# Raising and lowering operators for spin

1. Dec 6, 2007

### 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)?

2. Dec 6, 2007

### 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.

3. Dec 6, 2007

### Marco_84

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...

4. Dec 7, 2007

### 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!

Last edited: Dec 7, 2007