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So, take for instance the x and y pauli matrices $\sigma_1$ and $\sigma_2$, which have a representation in the $\sigma_3$ basis as

$$ \sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},\ \ \sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} $$

Now consider dynamic evolution given by a Hamiltonian like $H = \lambda\sigma_3$. In the Heisenberg picture, $\sigma_1$ and $\sigma_2$ become time-dependent operators, whose time derivatives are

$$ \frac{d \sigma_1}{d t} \propto \sigma_2,\ \ \ \frac{d \sigma_2}{d t} \propto \sigma_1 $$

Now, clearly the above matrix form for $\sigma_1$ and $\sigma_2$ (which I've always kinda taken as a definition) becomes inappropriate, since all its elements are constant numbers.

My question is: how do I integrate these equations to obtain the 4 time-dependent elements of each matrix?

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Integrating your equations is a bit silly. You just evaluate $$ \sigma_1(t) \equiv e^{-i\theta \sigma_3} \sigma_1 e^{i\theta \sigma_3}\\ \sigma_2(t) \equiv e^{-i\theta \sigma_3} \sigma_2 e^{i\theta \sigma_3} , $$ with $\theta\equiv t\lambda/\hbar$.

But Pauli vectors exponentiate trivially, $$ e^{i\theta \sigma_3} =\cos \theta +i \sigma_3 \sin\theta , $$ whence you just evaluate, $$ \sigma_1(t)= \sigma_1 \cos (2\theta) + \sigma_2\sin(2\theta) ,\\ \sigma_2(t)= \sigma_2 \cos (2\theta) - \sigma_1\sin(2\theta), $$ a uniform rotation, as you'd expect (why?).

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  • $\begingroup$ Thank you! For some reason I had forgotten about the time evolution operator. I take it the uniform rotation of $\sigma_1$ and $\sigma_2$ is the Heisenberg picture statement of the Larmor precession, right? $\endgroup$ Mar 23, 2021 at 16:55
  • $\begingroup$ Basically. Rotation round the z-axis. $\endgroup$ Mar 23, 2021 at 17:22

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