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Product of expectation values

  1. Oct 31, 2010 #1
    In quantum mechanics, when is this true
    [tex]
    \langle\psi|AB|\psi\rangle=\langle\psi | A|\psi\rangle\langle\psi |B|\psi\rangle
    [/tex]
    ? In probability theory, when the two variables are independent, the mean value of the product is the product of the mean values. What about QM?
     
  2. jcsd
  3. Oct 31, 2010 #2
    It is true if [itex]|\psi\rangle[/itex] is a normalized eigenstate of both A and B because then
    [tex]
    \langle \psi|AB| \psi\rangle=a_\psi b_\psi
    [/tex]
    ...
     
  4. Oct 31, 2010 #3
    The equation
    [tex]
    \langle \psi|AB|\psi\rangle = \langle \psi|A|\psi\rangle \langle \psi|B|\psi\rangle \ \forall \psi
    [/tex]
    also leads to [A,B]=0. But assuming that [A,B]=0 and that both are usual Hermitian observables does not seem to imply the above equation for general states, even though any state can be expanded as a linear combination of common eigenstates: this would require
    [tex]
    \sum_i a_ib_i|c_i|^2 = \sum_{i,j} a_i b_j |c_i|^2 |c_j|^2
    [/tex]
    for
    [tex]
    |\psi\rangle = \sum_i c_i |\psi_i \rangle.
    [/tex]
    So I don't have an answer to your question but I wrote anyway :) But
    [tex]
    \langle AB \rangle = \langle A \rangle \langle B \rangle
    [/tex]
    of course means that in the particular configuration, the operators are uncorrelated and there is e.g. no Heisenberg uncertainty in measuring both observables "simultaneously".
     
  5. Oct 31, 2010 #4
    Yeah, it is not enough for the operators to commute, because A commutes with A, but
    [tex]
    \langle\psi|A^2|\psi\rangle\neq\langle\psi | A|\psi\rangle\langle\psi |A|\psi\rangle
    [/tex]
     
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