I Where do these time derivatives of Pauli matrices come from?

[docnet]

TL;DR

Where do these time derivatives of Pauli matrices come from? These are in Susskind's Theoretical Minimum (which is not the best text to learn quantum mechanics from, I have heard). He introduces these equations as "the equations of motion" of the spin in a magnetic field. $$ \frac{d}{dt}\sigma_x=-i[\sigma_x,\sigma_z]\frac{\omega}{2}$$$$
\frac{d}{dt}\sigma_y=-i[\sigma_y,\sigma_z]\frac{\omega}{2}$$$$ \frac{d}{dt}\sigma_z=-i[\sigma_z,\sigma_z]\frac{\omega}{2}$$

Wolfgang Pauli's matrices are
$$\sigma_x=\begin{bmatrix}0& 1\\1 & 0\end{bmatrix},\quad \sigma_y=\begin{bmatrix}0& -i\\i & 0\end{bmatrix},\quad \sigma_z=\begin{bmatrix}1& 0\\0 & -1\end{bmatrix}$$
He introduces these equations as "the equations of motion" of the spin in a magnetic field.
$$ \frac{d}{dt}\sigma_x=-i[\sigma_x,\sigma_z]\frac{\omega}{2}$$$$
\frac{d}{dt}\sigma_y=-i[\sigma_y,\sigma_z]\frac{\omega}{2}$$$$ \frac{d}{dt}\sigma_z=-i[\sigma_z,\sigma_z]\frac{\omega}{2}$$
$$
\frac{d}{dt}\sigma_x=-\omega\sigma_y,\quad
\frac{d}{dt}\sigma_y=\omega\sigma_x,\quad
\frac{d}{dt}\sigma_z=0$$
"These are analogous to Saméon Poisson's brackets of angular momentum describing the rotational motion of a rigid body in classical physics. The $x$ and $y$ components of the spin precess around the $z$ axis, while the $z$ component of the spin does not change".

I am interested in how Susskind got the right hand side of the three equations.

They are not derivatives of the path in $\operatorname{SU(2)}$ $\gamma: q\longrightarrow I_2$ given by the rule $$\gamma(t)=q(t)=U\begin{pmatrix}e^{i\theta_1 t}&0\\0&e^{i\theta_2 t} \end{pmatrix}U^{-1}$$ with $\gamma(0)=I_2$ and $\gamma(1)=q$.

 

[PeroK]

I suspect those are operator equations which apply to the specific states in this particular scenario. I.e. if $|\alpha(t) \rangle$ is the state of the system at time $t$, then:
$$(\frac d {dt})(\sigma_x)|\alpha(t) \rangle = -i \frac \omega 2 [\sigma_x, \sigma_z] |\alpha(t) \rangle = -\omega \sigma_y |\alpha(t) \rangle$$
PS which, I guess, is another way of saying he's using the Heisenberg picture.

 

[vanhees71]

One should note that the spin of a charged particle (or a neutral particle being bound states of charged particles like the neutron) imply a magnetic moment,
$$\vec{\mu}=\frac{q}{2m} g \vec{s}.$$
Here $q$ is the charge, $m$ the mass the particle, and $g$ the socalled gyro-factor or Lande-factor. For an electron it's close to 2.

The potential for the forces on a dipole is
$$V(\vec{x},\vec{s})=-\vec{\mu} \cdot \vec{B}=-\frac{q g}{2m} \vec{s} \cdot \vec{B}.$$
For the equation of motion of the spin (in the Heisenberg picture of time evolution) you get
$$\dot{s}_a=\frac{1}{\mathrm{i} \hbar} [s_a,H]=-\frac{q g}{2m \mathrm{i} \hbar} B_b \cdot [ \sigma_a,\sigma_b] =-\frac{q g}{2 m \hbar} B_b \epsilon_{abc} s_c = -\frac{q g}{2m \hbar} \epsilon_{abc} B_b s_c= -\frac{q g}{2m \hbar} (\vec{B}\times \vec{s})_a$$
or
$$\dot{s}_a=\frac{q g}{2 m \hbar} \vec{s} \times \vec{B}=\vec{\mu} \times \vec{B},$$
as expected also from classical electromagnetism.

Note that for spin-1/2 particles $\vec{s}=\hbar \vec{\sigma}/2$.

 

[docnet]

$$\dot{s}_a=\frac{q g}{2 m \hbar} \vec{s} \times \vec{B}=\vec{\mu} \times \vec{B},$$
as expected also from classical electromagnetism.

Note that for spin-1/2 particles $\vec{s}=\hbar \vec{\sigma}/2$.

Okay. Question: Do the Hamiltonians with the Lie bracket operation form groups that are used in physics? (seriously, really curious.) Another question: Does this imply the the cross product in $\mathbb{R}^3$ corresponds to the Lie bracket of vector fields?

 

[vanhees71]

A lot of quantum theory is indeed best understood using the ideas of Lie groups and Lie algebras. E.g., the commutation relations between position and momentum result from the meaning of momentum as generators of spatial translations. In QT the Lie bracket is realized with self-adjoint operators describing observables via $\frac{1}{\mathrm{i}} [\hat{A},\hat{B}]$. The additional factor $1/\mathrm{i}$ ensures that the Lie bracket maps to another self-adjoint operator.

The commutation relations of angular momenta result from the fact that they are the generators of rotations, which in QT are represented by representations of the covering group SU(2) of the rotation group. Thus you have
$$[\hat{J}_a,\hat{J}_b]=\mathrm{i} \hbar \epsilon_{abc} \hat{J}_c.$$
That's valid also for orbital angular momentum and spin separately, because in non-relativistic physics the spin operators commute with orbital angular momentum.