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Suppose I find a $1/2$ spin particle in the eigenstate of the observable $\hat S_x$ relative to the eigenvalue $\hbar / 2$. I will use the short-hand notation $ \vert \uparrow_x \rangle$.

The goal is to express it in terms of the eigenstates of the observable $\hat S_z$: $\{ \vert \uparrow_z \rangle, \vert \downarrow_z \rangle\}$.

I have to solve the linear system: $$ \left\{ \begin{aligned} &\vert \uparrow_x \rangle = \alpha \vert \uparrow_z \rangle + \beta \vert \downarrow_z \rangle \\ &\vert \downarrow_x \rangle = \alpha' \vert \uparrow_z\rangle + \beta' \vert \downarrow_z \rangle \end{aligned}\right. $$

The condition of normalization $\langle \uparrow_x \vert \uparrow_x \rangle = \langle \downarrow_x \vert \downarrow_x \rangle = 1$ returns $|\alpha|^2 + |\beta|^2 = |\alpha'|^2 + |\beta'|^2 = 1 $, while the ortogonality, $\langle \uparrow_x \vert \downarrow_x \rangle = \langle \downarrow_x \vert \uparrow_x \rangle = 0$ implies $\bar{\alpha}\alpha' + \bar{\beta}\beta' = \bar{\alpha}'\alpha + \bar{\beta}'\beta = 0$.

Now, here comes the problem. Using the identity $[S_+, S_-] = 2 \hbar S_z$ I could write $$ [S_+, S_-]\vert \uparrow_x \rangle = 2 \hbar S_z \vert \uparrow_x \rangle $$ expanding the commutator $$ \begin{aligned} (S_+S_- - \require{cancel}\cancel{S_-S_+}) \vert \uparrow_x \rangle = \hbar ^2 \vert \uparrow_x \rangle \end{aligned} $$ I end up with $$\frac{\hbar}{2}\vert \uparrow_x \rangle = S_x\vert \uparrow_x \rangle = S_z\vert \uparrow_x \rangle $$

But it can't be! I have been checking it for quite a while, but I still don't know what went wrong.

ric.san
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    Maybe $\hat S_+$ vanishes only with $\vert \uparrow_z \rangle $, but I'm not sure – ric.san Sep 12 '21 at 17:16
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    @VladimirKalitvianski I think it's the same subspace, represented by different bases. They are not orthogonal one another – ric.san Sep 12 '21 at 17:41
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    Related : Understanding the Bloch sphere. The space of states of a spin-1/2 particle is a Hilbert space. Each one of the sets ${\vert \uparrow_z \rangle,\vert \downarrow_z \rangle}$,${\vert \uparrow_y \rangle,\vert \downarrow_y \rangle}$, ${\vert \uparrow_x \rangle,\vert \downarrow_x \rangle}$ is an orthonormal (eigen)basis. See equation (25) and Figure-04 in my answer in above link. – Frobenius Sep 12 '21 at 21:24
  • @Vladimir Kalitvianski : I apologize, but I must say that you have a very very wrong view about the Hilbert space of states of a spin-1/2 particle. I am not an expert but I suggest you to correct this view in order to avoid confusion about this stuff in the future. – Frobenius Sep 12 '21 at 22:20
  • I am stuck. What the relation between the 10 years-8 months user therein Vladimir Kalitvianski and the 3 months user herein Vladimir Kalitvianski ??? – Frobenius Sep 12 '21 at 22:29
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    @Frobenius: I am both of them, just seen from different computers. – Vladimir Kalitvianski Sep 13 '21 at 16:05
  • @Vladimir Kalitvianski : Thanks for the clarification. I got stuck because your comments about the Hilbert space of states of a spin-1/2 particle do not match the level of the 10 years-8 months user (Theoretical Physicist) we meet in answers herein. – Frobenius Sep 13 '21 at 16:37
  • I was wrong, but never mind - I am a bad user. – Vladimir Kalitvianski Sep 13 '21 at 17:18

1 Answers1

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You are mixing properties that are defined for $|\uparrow_z\rangle $ with those defined for $|\uparrow_x\rangle $. The right relations are $$S_z|\uparrow_z\rangle = \frac\hbar2 |\uparrow_z\rangle;$$ $$S_z|\downarrow_z\rangle = -\frac\hbar2 |\downarrow_z\rangle,$$ and thus $$S_z|\uparrow_x\rangle = \alpha\frac\hbar2 |\uparrow_z\rangle-\beta \frac\hbar2 |\downarrow_z\rangle .$$

The other one that is wrong is that you assumed $S_+|\uparrow_x\rangle =0$ which is not, the right condition is $S_+|\uparrow_z\rangle =0$ and $S_+|\downarrow_z\rangle =\hbar|\uparrow_z\rangle$ . Which leads to

$$S_+|\uparrow_x\rangle= \cancel{\alpha S_+|\uparrow_z\rangle} +\beta S_+|\downarrow_z\rangle=\beta\hbar|\uparrow_z\rangle.$$ The rest you can work it out if you work only on the $|\updownarrow_z\rangle$ basis.

Mauricio
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