## raising and lowering operators for spin

When we set the raising and lowering operators for spin to be $$S_{\pm} = S_x \pm i S_y$$, 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)?
 Recognitions: Science Advisor If we label the eigenstates of $S_z$ as $|{+}\rangle$ and $|{-}\rangle$, so that $S_z|{\pm}\rangle=\pm\frac12 |{\pm}\rangle$, then $$S_+|{-}\rangle=|{+}\rangle$$ $$S_-|{+}\rangle=|{-}\rangle$$ Also, $$S_+|{+}\rangle=0$$ $$S_-|{-}\rangle=0$$ That is, $S_+$ raises the value of $S_z$, and $S_-$ lowers it. That is how the raising and lowering operators are defined, and $S_x \pm i S_y$ is just what they work out to be.

 Quote by lion8172 When we set the raising and lowering operators for spin to be $$S_{\pm} = S_x \pm i S_y$$, 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)?
It is the right hand convention.
it is defined mathematically L=R x P... it is a "right handed representation".
We can of sure define that in left hand repr...

## raising and lowering operators for spin

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)

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