The spin 1/2 rotation matrix around the $z$-axis I worked out to be

$$ e^{i\theta S_z}=\begin{pmatrix} \exp\frac{i\theta}{2}&0\\ 0&\exp\frac{-i\theta}{2}\\ \end{pmatrix} $$

Is this taken to be anti-clockwise around the $z$-axis?

  • 2
    $\begingroup$ Well, how did you define $\theta$? $\endgroup$ – Kyle Kanos Dec 27 '13 at 17:37
  • $\begingroup$ I didn't. I just picked an arbitrary angle, used Pauli matices and expanded $\endgroup$ – user32462 Dec 27 '13 at 17:42
  • $\begingroup$ Perhaps you should try defining $\theta$; I would do it so that it is consistent with your work, so that you don't have to re-derive it. $\endgroup$ – Kyle Kanos Dec 27 '13 at 18:00
  • 2
    $\begingroup$ Just ask yourself what happens to the "vector" representing a complex $w$, when you multiply $w$ by $e^{i\alpha}$, $w$ being supposed in a $x,y$ plane, where the $z$-axis is at the usual place. $\endgroup$ – Trimok Dec 27 '13 at 18:27
  • 1
    $\begingroup$ Isn't this a hidden question about passive versus active transformation, or in this case, Schrodinger versus Heisenberg picture? $\endgroup$ – DanielSank Oct 9 '14 at 15:16

For your example, we have $e^{i\theta S_z}\mathbf{S}e^{-i\theta S_z}=\begin{pmatrix}\cos\theta & -\sin\theta&0\\\sin\theta & \cos\theta&0\\0&0&1\end{pmatrix}\mathbf{S}$, with $e^{i\theta S_z}=\begin{pmatrix}e^{i\frac{\theta }{2}} & 0\\ 0 & e^{-i\frac{\theta }{2}}\end{pmatrix}$ and $\mathbf{S}=\begin{pmatrix}S_x\\ S_y\\ S_z\end{pmatrix}$ representing the spin-1/2 operators.


In fact, for the most general spin rotation, we have $$U\mathbf{S}U^\dagger=A\mathbf{S}\rightarrow (1)$$, where $U$ represents the general spin rotation operator $U=e^{i\alpha S_z}e^{i\beta S_y}e^{i\gamma S_z}=\begin{pmatrix}\cos{\frac{\beta }{2}}e^{i\frac{\alpha + \gamma}{2}} & \sin{\frac{\beta }{2}}e^{i\frac{\alpha - \gamma}{2}}\\ -\sin{\frac{\beta }{2}}e^{i\frac{\gamma-\alpha}{2}} & \cos{\frac{\beta }{2}}e^{-i\frac{\alpha + \gamma}{2}}\end{pmatrix}\in SU(2)$, and $A=\begin{pmatrix}\cos\alpha \cos\beta\cos\gamma-\sin\alpha\sin\gamma& -\sin\alpha \cos\beta\cos\gamma-\cos\alpha\sin\gamma &\sin\beta\cos\gamma\\ \cos\alpha \cos\beta\sin\gamma+\sin\alpha\cos\gamma & -\sin\alpha \cos\beta\sin\gamma+\cos\alpha\cos\gamma&\sin\beta\sin\gamma\\-\cos\alpha\sin\beta&\sin\alpha\sin\beta&\cos\beta\end{pmatrix}$ $\in SO(3)$ with the three Euler angles $\alpha,\beta,\gamma$.

Eq.(1) gives the map from $SU(2)$ to $SO(3)$ and the relation $SO(3)\cong SU(2)/Z_2.$


$e^{i\theta S_x}=\begin{pmatrix}\cos{\frac{\theta }{2}} & i\sin{\frac{\theta }{2}}\\ i\sin{\frac{\theta }{2}} & \cos{\frac{\theta }{2}} \end{pmatrix},e^{i\theta S_y}=\begin{pmatrix}\cos{\frac{\theta }{2}} & \sin{\frac{\theta }{2}}\\ -\sin{\frac{\theta }{2}} & \cos{\frac{\theta }{2}}\end{pmatrix},e^{i\theta S_z}=\begin{pmatrix}e^{i\frac{\theta }{2}} & 0\\ 0 & e^{-i\frac{\theta }{2}}\end{pmatrix}.$

