# Vector wave-function spin

1. Nov 17, 2004

### RedX

$$\left(\begin{array}{cc}\grave{\psi_{x}}\\\grave{\psi_{y}}\end{array}\right)=(\left(\begin{array}{cc}1 & 0\\0 & 1\end{array}\right)-\frac{ie_{z}}{h}\left(\begin{array}{cc}L_{z} & 0\\0 & L_{z}\end{array}\right)-\frac{ie_{z}}{h}\left(\begin{array}{cc}0 & -ih\\ih & 0\end{array}\right))\left(\begin{array}{cc}\psi_{x}\\\psi_{y}\end{array}\right)$$

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. Nov 18, 2004

### santoshroy

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...
z`=z
x'=xcos(theta)+ysin(theta)
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
3. Nov 21, 2004

### RedX

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.