I Raising and Lowering Operators

  • Thread starter aliens123
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Why is it that the raising and lowering operators in a spin 1/2 system have a factor of $\hbar ?$

From Sakurai:
$$S_+ \equiv \hbar | + \rangle \langle - |, S_- \equiv \hbar | - \rangle \langle + |$$

"So the physical interpretation of $S_+$ is that it raises the component by one unit of $\hbar ? $ " (Page 22, 2017 edition).

But it seems like this does not raise the spin component by one unit of $\hbar. $ If I have a vector in the state $|- \rangle $ then
$$S_z |- \rangle = (-\hbar /2) | - \rangle$$
But now if I look at:
$$S_z S_+ |- \rangle = (\hbar^2 /2) | + \rangle$$
This didn't raise the spin component by one unit of $hbar.$ But if $S_+$ weren't defined with that factor of $\hbar,$ then it would have raised the factor by one unit of $hbar.$
 
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Just divide by the constant then, it’s defined that way since the raising and lowering operators take simple forms.
 

vanhees71

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Usually the raising- and loweing-operators of angular momentum operators are defined in terms of the angular-momentum components,
$$\hat{S}_{\pm}=\hat{S}_x \pm \mathrm{i} \hat{S}_y.$$
Thus it has dimensions of angular momentum, which is where the factors ##\hbar## come from in the OP.

Of course, you have to normalize the eigenvectors, which cancels one factor ##\hbar## again.
 

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