  • $\begingroup$ How have you defined $e^{i\theta S_z}\mathbf{S}e^{-i\theta S_z}$? The net result of this equation is multiplying a $2 \times 2$, $3 \times 1$, and $2 \times 2$ matrix with each other, and this is impossible. $\endgroup$ – Hunter Jan 31 '14 at 21:31
  • 3
    $\begingroup$ @Hunter No, what K-boy means is that you take $e^{i \theta S_z}$ (2x2) and act on it to each 2x2 component of the 3x1 column vector $\vec{S}$. You end up with a new column vector which is 3x1 with different 2x2 components. It is fine. $\endgroup$ – nervxxx Feb 1 '14 at 2:40
  • $\begingroup$ @nervxxx So the $3\times 3$ matrix (call it $R$) in Kai's first line is "actually" $R\otimes \mathrm{id}_2$, where $\mathrm{id}_2$ is the $2\times 2$ identity, right? $\endgroup$ – WetSavannaAnimal Aug 23 '15 at 12:22
  • $\begingroup$ @WetSavannaAnimalakaRodVance no, the mapping is this: let $U$ be an $SU(2)$ matrix (2x2). Then $U S_\mu U^\dagger = \sum_\nu R_{\mu \nu} S_\nu$, where $R$ is an $SO(3)$ matrix (3x3), and $\mu, \nu = x,y,z$. In math speak, the adjoint action of the Lie group $SU(2)$ gives an element of $SO(3)$ in the fundamental rep. In jargony terms, this gives rise to the oft-heard phrase: "SU(2) is the double cover of SO(3)", encapsulated in $SO(3) \cong SU(2)/Z_2$. $\endgroup$ – nervxxx Aug 24 '15 at 14:08

The three generators of right-handed spinor rotations are given by $\left\{- i\sigma_x,-i\sigma_y,-i\sigma_z\right\}$, see for instance Peskin & Schroeder page 44, and the rotation matrix for a spinor rotation over an angle $\phi$ around a unit vector $\hat{s}$ is given by:

$R~=~ \exp\left(-i\frac{\phi}{2}~\hat{s}\cdot\vec{\sigma}\right) ~=~ I\cos\frac{\phi}{2}+\left(-i\,\hat{s}\cdot\vec{\sigma}\right)\sin\frac{\phi}{2}$

Where $\vec{\sigma}=\{\sigma_x,\sigma_y,\sigma_z\}$ and $I$ is the unit matrix which is the same as $\sigma_o$. We can explicitly write the generators of (right-handed) rotation as follows starting from the definition of the Pauli matrices.

:\begin{align} \sigma_x = \begin{pmatrix} ~~0&~~1\\ ~~1&~~0~~ \end{pmatrix} && \sigma_y = \begin{pmatrix} ~~0&-i\\ ~~i&~~0~~ \end{pmatrix} && \sigma_z = \begin{pmatrix} ~~1&~~0\\ ~~0&-1~~ \end{pmatrix} \, \end{align}

:\begin{align} j_x = \begin{pmatrix} ~~0&-i\\ -i&~~0~~ \end{pmatrix} && j_y = \begin{pmatrix} ~~0&-1\\ ~~1&~~0~~ \end{pmatrix} && j_z = \begin{pmatrix} -i&~~0\\ ~~0&~~i~~ \end{pmatrix} \, \end{align}

The specific rotation matrix as given in the question above is a left-handed rotation since the right-handed rotation matrix is defined by:

$R~=~ \exp\left(\frac{\phi}{2} j_z\right) ~=~ I\cos\frac{\phi}{2}\phi~+~j_z\sin\frac{\phi}{2} ~=~ \begin{pmatrix} \exp-i\frac{\phi}{2}&0\\ ~~0&\exp i\frac{\phi}{2}~~ \end{pmatrix}$

Counter clockwise is right-handed if the rotation axis points towards you, but it is left-handed if the rotation-axis points away from you. It's up to your choice...


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.