- #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 would have raised the factor by one unit 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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