Noncommuting observables and indeterminacy principle

[microsansfil]

Hi all,

What is the link between noncommuting observables
{\displaystyle [{\hat {A}},{\hat {B}}]={\hat {A}}{\hat {B}}-{\hat {B}}{\hat {A}} \neq 0}
and indeterminacy principle (which is about inequality relation of standard deviation of the expectation value of observables A and B ) ?

If the observables commute we can find a complete set of simultaneous eigenvectors if not this implies that no quantum state can simultaneously be both A and B eigenstate. Why this implies that we cannot measure one without affecting statisticaly the other (uncertainty principle) ?

Is it possible to measure simultaneously the two incompatibles observables knowing that when a state is measured, it is projected onto an eigenstate in the basis of the relevant observable ?

best regards
Patrick

 

[vanhees71]

The link between the commutator and the uncertainty principle is given by the general uncertainty relation, which follows directly from the positive definiteness of the Hilbert-space scalar product:
$$\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle|,$$
where $\Delta A$ and $\Delta B$ are the standard deviations of observables $A$ and $B$. All expectation values have to be taken using the state the system is prepared in according to Born's rule, i.e.,
$$\langle A \rangle=\mathrm{Tr}(\hat{\rho} \hat{A}),$$
where $\hat{\rho}$ is the state (statistical operator).

The uncertainty relation tells you about the limitation of state preparations, not limitations of measurements. You can always measure any observable with arbitrary precision, no matter in which state the system is prepared in.

Of course this refers to measurements on ensembles, i.e., generally you cannot have a single simultaneous measurement of two (or more) incompatible observables, but you can repeatedly prepare the system in the same quantum state and then measure with arbitrary precision the one or the other observable repeatedly. The found expectation values and standard deviations obey the uncertainty relation, and in fact to verify it you have to measure the observables on ensembles of equally prepared systems in the above given sense with much higher accuracy than the standard deviations due to state preparation. If the measurement resolution is larger than the standard deviation due to the state preparation, you'll not have sufficient precision to resolve the uncertainty due to the state but only that due to the used measurement device.

 

[microsansfil]

Thank for the answer

Of course this refers to measurements on ensembles, i.e., generally you cannot have a single simultaneous measurement of two (or more) incompatible observables, but you can repeatedly prepare the system in the same quantum state and then measure with arbitrary precision the one or the other observable repeatedly. The found expectation values and standard deviations obey the uncertainty relation, and in fact to verify it you have to measure the observables on ensembles of equally prepared systems in the above given sense

1/ the concept of the ensemble does it mean :

successive measurements of A done on the same system (same quantum pur state ?) and measurements of B done on the same system ?
or
measurements of A done simultaneously on copies of the system (same quantum pur state ?) and measurements of B done simultaneously on copies of the system ?
or
other meaning ?2/ When we repeatedly prepare the system in the same quantum state, does A and B observable could be considered as independent and identically distributed random variables (IID) ?

Best regards
Patrick

 

[vanhees71]

What I meant is to prepare the system repeatedly in the same state and then measure repeatedly either A or B. Of course you can always prepare the same system again and again or always use a new one. E.g., for photons usually you can only do the latter since usually detecting them destroys them since usually the detection applies the photoelectric effect.

 

[microsansfil]

I am taking note that in fact the uncertainty relation can be deduced mathematically from the inequality relationship of Cauchy-Schwarz. This demonstration is clear, but its pratical setting up remains unclear to me.

If I prepare a system repeatedly in the same state and then I measure repeatedly A, I can calculate ΔA. Then an hour later I prepare the same experiment repeatedly in the same state and then I measure repeatedly B I can calculate ΔB. In this situation the relation

\Delta A \Delta B \geq \frac{1}{2} |\langle [\hat{A},\hat{B}] \rangle| is always verified ?

Best regards
Patrick

 

[vanhees71]

Yes, you summarized the meaning of the uncertainty relation in a very clear way. Also the excerpt from your source (textbook?) is very good. Which one is it?

 

[microsansfil]

Also the excerpt from your source (textbook?) is very good. Which one is it?

Jean-Louis Basdevant Jean Dalibard
Quantum Mechanics
https://books.google.fr/books/about/Quantum_Mechanics.html?id=wlgKaCItYrEC&redir_esc=y

Thank
Best regards
Patrick
I better understand why one say that quantum physics upset our common sense