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Rotations and Angular Momentum

This appendix summarizes the relation between the simplest angular momentum operators, represented by Pauli matrices, and rotations. Let us take for example the $z, y$ plane and a new rotated direction $Z$ , forming the angle $\theta$ with $z$ . How does the Pauli matrix, e.g. $\sigma_z$ , behave under this (passive) rotation of the axes? Equivalently, which is the matrix $\sigma_Z$ whose eigenvalues are spinors parallel and antiparallel to $Z$ (active rotations)? An easy way is to identify by eye that:

$\sigma_Z=\left(\begin{array} \cos\theta &-i\sin\theta\\ i\sin\theta & -\cos\theta\end{array}\right) = \sigma_z\cos\theta+\sigma_y\sin\theta$

evidently does the job. Its unnormalized eigenvectors are:

$|+\rangle = \left(\begin{array}i\sin\theta \\ \cos\theta-1\end{array}\right) \qquad\qquad |-\rangle = \left(\begin{array} \cos\theta-1 \\ -i\sin\theta \end{array}\right).$

The norm is $1/\sqrt{2\cos\theta(\cos\theta-1)}$ , but it diverges for $\theta=0$ and $\theta=\frac \pi 2$ , in which cases it is best to normalize the states afterwards. It is straightforward to check that in those two cases $|\pm\rangle$ are the eigenstates of $\sigma_z$ and of $\sigma_y$ , respectively.

A more elegant way of solving the problem is to recognize that the Pauli matrices are the generators of the rotation group in the spinor representation of dimension 2. This treatment is perhaps redundant for the present scope, but it serves to illustrate the concept.

Recalling that the rotated matrix is $A^\prime={\cal R}^{-1} A {\cal R}$ , the previous statement amounts to saying that infinitesimal rotations by $d\theta$ around, e.g., $x$ are produced by the rotation matrix ${\cal R}_x(d\theta)=1 -\frac {d\theta} 2\sigma_x$ , where $1$ stands for the unitary matrix. Hence finite rotations are produced by the rotation matrix:

${\cal R}_x(\theta) = e^{-i\sigma_x \theta/2}$

where the exponential operator must be defined by its eigenvalues $e^{\mp i\theta}$ on the two eigenvectors $|\pm\rangle$ of $\sigma_x$ . Using Euler equivalence in this basis we get $e^{-i\sigma_x \theta/2}=1\,cos\theta/2 -i\sigma_x\,\sin\theta/2$ . Writing this back on the basis of $\sigma_z$ :

$e^{-i\sigma_x\theta/2}=\left(\begin{array} \cos\theta/2 &i\sin\theta/2\\ -i\sin\theta/2 & \cos\theta/2\end{array}\right)$

and calculating $\sigma_Z=e^{i\sigma_x\theta/2}\sigma_ze^{-i\sigma_x\theta/2}$ we finally get again

$\sigma_Z=\left(\begin{array} \cos\theta & -i\sin\theta\\ i\sin\theta & -\cos\theta\end{array}\right)$

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Summary of Pauli Matrix Algebra

Pauli matrices obey the commutation relations

$\left[\sigma_x,\sigma_y\right]=2i\sigma_z\quad\quad \left[\sigma_y,\sigma_z\right]=2i\sigma_x\quad\quad \left[\sigma_z,\sigma_x\right]=2i\sigma_y$

Their behaviour is easily seen in the basis of σ_z , that is diagonal. If

$\sigma_z | \pm\rangle = \pm|\pm\rangle$

one can check that $\sigma_\pm=\sigma_x\pm i \sigma_y$ act as raise-operator and lower-operator, respectively. To show this, first note that they are Hermitian conjugates of each other, $\sigma_\pm^\dagger=\sigma_\mp$ . Then consider

$\sigma_\mp\sigma_\pm= \left(\sigma_x^2+\sigma_y^2\pm i \left[\sigma_x,\sigma_y\right]\right)= \left(2\mp2\sigma_z\right)$

and take the norm of vector $\sigma_\pm|\mp\rangle$ :

$\sqrt{\langle\mp|\sigma_\mp\sigma_\pm|\mp\rangle}= 2$

$\begin{eqnarray}\sigma_x=\left(\matrix{0 & 1\\1 & 0}\right) &\quad\quad& \sigma_y=\left(\matrix{0 & -i\\i & 0}\right)\\ \sigma_z=\left(\matrix{1 & 0\\0 & -1}\right)\end{eqnarray}$

Pauli matrices

$\sigma_\alpha=2S_\alpha,\quad\quad\alpha=x,y,z$

Pauli matrices are 2 times the spin S=1/2 matrices

$\sigma_x^2=\sigma_y^2=\sigma_z^2=1$

Simple to check

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PauliMatricesRotations

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