Probability of an entanglement state

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SUMMARY

The discussion centers on the probability of measuring specific states in an entangled quantum system, specifically the states |A> and |B> defined as |A>=(|0_{A}>+|1_{A}>)/√2 and |B>=(|0_B{}>+|1_B{}>)/√2. It is established that the probability of finding the combined state |AB> in either |0a>|0b> or |1a>|1b> is 1/2, as derived from the Born rule and the trace formula. The rephrased state |U> = (|0a>|0b> + |1a>|1b>)/√2 confirms this probability outcome.

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phyky
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given two system is entangled, |A>=(|[itex]0_{A}[/itex]>+|[itex]1_{A}[/itex]>)/√2, |B>=(|[itex]0_B{}[/itex]>+|[itex]1_B{}[/itex]>)/√2. entangle state |AB>what is the probability to find |[itex]0_A{}[/itex][itex]0_B{}[/itex]> and |[itex]1_A{}[/itex][itex]1_B{}[/itex]>. are there still 1/2 just like normal inner product formulation?
 
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phyky said:
|AB>=(|0a>+|1b>)/√2, what is the probability to find |0a> and |1b>. are there still 1/2?

You need to rephrase that - states with two particles have terms of the form |a>|b>.

Thanks
Bill
 
Well your rephrased question would be |U> = (|0a>|0b> + |1a>|1b>)/root 2. If you have an observation |0a>|0b> or |1a>|1b> then one or the other will be detected with probability 1/2. This is directly from the Born rule trace formula eg trace (|U><U| |1a>|1b>) = 1/2.

Thanks
Bill
 

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