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Physical interpretation of the rotation of a ket

  1. Jun 29, 2011 #1
    physical interpretation of the "rotation" of a ket

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

    No problem statement. I'm just having trouble imagining what it means to "rotate" a quantum -ket, especially since not all -kets are eigenstates of position. I know what the rotation operator is. I also can picture rotating a coordinate system in my head--that's easy of course. But what does it mean when a -ket is rotated?


    2. Relevant equations

    see attached .pdf. they are notes i've taken from shankar p. 306-306.

    3. The attempt at a solution

    see attached .pdf. they are notes i've taken from shankar p. 306-306.
     

    Attached Files:

  2. jcsd
  3. Jun 30, 2011 #2

    vela

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    Re: physical interpretation of the "rotation" of a ket

    Take the spin of an electron, for example. Suppose it's in the state
    [tex]\vert \alpha \rangle = \frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}[/tex]
    which is the +-eigenstate of Sx. If you rotate it by 90 degrees about the z-axis, you'd expect it to be in the +-eigenstate of Sy. The rotation operator about the z-axis for spin-1/2 is
    [tex]R(\phi)=\exp\left(-\frac{i S_z \phi}{\hbar}\right) = \begin{pmatrix}e^{-i\phi/2} & 0 \\ 0 & e^{i\phi/2}\end{pmatrix}[/tex]
    For [itex]\phi=\pi/2[/itex], you get
    [tex]\vert \alpha_R \rangle = R(\pi/2)\vert \alpha \rangle = \frac{1}{\sqrt{2}} \begin{pmatrix}e^{-i\pi/4} \\ e^{i\pi/4}\end{pmatrix} = \frac{e^{-i\pi/4}}{\sqrt{2}} \begin{pmatrix} 1 \\ i \end{pmatrix}[/tex]
    which is indeed the +-eigenstate of Sy.
     
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