raising and lowering operators for spin

lion8172

When we set the raising and lowering operators for spin to be [tex]S_{\pm} = S_x \pm i S_y[/tex], what convention are we following (i.e. why is the first term taken to be S_x and the second taken to be S_y)?

Avodyne

If we label the eigenstates of [itex]S_z[/itex] as [itex]|{+}\rangle[/itex] and [itex]|{-}\rangle[/itex], so that [itex]S_z|{\pm}\rangle=\pm\frac12 |{\pm}\rangle[/itex], then
[tex]S_+|{-}\rangle=|{+}\rangle[/tex]
[tex]S_-|{+}\rangle=|{-}\rangle[/tex]
Also,
[tex]S_+|{+}\rangle=0[/tex]
[tex]S_-|{-}\rangle=0[/tex]
That is, [itex]S_+[/itex] raises the value of [itex]S_z[/itex], and [itex]S_-[/itex] lowers it. That is how the raising and lowering operators are defined, and [itex]S_x \pm i S_y[/itex] is just what they work out to be.

Marco_84

It is the right hand convention.
think about the angular momentum...
it is defined mathematically L=R x P... it is a "right handed representation".
We can of sure define that in left hand repr...

ewilibrium

I think that is a Math method.
Example:
angular momentum: After "operate" with Math signals, Lz will be raised or lowered. (with operation L+=Lx+iLy)

--
If define [tex]L \pm = L_x \pm iL_y [/tex]
When comute them:
[tex][L_z ,L \pm ] = ... = \pm \hbar (L_x \pm iL_y )[/tex]
(Griffiths D.J_Quantum Mechanics...)
If define "left hand" the signal will become to convert.
+- become -+
That is not good!