, compatibility is by no means transitive, and this is one of the most interesting features of quantum theory. It gives rise to several no-go theorems regarding possible classical interpretations in terms of hidden variables (think of Kochen-Specker theorem for instance).

$\endgroup$

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.