(adsbygoogle = window.adsbygoogle || []).push({}); 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 [itex]S_{+}=\hbar \left|+\right\rangle \left\langle-\right|[/itex] and [itex]S_{-}=\hbar \left|-\right\rangle \left\langle+\right|[/itex].

Acting with the raising operator on, say, the spin down state, you get

[itex]S_{+} \left|-\right\rangle = \hbar \left|+\right\rangle [/itex]. The physical interpretation of this is that the raising operator increases the spin component by one unit of [itex]\hbar[/itex].

This makes sense to me but when I try to explicitly verify this I run into a misunderstanding.

Let's say I apply [itex]S_{+}[/itex] to [itex]\left|-\right\rangle[/itex] and get [itex]\hbar \left|+\right\rangle [/itex].

To then "measure" the eigenvalue of this spin-up state, would you not apply the [itex]S_{z}[/itex] operator, which would give another factor of [itex]\hbar[/itex]:

[itex]S_{z} S_{+} \left|-\right\rangle = S_{z} \hbar \left|+\right\rangle = \frac{\hbar^{2}}{2} \left|+\right\rangle[/itex]

Or is it a mistake to apply [itex]S_{z}[/itex] after applying the raising operator? If not, how does the extra factor of [itex]\hbar[/itex] disappear?

Thanks.

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# Spin 1/2-Raising and Lowering operators question

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