# Probability of an entanglement state

1. Feb 17, 2014

### phyky

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

Last edited: Feb 18, 2014
2. Feb 18, 2014

### bhobba

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

Thanks
Bill

3. Feb 18, 2014

### bhobba

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