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An eigenstates, eigenvectors and eigenvalues question

  1. May 11, 2013 #1
    Good evening :-)

    I have an exam on Wednesday and am working through some past papers. My uni doesn't give the model answers out, and I have come a bit stuck with one question. I have done part one, but not sure where to go from here, would be great if someone could point me in the right direction:

    S2) Show that the state vectors |Sx+> = [itex]\frac{1}{\sqrt{}2}[/itex] times a 2x1 matrix (1,1) and |Sy+> = [itex]\frac{1}{\sqrt{}2}[/itex] times a 2x1 matrix (1,-1) are eigenvectors of Sx = h/2 times a 2x2 matrix (0 1, 1 0) with respective eigenvalues plus and minus h/2...

    This I can do by using |A - λI| = 0, finding the eigenvalues, then using A.v=λv and setting up simutaneous quations to find the eigenvalues

    Part two... Of what operator is the state [itex]\frac{1}{sqrt{}2[/itex]}[/itex]/[itex](|Sx+> + |Sy+>) and eigenstate, and with what eigenvalue...

    Any help would be great and much appreciated
  2. jcsd
  3. May 11, 2013 #2
    My suggestion is to try to explicitly add up the two states in matrix notation...
    The answer should then be obvious to you :)

    The answer you gave to the first question is right. Nevertheless it is more time consuming then it was necessary and time is precious during exams :)
    You have to show that those are egenvectors with given eigenvalue, so you could simply show that

    [itex]S_x |S_x ± \rangle = ± \frac{h}{2}|S_x ± \rangle [/itex],

    without solving the egenvalue equation

    [itex]|S_x-\lambda I|=0[/itex].

    Last edited: May 11, 2013
  4. May 11, 2013 #3
    [itex]|S_x-\lambda I|=0[/itex]. is a lot easier and faster to solve ;)

    about the second question, it works in the same way:

    A(|Sx+> + |Sy+>) = λ (|Sx+> + |Sy+>)

    => (A-λI)(|Sx+> + |Sy+>) = 0

    A-λI = 0; Fill in λ = 1/sqrt(2) and the matrix will be (1/sqrt(2) 0; 0 1/sqrt(2) )

    It's been a while since I did this, I could be wrong ofcourse.. (and it looks a bit to easy, but hey?)

    edit: I think I'm wrong...
  5. May 11, 2013 #4
    Yes it is easy but no, it isn't faster.
    And if it were a bigger matrix (or worst a differential operator) it wouldn't be so easy to solve that equation while it would still be easy to let a matrix (or a differential operator) act on a vector (or a function).
    To solve an equation is (almost) always more difficult than checking one solution ^^

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