# Does the null operator correspond to an observable?

Sort of a silly question, but a worthwhile one to think about, I believe. On a recent test, my classmates and I were asked if it were possible to have three observables, A, B, and C, such that [A,B]=C.

Most of the students said no, but one student insisted that it could happen if A and B commuted, and thus C were the null operator. He claimed that as a hermitian operator, the null operator corresponded the the observable of performing no observation, and throwing your quantum state in the trash.

What do you think? Is that student correct? Should the rest of us get points off for not thinking of that possibility?

Edit: Since people are asking, no, the question was not about simultaneous observables. The question was about 3 unspecified observables, whose corresponding operators were such that the commutator of A and B was C.

My answer was no, three such observables could not exist, as the commutator of two hermitian operators is always anti-hermitian and thus has imaginary eigenvalues and cannot correspond to an observable.

• What was your answer? What did the teacher say was the correct answer? Sep 25, 2022 at 3:22
• Do you really mean “three observables”? Or do you mean “three simultaneous observables”? Sep 25, 2022 at 3:24
• If $0$ isn't an observable, observables aren't closed under addition. I'm not sure anyone wants that consequence.
– J.G.
Sep 25, 2022 at 6:38