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Minus sign in Bell derivation

  1. Jul 20, 2012 #1

    I am trying to reproduce Bell's calculation for the expectation value of paired spin measurements on particles in the singlet state. For unit vectors [itex] \hat{a} [/itex] and [itex] \hat{b} [/itex] we want to calculate

    [tex] P(a,b)=<\psi|(\hat{a}\cdot\vec{\sigma})(\hat{b} \cdot \vec{\sigma})|\psi>[/tex]

    where [itex]|\psi>[/itex] is the singlet state.

    Via the commutation and anticommutation relations for the Pauli matrices the enclosed operator is simply

    [tex](\hat{a}\cdot\hat{b})I + \imath\vec{\sigma}\cdot(\hat{a}\times\hat{b}).[/tex]

    As a scalar the dot product can be pulled from the bra-ket, leaving [itex](\hat{a}\cdot\hat{b})<\psi|I|\psi>=(\hat{a}\cdot \hat{b}) [/itex] since the singlet state is normalized. The cross product's expectation value turns out to vanish. Thus the final answer is

    [tex] P(a,b)=(\hat{a}\cdot\hat{b})=\cos(\theta). [/tex]

    The answer usually quoted, however, is [itex] -\cos(\theta) [/itex], and I can't figure out where the minus sign is coming from. Any ideas?
  2. jcsd
  3. Jul 21, 2012 #2
    Well, that would get you the minus sign. But I had thought the fact the spins were anti-parallel to be already encoded by the singlet state. It seems odd to me that you should have to insert this information again via the operator. Maybe I'm misunderstanding how [itex]\vec{\sigma}[/itex] is supposed to work?
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