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Homework Help: Sakurai, Chapter 1 Problems 23 & 24

  1. Dec 23, 2008 #1
    Problem 23:
    If a certain set of orthonormal kets, [tex] |1> |2> |3> [/tex], are used as the base kets, the operators A and B are represented by

    A = \left( \begin{array}{ccc} a & 0 & 0 \\
    0 & -a & 0 \\
    0 & 0 & -a \end{array} \right)

    B = \left( \begin{array}{ccc} b & 0 & 0 \\
    0 & 0 & -ib \\
    0 & ib & 0 \end{array} \right).


    A and B commute. Find a new set of orthonormal kets which are simultaneous eigenkets of both A and B. Specify the eigenvalues of A and B for each of the three eigenkets. Does your specification of eigenvalues completely characterize each eigenket?

    Problem 24:
    Prove that [tex] (1 / \sqrt{2})(1 + i\sigma_x) [/tex] acting on a two-component spinor can be regarded as the matrix representation of the rotation operator about the x-axis by angle [tex] -\pi / 2[/tex]. (The minus sign signifies that the rotation is clockwise.)
  2. jcsd
  3. Dec 23, 2008 #2


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    Hi quantumkiko! :wink:

    Show us what you've tried, and where you're stuck, and then we'll know how to help. :smile:
  4. Dec 23, 2008 #3
    Hi Tim!

    In problem 23, I don't know how to represent the simultaneous eigenkets [tex] |a, b> [/tex]. I just know how to solve the eigenvalues for each operator using the characteristic equation (some are degenerate). I also know that for two commuting observables, their simultaneous eigenkets form a complete set. Therefore, their simultaneous eigenkets are automatically orthogonal. That's all.

    For problem 24, I think we have to show that the result of letting the operator [tex] (1 / \\sqrt{2})(1 + i\\sigma_x) [/tex] act on a spinor is equivalent to a rotation operator acting on the same spinor. For a spinor of unit length, I used the matrix representation [tex] \left( \begin{array}{c} \cos \theta & \sin\theta \end{array} \right) [/tex] (I think this is where I was wrong.) Since the angle of rotation is [tex] -\pi / 2 [/tex], the rotation matrix will be given by,

    [tex] \left( \begin{array}{cc} cos(-\pi / 2) & sin(-\pi / 2) \\ -sin(-\pi / 2) & cos(-\pi / 2) \end{array} \right) = \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right) [/tex].

    If I let this operator act on the spinor, the resulting s
    Last edited: Dec 23, 2008
  5. Dec 23, 2008 #4


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    Hint: in problem 23, just look at the bottom right-hand 2x2 square of A …

    it's a multiple of the unit matrix!

    so its eigenkets are … ? :smile:
  6. Dec 23, 2008 #5
    It's eigenkets are [tex] \left( \begin{array}{c} 1 \\ 1 \end{array}\right) and \left( \begin{array}{c} -1 \\ -1 \end{array}\right) [/tex]?
  7. Dec 23, 2008 #6


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    waah! :cry:

    think … if C is the 2x2 unit matrix,

    for what vectors or kets V is CV = V? :biggrin:
  8. Dec 23, 2008 #7
    Oh, for all kets V! So how does that fit into finding the simultaneous eigenstates of A and B?
  9. Dec 23, 2008 #8


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    Well, there's one obvious simultaneous eigenstate …

    and once you've found the other two eigenstates of B, they're bound to be eigenstates of A also. :smile:

    (i'm logging out now for a few hours :wink:)
  10. Dec 23, 2008 #9
    I got it! The obvious one is [tex] \left( \begin{array}{c} 1 & 0 & 0 \end{array} \right) [/tex] while the others are [tex] \left( \begin{array}{c} 0 & 1/\sqrt{2} & i/\sqrt{2} \end{array} \right) [/tex] and [tex] \left( \begin{array}{c} 0 & -1/\sqrt{2} & -i/\sqrt{2} \end{array} \right) [/tex]. Thank you very much!

    Now how about Problem # 24?
  11. Dec 23, 2008 #10


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    erm … they're the same!! :redface:
    Le'ssee …
    Well … to prove it's a π/2 rotation …

    the obvious thing to do is to square it! :biggrin:
  12. Dec 23, 2008 #11
    Oh yeah, I should really get different eigenkets, not just multiples of one of the other. So the other two should be [tex]
    \left( \begin{array}{c} 0 & 1/\sqrt{2} & i/\sqrt{2} \end{array} \right)
    [/tex] and [tex]
    \left( \begin{array}{c} 0 & i/\sqrt{2} & 1/\sqrt{2} \end{array} \right)
    I was thinking that they won't be orthonormal, but I forgot that one of the [tex] i [/tex]'s changes sign when doing the inner product.

    I got Problem # 24 also. Thank you!
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