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Projecting onto an eigenstate when we make a measurement

  1. Feb 18, 2009 #1
    Suppose we prepare a system in some properly normalized superposition of the spherical harmonics: A|11> + B|10> + C|1-1>. One of the fundamental results of quantum mechanics is that, if we measure L_z, we will collapse the state of the system onto an eigenstate of the eigenvalue we measure. My question is this: Suppose we measure L_z and we get 0. Which eigenstate of L_z do we collapse to? There are infinitely many spherical harmonics for which m = 0!

    Thanks.
     
  2. jcsd
  3. Feb 18, 2009 #2
    If the original wave function is A|11> + B|10> + C|1-1> and it is normalizated, eigenfunction you can get when measurement of Lz is zero is l1 0>

    I hope my reply help you
     
  4. Feb 18, 2009 #3

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    The general rule is that when your measurement returns one of the eigenvalues, the state gets projected onto the corresponding eigenspace. That is, if there are n (orthonormal) eigenstates [itex]|\lambda_i>[/itex] corresponding to the eigenvalue [itex]\lambda[/itex], then if we define:

    [tex] P_\lambda = |\lambda_1><\lambda_1| + |\lambda_2><\lambda_2| + ... + |\lambda_n><\lambda_n| [/tex]

    Then if a measurement of [itex]|\psi>[/itex] returns [itex]\lambda[/itex], then the state collapses to [itex]P_\lambda |\psi>[/itex].

    If you have this degeneracy, it means there are other commuting observables you can still measure to distinguish between the different [itex]|\lambda_i>[/itex]. In your case, one such observable would be the total angular momentum, and your state is already in an eigenstate of this, so there's only one [itex]|\lambda_i>[/itex] possible, |1 0>.
     
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