Raising and lowering operators acting on spin 1 kets?

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I read in my notes that

S{-}|1> = sqrt(2)h(bar)|0>

and similar for all six products of using the raising and lowering operators on |1>, |0>, |-1>

I don't understand where the sqrt(2)h(bar) has come from?

Cheers

Philip
 
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J_{\pm}|j,m> = \hbar \sqrt{j(j+1) - m(m\pm 1 )}|j,m \pm 1 >
 
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philip041 said:
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Huh?
 

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