A Does the Zeno effect freeze all the commuting observables?

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The Zeno effect can freeze a two-level quantum system's state if the state cannot be separated into independent observables. If the observables commute and the state is separable, measuring one observable will not influence the other. However, in cases where the state is entangled, fast repeated measurements of one observable can indeed affect the other. This implies that the Zeno effect's impact depends on the system's state configuration. Understanding these interactions is crucial for quantum mechanics applications.
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Hi Pfs
Take a two level system up-down in its up level, There are other observables that commute
with that. is it true that applying the Zeno's effect by fast repeated measure of that
observable will also freeze the system on up?
thanks.
 
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Heidi said:
Hi Pfs
Take a two level system up-down in its up level, There are other observables that commute
with that. is it true that applying the Zeno's effect by fast repeated measure of that
observable will also freeze the system on up?
thanks.
It depends on the state of the system. If the state can be separated into a product state of each observable, then measuring one observable will not affect the other. However, of the state cannot be separated like that, then measuring one is equivalent to measuring the other, and the Zeno effect can freeze the other observable.
 

Thread 'Antilinear Operators'

An antilinear operator ##\hat{A}## can be considered as, ##\hat{A}=\hat{L}\hat{K}##, where ##\hat{L}## is a linear operator and ##\hat{K} c=c^*## (##c## is a complex number). In the Eq. (26) of the text https://bohr.physics.berkeley.edu/classes/221/notes/timerev.pdf the equality ##(\langle \phi |\hat{A})|\psi \rangle=[ \langle \phi|(\hat{A}|\psi \rangle)]^*## is given but I think this equation is not correct within a minus sign. For example, in the Hilbert space of spin up and down, having...
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