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Another probability question

  1. Aug 12, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the probability, [itex]P_{2a}(t)[/itex], that a measurement of the quantity [itex]A[/itex] in
    the state [itex]|\varphi (t)\rangle\right[/itex] will yield the value [itex]2a[/itex].


    2. Relevant equations
    [tex]\hat{A}|1\rangle\right = a(|1\rangle\right - i|2\rangle\right[/tex]
    [tex]\hat{A}|2\rangle\right = a(i|1\rangle\right + |2\rangle\right[/tex]
    [tex]\hat{A}|3\rangle\right = -2a(|3\rangle\right[/tex]

    [tex]A = \[ \left( \begin{array}{ccc}
    a & ia & 0 \\
    -ia & a & 0 \\
    0 & 0 & -2a\end{array} \right)\][/tex]

    [tex]|\varphi (t)\rangle\right = \[ \left( \begin{array}{ccc}
    cos(wt) \\
    0 \\
    -isin(wt) \end{array} \right)\][/tex]


    3. The attempt at a solution

    Well, I kinda suck at finding these probabilities. So I'm not sure what to do, since it asks for [itex]2a[/itex]. Is it just:
    [tex]P(2a) = \left|\langle\psi_j|\Psi\rangle\right|^2,[/tex]
    where [itex]\psi_j = \varphi[/itex] and [itex]\Psi = A|3\rangle\right[/itex], or am I just not getting it ?


    Regards.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 12, 2010 #2
    You need to find the eigenvectors and eigenvalues of the matrix, A, in terms of the basis vectors you are given. So basically you need to diagonalize the matrix you are given. This will give you 3 eigenvectors that are superpositions of the states: [itex]|1\rangle,|2\rangle,|3\rangle[/itex].

    Once you have the eigenvectors of 'A': [itex]|1'\rangle, |2'\rangle, |3'\rangle[/itex] (all of which you need to properly normalize) and their corresponding eigenvectors. Then you will choose the eigenvector [itex]|u'\rangle[/itex] with the eigenvalue, +2a. Since this will be the superposition state the wavefunction will collapse to once you measure the property A and find its value +2a.

    Finally, as you said before, the probability of finding this superposition state is:

    [tex]P(2a) = \frac{\left|\langle u'|\varphi(t)\rangle\right|^2}{\left|\langle \varphi(t)|\varphi(t) \rangle \right|^2}[/tex]
     
    Last edited: Aug 12, 2010
  4. Aug 14, 2010 #3
    So I get the eigenvectors to be:

    [itex][-i,1,0][/itex], [itex][0,0,1][/itex] and [itex][i,1,0][/itex].

    Normalized they will become:

    [itex][\frac{-i}{\sqrt{2}}, 0, 1][/itex], [itex][\frac{1}{\sqrt{2},0][/itex], and the last one can't be normalized, or am I wrong ?

    The eigenvalues is:

    [itex][0, -2a, 2a],[/itex]

    So I need to use the 3rd eigenvector ?

    And using the formula, I get that the probability must be:

    [tex]P_{2a} = \frac{cos^{2}(wt)}{4},[/tex]
    or am I way off ?


    Regards, and sorry for the late reply.
     
  5. Aug 14, 2010 #4
    You have the correct eigenvectors, but the normalized versions of them are wrong. You are correct about the eigenvalues and choosing the third eigenvector. But your probability is wrong. Double check the denominator in that probability.
     
  6. Aug 14, 2010 #5
    Think I screwed up the normalized eigenvectors. It should be:

    [itex][\frac{-i}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 0][/itex], [itex][0,0,1][/itex], and then again, the 3rd and last, I don't think I can normalize, since [itex]\sqrt{i^{2} + 1^{1} + 0} = \sqrt{0}[/itex]

    And I think I used the 3rd eigenvector instead of [itex]\varphi(t)[/itex] in the denominator, so that's why I got it wrong. It should of course be:

    [tex]P_{2a} = cos^{2}(wt)[/tex]

    Or maybe divided by 2 if I can normalize the eigenvector as the 1st one ?
     
    Last edited: Aug 14, 2010
  7. Aug 14, 2010 #6
    The third vector should normalized the same as the first.

    You want to multiply the vector by some normalization constant and solve for that constant. Remember that the magnitude of a vector is the dot product of itself with its complex conjugate. I believe you are leaving off the complex conjugate in the multiplication.
     
  8. Aug 14, 2010 #7
    Yup... I see now :)

    Once again, thank you very much.
     
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