# time evolution of Hamiltonian to generate the Bell pair

Consider two different Hamiltonians: $$H_1(t) = ZZ + \alpha(t)X_1 + \beta(t)X_2$$ and $$H_2(t) = XX + \alpha(t)Z_1 + \beta(t)Z_2$$, where $$\alpha(t)$$ and $$\beta(t)$$ are time-dependent functions. Starting from $$\lvert 00 \rangle$$, why does time-evolution of $$H_1(t)$$ can't generate the Bell pair $$= \frac{\lvert 00 \rangle + \lvert 11 \rangle}{\sqrt{2}}$$, but $$H_2(t)$$ can? (assuming that evolution time $$t = [0,T]$$ isn't too short)

Look at the matrix $$H_2$$: $$\begin{bmatrix} \alpha+\beta & 0 & 0 & 1 \\ 0 & \alpha-\beta & 1 & 0 \\ 0 & 1 & \beta-\alpha & 0 \\ 1 & 0 & 0 & -(\alpha+\beta) \end{bmatrix}.$$ You can easily divide this into two subspaces. The first is spanned by $$\{|00\rangle,|11\rangle\}$$ on which your initial state is fully supported. Thus you're only trying to find functions $$\alpha,\beta$$ such that $$\begin{bmatrix} \alpha+\beta & 1 \\ 1 & -(\alpha+\beta) \end{bmatrix}$$ evolves $$\begin{bmatrix} 1 \\ 0\end{bmatrix}$$ into $$\frac{1}{\sqrt{2}}\begin{bmatrix} 1 \\ 1\end{bmatrix}$$.
On the other hand, $$H_1=(H\otimes H)H_2(H\otimes H)$$ where $$H$$ is the Hadamard matrix. So, there is still a subspace structure, but it doesn't line up quite so nicely with our initial $$|00\rangle=\frac{1}{2}(|++\rangle+|--\rangle)+\frac12(|+-\rangle+|-+\rangle)$$ and final states $$(|00\rangle+|11\rangle)=\frac{1}{\sqrt{2}}(|++\rangle+|--\rangle).$$ In particular, the two states are supported by different amounts by the two subspaces. But the point of evolution under this subspace structure is that the amount within each subspace is preserved.
Having gained this insight, there's a more elegant way of expressing this. For $$H_2$$, there is an operator $$Z\otimes Z$$ that it commutes with, and both initial and final states have the same expectation with that operator.
For $$H_1$$, the equivalent commutation is $$X\otimes X$$. The initial and final states have different expectation values, and hence cannot evolve into each other: let $$|\psi\rangle=e^{-iH_it}|\phi\rangle$$ where $$O$$ is the operation that commutes with $$H_i$$. Then $$\langle\psi|O|\psi\rangle=\langle\phi|e^{iH_it}Oe^{-iH_it}|\phi\rangle=\langle\phi|e^{iH_it}e^{-iH_it}O|\phi\rangle=\langle\phi|O|\phi\rangle,$$ i.e. the expectation value of the state with the commuting operator is preserved.