# What does simultaneous measurement of a mutually commuting set of observables mean?

If I have a mutually commuting set of observables $$\{A, B, C, D,...\}$$ and when we talk of simultaneous measurements of these observables, do we mean a single apparatus is required for simultaneous measurement or a series of apparatuses are required?

Because "simultaneous" suggests that at the same instant knowing values of all observables.

## 2 Answers

A mutually commuting set of observables can be perfectly measured (in principle) simultaneously in the sense that they can be brought in the diagonal form to have the same eigenvector basis. "Simultaneous" here means the simultaneous sharing of the eigenvectors and not necessarily "simultaneous in time". Hence, it can be thought to be a simultaneous measurement in time also, with different apparatus. And hence 'an apparatus' that can share the measurement operators constructed from the same eigenbasis can perfectly measure (project) them all perfectly. Such an apparatus can always be constructed for these observables.

It might be confused with the position-momentum uncertainty relation where it is stated about the incompatible measurement 'in time'. But it essentially is the same phenomenon there because $$\hat{x}$$ and $$\hat{p}$$ do not commute and hence cannot be brought to the same eigenbasis at once. one can be measured making the other have great standard deviation and vice versa.

Also, refer to this question (precisely the same question asked on StackExchange Physics):

It depends a bit on the context, but the idea is that if two observables commute that means that measuring one will not affect measurement results of the other, and you can therefore think of the values of $$A$$ and $$B$$ in a given state as "simultaneously existing".

This means that you can characterize a state by the measurement outcomes you will get if you measure it in the different bases. For example, you might have a state $$\lvert\psi\rangle$$ which gives outcome "$$1$$" if you measure the observable $$A$$ and outcome "$$-1$$" if you measure the observable $$B$$, and therefore you can assign to the state the labels "$$1_A,-1_B$$", to represent this fact.

Note how this contrasts sharply with the case of noncommuting observables. If $$A$$ and $$B$$ do not commute, then measuring $$A$$ changes the results of measuring $$B$$, and vice versa. This means that you cannot meaningfully assign to a state labels representing what results you would get by measuring $$A$$ and $$B$$, because the order in which you do that changes the kind of results you would get.

• What if $A$ is degenerate? – RAMAN CHOUDHARY Feb 2 '19 at 10:36
• @RAMANCHOUDHARY what about it? – glS Feb 3 '19 at 18:35