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After studying the derivation of the generalised uncertainty principle, what I understand is:

Suppose we have two operators associated with observables $\hat{A}$ and $\hat{B}$. If I prepare a large number of quantum systems with identical states, $\left| \psi \right>$ and measure them using $\hat{A}$ on some of those systems and calculate to obtain the standard deviation of the measurements outcomes (call it $\Delta{A}$) and do the same by measuring the rest of the system using $\hat{B}$ to obtain standard deviation $\Delta{B}$, then the generalised uncertainty principle puts a constraint on the value of the product $\Delta{A}\Delta{B}$ if the operators do not commute. So, the standard deviation obtained from the probability distribution of the eigenvalues of both the operators are related by an inequality. In short, there really isn't a need to bring up 'physical' measurements to talk about the uncertainty principle - we can theoretically do all these.

I've heard how observables that commute can be precisely measured simultaneously while there will always be some uncertainty if operators do not commute. What does it mean to measure simultaneously mean here? In my above explanation, I never brought up the topic of simultaneous measurements at all; all I did was perform measurements on one part of the ensemble with one operator and then perform measurements on the remaining with the other operator.

What does it mean to simultaneously measure a state for two observables? If we mean simultaneously, don't we have to measure a single state first using an operator and then using another operator on the same state and not measuring a bunch of identical states for an observable and measuring a bunch of remaining identical states (not the one used for the previous observable) for another observable, like I mentioned in my interpretation of the uncertainty principle?

Tobias Fünke
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Pratham Hullamballi
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1 Answers1

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Operators that commute can be diagonalized by a single base. (the inverse is also true)

Let $[\hat{O_1},\hat{O_2}]=0$. Than there exists the base {$ |o_1o_2\rangle $} of simultaneous eigenstates.

Executing a measure of one, or both, of the observables means the initial state collapses into one of these eigenstates.

If i measure $\hat{O_1} $ and find the result $o_k$it means that my state is now

$|o_ko_2\rangle$

Where i don't necessarely know $o_2$ and have to execute another measure on the same system. I know i can do this because the system is in an eigenstate of $\hat{O_2} $ so there is determined value of this quantity that i can find.

In practice we can pack the multiple measures into one, as an example we can measure x, y and z with a particle detector. Theoretically we would have to diagonalize all the operators, find the simultaneous eigenstates and check against our state (ie. Do the inner product)

If the two observables did not commute than I could have only known that the final state is $|o_k\rangle$ which is not also an eigenstate of the second observable. Picking a set of commuting observables locks us in place, we can only know the value of these quantities.

Executing a simultameous measurement of two, or more, observables means being able to know the values of all the commuting observables because the measurement of one does not stop us from measuring another.

Riccardo Gandolfo
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