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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?

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

Comments:

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 $SO(3)\cong SU(2)/Z_2.$

Remarks:

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

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  • $\begingroup$ 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
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    $\begingroup$ @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
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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...

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