# On change representation of spin 1/2 particle

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.

• Maybe $\hat S_+$ vanishes only with $\vert \uparrow_z \rangle$, but I'm not sure Sep 12, 2021 at 17:16
• @VladimirKalitvianski I think it's the same subspace, represented by different bases. They are not orthogonal one another Sep 12, 2021 at 17:41
• 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. Sep 12, 2021 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. Sep 12, 2021 at 22:20
• @Frobenius: I am both of them, just seen from different computers. Sep 13, 2021 at 16:05

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.