# Heisenberg Equations of motion for two spins 1/2

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 $$$$e^{-A}Be^{A}=B+[A,B]+1/2[A,[A,B]]+...$$$$ which allows us to calculate $$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?

• Honestly, I am not familiar with it. Up until now, I have always worked with a single spin-1/2. Now that I need to work with two interacting spin-1/2, I need to get familiarised with new techniques. Thank you for pointing out this Rodrigues formula, I'll explore this. Commented Aug 24, 2023 at 8:44

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 $$\exp( it\mathcal {H}) \mathcal {O} \exp( -it\mathcal {H}) = \begin{pmatrix} 0&c+s&c-s&0\\ 0& 0& 0&c+s\\ 0& 0& 0&c-s\\ 0&0&0&0\\ \end{pmatrix}\\ = \cos(\gamma t/2)~~ \mathcal {O} - \sin(\gamma t/2) \left (\sigma_1^+ \sigma_2^z- \sigma_1^z \sigma_2^+\right ).$$

• Thanks Cosmas! Honestly, I was not expecting it to be constant... However, I can imagine it has been a fact of this reduced basis. It is possible to use this fact $\mathcal{O}(t)=\mathcal{O}$ in the 1+0 basis to now obtain $\mathcal{O}(t)$ in the 1/2x1/2 basis? Is there a unitary transformation that connects both operators? Commented Aug 24, 2023 at 14:41
• But this transformation is time-independent, right? Then, transforming the stationary $\mathcal{O}$ from the 1+0 basis to the 1/2x1/2 basis would leave the state stationary. But this is now what I expect, as when I work the Heisenberg equations of motion for $\mathcal{O}$ in the 1/2x1/2 basis, they give time-dependent equations of motion. What am I getting wrong? Commented Aug 24, 2023 at 15:05
• I do not seem to understand your point. You are right that the eigenvalues of [A,B] are all 0, but if the commutator itself [A,B] is not zero, then the equations of motion shouldn't be constant, right? Commented Aug 24, 2023 at 15:37
• I confess there is something off. I worked out the exponentials explicitly, and got a non-trivial time evolution of your nonhermitean $\mathcal{O}$, albeit quite simple. Commented Aug 24, 2023 at 16:15
• Your hamiltonian is hermitian but not rotationally invariant... so spins 0 and 1 do mix, unlike what I assumed in the answer, unjustifiably....Will need to drastically revise/upend the answer... Apologies. Commented Aug 24, 2023 at 16:36