# Commuting operators and their physical interpretation in QM

I'm studying Quantum Mechanics for the first time at the moment and I have a few questions in mind.

So recently, I saw a proof on that if two operators share the same eigenstates is equivalent to the two operators commuting. Furthermore, the physical interpretation of this is apparently that we can know two physical quantities without uncertainty if this holds. Now here comes my question:

1. can we interpret an operator as an act of "measuring",

2. and if we conform ourselves to the previous statement, that the order of measuring doesn't matter if the operators commute?

I can sort of see the answer to 2) intuitevly if one of the physical quantities measured is some scaling factor of the one we are looking for, i.e. velocity and momentum (scaling factor as mass)? We could also do the opposite order, meaning we measure momentum, from which velocity can be found. Maybe this is a bad example. since we are also limited by Heisenberg's uncertainty principle, but I hope you get the idea.

I hope I made myself clear enough.

1. can we interpret an operator as an act of "measuring",

Yes, there is a one-to-one mapping between mathematical operators (more precisely: self-adjoint operators) and physical observables. However, the act of measuring the observable $$A$$ on a physical state is not directly modeled as applying the operator $$A$$ to the state $$|\psi\rangle$$. It is actually more complicated (see for example Axioms of Quantum Mechanics):

• The measured value is any of the eigenvalues $$a_k$$ of $$A$$.
• The state collapses from $$|\psi\rangle$$ to the corresponding eigenvector $$|k\rangle$$ of $$A$$.
• The measurement outcome is not deterministic. It will yield eigenvalue $$a_k$$ and eigenvector $$|k\rangle$$ with probability $$|\langle k|\psi\rangle|^2$$.
1. and if we conform ourselves to the previous statement, that the order of measuring doesn't matter if the operators commute?

Yes. If two operators $$A$$ and $$B$$ commute, then you get the same results when you measure $$A$$ first and then $$B$$ as compared to the other way round. There are no accuracy restrictions when you measure both $$A$$ and $$B$$, and can you do this in any order: $$\Delta A\ \Delta B \ge 0.$$

If they don't commute (i.e. $$[A,B]\ne 0$$), then you have an uncertainty relation restricting the accuracies of the two measurements $$\Delta A\ \Delta B \ge \frac{1}{2}\Big|\langle[A,B]\rangle\Big|.$$ So if you measure $$A$$ precisely first (i.e. $$\Delta A=0$$) then you get a completely imprecise measurement for $$B$$ (i.e. $$\Delta B=\infty$$), and vice versa.

• Oh okay! So if they don't commute (and by extension of @alanf answer on Heisenbergs uncertainty principle) you would experience different results measuring whether you measure position vs momentum first? Commented Sep 4, 2023 at 8:42
• @Tanamas Yes, see my amendment. Commented Sep 4, 2023 at 10:12

I'm studying Quantum Mechanics for the first time at the moment and I have a few questions in mind.

1. can we interpret an operator as an act of "measuring",

No, not an "act," in the sense of a physical measurement action. The act of measuring is different from "acting" with an operator on a state.

An act of performing a measurement on a pure state collapses that state. Acting with an operator on a pure state does not collapse the state.

There is a relationship between the operator and the possible values of the measurement, but the act of measuring (i.e., performing the physical measurement) is different from these.

The eigenvalues of the operator are the possible results of the measurement. When the measurement is actually performed, the result will be one of the eigenvalues and the resulting state will be "collapsed" to the corresponding eigenstate of the operator (assuming the measured eigenvalue is not degenerate).

1. and if we conform ourselves to the previous statement, that the order of measuring doesn't matter if the operators commute?

The order of the act of measureing won't matter and also the order of operating the operators on a pure state won't matter. But, as discussed above, the act of measuring is not the same as the action of an operator on a state.

It is not the case that all operators represent measurements. A measurement is an interaction that produces a record that can be copied with arbitrarily high accuracy. This requirement roughly restricts the operators that can represent measurements to Hermitian operators (observables) and POVMs:

https://arxiv.org/abs/1212.3245

If two observables $$\hat{A}$$ and $$\hat{B}$$ don't commute, then there are limits to their mutual sharpness:

https://arxiv.org/abs/1511.04857

This is often called the uncertainty principle even though it has nothing to do with uncertainty. The uncertainty principle doesn't impose a limit on the accuracy with which you can measure momentum. Rather, because position and momentum observables don't commute, if you measure the momentum to some accuracy then there will be a limit to how sharp the resulting state can be in its position.

What matters is whether two observables commute, not whether they are related by scaling. For example, observables of two different systems will commute and so there won't be any limits on their mutual sharpness.

• When you use the word "sharp", do you use it in a sense of standard deviation mathematically speaking? Commented Sep 4, 2023 at 8:37
• @Tanamas No. See Section 2 of the paper at the second link. Commented Sep 4, 2023 at 8:54