compatibility of qunatum measurements


by sachi
Tags: compatibility, measurements, qunatum
sachi
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#1
Dec16-05, 05:41 PM
P: 76
The question is "two operators A,B do not commute". Is it true that
([A,B]psi) does not equal zero for any state psi?"

I'm not sure if there are any cases in which a state psi could produce this result (except for the obvious one where psi = 0 and there is no particle).

Thanks
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Physics Monkey
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#2
Dec16-05, 06:20 PM
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Here is a hint: think about the commutator [tex] [x^2 , p] [/tex]. Can you find a state that gives zero when you hit it with the commutator even though commutator isn't identically zero?
sachi
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#3
Dec17-05, 04:39 PM
P: 76
I've tried applying the commutator to an arbitrary psi and i get i*h bar * 2x* psi. The only case where I can think of this equalling zero is when psi = 0, which is the obvious case of having no particle.

Physics Monkey
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#4
Dec17-05, 04:41 PM
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compatibility of qunatum measurements


You can't think of any state that gives zero when you hit it with the position operator? Hint: think about position eigenstates.


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