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Spin 1/2--Raising and Lowering operators question
Hi,
Quick question regarding raising and lowering operators.
Sakurai (on pg 23 of Modern QM), gives the spin 1/2 raising and lowering operators S_{+}=\hbar \left|+\right\rangle \left\langle-\right| and S_{-}=\hbar \left|-\right\rangle \left\langle+\right|.
Acting with the raising operator on, say, the spin down state, you get
S_{+} \left|-\right\rangle = \hbar \left|+\right\rangle. The physical interpretation of this is that the raising operator increases the spin component by one unit of \hbar.
This makes sense to me but when I try to explicitly verify this I run into a misunderstanding.
Let's say I apply S_{+} to \left|-\right\rangle and get \hbar \left|+\right\rangle.
To then "measure" the eigenvalue of this spin-up state, would you not apply the S_{z} operator, which would give another factor of \hbar:
S_{z} S_{+} \left|-\right\rangle = S_{z} \hbar \left|+\right\rangle = \frac{\hbar^{2}}{2} \left|+\right\rangle
Or is it a mistake to apply S_{z} after applying the raising operator? If not, how does the extra factor of \hbar disappear?
Thanks.
Hi,
Quick question regarding raising and lowering operators.
Sakurai (on pg 23 of Modern QM), gives the spin 1/2 raising and lowering operators S_{+}=\hbar \left|+\right\rangle \left\langle-\right| and S_{-}=\hbar \left|-\right\rangle \left\langle+\right|.
Acting with the raising operator on, say, the spin down state, you get
S_{+} \left|-\right\rangle = \hbar \left|+\right\rangle. The physical interpretation of this is that the raising operator increases the spin component by one unit of \hbar.
This makes sense to me but when I try to explicitly verify this I run into a misunderstanding.
Let's say I apply S_{+} to \left|-\right\rangle and get \hbar \left|+\right\rangle.
To then "measure" the eigenvalue of this spin-up state, would you not apply the S_{z} operator, which would give another factor of \hbar:
S_{z} S_{+} \left|-\right\rangle = S_{z} \hbar \left|+\right\rangle = \frac{\hbar^{2}}{2} \left|+\right\rangle
Or is it a mistake to apply S_{z} after applying the raising operator? If not, how does the extra factor of \hbar disappear?
Thanks.
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