Given an operator $\mathcal{O}$ and a time-independent Hamiltonian $\mathcal{H}$, I can find the evolution of the operator as $\mathcal{O}(t)=e^{i \mathcal{H}t}\mathcal{O}e^{-i \mathcal{H}t}$.
For example, for a single spin-1/2 system under $\mathcal{H}=\Omega \sigma^x$ for $\mathcal{O}=\sigma^z$, I can find that $\sigma^z(t)=\cos(\Omega t)\sigma^z+\sin(\Omega t)\sigma^y$ where I used the BCH formula, which simplified the results to this nice closed formula. The BCH formula states that
\begin{equation}
e^{-A}Be^{A}=B+[A,B]+1/2[A,[A,B]]+…
\end{equation}
which allows us to calculate $e^{i \mathcal{H}t}\mathcal{O}e^{-i \mathcal{H}t}$ in terms of the commutators $[\mathcal{H},\mathcal{O}]$.
I am interested in the case of two spin-1/2 under $\mathcal{H}=-i\gamma/2 (\sigma_1^+\sigma_2^–\sigma_1^-\sigma_2^+)$, i.e. a Jaynes-Cummings interaction, for $\mathcal{O}=\sigma_1^++\sigma_2^+$.
In this case, brute force BCH formula does not give me a closed form, as $[A,B]=\sigma_1^+\sigma_2^z-\sigma_1^z\sigma_2^+$, $[A,[A,B]]=\sigma_1^++\sigma_2^+-2\sigma_1^+\sigma_2^z-2\sigma_1^z\sigma_2^+$, etc.
Is there any way to proceed, or is it known that for two spin-1/2, one cannot find a closed evolution for the Heisenberg equations of motion? Should I use instead a spin-1 and spin-0 representation?
Best Answer
You are misnaming the adjoint action lemma ("Hadamard's lemma", Campbell, 1897), useful to the CBH expansion.
It leads to $$ e^{it\Omega \sigma^x} \sigma^z e^{-it\Omega \sigma^x} =\sigma^z \cos(2t\Omega) +\sigma^y \sin (2t\Omega). $$
With apologies, this is a completely rewritten answer
... after the original, assuming rotational invariance of the hamiltonian, not there! You really don't need that adjoint action lemma, as your exponentials are straightforward to evaluate directly.
Namely, using the tensor product convention of sticking the 2-space matrices into the entries of the 1-space ones, $$ \mathcal{O}= \begin{pmatrix} 0&1&1&0\\ 0&0&0&1\\ 0&0&0&1\\ 0&0&0&0\\ \end{pmatrix}, $$ and $$ \mathcal{H}={\gamma\over 2} \begin{pmatrix} 0&0&0&0\\ 0&0&-i&0\\ 0&i&0&0\\ 0&0&0&0\\ \end{pmatrix}, \leadsto \\ \exp( it\mathcal {H})= \begin{pmatrix} 1&0&0&0\\ 0&\cos(\gamma t/2)&\sin(\gamma t/2)&0\\ 0&-\sin(\gamma t/2)&\cos(\gamma t/2)&0\\ 0&0&0&1\\ \end{pmatrix}, $$ so that $$ \exp( it\mathcal {H}) \mathcal {O} \exp( -it\mathcal {H}) = \begin{pmatrix} 0&c+s&c-s&0\\ 0& 0& 0&c+s\\ 0& 0& 0&c-s\\ 0&0&0&0\\ \end{pmatrix}\\ = \cos(\gamma t/2)~~ \mathcal {O} - \sin(\gamma t/2) \left (\sigma_1^+ \sigma_2^z- \sigma_1^z \sigma_2^+\right ). $$