Lecture4Notes - Two particles-Two apparatii(Entanglement...

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Two particles-Two apparatii (Entanglement) R.A.Harris Chem 120A, Spring 2005, Lecture 4 Suppose we have a source, which emits two particles simultaneously, 2 1 S Let us set up the following apparatus (see Figure 1 on next page). Particle 1 always goes to apparatus I, and particle 2 always goes to apparatus II. By experiment we find that when a particle goes through slit a its partner goes through a' . Similarly for b and b' . The apparatus on the left (II) has perfect slit checkers, i.e. we know which slit the particle has gone through when the slit checker is on. Thus, we must add probabilities when we use the slit checker. Let ψ a (1) be probability amplitude for particle 1 going through a . Let φ a’ (2) be the probability amplitude of particle 2 going through a' . Similarly for b and b' . First imagine that we have classical particles. The probability of a particle going through a and ending up at y and its partner going through a' and ending up at y' is P a (1)P a’ (2) . The y and y' are implicit. Similarly for b and b': P b (1)P b' (2) . The total probability for a particle being measured at y and another at y’ is P(y, y’) = P a (1)P a’ (2) + P b (1)P b' (2) Suppose we turn off the slit checker on apparatus II. Classically it makes no difference. Quantum mechanically we have to add amplitudes: Ψ (y,y’) = ψ a (1) φ a’ (2) + ψ b (1) φ b’ (2) Now P(y,y’) = | Ψ (y,y’) | 2 P(y,y’) = | ψ a (1) φ a’ (2) + ψ b (1) φ b’ (2)| 2 where Ψ (y,y’) is the total probability amplitude that particle 1 reaches the detector at y and particle 2 at y’ when particles 1 and 2 are correlated (entangled).
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There is an interference term. Now suppose we turn the slit checker on, so we can
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This note was uploaded on 09/29/2009 for the course CHEM 120A taught by Professor Whaley during the Spring '07 term at Berkeley.

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Lecture4Notes - Two particles-Two apparatii(Entanglement...

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