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Quantum Physics - Measurement/Eigenvalues(functions)

  1. Feb 21, 2010 #1
    1. The problem statement, all variables and given/known data

    For a certain system, an observable A has eigenvalues 1 and -1, with corresponding eigenfunctions [tex]u_+[/tex] and [tex]u_-[/tex]. Another observable B also has eigenvalues 1 and -1, but with corresponding eigenfunctions:

    [tex]v_+ = \frac{u_+ + u_1}{\sqrt{2}}[/tex]

    [tex]v_- = \frac{u_+ - u_1}{\sqrt{2}}[/tex]


    Find the possible results of a measurement of C = A+B


    2. Relevant equations


    3. The attempt at a solution


    Measured values are just the eigenvalues, in this case, the eigenvalues of C are just 2 and -2? but the answer is [tex]\pm\sqrt{2}[/tex]...I'm aware that the [tex]1/\sqrt{2}[/tex] in the eigenfunctions of B will make my answer "correct" but then they're not the eigenvalues anymore?
     
  2. jcsd
  3. Feb 21, 2010 #2

    gabbagabbahey

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    Why do you say this?
     
  4. Feb 21, 2010 #3
    if C = A + B, and A and B both have eigenvalues of 1 and -1, then the eigenvalues of C are 2 and -2?
     
  5. Feb 21, 2010 #4

    gabbagabbahey

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    No, why would you think this was true?
     
  6. Feb 21, 2010 #5
    then how do I do this question?
     
  7. Feb 21, 2010 #6

    gabbagabbahey

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    The same way one usually finds the eigenvalues of an operator...
     
  8. Feb 21, 2010 #7
    helpful. Much thanks!....
     
  9. Feb 21, 2010 #8
    does anyone have the patience to tell me how to do this problem?
     
  10. Feb 21, 2010 #9

    gabbagabbahey

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    If I gave you an operator in matrix form and asked you to calculate its eigenvalues, could you do it?
     
  11. Feb 21, 2010 #10
    yes...
     
  12. Feb 21, 2010 #11

    gabbagabbahey

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    Okay, so if you can put [itex]C[/itex] into matrix form, you can find its eigenvalues....do you see how to put [itex]C[/itex] into matrix form? How about putting [itex]A[/itex] into matrix form (you are given both its eigenvalues and eigenfunctions, so this should be trivial)?
     
  13. Feb 21, 2010 #12
    yes I can do put A in matrix form, for B do I need to use the transformation matrix and transform B into A basis?
     
  14. Feb 21, 2010 #13

    gabbagabbahey

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    Good, and what do you get when you do that?

    You can do it without a transformation matrix since you are given [itex]B[/itex]'s eigenfunctions in terms of [itex]A's[/itex] eigenfunctions.
     
  15. Feb 21, 2010 #14
    [itex]\begin{pmatrix} 1 & 1\\1 & -1 \end{pmatrix}[/itex] for A

    is there a systematic way of getting B or do I just write down the eigenvectors of B in terms of the eigenvectors of A and see which numbers i should put in?
     
  16. Feb 21, 2010 #15

    gabbagabbahey

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    That doesn't look right, how did you end up with this?
     
  17. Feb 21, 2010 #16
    [itex]\begin{pmatrix} 1 & 0\\0 & -1 \end{pmatrix}[/itex] for A

    sorry i messed up the typing
     
  18. Feb 21, 2010 #17

    gabbagabbahey

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    That's better, so I see you are representing the eigenfunctions of [itex]A[/itex] as

    [tex]u_{+}\to\begin{pmatrix}1 \\ 0 \end{pmatrix}[/tex] and [tex]u_{-}\to\begin{pmatrix} 0 \\ 1 \end{pmatrix}[/tex]

    correct?

    What does this make [itex]v_{\pm}[/itex] in this representation?
     
  19. Feb 21, 2010 #18
    yes

    my [itex]v_{\pm}[/itex] would be:

    [itex]\frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ \pm 1 \end{pmatrix}[/itex]



    so B would be [itex]\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}[/itex]

    but I had to just look at it and see what numbers i should assign for B, is there a better way of doing it?


    oh and i got the correct eigenvalues! thanks so much...i was really confused
     
    Last edited: Feb 21, 2010
  20. Feb 21, 2010 #19

    gabbagabbahey

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    Good.

    No, this isn't quite correct.

    Any operator [itex]F[/itex] can be decomposed in terms of its eigenvalues, [itex]\lambda_{i}[/itex] and corresponding eigenfunctions [itex]f_i[/itex] (provided they are orthogonal) according to the equation

    [tex]F=\sum_{i}\lambda_i f_i^\dagger f_i[/itex]

    (Where [itex]f_i^\dagger[/itex] is the adjoint of [itex]f_i[/itex])

    Does this look familiar? If so, you can use it to construct [itex]A[/itex] and [itex]B[/itex] from their eigenvalues/eigenvectors (this is what I had thought you had done to find [itex]A[/itex], but apparently you used some other method)
     
  21. Feb 21, 2010 #20

    for B i used the transformation matrix:

    [itex]\frac{1}{\sqrt{2}}\begin{pmatrix} 1 & 1 \\1 & -1 \end{pmatrix}[/itex]

    which means B =

    [itex]\frac{1}{2}\begin{pmatrix} 1 & 1 \\1 & -1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\0 & -1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\1 & -1 \end{pmatrix}[/itex]

    which is [itex]\begin{pmatrix} 0 & 1 \\1 & 0 \end{pmatrix}[/itex]
     
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