Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Vector wave-function spin

  1. Nov 17, 2004 #1
    [tex]\left(\begin{array}{cc}\grave{\psi_{x}}\\\grave{\psi_{y}}\end{array}\right)=(\left(\begin{array}{cc}1 & 0\\0 & 1\end{array}\right)-\frac{ie_{z}}{h}\left(\begin{array}{cc}L_{z} & 0\\0 & L_{z}\end{array}\right)-\frac{ie_{z}}{h}\left(\begin{array}{cc}0 & -ih\\ih & 0\end{array}\right))\left(\begin{array}{cc}\psi_{x}\\\psi_{y}\end{array}\right)[/tex]

    According to my book, the right hand side rotates a vector wave-function (psi_x and psi_y are both scalar functions of x and y) counterclockwise about the z axis by e_z. It seems to me that this must be a typo, and that instead, if you combine the first two matrices into a single operator L, and call the last matrix the operator S, then the transformation should be given by: J=L+SL, instead of J=L+S. I'm confused. thnx
    Last edited: Nov 18, 2004
  2. jcsd
  3. Nov 18, 2004 #2
    Yes it is a typo. Arguments are the following...
    1. Since Lz is responsible for the rotation so second term must contain Lz.
    2.It is a simple transformation which can be written as...
    y'=-xsin(theta)+y cos(theta)
    Now we are dealing with Q.M. where angular momentum is the generator of rotation. The cos and sine are replaced by Lz operators.
    >>> so the correct one is replace hbar within the second matrix by Lz .
    Last edited: Nov 18, 2004
  4. Nov 21, 2004 #3
    Actually, after rereading the problem, it is correct, to a first approximation in e_z, which is all that matters anyway for infinitismal operations.

    L is the generator of infinitismal rotatation for scalar wave functions, but for vector wave functions, you need J=L+S, where S rotates the vector (and L rotates the coordinates) and is the spin matrix.

    Although I don't understand how S in this example could be the spin matrix, as S_z should be diagonal in this basis.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook