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Measurements of spin

  1. May 8, 2010 #1
    I read in a book:

    For a qubit defined as: [tex]\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)[/tex]

    Since [tex]|0\rangle[/tex] and [tex]|1\rangle[/tex] are the eigenstates of [tex]\sigma_z[/tex] then measuring sigma_z will yield either [tex]|0\rangle[/tex] or [tex]|1\rangle[/tex]. Measuring [tex]\sigma_x[/tex] on the same qubit will give one of the eigenstates of [tex]\sigma_x[/tex], which are [tex]\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)[/tex] and [tex]\frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)[/tex].

    Only problem is I don't see how you can obtain these general/algebraic states? When I make a measurement on the qubit ie. ([tex](\frac{1}{\sqrt{2}}(\langle 0| +\langle1|))\sigma_x(\frac{1}{\sqrt{2}}(|0\rangle + |1\rangle))[/tex] I just obtain a number? How can I keep this algebraic structure and prove the above?
     
  2. jcsd
  3. May 10, 2010 #2
    What exactly is your question?

    In the last example you have calculated [tex]\langle \psi | \sigma_x | \psi \rangle[/tex] which is the expectation value of [tex]\sigma_x[/tex] for the state [tex]|\psi \rangle[/tex]. The expectation value is a number.
     
  4. May 10, 2010 #3

    Fredrik

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    The question appears to be "How do you express the eigenvectors of [itex]\sigma_x[/itex] as linear combinations of eigenvectors of [itex]\sigma_z[/itex]?". I don't have time to answer that right now, so I'll leave it for someone else.
     
  5. May 10, 2010 #4
    Yes exactly this!
     
  6. May 10, 2010 #5
    The representation of [tex]\sigma_x[/tex] in the basis of the eigenvectors of [tex]\sigma_z[/tex] is
    [tex]\left[ \begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right][/tex]
    So if you diagonalize that you will get the eigenvectors, which will be (1,1) and (1,-1). The eigenvectors for [tex]\sigma_z[/tex] in this basis are of course (1,0) and (0,1)
     
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