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The interference term. Confused.

  1. Jun 27, 2011 #1
    Can someone explain what the difference is between the expectation value of an observable in a pure vs a mixed state. The equations are identical. For example with spin up and down [tex]|A|^2<S_u|O|S_u> + |B|^2<S_d|O|S_d>[/tex]
  2. jcsd
  3. Jun 27, 2011 #2


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    What you've quoted is <O> for the case where the system is in an incoherent mixture of |Su> and |Sd>. This would be a classical mixture, like heads and tails on a coin, and is represented by a density matrix rather than a wavefunction.

    More typically what we mean by a mixed state is a coherent mixture, |S> = A |Su> + B |Sd>, in which case the expectation value <S|O|S> will need to include interference terms A*B <Su|O|Sd> and B*A <Sd|O|Su>.
  4. Jun 27, 2011 #3
    If you could quickly reference page 358 of this pdf http://www.physics.sfsu.edu/~greensit/book.pdf [Broken]

    It shows that your equation is equivalent to mine...
    Last edited by a moderator: May 5, 2017
  5. Jun 27, 2011 #4


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    That's true only if your states are eigenstates of O (so the interference terms vanish).

    This means, you can't distinguish between a coherent superposition and an incoherent mixture by measuring in one basis alone. But you don't get identical expectation values in all bases!

    Consider a superposition of 50:50 spin up and spin down along a given axis. If you change your measurement axis (corresponding to another basis) you get other ratios. In the case of incoherent mixing, the ratio is 50:50 regardless of basis. You can verify this by an easy calculation of the corresponding density matrices.
  6. Jun 27, 2011 #5
    What are the terms that vanish. If the system is mixed or pure it does not have any extra terms.

    Edit: I see the term now, I looked right over it, thanks!
    Last edited: Jun 27, 2011
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