I have a question regarding the rotation of spinors in a spin-1/2 system.

We have a Spin generator $\hat{S}$ for rotations of spinors. A rotation around the axis $\vec{n}$ with the angle $\phi$ is generated by the operator:

$$ D_{\vec{n}}(\phi) = \exp(-i\phi \hat{S}\cdot \vec{n}) $$

This operator can also be written, for a rotation about $z$ e.g. as:

$$D_z(\phi) = \cos\left(\frac{\phi}{2}\right) - i \sigma_z \sin\left(\frac{\phi}{2}\right) $$

Here, $\hat{S}_i = \sigma_i/2$ and $\sigma_i$ are the pauli matrices.

$\sigma_1 = \begin{bmatrix}0 & 1 \\ 1& 0 \end{bmatrix}$ $\sigma_2 = \begin{bmatrix}0&-i\\ i&0\end{bmatrix}$ and $\sigma_3 = \begin{bmatrix} 1&0\\ 0&-1\end{bmatrix}$

Then we have given two states with a spin into the direction of the z-axis:

$\vert S_z= + \frac{1}{2} \rangle = \begin{bmatrix}1\\ 0\end{bmatrix} = \vert {\uparrow}\rangle $


$\vert S_z = - \frac{1}{2} \rangle = \begin{bmatrix}0\\ 1\end{bmatrix} = \vert {\downarrow}\rangle $

Now my question is:

With which rotation $D_\vec{n}(\phi)$ can the eigenstate $\vert S_x = +\frac{1}{2}\rangle$ be obtained using $\vert\uparrow\rangle $?

How can I calculate that?


2 Answers 2


The simplest way to think about it is by thinking of spin as a classical vector.

What kind of rotation would take a vector completely along $\hat z$ to the $\hat x$ axis? Clearly, this would be a rotation in the $xz$ plane, i.e. a rotation about $\hat y$.

The same argument will work for spin. You might care to reflect on the relation between the classical angle and the angle of rotation in spin space, remembering that $\vert +\rangle$ and $\vert -\rangle$ are orthogonal vector in spin space but antiparallel in ordinary space. That’s why $\phi$ is mutiplied by $1/2$ in your expressions.

  • $\begingroup$ Thank you very much! Now I have the following result: $D_y |\uparrow> = cos(\phi / 2) - i \begin{bmatrix}0 & -i\\ i & 0\end{bmatrix} sin(\phi /2) |\phi> = cos(\phi / 2) - \begin{bmatrix}0 & 1\\ -1 & 0\end{bmatrix} sin(\phi / 2) |\uparrow>$ And how do I now show, that this is $|S_x = +1/2>$? $\endgroup$
    – Armani42
    Commented Oct 21, 2017 at 12:55
  • $\begingroup$ @Armani42 You want to be sure that the $\cos(\phi/2)$ term is multiplied by the unit matrix matrix. There remains for you to find the eigenstates of $S_x$ as combinations of eigenstates of $S_z$, and find the angle $\phi$ that will produce these eigenstates. Can you guess $\phi$ from classical arguments? $\endgroup$ Commented Oct 21, 2017 at 12:58
  • $\begingroup$ Oh okay, and the Angle is than because spins rotate half as fast: $\phi$= $\pi$ ? $\endgroup$
    – Armani42
    Commented Oct 21, 2017 at 13:01
  • $\begingroup$ @Armani42 why do you try it and see if the result is an eigenstate of $S_x$? $\endgroup$ Commented Oct 21, 2017 at 13:02
  • $\begingroup$ Okay now I tried and I get: $D_y |\uparrow> = \begin{bmatrix} cos(\phi/2) & -sin(\phi/2) \\ sin(\phi/2) & cos(\phi/2)\end{bmatrix} |\uparrow>$ And $D_y(\pi) = \begin{bmatrix} 0 & -1 \\ 1 & 0\end{bmatrix} \begin{bmatrix}1 \\ 0\end{bmatrix} = \begin{bmatrix}0 \\ 1\end{bmatrix}$ But I can't take an eigenvalue of that :/ $\endgroup$
    – Armani42
    Commented Oct 21, 2017 at 13:09

When you rotate spin operators i.e. $$D_{\vec{n}}(\phi) \cdot (\hat{S} \cdot \vec{m}) \cdot D^{\dagger}_{\vec{n}}(\phi),$$ it is the same as if you would rotate (right-hand) the vector $\vec{m}$ around the vector $\vec{n}$ at angle $\phi$. It is a consequence of algebraic properties of spin operators (they span $su(2)$ Lie algebra). Following this analogy, $\hat{S}_{x,y,z}$ can be pictured as three orthogonal versors of the 3D Euclidean space. If you rotate versor $e_{x}$ at angle $\phi = -\pi/2$ around $e_y$ you will get $e_z$. So, in terms of spin operators: $$\hat{S}_z = e^{i\frac{\pi}{2}\hat{S}_y} \cdot \hat{S}_x \cdot e^{-i\frac{\pi}{2}\hat{S}_y}.$$

Now your initial state is an eigenstate of $\hat{S}_z$ i.e. $$\hat{S}_z |s_z=+1/2\rangle = \frac{1}{2} |s_z=+1/2\rangle.$$ If you use previous relation you can write $$\hat{S}_x e^{-i\frac{\pi}{2}\hat{S}_y}|s_z=+1/2\rangle = \frac{1}{2} e^{-i\frac{\pi}{2}\hat{S}_y}|s_z=+1/2\rangle,$$ and according to definition of eigenstate $$|s_x=+1/2\rangle := e^{-i\frac{\pi}{2}\hat{S}_y}|s_z=+1/2\rangle$$


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