How Does the General Uncertainty Principle Apply to Non-Commuting Operators?

AI Thread Summary
The discussion centers on the application of the uncertainty principle to non-commuting operators, specifically in the context of Hermitian observables. Participants confirm that since the operators do not commute, the uncertainty principle is applicable, and they explore the implications of a "minimal uncertainty state." There is a focus on calculating variances and understanding the relationship between the operators, with a realization that the operators resemble Pauli spin matrices. The conversation highlights the importance of recognizing the physical context behind mathematical representations in quantum mechanics. Overall, the discussion enhances understanding of the uncertainty principle in relation to specific quantum states.
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Homework Statement



[PLAIN]http://img219.imageshack.us/img219/306/prblem.png

(oops, that line should end with "just before measurement? (Express your answer in terms of the variances of the two operators)" )

The Attempt at a Solution



[PLAIN]http://img641.imageshack.us/img641/5214/answerm.png

Does this make sense? I know that each operator is Hermitian, so it is an observable. I also know that they are incompatible because they do not commute. So the uncertainty principle must hold. However, the wording of the question makes me question myself... it merely gives the system's state directly before the measurement. Does time dependence matter?

Also, I can calculate the deviation of each operator by themselves, and multiply them together to get the same as the other side of the uncertainty inequality. I suppose the question asks for an answer in terms of the variance of the two operators, so perhaps it is not asking me to confirm both sides.

Any insight would be appreciated.

(sorry for the odd format, I'm still learning the ropes at making equations on the computer)
 
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agooddog said:
Also, I can calculate the deviation of each operator by themselves, and multiply them together to get the same as the other side of the uncertainty inequality.

So, you have just noticed in this way that the state given is a "minimal uncertainty state" for you pair of observables - you have equality and it can not be any better. For position and momentum observables these are Gaussian (bell shaped) states. For spin components Sx and Sy these are eigenstates of Sz.

Knowing this what will be the other state for which you will also have equality rather than inequality? Can you guess?
 
Gut instinct tells me it will be ( 0 1 ) (transposed, of course)

Ah, I did not notice that these represent spins, but merely saw them as (somewhat) arbitrary operators that the Prof made up. I feel a little dumb now realizing how similar they are to the Pauli spin matrices. Although I am glad because I understand the linear algebra behind this all a bit better after struggling through that without being able to picture the physical situation in my mind.

Thanks for the quick feedback!
 
agooddog said:
Gut instinct tells me it will be ( 0 1 ) (transposed, of course)

Not a bad instinct!
 
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