- #1

aliens123

- 75

- 5

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

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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