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I Non-commuting operators on the same eigenfunctions

  1. Apr 3, 2016 #1
    In Griffiths chapter 4 (pg. 179-180) there is an example (Ex. 4.3) that details the expectation value of ## S_x ##, ##S_y##, and ##S_z## of a spin 1/2 particle in a magnetic field. In this example, they find an eigenvector of ## H## (which commutes with ## S_z##) but then use this same eigenvector to compute the expectation value of both ##S_x## and ##S_y##, too. But the three spin operators do not commute. They have different eigenvectors. So what exactly is the significance of computing the expectation values of the two other spin components with this eigenvector that is not an eigenvector of those 2 operators themselves?

    I seem to be missing something very fundamental here. What exactly does it mean for an operator to act on a state that is not its own eigenvector (i.e. what is the physical meaning of this value)?
     
  2. jcsd
  3. Apr 3, 2016 #2

    A. Neumaier

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    An operator ##A## on a finite-dimensional Hilbert space acts on all vectors, not only on the eigenvectors. The eigenvectors are just the special vectors ##\psi## for which the image ##A\psi## is parallel to ##\psi##.
     
    Last edited: Apr 4, 2016
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