Raising and lowering operators for spin

[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)?

 

[Avodyne]

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.

 

[Marco_84]

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.
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 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!